Triangles: Lengths, Angles and Areas, Between the Challenge and the Olympiad

Problems

Problem 1

In a triangle $ABC$ , $X$ and $Y$ are points on the segments $AB$ and $AC$ , respectively, such that $AX : XB = 1 : 2$ and $AY : YC = 2 : 1$ . If the area of triangle $AXY$ is $10$ then what is the area of triangle $ABC$ ?

Problem 2

From a square with sides of length $5$ , triangular pieces from the four corners are removed to form a regular octagon. Find the area removed to the nearest integer?

Problem 3

In rectangle $ABCD$ , $AB = 5$ and $BC = 3$ . Points $F$ and $G$ are on line segment $CD$ so that $DF = 1$ and $GC = 2$ . Lines $AF$ and $BG$ intersect at $E$ . What is the area of $\triangle AEB$ ?

Problem 4

Let $AD$ and $BC$ be the parallel sides of a trapezium $ABCD$ . Let $P$ and $Q$ be the midpoints of the diagonals $AC$ and $BD$ . If $AD = 16$ and $BC = 20$ , what is the length of $PQ$ ?

Problem 5

Let $ABCD$ be a convex quadrilateral with perpendicular diagonals. If $AB = 20, BC = 70$ and $CD = 90$ , then what is the value of $DA$ ?

Problem 6

Let $XOY$ be a triangle with $\angle XOY = 90^\circ$ . Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$ , respectively. Suppose that $XN = 19$ and $YM = 22$ . What is $XY$ ?

Problem 7

In a quadrilateral $ABCD$ , it is given that $AB = AD = 13, BC = CD = 20, BD = 24$ . If $r$ is the radius of the circle inscribable in the quadrilateral, then what is the integer closest to $r$ ?

Problem 8

Let $ABCD$ be a trapezium in which $AB \parallel CD$ and $AB = 3CD$ . Let $E$ be the midpoint of the diagonal $BD$ . If $[ABCD] = n \times [CDE]$ , what is the value of $n$ ? (Here $[\Gamma]$ denotes the area of the geometrical figure $\Gamma$ .)

Problem 9

Let $ABCD$ be a rectangle in which $AB + BC + CD = 20$ and $AE = 9$ where $E$ is the mid-point of the side $BC$ . Find the area of the rectangle.

Problem 10

In a triangle $ABC$ , let $E$ be the midpoint of $AC$ and $F$ be the midpoint of $AB$ . The medians $BE$ and $CF$ intersect at $G$ . Let $Y$ and $Z$ be the midpoints of $BE$ and $CF$ respectively. If the area of triangle $ABC$ is $480$ , find the area of triangle $GYZ$ .

Problem 11

Let $ABCD$ be a quadrilateral with $\angle ADC = 70^{\circ}, \angle ACD = 70^{\circ}$ , $\angle ACB = 10^{\circ}$ and $\angle BAD = 110^{\circ}$ . The measure of $\angle CAB$ (in degrees) is:

Problem 12

In trapezium $ABCD$ , it is given that $AB$ is parallel to $CD$ . Assume that $AB = 3CD$ , $CD = DA$ and $\angle CDA = 120^{\circ}$ . If the largest angle of $ABCD$ is $x^{\circ}$ and the smallest angle is $y^{\circ}$ , what is the value of $x/y$ ?

Problem 13

$ABCD$ is a square and $AB = 1$ . Equilateral triangles $AYB$ and $CXD$ are drawn such that $X$ and $Y$ are inside the square. What is the length of $XY$ ?

Problem 14

In a triangle $ABC$ , let $H$ , $I$ and $O$ be the orthocentre, incentre and circumcentre, respectively. If the points $B$ , $H$ , $I$ , $C$ lie on a circle, what is the magnitude of $\angle BOC$ in degrees?

Problem 15

Let $ABC$ be an equilateral triangle. Let $P$ and $S$ be points on $AB$ and $AC$ , respectively, and let $Q$ and $R$ be points on $BC$ such that $PQRS$ is a rectangle. If $PQ = \sqrt{3} PS$ and the area of $PQRS$ is $28\sqrt{3}$ , what is the length of $PC$ ?

Problem 16

In a triangle $ABC$ with $\angle BCA = 90^\circ$ , the perpendicular bisector of $AB$ intersects segments $AB$ and $AC$ at $X$ and $Y$ , respectively. If the ratio of the area of quadrilateral $BXYC$ to the area of triangle $ABC$ is $13 : 18$ and $BC = 12$ then what is the length of $AC$ ?

Problem 17

In rectangle $ABCD$ , $AB = 8$ and $BC = 20$ . Let $P$ be a point on $AD$ such that $\angle BPC = 90^\circ$ . If $r_1, r_2, r_3$ are the radii of the incircles of triangles $APB$ , $BPC$ and $CPD$ , what is the value of $r_1 + r_2 + r_3$ ?

Problem 18

In a rectangle $ABCD$ , $E$ is the midpoint of $AB$ ; $F$ is a point on $AC$ such that $BF$ is perpendicular to $AC$ ; and $FE$ perpendicular to $BD$ . Suppose $BC = 8\sqrt{3}$ . Find $AB$ .

Problem 19

Suppose the altitudes of a triangle are 10, 12 and 15. What is its semi-perimeter?

Problem 20

Let $ABCD$ be a rectangle and let $E$ and $F$ be points on $CD$ and $BC$ respectively such that area $(ADE) = 16$ , area $(CEF) = 9$ and area $(ABF) = 25$ . What is the area of triangle $AEF$ ?

Problem 21

In a triangle $ABC$ , the median from $B$ to $CA$ is perpendicular to the median from $C$ to $AB$ . If the median from $A$ to $BC$ is 30, determine $(BC^2 + CA^2 + AB^2)/100$ .

Problem 22

Triangles $ABC$ and $DEF$ are such that $\angle A = \angle D$ , $AB = DE = 17$ , $BC = EF = 10$ and $AC - DF = 12$ . What is $AC + DF$ ?

Problem 23

Let $ABC$ be an acute-angled triangle and let $H$ be its orthocentre. Let $G_1$ , $G_2$ and $G_3$ be the centroids of the triangles $HBC$ , $HCA$ and $HAB$ , respectively. If the area of triangle $G_1G_2G_3$ is 7 units, what is the area of triangle $ABC$ ?

Problem 24

Let $ABC$ be a triangle and let $\Omega$ be its circumcircle. The internal bisectors of angles $A$ , $B$ and $C$ intersect $\Omega$ at $A_1$ , $B_1$ , and $C_1$ , respectively, and the internal bisectors of angles $A_1$ , $B_1$ and $C_1$ of the triangle $A_1 B_1 C_1$ intersect $\Omega$ at $A_2$ , $B_2$ and $C_2$ , respectively. If the smallest angle of triangle $ABC$ is $40^\circ$ , what is the magnitude of the smallest angle of triangle $A_2 B_2 C_2$ in degrees?

Problem 25

Let $ABC$ be a triangle with $AB = AC$ . Let $D$ be a point on the segment $BC$ such that $BD = 48\tfrac{1}{61}$ and $DC = 61$ . Let $E$ be a point on $AD$ such that $CE$ is perpendicular to $AD$ and $DE = 11$ . Find $AE$ .

Problem 26

Let $ABC$ be a triangle with $AB = 5$ , $AC = 4$ , $BC = 6$ . The internal angle bisector of $C$ intersects the side $AB$ at $D$ . Points $M$ and $N$ are taken on sides $BC$ and $AC$ , respectively, such that $DM \parallel AC$ and $DN \parallel BC$ . If $(MN)^2 = \dfrac{p}{q}$ where $p$ and $q$ are relatively prime positive integers then what is the sum of the digits of $|p - q|$ ?

Problem 27

Let $ABCD$ be a parallelogram. Let $E$ and $F$ be midpoints of $AB$ and $BC$ respectively. The lines $EC$ and $FD$ intersect in $P$ and form four triangles $APB$ , $BPC$ , $CPD$ and $DPA$ . If the area of the parallelogram is 100 sq. units, what is the maximum area in sq. units of a triangle among these four triangles?

Problem 28

Three parallel lines $L_1, L_2, L_3$ are drawn in the plane such that the perpendicular distance between $L_1$ and $L_2$ is 3 and the perpendicular distance between $L_2$ and $L_3$ is also 3. A square $ABCD$ is constructed such that $A$ lies on $L_1$ , $B$ lies on $L_3$ and $C$ lies on $L_2$ . Find the area of the square.

Problem 29

In parallelogram $ABCD$ the longer side is twice the shorter side. Let $XYZW$ be the quadrilateral formed by the internal bisectors of the angles of $ABCD$ . If the area of $XYZW$ is 10, find the area of $ABCD$ .

Problem 30

A triangle $ABC$ with $AC = 20$ is inscribed in a circle $\omega$ . A tangent $t$ to $\omega$ is drawn through $B$ . The distance of $t$ from $A$ is $25$ and that from $C$ is $16$ . If $S$ denotes the area of the triangle $ABC$ , find the largest integer not exceeding $S/20$ .

Problem 31

In a parallelogram $ABCD$ , a point $P$ on the segment $AB$ is taken such that $\dfrac{AP}{AB} = \dfrac{61}{2022}$ and a point $Q$ on the segment $AD$ is taken such that $\dfrac{AQ}{AD} = \dfrac{61}{2065}$ . If $PQ$ intersects $AC$ at $T$ , find $\dfrac{AC}{AT}$ to the nearest integer.

Problem 32

Two sides of an integer sided triangle have lengths $18$ and $x$ where $x < 100$ . If there are exactly 35 possible integer values $y$ such that $18, x, y$ are the sides of a non-degenerate triangle, find the number of possible integer values $x$ can have.

Problem 33

Consider a square $ABCD$ of side length $16$ . Let $E, F$ be points on $CD$ such that $CE = EF = FD$ . Let the line $BF$ and $AE$ meet in $M$ . The area of $\triangle MAB$ is:

Problem 34

In a convex quadrilateral $ABCD$ , the lengths of the diagonals are 12 and 16 and the line segments joining the midpoints of the opposite sides are of equal length. What is the maximum possible area of the quadrilateral $ABCD$ ?

Problem 35

In an isosceles triangle $ABC$ , with $\angle ACB = 90^{\circ}$ . The point $D$ is on the side $BC$ such that $\angle ADC = 75^{\circ}$ . If the area of triangle $ADC$ is $81$ , what is the length of segment $BD$ ?

Problem 36

Let $ABCD$ be a rectangle and let $E$ be a point on $BD$ such that $AE$ is perpendicular to $BD$ . If $AE = 12$ and $CE = \sqrt{193}$ , compute the area of the rectangle $ABCD$ .

Problem 37

$O$ and $I$ are the circumcentre and incentre of $\triangle ABC$ respectively. Suppose $O$ lies in the interior of $\triangle ABC$ and $I$ lies on the circle passing through $B$ , $O$ , and $C$ . What is the magnitude of $\angle BAC$ in degrees?

Problem 38

Let $A_1, B_1, C_1, D_1$ be the midpoints of the sides of a convex quadrilateral $ABCD$ and let $A_2, B_2, C_2, D_2$ be the midpoints of the sides of the quadrilateral $A_1B_1C_1D_1$ . If $A_2B_2C_2D_2$ is a rectangle with sides 4 and 6, then what is the product of the lengths of the diagonals of $ABCD$ ?

Problem 39

Let $ABCD$ be a convex quadrilateral with $\angle DAB = \angle BDC = 90^\circ$ . Let the incircles of triangles $ABD$ and $BCD$ touch $BD$ at $P$ and $Q$ , respectively, with $P$ lying in between $B$ and $Q$ . If $AD = 999$ and $PQ = 200$ then what is the sum of the radii of the incircles of triangles $ABD$ and $BDC$ ?

Problem 40

In a triangle $ABC$ , let $I$ denote the incentre. Let the lines $AI, BI$ and $CI$ intersect the incircle at $P, Q$ and $R$ , respectively. If $\angle BAC = 40^\circ$ , what is the value of $\angle QPR$ in degrees?

Problem 41

In acute-angled triangle $ABC$ , let $D$ be the foot of the altitude from $A$ , and $E$ be the midpoint of $BC$ . Let $F$ be the midpoint of $AC$ . Suppose $\angle BAE = 40^\circ$ . If $\angle DAE = \angle DFE$ , what is the magnitude of $\angle ADF$ in degrees?

Problem 42

Let $P$ be an interior point of a triangle $ABC$ whose sidelengths are 26, 65, 78. The line through $P$ parallel to $BC$ meets $AB$ in $K$ and $AC$ in $L$ . The line through $P$ parallel to $CA$ meets $BC$ in $M$ and $BA$ in $N$ . The line through $P$ parallel to $AB$ meets $CA$ in $S$ and $CB$ in $T$ . If $KL$ , $MN$ , $ST$ are of equal lengths, find this common length.

Problem 43

In a triangle $ABC$ , right-angled at $A$ , the altitude through $A$ and the internal bisector of $\angle A$ have lengths 3 and 4, respectively. Find the length of the median through $A$ .

Problem 44

Let $ABC$ be a triangle with sides $51$ , $52$ , $53$ . Let $\Omega$ denote the incircle of $\triangle ABC$ . Draw tangents to $\Omega$ which are parallel to the sides of $ABC$ . Let $r_1, r_2, r_3$ be the inradii of the three corner triangles so formed. Find the largest integer that does not exceed $r_1 + r_2 + r_3$ .

Problem 45

The sides $x$ and $y$ of a scalene triangle satisfy $x + \dfrac{2\Delta}{x} = y + \dfrac{2\Delta}{y}$ , where $\Delta$ is the area of the triangle. If $x = 60$ , $y = 63$ , what is the length of the largest side of the triangle?

Problem 46

Let $x, y, z$ be positive real numbers such that $x^2 + y^2 = 49$ , $y^2 + yz + z^2 = 36$ and $x^2 + \sqrt{3}xz + z^2 = 25$ . If the value of $2xy + \sqrt{3}yz + zx$ can be written as $p\sqrt{q}$ where $p, q$ are integers and $q$ is not divisible by square of any prime number, find $p + q$ .

Problem 47

In a trapezium $ABCD$ , the internal bisector of angle $A$ intersects the base $BC$ (or its extension) at the point $E$ . Inscribed in the triangle $ABE$ is a circle touching the side $AB$ at $M$ and side $BE$ at the point $P$ . Find the angle $DAE$ in degrees, if $AB : MP = 2$ .

Problem 48

Let $ABC$ be a triangle and let $D$ be a point on the segment $BC$ such that $AD = BC$ . Suppose $\angle CAD = x^{\circ}$ , $\angle ABC = y^{\circ}$ and $\angle ACB = z^{\circ}$ and $x, y, z$ are in an arithmetic progression in that order where the first term and the common difference are positive integers. Find the largest possible value of $\angle ABC$ in degrees.

Problem 49

The ex-radii of a triangle are $10\frac{1}{2}, 12$ and 14. If the sides of the triangle are the roots of the cubic $x^3 - px^2 + qx - r = 0$ , where $p, q, r$ are integers, find the integer nearest to $\sqrt{p + q + r}$ .

Problem 50

In a triangle $ABC$ , $\angle BAC = 90^{\circ}$ . Let $D$ be the point on $BC$ such that $AB + BD = AC + CD$ . Suppose $BD : DC = 2 : 1$ . If $\dfrac{AC}{AB} = \dfrac{m + \sqrt{p}}{n}$ , where $m, n$ are relatively prime positive integers and $p$ is a prime number, determine the value of $m + n + p$ .

Problem 51

Let $P$ be a point in the interior of a triangle $ABC$ and let $AP, BP, CP$ meet the sides $BC, CA, AB$ in $D, E, F$ respectively. If $\frac{BP}{PE} = \frac{5}{2}, \quad \frac{CP}{PF} = \frac{7}{3}, \text{ and } \frac{AP}{PD} = \frac{p}{q}$ where p, q are natural numbers and $\gcd(p,q) = 1$ , find $p + q$ .

Problem 52

In triangle $ABC$ , it is given that $\angle CAB = 50^{\circ}$ and $\angle ABC = 70^{\circ}$ . Points $D$ and $E$ are chosen on sides $BC$ and $AC$ respectively such that $\angle ABE = \angle DAB = 30^{\circ}$ . If $\angle DEB = x^{\circ}$ , what is the value of $x$ ?

Problem 53

Consider the areas of the four triangles obtained by drawing the diagonals $AC$ and $BD$ of a trapezium $ABCD$ . The product of these areas, taken two at a time, are computed. If among the six products so obtained, two products are 1296 and 576, determine the square root of the maximum possible area of the trapezium to the nearest integer.

Problem 54

Let $D$ be an interior point of the side $BC$ of a triangle $ABC$ . Let $I_1$ and $I_2$ be the incentres of triangles $ABD$ and $ACD$ respectively. Let $AI_1$ and $AI_2$ meet $BC$ in $E$ and $F$ respectively. If $\angle BI_1E = 60^\circ$ , what is the measure of $\angle CI_2F$ in degrees?

Problem 55

In a triangle $ABC$ , the median $AD$ (with $D$ on $BC$ ) and the angle bisector $BE$ (with $E$ on $AC$ ) are perpendicular to each other. If $AD = 7$ and $BE = 9$ , find the integer nearest to the area of triangle $ABC$ .

Problem 56

In triangle $ABC$ , let $P$ and $R$ be the feet of the perpendiculars from $A$ onto the external and internal bisectors of $\angle ABC$ , respectively; and let $Q$ and $S$ be the feet of the perpendiculars from $A$ onto the internal and external bisectors of $\angle ACB$ , respectively. If $PQ = 7$ , $QR = 6$ and $RS = 8$ , what is the area of triangle $ABC$ ?

Problem 57

The incircle $\Gamma$ of a scalene triangle $ABC$ touches $BC$ at $D$ , $CA$ at $E$ and $AB$ at $F$ . Let $r_A$ be the radius of the circle inside $ABC$ which is tangent to $\Gamma$ and the sides $AB$ and $AC$ . Define $r_B$ and $r_C$ similarly. If $r_A = 16$ , $r_B = 25$ and $r_C = 36$ , determine the radius of $\Gamma$ .

Problem 58

Given $\Delta ABC$ with $\angle B = 60^{\circ}$ and $\angle C = 30^{\circ}$ , let $P, Q, R$ be points on sides $BA, AC, CB$ respectively such that $BPQR$ is an isosceles trapezium with $PQ \parallel BR$ and $BP = QR$ . Find the minimum possible value of $\dfrac{2[ABC]}{[BPQR]}$ where $[S]$ denotes the area of any polygon $S$ .

Problem 59

In a triangle $ABC$ , a point $P$ in the interior of $ABC$ is such that $\angle BPC - \angle BAC = \angle CPA - \angle CBA = \angle APB - \angle ACB.$ Suppose $\angle BAC = 30^{\circ}$ and $AP = 12$ . Let $D, E, F$ be the feet of perpendiculars from $P$ on to $BC, CA, AB$ respectively. If $m\sqrt{n}$ is the area of the triangle $DEF$ where $m, n$ are integers with $n$ prime, then what is the value of the product $mn$ ?

Problem 60

Let $ABC$ be a triangle, $D$ be the midpoint of side $BC$ , $O$ be the circumcentre and $H$ be the orthocentre. If the triangle $ODH$ is equilateral with side length equal to 6 and the area of the triangle $ABC$ can be written as $a\sqrt{b}$ , where $a, b$ are positive integers and $b$ is not divisible by the square of any prime, find $a + b$ .

Problem 61

Let $ABC$ be an isosceles triangle with sides $13, 13$ and $10$ . The tangents to the incircle, drawn parallel to the sides intersect the sides in points $D, E, F, G, H, K$ which form a hexagon. If the area of the hexagon $DEFGHK$ is $m + \dfrac{n}{l}$ , where $m, n, l$ are positive integers with $n < l$ and $\gcd(n,l) = 1$ , what is $m + n + l$ ?

Solutions

Solution: PRMO 2014, Q10

Key idea

Two triangles sharing the angle at $A$ have areas in the ratio of the products of the sides around it.

Triangles $AXY$ and $ABC$ share the angle at $A$ , so $\frac{[AXY]}{[ABC]} = \frac{AX \cdot AY}{AB \cdot AC} = \frac13 \times \frac23 = \frac29,$ using $AX : XB = 1 : 2$ , which makes $AX = \tfrac13 AB$ , and $AY : YC = 2 : 1$ , which makes $AY = \tfrac23 AC$ .

Hence $[ABC] = \frac92 \times 10 = 45.$

Answer 45

Solution: PRMO 2019, Q1

Key idea

Each corner triangle is right-angled and isosceles, because the octagon’s $135^{\circ}$ angle leaves $45^{\circ}$ on either side, so one equation between the cut and what survives of the side fixes everything.

Let the piece cut from a corner have legs of length $x$ along the two sides of the square. The octagon is regular, so its interior angle at every vertex is $135^{\circ}$ , and at a vertex lying on a side of the square the remaining angle is $180^{\circ} - 135^{\circ} = 45^{\circ}$ . Each corner triangle therefore has a right angle at the square’s corner and two angles of $45^{\circ}$ , which makes it isosceles with legs $x$ and hypotenuse $x\sqrt2$ .

Two kinds of side now appear in the octagon: the four hypotenuses, of length $x\sqrt2$ , and the four leftovers of the square’s sides, of length $5 - 2x$ . Regularity says these are equal: $5 - 2x = x\sqrt2, \qquad \text{so} \qquad x = \frac{5}{2 + \sqrt2}.$

The area removed is four half-squares of side $x$ : $4 \cdot \frac{x^2}{2} = 2x^2 = \frac{50}{(2+\sqrt2)^2} = \frac{50}{6 + 4\sqrt2} = \frac{25}{3 + 2\sqrt2}.$ That last denominator rationalises beautifully, since $(3+2\sqrt2)(3-2\sqrt2) = 9 - 8 = 1$ , so the area removed is exactly $25(3 - 2\sqrt2) = 75 - 50\sqrt2.$ The nearest integer is now settled without a calculator. Saying the area lies strictly between $3.5$ and $4.5$ is the same as saying $50\sqrt2$ lies strictly between $70.5$ and $71.5$ , which is the same as $1.41 < \sqrt2 < 1.43.$ Both halves follow from squaring, since $1.41^2 = 1.9881 < 2$ and $1.43^2 = 2.0449 > 2$ . So the nearest integer is $4$ .

The rationalisation is worth the trouble. A decimal value of $x$ carried through the squaring loses precision exactly where the question is delicate, and $75 - 50\sqrt2$ leaves no doubt about which side of $4.5$ the answer falls.

Answer 4

Solution: PRMO 2012, Q8

Key idea

Put the rectangle in coordinates; the two lines meet above the rectangle, at height $5$ .

Take $A = (0,0)$ , $B = (5,0)$ , $C = (5,3)$ and $D = (0,3)$ . Then $DF = 1$ puts $F$ at $(1,3)$ and $GC = 2$ puts $G$ at $(3,3)$ .

The line $AF$ has slope $3$ , so it is $y = 3x$ . The line $BG$ runs from $(5,0)$ to $(3,3)$ , with slope $-\tfrac32$ , so it is $y = -\tfrac32(x - 5)$ . Setting the two equal, $3x = -\tfrac32 x + \tfrac{15}{2}, \qquad \tfrac92 x = \tfrac{15}{2}, \qquad x = \tfrac53,$ and then $y = 5$ . So $E = \left(\tfrac53, 5\right)$ , above the rectangle.

Triangle $AEB$ has the base $AB = 5$ along the $x$ -axis and height $5$ , so its area is $\tfrac12 \times 5 \times 5 = \frac{25}{2}.$

Answer $25/2$

Solution: PRMO 2013, Q8

Key idea

Both midpoints lie on the midline of the trapezium, and their positions along it differ by half the difference of the parallel sides.

Put $B$ at the origin with $BC$ along the $x$ -axis, so $B = (0,0)$ , $C = (20, 0)$ , and the other two vertices sit at height $h$ : $A = (p, h)$ and $D = (p + 16, h)$ , since $AD = 16$ is parallel to $BC$ .

The two midpoints are $P = \frac{A + C}{2} = \left(\frac{p + 20}{2},\ \frac h2\right), \qquad Q = \frac{B + D}{2} = \left(\frac{p + 16}{2},\ \frac h2\right).$ They have the same height, so $PQ = \frac{(p+20) - (p+16)}{2} = \frac{20 - 16}{2} = 2.$

The horizontal offset $p$ cancelled, as did the height: in any trapezium the segment joining the midpoints of the diagonals lies along the midline and has length half the difference of the parallel sides.

Answer 2

Solution: PRMO 2014, Q3

Key idea

Perpendicular diagonals make the two pairs of opposite sides satisfy $AB^2 + CD^2 = BC^2 + DA^2$ , since every side is a hypotenuse over the same two axes.

the two shaded triangles use all four legs, so AB^2 + CD^2 and BC^2 + DA^2 are both p^2 + q^2 + r^2 + s^2 — the two shaded triangles use all four legs, so $AB^2 + CD^2$
and $BC^2 + DA^2$ are both $p^2 + q^2 + r^2 + s^2$

Let the diagonals meet at $O$ , and take them as the coordinate axes, so $A = (-p, 0), \quad C = (q, 0), \quad B = (0, r), \quad D = (0, -s)$ with $p, q, r, s > 0$ . Each side is then the hypotenuse of a right triangle at $O$ : $AB^2 = p^2 + r^2, \quad BC^2 = q^2 + r^2, \quad CD^2 = q^2 + s^2, \quad DA^2 = p^2 + s^2.$

Adding the first and third and comparing with the second and fourth, both totals are $p^2 + q^2 + r^2 + s^2$ : $AB^2 + CD^2 = BC^2 + DA^2.$ So $DA^2 = 20^2 + 90^2 - 70^2 = 400 + 8100 - 4900 = 3600, \qquad DA = 60.$

Answer 60

Solution: PRMO 2014, Q15

Key idea

The two medians to the legs give $4a^2 + b^2$ and $a^2 + 4b^2$ , whose sum is $5(a^2+b^2)$ , and $XY^2 = 4(a^2+b^2)$ .

Let $OX = 2a$ and $OY = 2b$ , so $M = (a, 0)$ and $N = (0, b)$ in coordinates with the right angle at the origin, $X = (2a, 0)$ and $Y = (0, 2b)$ .

Pythagoras on the two given segments: $XN^2 = 4a^2 + b^2 = 361, \qquad YM^2 = a^2 + 4b^2 = 484.$ Adding them, $5\left(a^2 + b^2\right) = 845, \qquad a^2 + b^2 = 169.$ And the hypotenuse is $XY^2 = (2a)^2 + (2b)^2 = 4 \times 169 = 676, \qquad XY = 26.$

Adding the two equations rather than solving them is what keeps this short: the answer needs only the symmetric combination $a^2 + b^2$ , and neither $a$ nor $b$ is rational on its own.

Answer 26

Solution: PRMO 2018, Q2

Key idea

A quadrilateral with an inscribed circle has area $rs$ , and here the diagonal $BD$ splits the kite into two isosceles triangles whose areas are immediate.

The question hands us the inscribed circle, so there is nothing to establish about its existence. The given lengths do satisfy the identity that any tangential quadrilateral must satisfy, as they had to. The two tangents drawn to a circle from a point outside it have equal length, so writing $w, x, y, z$ for the tangent lengths from $A, B, C, D$ , $AB + CD = (w+x) + (y+z) = (w+z) + (x+y) = AD + BC,$ and here both sides come to $13 + 20 = 33$ . That is a necessary condition and no more; on its own it would not produce a circle, which is why it is worth having been given one.

For any tangential polygon, joining the centre to each vertex cuts it into triangles, each with a side of the polygon as base and the inradius $r$ as height. Adding the areas, $[ABCD] = \tfrac12 r \times (\text{perimeter}) = rs, \qquad s = \tfrac{66}{2} = 33.$

So all that remains is the area. The diagonal $BD = 24$ splits the quadrilateral into triangles $ABD$ and $CBD$ , both isosceles with base $24$ . Their heights come from Pythagoras on half the base: $\sqrt{13^2 - 12^2} = 5, \qquad \sqrt{20^2 - 12^2} = 16,$ so the two areas are $\tfrac12 \cdot 24 \cdot 5 = 60$ and $\tfrac12 \cdot 24 \cdot 16 = 192$ , totalling $252$ . Hence $r = \frac{252}{33} = \frac{84}{11} = 7.636\ldots,$ and the nearest integer is $8$ .

Answer 8

Solution: IOQM 2020, Q1

Key idea

Triangles standing on the same base have areas in the ratio of their heights, and a midpoint halves one of those heights.

Let us call the shorter parallel side $CD = a$ , so that $AB = 3a$ , and let $h$ be the distance between the two parallel lines. The trapezium’s area is then the familiar average of the parallel sides times the height, $[ABCD] = \tfrac12 (3a + a)h = 2ah,$ and everything now depends on getting a handle on the small triangle $CDE$ .

It helps to reach it in two steps rather than one. Consider first the triangle $CDB$ , which stands on the base $CD$ of length $a$ with its apex $B$ sitting on the other parallel line, so that its height is the full $h$ and its area is $\tfrac12 ah$ . The triangle we actually want, $CDE$ , stands on that same base $CD$ , and its apex $E$ is the midpoint of $BD$ . Since $D$ lies on the line $CD$ already, moving from $B$ to the midpoint of $BD$ moves us exactly halfway down to that line, so the height of $CDE$ above $CD$ is $h/2$ and $[CDE] = \tfrac12 \cdot a \cdot \tfrac{h}{2} = \tfrac{ah}{4}.$

Comparing the two areas, $n = 2ah \div \tfrac{ah}{4} = 8$ . Notice that both $a$ and $h$ vanished from the answer, which is the signal that the ratio was never going to depend on the particular trapezium we drew.

Answer 8

Solution: IOQM 2020, Q4

Key idea

The quadratic that the conditions produce is, after a single rearrangement, precisely the area that was asked for, so the sides themselves never need to be found.

Write $AB = CD = a$ for the two equal sides of the rectangle and $BC = b$ for the third, so that the condition $AB + BC + CD = 20$ becomes $2a + b = 20$ . Because $E$ is the midpoint of $BC$ and the rectangle has a right angle at $B$ , Pythagoras in the triangle $ABE$ gives $a^2 + \left(\frac{b}{2}\right)^2 = 81.$ Substituting $b = 20 - 2a$ , so that $b/2 = 10 - a$ , and expanding, $a^2 + (10-a)^2 = 81, \qquad 2a^2 - 20a + 100 = 81, \qquad 2a^2 - 20a + 19 = 0.$

Solving that for $a$ with the quadratic formula produces an unpleasant surd. Look instead at what the question actually asks for. The area of the rectangle is $ab = a(20 - 2a) = 20a - 2a^2 = -\left(2a^2 - 20a\right),$ and the equation above tells us that $2a^2 - 20a = -19$ . The area is therefore $19$ exactly.

The lesson generalises well beyond this problem. The side lengths here are irrational, yet their product is a whole number, and a question that asks only for the product is often built so that the product falls out of the relation directly. Reading the target carefully before grinding through the algebra saves the grinding.

Answer 19

Solution: IOQM 2023, Q5

Key idea

Both $Y$ and $Z$ lie on the segments $GB$ and $GC$ themselves, and each is at one sixth of the median from $G$ , so triangle $GYZ$ is a scaled copy of triangle $GBC$ with ratio $1/4$ .

The intersection of the medians is called the centroid. To justify the ratio we need, read $A$ , $B$ and $C$ as coordinate pairs. The midpoints are $E=(A+C)/2$ and $F=(A+B)/2$ . The point with position vector $g=(A+B+C)/3$ satisfies $g-B=\frac23(E-B), \qquad g-C=\frac23(F-C).$ It therefore lies on both medians, so it is $G$ , and lies two thirds of the way along each median from the vertex. Also its perpendicular height above $BC$ is one third of the height of $A$ : the heights of $B$ and $C$ above that line are zero, and taking an average of coordinates averages that perpendicular component too. Thus $[GBC]=[ABC]/3$ .

The median ratios now give $GB = \tfrac23 BE, \qquad GC = \tfrac23 CF.$ Now locate $Y$ , the midpoint of $BE$ . Measuring from $B$ we have $BY = \tfrac12 BE$ while $BG = \tfrac23 BE$ , and since $\tfrac12 < \tfrac23$ the point $Y$ lies between $B$ and $G$ . In particular $Y$ is on the ray $GB$ , and $GY = BG - BY = \left(\tfrac23 - \tfrac12\right)BE = \tfrac16 BE.$ The same computation on the other median puts $Z$ on the ray $GC$ with $GZ = \tfrac16 CF$ .

Triangles $GYZ$ and $GBC$ therefore share the angle at $G$ , and their sides about that angle are in the ratios $\frac{GY}{GB} = \frac{\tfrac16 BE}{\tfrac23 BE} = \frac14, \qquad \frac{GZ}{GC} = \frac14.$ Two triangles with a common angle have areas in the ratio of the products of the enclosing sides, so $[GYZ] = \tfrac14 \cdot \tfrac14 \cdot [GBC] = \tfrac{1}{16}[GBC].$

Finally, the three triangles $GBC$ , $GCA$ , $GAB$ have equal areas, each one third of $[ABC]$ , so $[GBC] = 160$ and $[GYZ] = \frac{160}{16} = 10.$

Answer 10

Solution: IOQM 2024, Q4

Key idea

Triangle $ACD$ is isosceles, which hands over $\angle DAC = 40^{\circ}$ , and the angle at $A$ splits into that piece and the one asked for.

In triangle $ACD$ the angles at $D$ and at $C$ are both $70^{\circ}$ , so the third one is $\angle DAC = 180^{\circ} - 70^{\circ} - 70^{\circ} = 40^{\circ}.$

Now look at the angle $\angle BAD$ , which the problem gives as $110^{\circ}$ . Since $C$ lies inside that angle, it splits into two parts, $\angle BAD = \angle BAC + \angle CAD,$ and therefore $\angle CAB = 110^{\circ} - 40^{\circ} = 70^{\circ}.$

The given $\angle ACB = 10^{\circ}$ is not needed for the answer. It fixes the shape of the quadrilateral, and a quick check confirms consistency: in triangle $ABC$ the angles would then be $70^{\circ}$ , $10^{\circ}$ and $100^{\circ}$ .

Answer 70

Solution: IOQM 2026, Q8

Key idea

The line through $C$ parallel to $DA$ cuts off a rhombus, and what remains is a triangle with sides $1$ and $2$ around a $60^{\circ}$ angle, which is half of an equilateral triangle.

Take $CD = DA = 1$ , so $AB = 3$ . Since $AB$ is parallel to $CD$ , the angles at $D$ and $A$ add to $180^{\circ}$ , and $\angle DAB = 60^{\circ}$ .

Mark $E$ on $AB$ with $AE = 1$ . Then $AE$ and $DC$ are parallel and equal, so $AECD$ is a parallelogram, and since $DA = DC$ it is a rhombus. So $EC = 1$ , and $EC$ is parallel to $DA$ , which makes $\angle CEB = \angle DAB = 60^{\circ}$ . The rest of the long side is $EB = 2$ .

Let $M$ be the midpoint of $EB$ . Triangle $CEM$ has $EC = EM = 1$ and a $60^{\circ}$ angle between them, so it is equilateral and $MC = 1$ . Now $MC = MB = 1$ , and triangle $MCB$ is isosceles with apex angle $\angle CMB = 180^{\circ} - 60^{\circ} = 120^{\circ}$ , so its base angles are $30^{\circ}$ . In particular $\angle ABC = 30^{\circ}$ .

The angles at $B$ and $C$ also add to $180^{\circ}$ , so $\angle BCD = 150^{\circ}$ . The four angles are $60^{\circ}$ , $30^{\circ}$ , $150^{\circ}$ and $120^{\circ}$ , the largest is $x = 150$ and the smallest $y = 30$ , and $\frac{x}{y} = \frac{150}{30} = 5.$

Answer 5

Solution: PRMO 2012, Q10

Key idea

Both apexes sit on the vertical centre line, one at height $\sqrt3/2$ and the other the same distance below the top.

Put $A = (0,0)$ , $B = (1,0)$ , $C = (1,1)$ , $D = (0,1)$ .

The equilateral triangle on $AB$ with its apex inside has $Y$ directly above the midpoint of $AB$ , at height $\tfrac{\sqrt3}{2}$ : $Y = \left(\tfrac12,\ \tfrac{\sqrt3}{2}\right).$ The equilateral triangle on $CD$ with its apex inside has $X$ the same distance below $CD$ : $X = \left(\tfrac12,\ 1 - \tfrac{\sqrt3}{2}\right).$

Both lie on the vertical line $x = \tfrac12$ , so $XY = \frac{\sqrt3}{2} - \left(1 - \frac{\sqrt3}{2}\right) = \sqrt3 - 1.$

That the two apexes overlap vertically, rather than missing each other, is because $\tfrac{\sqrt3}{2} = 0.866$ exceeds $\tfrac12$ : each triangle reaches past the centre of the square.

Answer $\sqrt3-1$

Solution: PRMO 2013, Q9

Key idea

$\angle BIC = 90^{\circ} + \tfrac A2$ , while $\angle BHC$ is $A$ or $180^{\circ} - A$ according to which side of $BC$ the orthocentre lies on. Concyclicity asks for equal angles in the first case and supplementary ones in the second, and both give the same $A$ .

Start with the incentre, which is the easy one. It lies inside the triangle, so it is on the same

side of $BC$ as $A$ , and the angles at $B$ and $C$ inside triangle $BIC$ are halved: $\angle BIC = 180^{\circ} - \frac{B + C}{2} = 90^{\circ} + \frac{A}{2}.$

The orthocentre needs more care, because where it sits depends on the shape of the triangle. The line $BH$ is perpendicular to $CA$ and the line $CH$ is perpendicular to $AB$ , whatever the triangle looks like. Rotating both of a pair of lines through $90^{\circ}$ does not change the angle between them, so the angle between the lines $BH$ and $CH$ equals the angle between the lines $CA$ and $AB$ . Two lines make two angles, supplementary to each other, so $\angle BHC = A \qquad \text{or} \qquad \angle BHC = 180^{\circ} - A,$ and which one holds is decided by which side of $BC$ the point $H$ falls on. If $B$ or $C$ is a right angle then $H$ is that vertex itself and $B$ , $H$ , $I$ , $C$ are not four points at all, so that case is out; otherwise $H$ is off the line $BC$ .

Both cases give the same angle

Suppose first that $H$ is on the same side of $BC$ as $A$ , hence on the same side as $I$ . Then $H$ and $I$ lie on the same arc over the chord $BC$ , so they subtend equal angles there, and the angle at $H$ is the one measuring $180^{\circ} - A$ : $180^{\circ} - A = 90^{\circ} + \frac A2, \qquad \frac{3A}{2} = 90^{\circ}, \qquad A = 60^{\circ}.$

Suppose instead that $H$ is on the far side of $BC$ from $A$ . Then $BHCI$ is a cyclic quadrilateral with $H$ and $I$ at opposite vertices, so those two angles are supplementary, and now the angle at $H$ is the one measuring $A$ : $A + 90^{\circ} + \frac A2 = 180^{\circ}, \qquad \frac{3A}{2} = 90^{\circ}, \qquad A = 60^{\circ}.$

Both readings land on the same place, so the case split costs nothing. It is not idle, though: the second case really does occur. A triangle with $A = 60^{\circ}$ and $B = 100^{\circ}$ is obtuse at $B$ , its orthocentre falls beyond $BC$ , and $B$ , $H$ , $I$ , $C$ are concyclic all the same.

Finally the central angle over $BC$ is twice the inscribed one: $\angle BOC = 2A = 120^{\circ}.$

At $A = 60^{\circ}$ something pretty happens: $H$ , $I$ and $O$ all lie on one circle through $B$ and $C$ , namely the circle centred at the midpoint of arc $BC$ . When the triangle is acute this shows itself as $\angle BHC = \angle BIC = \angle BOC = 120^{\circ}$ , three equal angles on the same arc. When it is obtuse at $B$ or $C$ , the orthocentre changes sides and its angle drops to $60^{\circ}$ , which is exactly what a point on the other arc of the same circle should give.

Answer 120

Solution: PRMO 2013, Q12

Key idea

The area fixes both sides of the rectangle, and its height then fixes where the upper left corner $P$ meets $AB$ ; after that $PC$ is one application of Pythagoras.

Write $PS = w$ for the horizontal side, parallel to $BC$ , and $PQ = h = \sqrt3\,w$ for the vertical one. The area gives $wh = \sqrt3\,w^2 = 28\sqrt3, \qquad w^2 = 28, \qquad w = 2\sqrt7, \quad h = 2\sqrt{21}.$

Now place the triangle: $B = (0,0)$ , $C = (s, 0)$ and $A = \left(\tfrac s2, \tfrac{s\sqrt3}{2}\right)$ , where $s$ is the side. Then $Q = (q, 0)$ , $P = (q, h)$ , $S = (q + w, h)$ and $R = (q+w, 0)$ .

The side $AB$ runs from the origin with slope $\sqrt3$ , so $P$ on it means $h = \sqrt3 q$ , that is $q = 2\sqrt7$ . The side $AC$ is the line $y = \sqrt3(s - x)$ , so $S$ on it means $2\sqrt{21} = \sqrt3\left(s - 4\sqrt7\right), \qquad s - 4\sqrt7 = 2\sqrt7, \qquad s = 6\sqrt7.$

Finally, $P = \left(2\sqrt7,\ 2\sqrt{21}\right)$ and $C = \left(6\sqrt7,\ 0\right)$ , so $PC^2 = \left(4\sqrt7\right)^2 + \left(2\sqrt{21}\right)^2 = 112 + 84 = 196, \qquad PC = 14.$

Answer 14

Solution: PRMO 2013, Q19

Key idea

Triangle $AXY$ is similar to triangle $ACB$ , with ratio $AX/AC$ , so the area ratio $\tfrac5{18}$ says $(c/2b)^2 = \tfrac5{18}$ .

Write $b = AC$ and $c = AB$ , so $b^2 + 144 = c^2$ by Pythagoras at $C$ .

The quadrilateral $BXYC$ is what remains of the triangle after removing triangle $AXY$ , so $\frac{[AXY]}{[ABC]} = 1 - \frac{13}{18} = \frac{5}{18}.$

Now compare the two triangles. They share the angle at $A$ , and $\angle AXY = 90^{\circ} = \angle ACB$ , so triangle $AXY$ is similar to triangle $ACB$ , matching $X$ with $C$ and $Y$ with $B$ . The ratio of similarity is $AX/AC$ , and $X$ is the midpoint of $AB$ , so $AX = c/2$ . Hence $\left(\frac{c}{2b}\right)^2 = \frac{5}{18}, \qquad 18c^2 = 20 b^2, \qquad 18\left(b^2 + 144\right) = 20b^2.$ So $2b^2 = 2592$ , that is $b^2 = 1296$ and $AC = 36.$

Answer 36

Solution: PRMO 2015 Part A, Q11

Key idea

In a right triangle the inradius is $\tfrac12(\text{sum of legs} - \text{hypotenuse})$ , and the legs of the middle triangle are precisely the hypotenuses of the two outer ones, so those lengths cancel and the position of $P$ never has to be found.

For a right triangle with legs $p$ and $q$ and hypotenuse $h$ , the inradius is $r = \tfrac12(p + q - h)$ . This is standard, and it comes straight from equal tangent lengths: the two tangents from the right-angle vertex both have length $r$ , so the remaining pieces of the legs are $p - r$ and $q - r$ , and these are also the tangent lengths onto the hypotenuse, giving $(p-r) + (q-r) = h$ .

Each of the three triangles here has a right angle. Triangle $APB$ is right-angled at $A$ , triangle $CPD$ at $D$ , and triangle $BPC$ at $P$ by hypothesis. Applying the formula to each, $r_1 = \tfrac12(AP + AB - PB), \qquad r_2 = \tfrac12(PB + PC - BC),$ $r_3 = \tfrac12(PD + DC - PC).$ Adding the three, the terms $PB$ and $PC$ each appear once positively and once negatively, so they vanish without ever being computed: $r_1 + r_2 + r_3 = \tfrac12\bigl(AP + PD + AB + DC - BC\bigr).$ Since $P$ lies on $AD$ we have $AP + PD = AD = BC = 20$ , and $AB = DC = 8$ , so $r_1 + r_2 + r_3 = \tfrac12(20 + 8 + 8 - 20) = 8.$

It is worth seeing what has happened. The answer does not depend on where $P$ sits on $AD$ , only on the rectangle. The condition $\angle BPC = 90^{\circ}$ was needed to make the middle triangle right-angled so the formula applies, but the two possible positions of $P$ , namely $AP = 4$ and $AP = 16$ , give the same total.

Answer 8

Solution: PRMO 2017, Q13

Key idea

The midpoint of the hypotenuse is equally far from the three vertices of a right triangle. Here that makes $EAF$ isosceles, and the second perpendicularity relates its angles to the rectangle’s diagonals.

The condition $BF\perp AC$ makes triangle $AFB$ right-angled at $F$ . Its hypotenuse is $AB$ , whose midpoint is $E$ . The circle with diameter $AB$ passes through $F$ , so $EA=EF=EB.$ This gives an isosceles triangle before we do any length work.

The two diagonals of a rectangle make equal acute angles with its horizontal sides: right triangles $ABC$ and $BAD$ are congruent, with matching legs $AB$ and $BC=AD$ . Write $\theta=\angle CAB=\angle ABD$ for that common angle. Since $AF$ lies on $AC$ , $\angle EAF=\theta$ . The equal lengths $EA=EF$ give $\angle AFE=\theta$ , so $\angle AEF=180^\circ-2\theta, \qquad \angle FEB=2\theta,$ using the straight line $AEB$ .

On the other hand, $FE\perp BD$ and $EB$ lies on the horizontal side $AB$ , so $\angle FEB=90^\circ-\angle ABD=90^\circ-\theta.$ Equating the two expressions, $2\theta=90^\circ-\theta, \qquad \theta=30^\circ.$

Finally, in right triangle $ABC$ , $\frac{BC}{AB}=\tan30^\circ=\frac1{\sqrt3}, \qquad AB=\sqrt3\,BC=\sqrt3\cdot8\sqrt3=24.$

Answer 24

Solution: PRMO 2017, Q17

Key idea

The altitudes fix the shape, and Heron’s formula then fixes the size, but the semi-perimeter comes out as $60/\sqrt7$ , which is not the whole number the answer sheet demands.

This question was discounted by the organisers, and working it through shows why.

Altitudes and sides are inversely proportional, since twice the area equals each side times its altitude. So $a : b : c = \frac{1}{10} : \frac{1}{12} : \frac{1}{15} = 6 : 5 : 4,$ multiplying through by $60$ . Write $a = 6t$ , $b = 5t$ , $c = 4t$ .

Now match the two expressions for the area. On one hand, $\Delta = \tfrac12 \cdot a \cdot 10 = 30t.$ For the next area computation, here is the needed formula. If $a,b,c$ are a triangle’s sides, $\theta$ is the angle between $b$ and $c$ , and $s=(a+b+c)/2$ , the cosine rule gives $2bc\cos\theta=b^2+c^2-a^2$ . Hence $16\Delta^2=4b^2c^2\sin^2\theta=4b^2c^2-(b^2+c^2-a^2)^2.$ Factoring the difference of squares, $16\Delta^2=((b+c)^2-a^2)(a^2-(b-c)^2)$ $=(a+b+c)(b+c-a)(a+c-b)(a+b-c)=16s(s-a)(s-b)(s-c).$ Since area is positive, $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$ . This is Heron’s formula.

Here the semi-perimeter is $s=15t/2$ , so the second expression for the area is $\Delta = \sqrt{\frac{15t}{2}\cdot\frac{3t}{2}\cdot\frac{5t}{2}\cdot\frac{7t}{2}} = \frac{t^2}{4}\sqrt{1575} = \frac{15\sqrt7\,t^2}{4}.$ Equating the two, $30t = \tfrac{15\sqrt7}{4}t^2$ , so $t = \tfrac{8}{\sqrt7}$ and $s = \frac{15t}{2} = \frac{60}{\sqrt7} = \frac{60\sqrt7}{7} = 22.677\ldots$

The triangle exists and is perfectly ordinary; the trouble is that the paper asks for an answer between $0$ and $99$ to be written as an integer, and this semi-perimeter is irrational. The question needed altitudes chosen to make $\sqrt{1575}$ rational, and these are not they.

Answer none

Solution: PRMO 2017, Q25

Key idea

Writing the three given areas in coordinates and expanding the third gives $wh + ef = 100$ with $ef = 1600/wh$ , a quadratic in the area of the rectangle.

Put $A = (0,0)$ , $B = (w, 0)$ , $C = (w, h)$ , $D = (0, h)$ , and let $E = (e, h)$ on $CD$ and $F = (w, f)$ on $BC$ .

The three given areas are right triangles with legs along the sides: $[ADE] = \tfrac12 he = 16, \qquad [ABF] = \tfrac12 wf = 25,$ $[CEF] = \tfrac12 (w-e)(h-f) = 9.$ So $he = 32$ , $wf = 50$ and $(w-e)(h-f) = 18$ .

Expand the third: $wh - wf - eh + ef = 18, \qquad\text{so}\qquad wh + ef = 100,$ once $wf = 50$ and $eh = 32$ are put in. And $ef$ is determined by the other two, since $e = 32/h$ and $f = 50/w$ give $ef = 1600/(wh)$ . Writing $S = wh$ for the area of the rectangle, $S + \frac{1600}{S} = 100, \qquad S^2 - 100S + 1600 = 0, \qquad S = 80 \ \text{ or } \ 20.$ The three triangles already occupy $16 + 25 + 9 = 50$ , so $S = 20$ is impossible and $S = 80$ .

The fourth piece of the rectangle is the triangle $AEF$ , so $[AEF] = 80 - 50 = 30.$

Answer 30

Solution: PRMO 2018, Q10

Key idea

The centroid cuts each median in the ratio $2 : 1$ , so perpendicular medians from $B$ and $C$ give $\tfrac49(m_b^2 + m_c^2) = a^2$ , which is the relation $b^2 + c^2 = 5a^2$ .

the right angle at G makes BGC a right triangle, so \tfrac49\left(m_b^2 + m_c^2\right) = a^2 — the right angle at $G$ makes $BGC$ a right triangle,
so $\tfrac49\left(m_b^2 + m_c^2\right) = a^2$

Write $a=BC$ , $b=CA$ and $c=AB$ , and let $m_a$ , $m_b$ and $m_c$ be the lengths of the medians from $A$ , $B$ and $C$ . Let $G$ be the centroid. It lies on both medians and divides each in the ratio $2:1$ from the vertex, so $BG = \tfrac23 m_b$ and $CG = \tfrac23 m_c$ . The two medians are perpendicular at $G$ , so triangle $BGC$ has a right angle there and $BG^2 + CG^2 = BC^2 = a^2, \qquad \text{that is} \qquad \tfrac49\left(m_b^2 + m_c^2\right) = a^2.$

The median formula comes from adding the cosine rule in the two triangles on either side of a midpoint. Let $E$ be the midpoint of $CA$ . Since $EA=EC=b/2$ and the two angles at $E$ are supplementary, $a^2=m_b^2+\frac{b^2}{4}-bm_b\cos\angle BEC, \qquad c^2=m_b^2+\frac{b^2}{4}+bm_b\cos\angle BEC.$ Adding cancels the angle and gives $4m_b^2=2a^2+2c^2-b^2$ . Applying the same argument to the median from $C$ gives $4m_c^2 = 2a^2 + 2b^2 - c^2$ . Adding, $4\left(m_b^2 + m_c^2\right) = 4a^2 + b^2 + c^2,$ so the perpendicularity condition reads $\tfrac19(4a^2 + b^2 + c^2) = a^2$ , that is $b^2 + c^2 = 5a^2.$

The third median now follows at once. From $4m_a^2 = 2b^2 + 2c^2 - a^2 = 10a^2 - a^2 = 9a^2$ we get $m_a = \tfrac32 a$ , and $m_a = 30$ gives $a = 20$ . Hence $a^2 = 400$ , $b^2 + c^2 = 2000$ , and $\frac{a^2 + b^2 + c^2}{100} = \frac{2400}{100} = 24.$

Answer 24

Solution: PRMO 2018, Q17

Key idea

Both third sides satisfy the same quadratic from the cosine rule, so they are its two roots, and Vieta gives their product without ever finding the angle.

Write $x = AC$ and $y = DF$ . The cosine rule in each triangle, with the equal angle $\theta = \angle A = \angle D$ , gives $10^2 = 17^2 + x^2 - 2 \cdot 17 x\cos\theta, \qquad 10^2 = 17^2 + y^2 - 2 \cdot 17 y\cos\theta.$ So $x$ and $y$ are both roots of the same quadratic $t^2 - (34\cos\theta)\,t + 189 = 0,$ where $189 = 17^2 - 10^2$ . Since $x - y = 12$ they are different numbers, so they are the two roots, and Vieta gives $xy = 189.$

Now one identity finishes it: $(x+y)^2 = (x-y)^2 + 4xy = 144 + 756 = 900,$ so $AC + DF = x + y = 30$ .

The angle never had to be found, and indeed it is determined only afterwards, by $\cos\theta = 30/34$ . Two triangles with two sides and a non-included angle equal are the classic ambiguous case, and this problem is that ambiguity turned into an exercise.

Answer 30

Solution: PRMO 2018, Q21

Key idea

Averaging the three points of each small triangle makes the differences $G_1 - G_2$ exactly one third of $B - A$ , so the two triangles are similar in the ratio $1 : 3$ and $H$ drops out entirely.

Treat the points as position vectors. Nothing here needs any theory of vectors: reading every letter below as a pair of coordinates and doing the arithmetic separately on the two coordinates gives exactly the same argument twice over.

The centroid of a triangle is the average of its vertices, so $G_1 = \frac{H + B + C}{3}, \qquad G_2 = \frac{H + C + A}{3}, \qquad G_3 = \frac{H + A + B}{3}.$ Subtracting in pairs, the $H$ and the shared vertex cancel: $G_1 - G_2 = \frac{B - A}{3}, \qquad G_2 - G_3 = \frac{C - B}{3}, \qquad G_3 - G_1 = \frac{A - C}{3}.$

So the triangle $G_1G_2G_3$ has every side one third of the corresponding side of $ABC$ , and is therefore similar to it with ratio $\tfrac13$ . Areas scale by the square, $[ABC] = 9\,[G_1G_2G_3] = 9 \times 7 = 63.$

Notice what was never used: that $H$ is the orthocentre, or that the triangle is acute. Any point $H$ whatever gives the same three centroids’ triangle, since $H$ cancels in every difference. The orthocentre is there to make the configuration concrete, not because the answer depends on it.

Answer 63

Solution: PRMO 2019, Q10

Key idea

Bisecting the angles and re-inscribing sends each angle $A$ to $90^{\circ} - A/2$ , so doing it twice sends $A$ to $45^{\circ} + A/4$ , an increasing map that carries the smallest angle to the smallest angle.

First find how the angles of $A_1B_1C_1$ relate to those of $ABC$ . The bisector from $A$ meets $\Omega$ again at the midpoint $A_1$ of the arc $BC$ not containing $A$ , and similarly for $B_1$ and $C_1$ . Write the arcs of $\Omega$ subtended by the sides as usual: arc $BC$ not containing $A$ measures $2A$ , arc $CA$ not containing $B$ measures $2B$ , and arc $AB$ not containing $C$ measures $2C$ .

The inscribed angle of triangle $A_1B_1C_1$ at $A_1$ subtends the arc from $B_1$ to $C_1$ that does not contain $A_1$ , and that arc passes through $A$ . It is made of half of arc $CA$ , namely $B$ , together with half of arc $AB$ , namely $C$ . So the arc measures $B + C$ and the inscribed angle is half of it: $\angle A_1 = \frac{B+C}{2} = \frac{180^{\circ} - A}{2} = 90^{\circ} - \frac{A}{2}.$

Applying the same formula a second time to the triangle $A_1B_1C_1$ , $\angle A_2 = 90^{\circ} - \frac{\angle A_1}{2} = 90^{\circ} - \frac{90^{\circ} - A/2}{2} = 45^{\circ} + \frac{A}{4}.$

Now the bookkeeping. One application reverses the order of the angles, since $A \mapsto 90^{\circ} - A/2$ is decreasing, but two applications restore it: $A \mapsto 45^{\circ} + A/4$ increases with $A$ . So the smallest angle of $A_2B_2C_2$ comes from the smallest angle of $ABC$ , and with that angle equal to $40^{\circ}$ , $45^{\circ} + \frac{40^{\circ}}{4} = 55^{\circ}.$

The map $A \mapsto 45^{\circ} + A/4$ also explains what repeated bisection does in the long run: every angle is dragged towards the fixed point $60^{\circ}$ , four times closer at each step, so the triangles rapidly become equilateral.

Answer 55

Solution: IOQM 2020, Q7

Key idea

The perpendicular $CE$ and the altitude from the apex create two right triangles that share the angle at $D$ , and the similarity between them converts the unknown length $AD$ into a product of two lengths already in hand.

$both shaded triangles are right-angled and share the angle at D , so DE \cdot DA = DM \cdot DC$

both shaded triangles are right-angled and share the angle at

D

,
so

DE \cdot DA = DM \cdot DC

The given number $48\tfrac{1}{61}$ is so deliberately ugly that it is worth treating as a promise: whoever set the problem intended it to cancel, and our job is to find the relation in which it does.

Begin by letting $M$ be the midpoint of $BC$ . Because the triangle is isosceles with $AB = AC$ , the segment $AM$ is the altitude from $A$ , so the angle at $M$ in triangle $AMD$ is a right angle. We now have two right triangles in the figure, $DEC$ with its right angle at $E$ , since $CE \perp AD$ , and $DMA$ with its right angle at $M$ . They share the angle at $D$ , which makes them similar, and similarity gives $\frac{DE}{DM} = \frac{DC}{DA}, \qquad \text{or equivalently} \qquad DE \cdot DA = DM \cdot DC.$ Every quantity in that relation except $DA$ is either given or computable, which is exactly what we want.

Computing $DM$ is where the awkward fraction earns its place. Writing the given length over a common denominator, $BD = 48\tfrac{1}{61} = \tfrac{2929}{61}$ , and since $DC = 61 = \tfrac{3721}{61}$ , $BC = \frac{2929 + 3721}{61} = \frac{6650}{61}, \qquad BM = \frac{3325}{61},$ $DM = BM - BD = \frac{3325 - 2929}{61} = \frac{396}{61}.$

The promised cancellation now arrives, since $DM \cdot DC = \tfrac{396}{61} \cdot 61 = 396$ . With $DE = 11$ we get $DA = 396/11 = 36$ , and because $E$ lies between $D$ and $A$ on the segment, $AE = DA - DE = 36 - 11 = 25.$ No trigonometry was needed anywhere, and the fraction that looked so hostile was the friendliest thing in the problem.

Answer 25

Solution: IOQM 2020, Q9

Key idea

The two parallels make $DMCN$ a parallelogram, but the angle bisector makes it something stronger, a rhombus, and the length asked for is its second diagonal.

The two parallel conditions are there to build a quadrilateral, so let us build it. Since $DM \parallel AC$ and $DN \parallel BC$ , the quadrilateral $DMCN$ has both pairs of opposite sides parallel and is therefore a parallelogram. That alone would not be enough to finish, and the extra ingredient is the bisector: $CD$ is a diagonal of this parallelogram, and it bisects the angle at $C$ . Here $DM\parallel AC$ gives $\angle MDC=\angle DCA=\angle DCM$ , so triangle $DMC$ is isosceles and $DM=CM$ . Opposite sides of a parallelogram are equal, so all four sides are equal, $CM = CN = DM = DN.$

Finding that common side is now a short piece of standard work. The angle bisector theorem in triangle $ABC$ gives $\dfrac{AD}{DB} = \dfrac{CA}{CB} = \dfrac{4}{6}$ , so on the side $AB = 5$ we have $AD = 2$ and $DB = 3$ . Because $DM \parallel AC$ , the triangle $BDM$ is similar to the triangle $BAC$ , and therefore $\frac{BM}{BC} = \frac{BD}{BA} = \frac{3}{5}, \qquad BM = \frac{18}{5}, \qquad CM = 6 - \frac{18}{5} = \frac{12}{5}.$

What remains is the diagonal $MN$ of the rhombus, which is the third side of the triangle $MCN$ whose two equal sides are $CM = CN = \tfrac{12}{5}$ and whose included angle is $\angle ACB$ . The cosine rule in the original triangle supplies that angle, $\cos C = \frac{6^2 + 4^2 - 5^2}{2 \cdot 6 \cdot 4} = \frac{27}{48} = \frac{9}{16},$ and applying the cosine rule once more, now in triangle $MCN$ , $MN^2 = 2\left(\frac{12}{5}\right)^2 (1 - \cos C) = \frac{288}{25} \cdot \frac{7}{16} = \frac{126}{25}.$

Since $126$ and $25$ share no factor, we have $p = 126$ and $q = 25$ , so $|p - q| = 101$ and the sum of its digits is $1 + 0 + 1 = 2$ .

Answer 2

Solution: IOQM 2020, Q19

Key idea

For any point inside a parallelogram, the two triangles on opposite sides have areas adding to half the whole, so the four areas fall into two pairs each summing to fifty.

Start with the structural fact, because it organises everything that follows. If $P$ is any point inside the parallelogram and $h$ is its distance from $AB$ , then its distance from the opposite side $CD$ is $H - h$ , where $H$ is the distance between those two parallel sides. Hence $[APB] + [CPD] = \tfrac12 \cdot AB \cdot h + \tfrac12 \cdot CD \cdot (H-h) = \tfrac12 \cdot AB \cdot H,$ using $AB = CD$ , and that is half the parallelogram, namely $50$ . The same argument applied to the other pair of sides gives $[BPC] + [DPA] = 50$ . So the four areas split into two pairs, each pair totalling $50$ , and we only need to locate $P$ well enough to split each pair.

To locate $P$ , use the two sides at $A$ as a frame: take $A$ as origin and write each point as $\lambda\,\overrightarrow{AB} + \mu\,\overrightarrow{AD}$ , recorded as $(\lambda, \mu)$ . A point with second coordinate $\mu$ lies at the fraction $\mu$ of the way from $AB$ to $CD$ , so $[APB] = \tfrac12 \cdot AB \cdot \mu H = \mu \cdot \tfrac12[ABCD]$ ; in the same way $[DPA] = \lambda \cdot \tfrac12 [ABCD]$ . In this frame $B = (1,0)$ , $C = (1,1)$ , $D = (0,1)$ , and the midpoints are $E = \left(\tfrac12, 0\right)$ and $F = \left(1, \tfrac12\right)$ , exactly as if the parallelogram were a unit square. Parametrising the line $EC$ as $\left(\tfrac12 + \tfrac{t}{2},\, t\right)$ and the line $FD$ as $(1-s,\, \tfrac12 + \tfrac{s}{2})$ , equating the two coordinates gives $\frac12+\frac t2=1-s, \qquad t=\frac12+\frac s2.$ The first says $t=1-2s$ . Substituting into the second, $1-2s=\frac12+\frac s2, \qquad \frac52s=\frac12, \qquad s=\frac15, \qquad t=\frac35.$ Hence $P = \left(\tfrac45,\ \tfrac35\right).$

With $\lambda = \tfrac45$ and $\mu = \tfrac35$ , the area $[APB]$ is $\tfrac{3}{10}$ of the parallelogram and $[DPA]$ is $\tfrac{4}{10}$ , and the pairs found at the start supply the other two: $[CPD] = \tfrac12 - \tfrac{3}{10} = \tfrac{2}{10}$ and $[BPC] = \tfrac12 - \tfrac{4}{10} = \tfrac{1}{10}$ . For a parallelogram of area $100$ the four triangles have areas $30$ , $10$ , $20$ and $40$ , so the largest is $40$ .

Answer 40

Solution: IOQM 2021 Part A, Q1

Key idea

Write the side as a vector rather than an angle. Turning that vector through a right angle to reach the next vertex converts the two given distances into the two components, and the area is the sum of their squares.

Put $A$ at the origin with $L_1$ as the $x$ -axis, so that $L_2$ is the line $y = 3$ and $L_3$ is the line $y = 6$ . Writing the side as a vector keeps the work to two coordinates, with no angle anywhere and no trigonometry.

Write the side $AB$ as the vector $(p, 6)$ . The second coordinate is forced, because $B$ lies on $L_3$ and $A$ is on $L_1$ , so $B$ is six units higher. After reflecting the figure in the vertical axis if necessary, take the turn from $AB$ to $BC$ anticlockwise. Reflection preserves lengths and area. To get from $B$ to $C$ we turn that vector through a right angle, and a quarter turn sends $(p, 6)$ to $(-6, p)$ . Hence $C = B + (-6,\, p) = (p - 6,\; 6 + p).$

Now impose the one condition not yet used, that $C$ lies on $L_2$ . That says $6 + p = 3$ , so $p = -3$ , and no angle was ever needed. The area of the square is the squared length of its side, $AB^2 = p^2 + 6^2 = 9 + 36 = 45.$

It is worth noticing that the answer depends only on $p^2$ , so the sign of $p$ , which is to say which way the square leans, makes no difference.

Answer 45

Solution: IOQM 2021 Part A, Q5

Key idea

In any parallelogram the four internal bisectors meet at right angles, so $XYZW$ is a rectangle, and its diagonal is the difference of the two side lengths.

Adjacent angles of a parallelogram add to $180^{\circ}$ , so their halves add to $90^{\circ}$ . In the triangle cut off at a corner by two adjacent bisectors, the third angle is therefore a right angle, and since this happens at all four corners the quadrilateral $XYZW$ is a rectangle.

Let $AB = 2s$ and $AD = s$ , with $\angle DAB = \theta$ . As in the figure, $Y$ is where the bisectors from $A$ and $B$ meet, $X$ where those from $A$ and $D$ meet, and $Z$ where those from $B$ and $C$ meet; the angle at each is a right angle, by the paragraph above. Both $Y$ and $X$ lie on the bisector from $A$ , which makes the angle $\tfrac{\theta}{2}$ with $AB$ and with $AD$ . In the right triangle $AYB$ this gives $AY = 2s\cos\tfrac{\theta}{2}$ , and in the right triangle $AXD$ it gives $AX = s\cos\tfrac{\theta}{2}$ , so $XY = AY - AX = (2s - s)\cos\tfrac{\theta}{2} = s\cos\tfrac{\theta}{2}.$ In the same way $Y$ and $Z$ lie on the bisector from $B$ . The right triangle $AYB$ gives $BY = 2s\sin\tfrac{\theta}{2}$ , and in the right triangle $BZC$ the angle at $B$ is $90^{\circ} - \tfrac{\theta}{2}$ , so $BZ = s\sin\tfrac{\theta}{2}$ and $YZ = BY - BZ = (2s - s)\sin\tfrac{\theta}{2} = s\sin\tfrac{\theta}{2}.$ So the rectangle’s area is $s^2 \sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2} = \tfrac12 s^2 \sin\theta$ . The parallelogram itself has area $2s^2 \sin\theta$ .

The ratio is therefore $\frac{[ABCD]}{[XYZW]} = \frac{2s^2 \sin\theta}{\tfrac12 s^2 \sin\theta} = 4,$ independent of both the size and the angle, which is why the problem could give a bare number. With $[XYZW] = 10$ we get $[ABCD] = 40$ .

Answer 40

Solution: IOQM 2022, Q1

Key idea

The distance from a vertex to the tangent at another vertex is that chord squared divided by twice the circumradius. Both distances then carry a factor of $R$ , and it cancels out of the area formula entirely.

The key is a single fact about a tangent, and it is worth deriving rather than quoting. Let the tangent at $B$ be $t$ , and drop a perpendicular from $A$ to $t$ , of length $d_A$ . If $\theta$ is the angle between the tangent $t$ and the chord $BA$ , then $d_A = AB\sin\theta$ . But the tangent-chord angle equals the inscribed angle in the alternate segment, which is $\angle ACB$ , and the extended sine rule gives $AB = 2R\sin(\angle ACB) = 2R\sin\theta$ . Substituting $\sin\theta = AB/(2R)$ , $d_A = \frac{AB^2}{2R}, \qquad \text{and likewise} \qquad d_C = \frac{CB^2}{2R}.$

Now put in the numbers. From $d_A = 25$ we get $AB^2 = 50R$ , and from $d_C = 16$ we get $CB^2 = 32R$ . Multiplying, $AB \cdot CB = \sqrt{50R \cdot 32R} = \sqrt{1600R^2} = 40R.$ That product is linear in $R$ , which is exactly what the area formula wants.

The area of a triangle in terms of its sides and circumradius is $S = \dfrac{abc}{4R}$ , so $S = \frac{AB \cdot CB \cdot CA}{4R} = \frac{40R \cdot 20}{4R} = 200,$ and $R$ has vanished, meaning the area is the same for every circle in which this configuration can be drawn. Hence $S/20 = 10$ , and the largest integer not exceeding it is $10$ .

Answer 10

Solution: IOQM 2022, Q2

Key idea

Use $AB$ and $AD$ as a coordinate frame. Points of the diagonal have equal coordinates, and points of the line $PQ$ satisfy an intercept equation, so the two conditions meet in one line of algebra.

Take $A$ as origin and use the two sides as basis vectors, writing a point as $\lambda \, \overrightarrow{AB} + \mu \, \overrightarrow{AD}$ . This is legitimate because the two sides of a parallelogram are independent directions, and it makes both conditions easy to state.

The diagonal $AC$ runs to $\overrightarrow{AB} + \overrightarrow{AD}$ , so every point of it has $\lambda = \mu$ . Write the common value at $T$ as $\tau$ , so that $AT / AC = \tau$ and the answer we want is $1/\tau$ .

The line $PQ$ joins $P = \tfrac{61}{2022}\overrightarrow{AB}$ to $Q = \tfrac{61}{2065}\overrightarrow{AD}$ . A line meeting the axes of a frame at those two points can be written as $(\lambda,\mu)=((1-t)61/2022,t61/2065)$ . Dividing the two coordinates by their intercepts and adding eliminates $t$ , giving the equation $\frac{\lambda}{61/2022} + \frac{\mu}{61/2065} = 1.$ Putting $\lambda = \mu = \tau$ and factoring, $\tau\left(\frac{2022}{61} + \frac{2065}{61}\right) = 1, \qquad \tau \cdot \frac{4087}{61} = 1, \qquad \tau = \frac{61}{4087}.$

Therefore $\dfrac{AC}{AT} = \dfrac{4087}{61} = 67$ exactly, since $61 \times 67 = 4087$ . The question asked for the nearest integer as a courtesy; the value is a whole number on the nose, which is the usual sign that the ugly-looking $2022$ and $2065$ were chosen for it.

Answer 67

Solution: IOQM 2022, Q9

Key idea

The triangle inequality gives $y$ a window of width $2\min(18,x)$ , so the count $35$ pins down the minimum rather than $x$ itself.

For $18$ , $x$ and $y$ to form a non-degenerate triangle we need $|18 - x| < y < 18 + x$ . The number of integers strictly between those bounds is $(18 + x - 1) - (|18 - x| + 1) + 1 = 18 + x - |18-x| - 1 = 2\min(18, x) - 1,$ where the last step uses $t + s - |t - s| = 2\min(t,s)$ .

Setting this equal to $35$ gives $\min(18, x) = 18$ , which says nothing about $x$ except that $x \geq 18$ . That is the point of the problem: the count saturates once $x$ passes $18$ , so a whole range of $x$ produces exactly $35$ values of $y$ .

With the given restriction $x < 100$ , the admissible values are $x = 18, 19, \ldots, 99$ , and there are $99 - 18 + 1 = 82$ of them.

Answer 82

Solution: IOQM 2024, Q12

Key idea

Coordinates turn the two lines into $y = \tfrac32 x$ and $y = -\tfrac32(x-16)$ , whose crossing height is all the area formula needs.

Put $A = (0,0)$ , $B = (16,0)$ , $C = (16,16)$ and $D = (0,16)$ , so that $CD$ is the top side. The points $E$ and $F$ cut it into three equal parts, with $E$ nearer $C$ : $E = \left(\tfrac{32}{3},\, 16\right), \qquad F = \left(\tfrac{16}{3},\, 16\right).$

The line $AE$ passes through the origin with slope $16 \div \tfrac{32}{3} = \tfrac32$ , so it is $y = \tfrac32 x$ . The line $BF$ passes through $(16,0)$ with slope $\frac{16 - 0}{\tfrac{16}{3} - 16} = \frac{16}{-\tfrac{32}{3}} = -\frac32,$ so it is $y = -\tfrac32(x - 16)$ . Setting the two equal, $\tfrac32 x = -\tfrac32 x + 24$ , so $x = 8$ and $y = 12$ .

The triangle $MAB$ has the side $AB$ of length $16$ lying along the $x$ -axis, and $M$ is at height $12$ above it, so $[MAB] = \tfrac12 \cdot 16 \cdot 12 = 96.$

The symmetry of the picture is worth noticing: $E$ and $F$ are placed symmetrically about the vertical midline, so $M$ had to land on that midline, and only its height was ever in question.

Answer 96

Solution: IOQM 2025 Part SEP, Q28

Key idea

The midpoints of the four sides form a parallelogram whose sides are parallel to the diagonals, and the two segments in the question are its diagonals, so they are equal exactly when that parallelogram is a rectangle, which happens exactly when the diagonals of $ABCD$ are perpendicular.

PQ \parallel AC and QR \parallel BD , each half as long — $PQ \parallel AC$ and $QR \parallel BD$ , each half as long

Let $P, Q, R, S$ be the midpoints of $AB, BC, CD, DA$ . In triangle $ABC$ the segment $PQ$ joins the midpoints of two sides, so $PQ$ is parallel to $AC$ and half its length; the same argument in triangle $ACD$ gives $SR$ parallel to $AC$ and half its length. So $PQRS$ is a parallelogram, with one pair of sides parallel to $AC$ and of length $6$ , the other pair parallel to $BD$ and of length $8$ .

The two segments joining midpoints of opposite sides of $ABCD$ are $PR$ and $QS$ , which are precisely the diagonals of this parallelogram. A parallelogram has equal diagonals exactly when it is a rectangle, and $PQRS$ is a rectangle exactly when its two side directions, namely those of $AC$ and $BD$ , are perpendicular.

For a quadrilateral with perpendicular diagonals the area is half the product of the diagonals, since each of the four small triangles has one leg along each diagonal. Hence $[ABCD] = \tfrac12 \cdot 12 \cdot 16 = 96.$ The condition forces this value rather than merely bounding it, so $96$ is both the maximum and the only possibility.

Answer 96

Solution: IOQM 2026, Q13

Key idea

The work is in the $15^{\circ}$ angle at $A$ . One fold turns it into a $30^{\circ}$ - $60^{\circ}$ - $90^{\circ}$ triangle and gives $CD$ exactly, and then $BD^2$ turns out to be four times the area of $ADC$ .

Let $CA = CB = a$ . Triangle $ACD$ has a right angle at $C$ and $\angle ADC = 75^{\circ}$ , so $\angle CAD = 15^{\circ}$ .

To find $CD$ without tables, mark $F$ on $AC$ with $\angle FDA = 15^{\circ}$ . Triangle $AFD$ has two angles of $15^{\circ}$ , so $FA = FD$ . Its exterior angle at $F$ is $\angle DFC = 15^{\circ} + 15^{\circ} = 30^{\circ}$ , so $FCD$ is a $30^{\circ}$ - $60^{\circ}$ - $90^{\circ}$ triangle, with $FD = 2\,CD$ and $FC = \sqrt3\,CD$ . Hence $a = AF + FC = (2 + \sqrt3)\,CD, \qquad CD = \frac{a}{2 + \sqrt3} = (2 - \sqrt3)\,a.$

The area of $ADC$ is $\tfrac12 \cdot a \cdot CD = \tfrac12 (2 - \sqrt3)a^2 = 81$ , so $(2 - \sqrt3)a^2 = 162$ . And $BD = a - CD = (\sqrt3 - 1)a$ , so $BD^2 = (4 - 2\sqrt3)\,a^2 = 2(2 - \sqrt3)\,a^2 = 324,$ and $BD = 18$ .

Answer 18

Solution: IOQM 2026, Q24

Key idea

The foot of the perpendicular from $C$ to $BD$ is the image of $E$ under the half-turn about the centre, so $CE$ is the hypotenuse of a right triangle with legs $12$ and the gap between the two feet. That gives $BD$ , and the area is $BD \times AE$ .

This question was discounted by the organisers, who gave every candidate full marks. Its answer is $300$ , which does not fit the two-digit answer sheet: the words asking for the sum of the digits of the area were lost in review. The mathematics is worth doing all the same.

Let $BE = p$ and $ED = q$ . The angle at $A$ is a right angle and $AE$ is the altitude to the hypotenuse $BD$ , so triangles $ABE$ and $DAE$ are similar and $BE/AE = AE/ED$ , that is $pq = AE^2 = 144.$

A half-turn about the centre of the rectangle swaps $A$ with $C$ and $B$ with $D$ . It carries $E$ to the foot $F$ of the perpendicular from $C$ to $BD$ , so $CF = 12$ and $DF = p$ , which puts $F$ at distance $q$ from $B$ . So $EF = |p - q|$ , and in the right triangle $CFE$ , $193 = CE^2 = 12^2 + (p - q)^2, \qquad (p - q)^2 = 49.$ Then $(p + q)^2 = (p - q)^2 + 4pq = 49 + 576 = 625$ , so $BD = 25$ , and $\text{area} = 2 \times \tfrac12 \cdot BD \cdot AE = 25 \times 12 = 300.$ The rectangle is $20$ by $15$ . The intended answer, the sum of the digits of $300$ , would have been $3$ .

Answer none

Solution: PRMO 2012, Q14

Key idea

$\angle BOC = 2A$ and $\angle BIC = 90^{\circ} + \tfrac A2$ , and the two are equal exactly when $B$ , $O$ , $I$ , $C$ are concyclic.

Both $O$ and $I$ lie inside the triangle, hence on the same side of $BC$ as $A$ . So if they lie on a common circle through $B$ and $C$ , they subtend equal angles on the chord $BC$ : $\angle BOC = \angle BIC.$

The central angle is twice the inscribed one, $\angle BOC = 2A$ , and the incentre satisfies $\angle BIC = 180^{\circ} - \frac{B + C}{2} = 90^{\circ} + \frac A2.$ Setting them equal, $2A = 90^{\circ} + \frac A2, \qquad \frac{3A}{2} = 90^{\circ}, \qquad A = 60^{\circ}.$

At $A = 60^{\circ}$ both angles are $120^{\circ}$ , and one may check that the triangle is then acute for a range of shapes, so $O$ really can lie inside as the question requires.

Answer 60

Solution: PRMO 2013, Q15

Key idea

Halving twice: the sides of $A_1B_1C_1D_1$ are half the diagonals of $ABCD$ , and the sides of $A_2B_2C_2D_2$ are half the diagonals of $A_1B_1C_1D_1$ .

The midpoint quadrilateral of any quadrilateral is a parallelogram whose sides are parallel to the diagonals of that quadrilateral and half as long: that is Varignon’s theorem, and it follows from the midline of each of the four triangles cut off by a diagonal.

There are three quadrilaterals here and it pays to keep their diagonals apart. Applying Varignon to $ABCD$ , the sides of $A_1B_1C_1D_1$ are half the diagonals of $ABCD$ , namely $\frac{AC}{2} \quad \text{and} \quad \frac{BD}{2}.$ Applying it again to $A_1B_1C_1D_1$ , the sides of $A_2B_2C_2D_2$ , which are the given $4$ and $6$ , are half the diagonals of $A_1B_1C_1D_1$ . So the diagonals of $A_1B_1C_1D_1$ have lengths $8$ and $12$ . The diagonals of $A_2B_2C_2D_2$ never enter the argument.

Now use the shape. A parallelogram has perpendicular diagonals exactly when it is a rhombus, and $A_2B_2C_2D_2$ is a rectangle exactly when the diagonals of $A_1B_1C_1D_1$ are perpendicular. So $A_1B_1C_1D_1$ is a rhombus, that is $\frac{AC}{2} = \frac{BD}{2} = t \quad \text{for some } t.$

The diagonals of this rhombus are perpendicular and bisect each other, so one of the four right triangles has legs $8/2=4$ and $12/2=6$ and hypotenuse $t$ . Pythagoras gives $t^2=4^2+6^2=52, \qquad 4t^2=208.$ Since $AC = BD = 2t$ , the product of the diagonals of $ABCD$ is $AC \cdot BD = 4t^2 = 208.$

The answer is the sum of the squares of the two given sides, doubled twice and halved twice, and it is no coincidence that it equals $8^2 + 12^2$ exactly.

Answer 208

Solution: PRMO 2014, Q12

Key idea

In the first triangle the distance from $D$ to the touch point is $AD - r_1$ , and in the second the touch point is at distance $r_2$ from $D$ , so $PQ = AD - r_1 - r_2$ .

Write $r_1$ and $r_2$ for the two inradii and $m = BD$ .

The first triangle

Triangle $ABD$ has its right angle at $A$ , so $m$ is its hypotenuse and $r_1 = \frac{AB + AD - m}{2}.$ The tangent length from $D$ is the semi-perimeter minus the opposite side $AB$ : $DP = \frac{AB + AD + m}{2} - AB = \frac{AD + m - AB}{2}.$ Adding the two, everything cancels but $AD$ : $DP + r_1 = AD.$

The second triangle

Triangle $BDC$ has its right angle at $D$ , and the two tangent lengths from the right-angle vertex of any triangle both equal the inradius. So $DQ = r_2.$

Putting them together

Along $BD$ the order is $B$ , $P$ , $Q$ , $D$ , so $PQ = DP - DQ = (AD - r_1) - r_2.$ Hence $r_1 + r_2 = AD - PQ = 999 - 200 = 799.$

Notice how little was used: not the lengths $AB$ or $DC$ , not $BD$ , only the two right angles and the two given numbers. The configuration has a free parameter, and the sum of the radii does not feel it.

Answer 799

Solution: PRMO 2014, Q16

Key idea

$P$ , $Q$ , $R$ lie on the incircle along the directions $IA$ , $IB$ , $IC$ , so the arc $QR$ subtends the central angle $\angle BIC = 90^{\circ} + \tfrac12 \angle A$ , and the inscribed angle at $P$ is half of it.

One point about the reading first. The paper says the lines $AI$ , $BI$ , $CI$ meet the incircle, and each of those lines meets it twice, once on the side of $I$ towards the vertex and once on the opposite side. Throughout we take $P$ , $Q$ , $R$ to be the intersections lying on the rays from $I$ towards $A$ , $B$ , $C$ respectively, which is the reading the official answer uses. It matters: the other choice for $P$ puts it on the short arc $QR$ instead of the long one and turns the answer into $125^{\circ}$ .

The three points lie on the incircle, which is centred at $I$ , so $I$ is the centre of the circle through $P$ , $Q$ , $R$ and the angle $\angle QPR$ is an inscribed angle in it.

The central angle standing on the same arc $QR$ is the angle between the radii $IQ$ and $IR$ , that is the angle between the lines $IB$ and $IC$ : $\angle BIC = 180^{\circ} - \frac{\angle B}{2} - \frac{\angle C}{2} = 180^{\circ} - \frac{180^{\circ} - \angle A}{2} = 90^{\circ} + \frac{\angle A}{2}.$ With $\angle A = 40^{\circ}$ this is $110^{\circ}$ .

The point $P$ lies in the direction of $IA$ , which is on the other side of the circle from the arc $QR$ just measured, so the inscribed angle theorem gives $\angle QPR = \frac{110^{\circ}}{2} = 55^{\circ}.$

Only $\angle A$ mattered. The other two angles fix where $Q$ and $R$ sit individually, but not the arc between them.

Answer 55

Solution: PRMO 2015 Part A, Q16

Key idea

$F$ is the midpoint of the hypotenuse of right triangle $ADC$ , so $FA = FD$ and $\angle ADF = \angle DAC$ ; meanwhile the given equality of angles says exactly $\angle DAE = |\angle B - \angle C|$ , which is what turns $\angle BAE$ into $\angle DAC$ .

Write $B$ and $C$ for the angles of the triangle at those vertices, and note first that $AD$ being an altitude makes $\angle ADB = \angle ADC = 90^{\circ}$ . In right triangle $ABD$ this gives $\angle BAD = 90^{\circ} - B$ , and in right triangle $ACD$ it gives $\angle DAC = 90^{\circ} - C$ . Since the triangle is acute, $D$ lies strictly between $B$ and $C$ ; first consider $B>C$ , which puts $D$ between $B$ and the midpoint $E$ . The other orders are treated below.

The quantity we are asked for is easy to convert. In the right triangle $ADC$ the point $F$ is the midpoint of the hypotenuse $AC$ , so it is equidistant from all three vertices, and in particular $FA = FD$ . Triangle $AFD$ is therefore isosceles and $\angle ADF = \angle DAF = \angle DAC = 90^{\circ} - C.$ So the whole problem is to find $C$ .

Now for the given condition, and the useful half of it is the right-hand side. Two facts about $F$ and $E$ pin down $\angle DFE$ . Since $E$ and $F$ are the midpoints of $BC$ and $CA$ , the segment $FE$ is a midline of the triangle and is parallel to $AB$ , so the angle it makes with line $BC$ at $E$ is $\angle FEC = B$ . And since $FD = FC$ , triangle $FDC$ is isosceles with $\angle FDC = \angle FCD = C$ . Looking now at triangle $FDE$ , whose vertices $D$ and $E$ both lie on $BC$ , its angle at $D$ is $C$ and its angle at $E$ is $180^{\circ} - B$ , so $\angle DFE = 180^{\circ} - C - (180^{\circ} - B) = B - C.$

The hypothesis $\angle DAE = \angle DFE$ therefore reads $\angle DAE = B - C$ . Finally, $E$ lies on the far side of $D$ from $B$ , so the angle at $A$ splits as $\angle BAE = \angle BAD + \angle DAE = (90^{\circ} - B) + (B - C) = 90^{\circ} - C.$ The two occurrences of $B$ cancel, which is the point of the whole configuration. Comparing with the expression found earlier, $\angle ADF = 90^{\circ} - C = \angle BAE = 40^{\circ}.$

If $B<C$ , the order along $BC$ is $B,E,D,C$ . Triangle $FDE$ now has angles $\angle FDE=180^\circ-C$ and $\angle FED=B$ , so $\angle DFE=C-B$ . The given equality makes $\angle DAE=C-B$ , and this time $\angle BAE=\angle BAD-\angle DAE=(90^\circ-B)-(C-B)=90^\circ-C.$ The earlier identity $\angle ADF=90^\circ-C$ still applies, so the answer is again $40^\circ$ . If $B=C$ , then $D=E$ and both angles in the given equality are zero. Thus $\angle BAE=\angle BAD=90^\circ-B=90^\circ-C=\angle ADF$ , with the same answer.

Nothing here needed the individual angles of the triangle, only the combination $90^{\circ} - C$ , and the given condition was designed precisely to hand that combination over.

Answer 40

Solution: PRMO 2017, Q24

Key idea

The three equal lengths force $\ell = 2/(1/a + 1/b + 1/c)$ , which here is $30$ ; but the segment parallel to a side is shorter than that side, and the shortest side is $26$ .

This question was discounted by the organisers. The formula it wants is easy to derive, and the derivation is exactly what shows the configuration cannot exist.

Let $\alpha$ be the distance from $P$ to $BC$ divided by the height from $A$ , and define $\beta$ and $\gamma$ in the same way with $CA$ and $AB$ . Triangles $PBC$ and $ABC$ share the base $BC$ , so $\alpha = [PBC]/[ABC]$ , and likewise $\beta = [PCA]/[ABC]$ and $\gamma = [PAB]/[ABC]$ . For an interior point the three triangles $PBC$ , $PCA$ , $PAB$ fill $ABC$ exactly, so $\alpha + \beta + \gamma = 1,$ with all three positive.

The line through $P$ parallel to $BC$ cuts off at $A$ a triangle similar to $ABC$ , and its ratio of similarity is the ratio of the distances from $A$ , namely $1 - \alpha$ . Hence $KL = a(1-\alpha), \qquad MN = b(1-\beta), \qquad ST = c(1-\gamma),$ by the same argument at each vertex. If all three are equal to $\ell$ then $\frac{\ell}{a} + \frac{\ell}{b} + \frac{\ell}{c} = (1-\alpha)+(1-\beta)+(1-\gamma) = 2,$ so $\ell = \frac{2}{\tfrac1a + \tfrac1b + \tfrac1c} = \frac{2abc}{ab+bc+ca}.$ With $a = 26$ , $b = 65$ , $c = 78$ this gives $abc = 131820$ and $ab+bc+ca = 8788$ , so $\ell = \frac{263640}{8788} = 30.$

And there is the defect. For $P$ to be interior we need $\alpha > 0$ , that is $\ell < a$ for every side, so $\ell$ must be smaller than the shortest side. Here the shortest side is $26$ and $\ell = 30$ exceeds it, which means $\alpha = 1 - \tfrac{30}{26} < 0$ : the point $P$ would have to lie outside the triangle, against the hypothesis.

So $30$ is the answer to a question about a configuration that does not exist for this triangle, and the discounting is right. A triangle admits such a point exactly when $2abc/(ab+bc+ca)$ is less than its shortest side.

Answer none

Solution: PRMO 2018, Q13

Key idea

Both the altitude and the bisector from the right angle have short formulas in the legs, and the two of them turn $b^2 + c^2 = a^2$ into a single linear equation for the hypotenuse.

Let the legs be $b = AC$ and $c = AB$ and the hypotenuse $a = BC$ ; the median from $A$ is $\tfrac{a}{2}$ , since in a right triangle the midpoint of the hypotenuse is the circumcentre.

The altitude from the right angle has length $h = bc/a$ , because the area is both $\tfrac12 bc$ and $\tfrac12 a h$ . So $h = 3$ gives $bc = 3a.$

The internal bisector from $A$ splits the right angle into two $45^{\circ}$ angles, so comparing areas again, $\tfrac12 bc \sin 90^{\circ} = \tfrac12 bt \sin 45^{\circ} + \tfrac12 ct\sin 45^{\circ}, \qquad \text{giving} \qquad t = \frac{\sqrt2\, bc}{b + c}.$ So $t = 4$ gives $bc = 2\sqrt2\,(b+c)$ , and with $bc = 3a$ , $b + c = \frac{3a}{2\sqrt2}.$

Now square and use Pythagoras. Since $b^2 + c^2 = a^2$ , $(b+c)^2 = a^2 + 2bc, \qquad \frac{9a^2}{8} = a^2 + 6a, \qquad \frac{a^2}{8} = 6a,$ so $a = 48$ and the median is $\tfrac{a}{2} = 24$ .

Answer 24

Solution: PRMO 2019, Q28

Key idea

Each corner triangle is similar to $ABC$ with ratio $(h - 2r)/h$ for the corresponding altitude, and the reciprocals of the three altitudes sum to $1/r$ , so the three small inradii add up to exactly $r$ .

Let $r$ be the inradius of $ABC$ , and let $h_a$ be the altitude from $A$ . The tangent to $\Omega$ parallel to $BC$ lies at distance $2r$ from $BC$ , on the far side of the incircle, so it cuts off at $A$ a triangle whose sides are parallel to those of $ABC$ . That corner triangle is therefore similar to $ABC$ , and the ratio of similarity is the ratio of the distances from $A$ to the two parallel lines: $\frac{h_a - 2r}{h_a}.$ Inradius scales the same way, so $r_1 = r\left(1 - \frac{2r}{h_a}\right),$ and likewise for the other two corners.

Adding the three, $r_1 + r_2 + r_3 = 3r - 2r^2\left(\frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c}\right).$ The bracket is the classical identity. Writing $\Delta$ for the area and $s$ for the semi-perimeter, each altitude satisfies $h_a = 2\Delta/a$ , so $\frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c} = \frac{a + b + c}{2\Delta} = \frac{2s}{2rs} = \frac{1}{r},$ using $\Delta = rs$ . Substituting, $r_1 + r_2 + r_3 = 3r - 2r^2 \cdot \frac{1}{r} = r.$

So the answer is the inradius of the original triangle, and nothing about the three corner triangles has to be computed separately. With sides $51, 52, 53$ the semi-perimeter is $s = 78$ , and Heron’s formula gives $\Delta = \sqrt{78 \times 27 \times 26 \times 25} = \sqrt{1{,}368{,}900} = 1170,$ so $r = \frac{\Delta}{s} = \frac{1170}{78} = 15.$

The sum is exactly $15$ , not merely close to it, so the largest integer not exceeding it is $15$ . That exactness is worth insisting on: computed in decimals the three inradii add to $14.999999\ldots$ , and a floor taken at that point would return $14$ . The identity above is what makes the answer safe.

Answer 15

Solution: IOQM 2020, Q16

Key idea

Rearranged, the given condition says $2\Delta = xy$ , and that is exactly the statement that the angle between those two sides is a right angle.

The equation relates two sides and the area, and the useful move is to gather the two sides on one side of it and the area on the other: $x - y = \frac{2\Delta}{y} - \frac{2\Delta}{x} = 2\Delta \cdot \frac{x - y}{xy}.$ Because the triangle is scalene we know $x \neq y$ , so we may divide both sides by $x - y$ , and this is precisely where the scalene condition earns its place in the problem. What survives is $1 = \frac{2\Delta}{xy}, \qquad \text{that is} \qquad \Delta = \tfrac12 xy.$

Now recall that a triangle with sides $x$ and $y$ enclosing an angle $\theta$ has area $\tfrac12 xy \sin\theta$ . Comparing this with what we have just derived forces $\sin\theta = 1$ , so $\theta = 90^{\circ}$ and the sides $x$ and $y$ are the two legs of a right angle. The condition that looked analytic was a disguised statement about shape.

The third side is then the hypotenuse, and with $x = 60$ and $y = 63$ , $\sqrt{60^2 + 63^2} = \sqrt{3600 + 3969} = \sqrt{7569} = 87.$ Since the hypotenuse is necessarily the longest side, the largest side of the triangle is $87$ .

Answer 87

Solution: IOQM 2021 Part A, Q6

Key idea

Each equation is a cosine rule in disguise, for angles of $90^{\circ}$ , $120^{\circ}$ and $150^{\circ}$ . Those sum to $360^{\circ}$ , so the three quantities are distances from an interior point to the vertices of a $5$ - $6$ - $7$ triangle, and the expression asked for is four times its area.

Look at the three equations as instances of $u^2 + v^2 - 2uv\cos\theta$ . The first, $x^2 + y^2 = 49$ , has no cross term, so $\theta = 90^{\circ}$ and the opposite side is $7$ . The second, $y^2 + yz + z^2 = 36$ , has cross term $+yz$ , which matches $-2\cos\theta = 1$ , so $\theta = 120^{\circ}$ and the opposite side is $6$ . The third has cross term $\sqrt3 xz$ , giving $\cos\theta = -\tfrac{\sqrt3}{2}$ , so $\theta = 150^{\circ}$ and the opposite side is $5$ .

Now count the angles: $90 + 120 + 150 = 360$ . Three angles at a point. So there is a point $P$ with $PA = x$ , $PB = y$ , $PC = z$ , the three angles at $P$ as above, and the triangle $ABC$ having sides $7$ , $6$ and $5$ opposite them.

The area of $ABC$ is the sum of the three little triangles at $P$ : $[ABC] = \tfrac12 xy\sin 90^{\circ} + \tfrac12 yz\sin 120^{\circ} + \tfrac12 zx\sin 150^{\circ},$ which evaluates to $[ABC] = \tfrac12 xy + \tfrac{\sqrt3}{4}yz + \tfrac14 zx.$ Multiplying by four gives exactly the expression we were asked for: $2xy + \sqrt3\,yz + zx = 4\,[ABC].$

Heron’s formula finishes it. With sides $5, 6, 7$ the semiperimeter is $9$ , so $[ABC] = \sqrt{9 \cdot 4 \cdot 3 \cdot 2} = \sqrt{216} = 6\sqrt6$ , and the expression equals $24\sqrt6$ . Hence $p = 24$ , $q = 6$ , and $p + q = 30$ .

Answer 30

Solution: IOQM 2022, Q3

Key idea

The bisector and the parallel sides force triangle $ABE$ to be isosceles, after which $MP$ is the base of the small isosceles triangle cut off at $B$ by the two tangent lengths.

First establish the shape. Since $AD \parallel BC$ in the trapezium, the angles $\angle DAE$ and $\angle AEB$ are alternate angles and so are equal. But $AE$ bisects angle $A$ , so $\angle DAE = \angle BAE$ as well. Hence $\angle BAE = \angle AEB$ , and triangle $ABE$ is isosceles with $BA = BE$ . Everything now happens inside that triangle.

Write $\beta = \angle ABE$ and scale so that $BA = BE = 1$ , so $AE = 2\sin\tfrac{\beta}{2}$ . The incircle touches the two equal sides at $M$ and $P$ , and the tangent lengths from a vertex are equal, so $BM = BP = s - AE$ where $s$ is the semiperimeter. Computing, $s = \frac{1 + 1 + 2\sin\frac{\beta}{2}}{2} = 1 + \sin\frac{\beta}{2}, \qquad BM = BP = 1 - \sin\frac{\beta}{2}.$

So $BMP$ is itself isosceles with apex angle $\beta$ at $B$ and legs $1 - \sin\tfrac{\beta}{2}$ , giving $MP = 2\left(1 - \sin\tfrac{\beta}{2}\right)\sin\tfrac{\beta}{2}.$

Now impose $AB : MP = 2$ , that is $MP = \tfrac12$ . Writing $u = \sin\tfrac{\beta}{2}$ , $2u(1-u) = \tfrac12 \quad\Longrightarrow\quad 4u^2 - 4u + 1 = 0 \quad\Longrightarrow\quad (2u-1)^2 = 0,$ so $u = \tfrac12$ and $\beta = 60^{\circ}$ . The repeated root is worth noticing: the condition is exactly extremal, so this configuration is the unique one of its kind.

Finally $\angle DAE = \angle AEB = \tfrac{180^{\circ} - \beta}{2} = 60^{\circ}$ .

Answer 60

Solution: IOQM 2022, Q13

Key idea

Because $x$ , $y$ , $z$ are in arithmetic progression, $x + z = 2y$ . Feeding that into the sine rule collapses the condition $AD = BC$ all the way down to $2y + z = 180^{\circ}$ .

Write $y = x + d$ and $z = x + 2d$ with $x, d$ positive integers. The progression gives us one free gift before any geometry: $x + z = 2x + 2d = 2y$ .

Now express both lengths through the sine rule. In triangle $ACD$ the angles are $x$ at $A$ , $z$ at $C$ , and therefore $180^{\circ} - x - z$ at $D$ , so $AD = \frac{AC \sin z}{\sin(x+z)} = \frac{AC \sin z}{\sin 2y},$ using $x + z = 2y$ . In triangle $ABC$ the angle at $A$ is $180^{\circ} - y - z$ , so $BC = \frac{AC \sin(y+z)}{\sin y}.$

Setting $AD = BC$ and cancelling $AC$ , $\frac{\sin z}{\sin 2y} = \frac{\sin(y+z)}{\sin y} \quad\Longrightarrow\quad \sin z \sin y = \sin 2y \, \sin(y+z).$ Replacing $\sin 2y = 2 \sin y \cos y$ and cancelling $\sin y$ , which is not zero, $\sin z = 2 \cos y \sin(y+z) = \sin(2y + z) + \sin z,$ where the last step is the product-to-sum identity. So $\sin(2y+z) = 0$ , and since $2y + z$ lies strictly between $0^{\circ}$ and $360^{\circ}$ , $2y + z = 180^{\circ}.$

Everything is now arithmetic. Substituting the progression, $2(x+d) + (x+2d) = 3x + 4d = 180$ . For $x$ to be an integer we need $3 \mid 4d$ , hence $3 \mid d$ , so write $d = 3k$ and get $x = 60 - 4k$ and $y = 60 - k$ .

To make $y$ as large as possible we take $k$ as small as possible, and $k = 1$ gives $x = 56$ , $y = 59$ , $z = 62$ . Check that this is a genuine triangle: the angle at $A$ is $180 - 59 - 62 = 59^{\circ}$ , and $\angle CAD = 56^{\circ}$ is indeed less than that, so $D$ falls properly inside $BC$ . The largest possible $\angle ABC$ is $59^{\circ}$ .

The relation $2y + z = 180^{\circ}$ is worth reading geometrically, because it says more than it looks. The angle at $A$ is $180^{\circ} - y - z$ , so $2y + z = 180^{\circ}$ says exactly that $\angle BAC = \angle ABC$ , which is to say $BC = CA$ : the condition $AD = BC$ , given the progression, is the condition that the triangle is isosceles at $C$ . And then $AD = BC = CA$ makes triangle $ACD$ isosceles as well, so $\angle ADC = \angle ACD = z$ and $x + 2z = 180^{\circ}$ , which is the same equation arrived at from the other end.

Answer 59

Solution: IOQM 2023, Q13

Key idea

The reciprocals of the exradii add to the reciprocal of the inradius, which gives $\rho = 4$ at once, and then each $s-a$ is a fixed multiple of $s$ , so the triangle is forced to be $13, 14, 15$ .

Write $\Delta$ for the area and $s$ for the semiperimeter. The excircle opposite side $a=BC$ is the circle outside the triangle touching $BC$ and the extensions of $AB$ and $AC$ . Its centre $I_A$ has perpendicular distance $r_a$ from each of those three lines. The areas of triangles $I_AAB$ and $I_AAC$ , less the area of triangle $I_ABC$ , give $[ABC]$ : $\Delta=\frac12r_a(c+b-a)=r_a(s-a).$ Thus $r_a=\Delta/(s-a)$ ; the other two exradii follow by changing the vertex. The inradius is $\rho=\Delta/s$ , by the area decomposition already used in Solution 11.7. Adding the reciprocals of the three exradii, $\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c} = \frac{(s-a)+(s-b)+(s-c)}{\Delta} = \frac{3s - 2s}{\Delta} = \frac{s}{\Delta} = \frac1\rho.$ With the given values, $\frac1\rho = \frac{2}{21} + \frac{1}{12} + \frac{1}{14} = \frac{8 + 7 + 6}{84} = \frac{21}{84} = \frac14,$ so $\rho = 4$ and $\Delta = 4s$ .

Now each side follows. From $s - a = \Delta/r_a = 4s/r_a$ we get $s-a = \frac{4s}{21/2} = \frac{8s}{21}, \qquad s-b = \frac{4s}{12} = \frac{7s}{21}, \qquad s-c = \frac{4s}{14} = \frac{6s}{21},$ and therefore $a = \tfrac{13s}{21}$ , $b = \tfrac{14s}{21}$ , $c = \tfrac{15s}{21}$ . The sides are in the ratio $13 : 14 : 15$ , and only the scale remains to be found.

Write $a = 13t$ , $b = 14t$ , $c = 15t$ , so that $s = 21t$ . Heron’s formula gives $\Delta = \sqrt{21t \cdot 8t \cdot 7t \cdot 6t} = \sqrt{7056\,t^4} = 84t^2,$ while we already know $\Delta = 4s = 84t$ . Comparing, $t = 1$ and the sides are $13$ , $14$ , $15$ .

The cubic with these roots has $p = 42, \qquad q = 13\cdot14 + 14\cdot15 + 15\cdot13 = 587,$ $r = 13\cdot14\cdot15 = 2730,$ so $p + q + r = 3359$ . The nearest integer to $\sqrt{3359}$ is $58$ : the root is below $58$ because $58^2 = 3364 > 3359$ , and above $57.5$ because $115^2 = 13225$ is less than $4 \cdot 3359 = 13436$ .

Answer 58

Solution: IOQM 2024, Q22

Key idea

The equal-sum condition says $BD - DC$ equals $AC - AB$ , and with $BD : DC = 2 : 1$ that turns into $BC = 3(AC - AB)$ , which Pythagoras converts into a quadratic in the ratio.

Write $c = AB$ , $b = AC$ and $a = BC$ , so $a^2 = b^2 + c^2$ . From $BD : DC = 2 : 1$ we have $BD = \tfrac{2a}{3}$ and $DC = \tfrac a3$ , and the given condition $AB + BD = AC + CD$ reads $c + \frac{2a}{3} = b + \frac{a}{3}, \qquad \text{that is} \qquad a = 3(b - c).$ In particular $b > c$ .

Now substitute into Pythagoras: $9(b-c)^2 = b^2 + c^2 \quad\Longrightarrow\quad 4b^2 - 9bc + 4c^2 = 0.$ Dividing by $c^2$ and writing $t = b/c$ , $4t^2 - 9t + 4 = 0, \qquad t = \frac{9 \pm \sqrt{81-64}}{8} = \frac{9 \pm \sqrt{17}}{8}.$ Since $b > c$ we need $t > 1$ , which selects the plus sign: $\frac{AC}{AB} = \frac{9 + \sqrt{17}}{8}.$

Here $m = 9$ and $n = 8$ are coprime and $p = 17$ is prime, as required, so $m + n + p = 9 + 8 + 17 = 34.$

Answer 34

Solution: IOQM 2025 Part SEP, Q28

Key idea

Write $x, y, z$ for the areas of $PBC$ , $PCA$ , $PAB$ ; then each of the three ratios along the three segments is one of $\dfrac{y+z}{x}$ , $\dfrac{z+x}{y}$ , $\dfrac{x+y}{z}$ .

Each segment ratio compares the two adjacent areas with the opposite area.

The three segments from the vertices have no common measure, since nothing fixes the shape of the triangle. They do, however, cut the triangle into three pieces around $P$ , and those three areas are the natural quantities, because every ratio in the question turns out to be a ratio of them.

Let $x = [PBC]$ , $y = [PCA]$ and $z = [PAB]$ . Triangles $APB$ and $DPB$ share the vertex $B$ and have bases $AP$ and $PD$ on one line, so their areas are in the ratio $AP : PD$ ; the same holds with $C$ in place of $B$ . Adding the two, $\frac{AP}{PD} = \frac{[APB]+[APC]}{[DPB]+[DPC]} = \frac{z+y}{x},$ since the triangles $DPB$ and $DPC$ together make up $PBC$ . Cyclically, $\frac{BP}{PE} = \frac{z+x}{y}, \qquad \frac{CP}{PF} = \frac{x+y}{z}.$

Now use the data. Writing $T = x+y+z$ , the first given ratio says $\dfrac{T-y}{y} = \dfrac52$ , so $\dfrac Ty = \dfrac72$ and $y = \dfrac{2T}{7}$ . The second says $\dfrac{T-z}{z} = \dfrac73$ , so $z = \dfrac{3T}{10}$ . Hence $x = T\left(1 - \frac27 - \frac{3}{10}\right) = T \cdot \frac{70 - 20 - 21}{70} = \frac{29T}{70},$ and $\frac{AP}{PD} = \frac{T - x}{x} = \frac{T}{x} - 1 = \frac{70}{29} - 1 = \frac{41}{29}.$

Since $\gcd(41,29) = 1$ , we get $p + q = 41 + 29 = 70$ .

Answer 70

Solution: IOQM 2026, Q30

Key idea

The sine rule in the two triangles standing on $AB$ gives $BE = 2\,BD\cos 40^{\circ}$ . That says the foot of the perpendicular from $D$ to $BE$ is the midpoint of $BE$ , so $DB = DE$ , and the angle at $E$ copies the angle at $B$ .

The angle $ABC$ is $70^{\circ}$ and $\angle ABE = 30^{\circ}$ , so $\angle EBC = 40^{\circ}$ .

In triangle $ABD$ the angles at $A$ and $B$ are $30^{\circ}$ and $70^{\circ}$ , so $\angle ADB = 80^{\circ}$ , and the sine rule gives $BD = AB \sin 30^{\circ}/\sin 80^{\circ}$ . In triangle $ABE$ the angles at $A$ and $B$ are $50^{\circ}$ and $30^{\circ}$ , so $\angle AEB = 100^{\circ}$ , and $BE = AB \sin 50^{\circ}/\sin 100^{\circ}$ . Since $\sin 100^{\circ} = \sin 80^{\circ}$ , dividing gives $\frac{BE}{BD} = \frac{\sin 50^{\circ}}{\sin 30^{\circ}} = 2\sin 50^{\circ} = 2\cos 40^{\circ}.$

Drop the perpendicular from $D$ to $BE$ , with foot $M$ . The angle $DBM$ is $\angle EBC = 40^{\circ}$ , so $BM = BD\cos 40^{\circ} = \tfrac12 BE$ , and $M$ is the midpoint of $BE$ . So $D$ lies on the perpendicular bisector of $BE$ , and $DB = DE$ . In the isosceles triangle $DBE$ the base angles are equal: $\angle DEB = \angle DBE = 40^{\circ}.$

Answer 40

Solution: PRMO 2017, Q30

Key idea

In a trapezium the two triangles on the parallel sides have areas $p$ and $q$ and the other two both have area $\sqrt{pq}$ , so the total is $(\sqrt p + \sqrt q)^2$ and only three distinct products exist.

the two triangles on the parallel sides carry s^2 and t^2 , and the other two both carry st — the two triangles on the parallel sides carry $s^2$ and $t^2$ ,
and the other two both carry $st$

Let the diagonals meet at $O$ , and write $p = [AOB]$ and $q = [COD]$ for the triangles on the two parallel sides. Triangles $AOB$ and $COD$ are similar, and triangles $BOC$ and $AOD$ have equal areas, because $[ABC] = [ABD]$ when $AB \parallel CD$ and one subtracts the common part $[AOB]$ . Writing $m$ for that common area, the standard relation $m^2 = pq$ holds, since $\frac{m}{p} = \frac{OC}{OA} = \frac{q}{m}.$

Put $s = \sqrt p$ and $t = \sqrt q$ , so $m = st$ and the four areas are $s^2$ , $t^2$ , $st$ , $st$ . The total area is $s^2 + t^2 + 2st = (s+t)^2.$

Now the products, taken two at a time. Only three distinct values arise: $pq = s^2t^2, \qquad pm = s^3t, \qquad qm = st^3,$ each occurring twice, the first as $p \cdot q$ and again as $m \cdot m$ .

Two of the six are $1296$ and $576$ , so the pair we are given is one of these three values in some order. Take the possibilities in turn, always maximising $(s+t)^2$ .

$s^2t^2 = 1296$ and $s^3t = 576$ . Then $st = 36$ and $s^2 (st) = 576$ , so $s^2 = 16$ , $s = 4$ and $t = 9$ . The total area is $(4+9)^2 = 169$ .
$s^2t^2 = 576$ and $s^3t = 1296$ . Then $st = 24$ and $s^2 (st) = 1296$ , so $s^2 = 54$ and $t^2 = (st)^2 / s^2 = 576/54 = \tfrac{32}{3}$ . The total is $s^2 + 2st + t^2 = 54 + 48 + \tfrac{32}{3} = \tfrac{338}{3}$ .
$s^3t = 1296$ and $st^3 = 576$ . Multiplying, $(st)^4 = 1296 \cdot 576 = 864^2$ , so $st = \sqrt{864} = 12\sqrt6$ ; dividing, $s^2/t^2 = \tfrac94$ , so $s^2 = \tfrac32 st$ and $t^2 = \tfrac23 st$ . The total is $s^2 + 2st + t^2 = \left(\tfrac32 + 2 + \tfrac23\right) st = \tfrac{25}{6} \cdot 12\sqrt6 = 50\sqrt6$ .

Swapping the roles of $s$ and $t$ changes nothing, since the total is symmetric.

Both runners-up fall short of $169$ , and exactly: $\tfrac{338}{3} < 169$ because $338 < 507$ , and $50\sqrt6 < 169$ because $(50\sqrt6)^2 = 15000$ while $169^2 = 28561$ . The largest total area is therefore $169$ , and its square root is exactly $13.$

The first case is the one that lands on whole numbers, and that is no accident: $1296 = 36^2$ and $576 = 24^2$ were chosen so that $st$ and $s$ come out integral.

Answer 13

Solution: PRMO 2018, Q29

Key idea

The angle an incentre subtends at two vertices is $90^{\circ}$ plus half the third angle, so $\angle BI_1E$ is $90^{\circ} - \tfrac12\angle ADB$ , and the two angles at $D$ add to $180^{\circ}$ .

Since $I_1$ is the incentre of triangle $ABD$ , it lies on the bisector from $A$ , which meets $BD$ at $E$ . So $A$ , $I_1$ , $E$ are collinear with $I_1$ between $A$ and $E$ , and $\angle BI_1E = 180^{\circ} - \angle BI_1A.$

Now the standard incentre formula. In any triangle the incentre $I$ satisfies $\angle BIA = 90^{\circ} + \tfrac12\angle C$ , because the angles at $A$ and $B$ inside triangle $ABI$ are half of the triangle’s, so $\angle BIA = 180^{\circ} - \tfrac12\angle A - \tfrac12\angle B = 180^{\circ} - \tfrac12\left(180^{\circ} - \angle C\right) = 90^{\circ} + \tfrac12\angle C.$ Applying it in triangle $ABD$ , whose third angle is $\angle ADB$ , $\angle BI_1E = 180^{\circ} - \left(90^{\circ} + \tfrac12\angle ADB\right) = 90^{\circ} - \tfrac12 \angle ADB.$

The hypothesis $\angle BI_1E = 60^{\circ}$ therefore gives $\angle ADB = 60^{\circ}$ . Since $D$ lies inside $BC$ , the two angles at $D$ are supplementary, so $\angle ADC = 120^{\circ}$ , and the same formula in triangle $ACD$ gives $\angle CI_2F = 90^{\circ} - \tfrac12 \angle ADC = 90^{\circ} - 60^{\circ} = 30^{\circ}.$

Everything else about the triangle is irrelevant: the answer depends only on the angle at $D$ , and the two halves of the picture are linked by nothing more than the straight line $BC$ .

Answer 30

Solution: PRMO 2019, Q29

Key idea

A bisector that is also perpendicular to a median makes an isosceles triangle, so $BA = BD$ and the foot is the midpoint of $AD$ ; that fixes the ratio $AE : EC$ and hence the fraction of $BE$ that lies inside the triangle $ABD$ .

Let $F$ be the point where $BE$ meets $AD$ .

The isosceles triangle

Compare triangles $ABF$ and $DBF$ . They share $BF$ ; the angles at $F$ are both right angles, since $AD \perp BE$ ; and the angles at $B$ are equal, because $BF$ lies along the bisector of $\angle ABC$ while $BA$ and $BD$ lie along the two arms, $D$ being a point of $BC$ . So the two triangles are congruent, and $BA = BD, \qquad AF = FD.$ In particular $F$ is the midpoint of $AD$ , so $AF=FD=7/2$ . Write $a=BC$ and $c=BA$ . Since $D$ is the midpoint of $BC$ , $c = BA = BD = \tfrac{a}{2}, \qquad \text{that is} \qquad a = 2c.$

How far along $BE$ the foot lies

The angle bisector from $B$ divides $AC$ in the ratio of the adjacent sides, so $\frac{AE}{EC} = \frac{BA}{BC} = \frac{c}{a} = \frac12.$ Now take $B$ as the origin and write $A$ and $C$ for the position vectors of those vertices. Then $D = \tfrac12 C$ and $E = A + \tfrac13(C - A) = \tfrac13(2A + C)$ .

The point $F$ lies on $BE$ , so $F = \tfrac{t}{3}(2A + C)$ for some $t$ ; and it lies on $AD$ , so $F = (1-u)A + \tfrac{u}{2}C$ for some $u$ . Since $A$ and $C$ are not parallel, the coefficients must match: $\frac{2t}{3} = 1 - u, \qquad \frac{t}{3} = \frac{u}{2}.$ The second gives $u = \tfrac{2t}{3}$ , and substituting into the first gives $\tfrac{4t}{3} = 1$ , so $t = \frac34, \qquad u = \frac12.$ The value $u = \tfrac12$ confirms independently that $F$ is the midpoint of $AD$ , and $t = \tfrac34$ is what we were after: $BF = \tfrac34 \, BE = \tfrac34 \times 9 = \frac{27}{4}.$

The area

Since $AD$ is a median, it halves the triangle, so $[ABC] = 2\,[ABD]$ . And in triangle $ABD$ the segment $AD$ is a base with $BF$ as its perpendicular height. Hence $[ABC] = 2 \cdot \tfrac12 \cdot AD \cdot BF = 7 \times \frac{27}{4} = \frac{189}{4} = 47.25,$ and the nearest integer is $47$ .

Notice that the two given lengths enter separately and linearly: the area is $\tfrac34 \cdot AD \cdot BE$ for any triangle in which a median and a bisector cross at right angles.

Answer 47

Solution: IOQM 2020, Q22

Key idea

Reflecting $A$ in a bisector of $B$ sends it to a point of line $BC$ , and the foot of the perpendicular is the midpoint of that journey. So all four feet lie on the midline, where the three given lengths read off as $AB$ , $AC$ and the semiperimeter.

The four feet look unrelated until one notices what a perpendicular from $A$ to a bisector really is. Dropping a perpendicular from $A$ to a line and continuing the same distance beyond reflects $A$ in that line, and the foot is the midpoint of $A$ and its reflection. Now, reflecting $A$ in the internal bisector of angle $B$ maps the ray $BA$ onto the ray $BC$ , so the image is the point of line $BC$ at distance $AB$ from $B$ . Reflecting in the external bisector instead sends $A$ to the point of line $BC$ at the same distance $AB$ from $B$ but on the opposite side.

Each foot is therefore the midpoint of a segment joining $A$ to a point of line $BC$ , which places it halfway between $A$ and that line, on the midline parallel to $BC$ . The same reasoning at vertex $C$ places $Q$ and $S$ on that same midline. All four points are collinear, which is why the problem could give us three consecutive gaps along a line.

Put coordinates on it. Take $B$ at the origin and $C$ at $(a, 0)$ , and write $c = AB$ and $b = AC$ . The four reflections land on the $x$ -axis at $\pm c$ from $B$ and at $a \pm b$ from the origin, so the four feet have $x$ -coordinates $\frac{A_x - c}{2}, \quad \frac{A_x + c}{2}, \quad \frac{A_x + a - b}{2}, \quad \frac{A_x + a + b}{2},$ all at the same height. Taking differences, the distance between the two feet belonging to vertex $B$ is exactly $c$ , the distance between the two belonging to $C$ is exactly $b$ , and the distance from the leftmost to the rightmost is $\tfrac{a+b+c}{2}$ , the semiperimeter.

The order is $P,Q,R,S$ : the successive differences of their horizontal coordinates are $(a+c-b)/2$ , $(b+c-a)/2$ and $(a+b-c)/2$ , each positive by the triangle inequalities. Reading the given data in that order, we get $c = PR = 13, \qquad b = QS = 14, \qquad s = PS = 7 + 6 + 8 = 21,$ and hence $a = 2s - b - c = 42 - 27 = 15$ . Heron’s formula finishes it: $[ABC] = \sqrt{21 \cdot 6 \cdot 7 \cdot 8} = \sqrt{7056} = 84.$

Answer 84

Solution: IOQM 2020, Q23

Key idea

Each small circle sits on a bisector with the incentre, so tangency gives $\sin\tfrac{A}{2} = \tfrac{r - r_A}{r + r_A}$ . Half-angle algebra turns that into $\sqrt{r_A/r} = \tan\left(45^\circ - \tfrac{A}{4}\right)$ , and those three angles sum to a right angle.

Fix attention on the circle at vertex $A$ . It touches both $AB$ and $AC$ , so its centre lies on the bisector of angle $A$ , which is where the incentre lies too. Along that bisector, a circle of radius $\rho$ inscribed in the angle sits at distance $\rho / \sin\tfrac{A}{2}$ from $A$ , since the perpendicular from the centre to a side makes a right triangle with the half-angle at $A$ . The two circles touch each other externally, so the gap between their centres equals the sum of their radii: $\frac{r}{\sin\frac{A}{2}} - \frac{r_A}{\sin\frac{A}{2}} = r + r_A, \qquad \text{so} \qquad \sin\frac{A}{2} = \frac{r - r_A}{r + r_A}.$

Solving for the ratio gives $\dfrac{r_A}{r} = \dfrac{1 - \sin\frac{A}{2}}{1 + \sin\frac{A}{2}}$ , and the same holds at $B$ and at $C$ . So the three radii are known, but they are known one at a time, and the answer wants them combined.

Here is what to aim for. The one thing the three angles of a triangle satisfy is $A + B + C = 180^\circ$ , so any hope of combining the three ratios rests on turning each of them back into an angle and using that sum. A ratio becomes an angle when it becomes a tangent, and the shape $\tfrac{1-\sin}{1+\sin}$ is exactly a squared tangent in disguise. Writing $\sin\frac{A}{2} = \cos\left(90^\circ - \frac{A}{2}\right)$ and using $\dfrac{1-\cos\varphi}{1+\cos\varphi} = \tan^2\dfrac{\varphi}{2}$ with $\varphi = 90^\circ - \frac{A}{2}$ , we get $\frac{r_A}{r} = \tan^2\left(45^\circ - \frac{A}{4}\right), \qquad \text{so} \qquad \sqrt{\frac{r_A}{r}} = \tan\left(45^\circ - \frac{A}{4}\right).$

The quarter of an angle looks like an odd thing to arrive at, but it is precisely what makes the sum work. Set $u = 45^\circ - \tfrac{A}{4}$ and define $v$ , $w$ likewise from $B$ and $C$ . Since $A + B + C = 180^\circ$ , $u + v + w = 135^\circ - \frac{A+B+C}{4} = 135^\circ - 45^\circ = 90^\circ.$ Since $u+v+w=90^\circ$ , the tangent addition formula gives $\frac{\tan u+\tan v}{1-\tan u\tan v}=\cot w=\frac{1}{\tan w}.$ Here $u,v,w$ are positive and their sum is a right angle, so the denominators are non-zero. Multiplying through and moving the last term, $\tan u\tan w+\tan v\tan w=1-\tan u\tan v,$ hence $\tan u\tan v+\tan v\tan w+\tan w\tan u=1$ . Translating that identity back through $\tan u = \sqrt{r_A/r}$ and its companions, $\frac{\sqrt{r_A r_B} + \sqrt{r_B r_C} + \sqrt{r_C r_A}}{r} = 1,$ that is, $r = \sqrt{r_A r_B} + \sqrt{r_B r_C} + \sqrt{r_C r_A}.$

The given radii are all perfect squares, which is the setter’s kindness: $r = \sqrt{16 \cdot 25} + \sqrt{25 \cdot 36} + \sqrt{36 \cdot 16} = 20 + 30 + 24 = 74.$

Answer 74

Solution: IOQM 2022, Q12

Key idea

One parameter controls the whole trapezium. Writing both areas in terms of it turns the question into maximising a quadratic.

The angles are $\angle B = 60^{\circ}$ and $\angle C = 30^{\circ}$ , so $\angle A = 90^{\circ}$ . Put $B$ at the origin with $C$ at $(1,0)$ , so $AB = \tfrac12$ and $A = \left(\tfrac14, \tfrac{\sqrt3}{4}\right)$ , and the triangle has area $\tfrac{\sqrt3}{8}$ .

Let $BP = u$ along $BA$ . Since $AB=1/2$ and the trapezium is non-degenerate, $0<u<1/2$ . Since $PQ \parallel BR$ , that is parallel to $BC$ , the point $Q$ sits at the same height as $P$ , namely $\tfrac{\sqrt3}{2}u$ . Following the line $AC$ down to that height puts $Q$ at horizontal position $1 - \tfrac{3u}{2}$ . Finally $R$ lies on $BC$ with $QR = BP = u$ , and since $Q$ is at height $\tfrac{\sqrt3}{2}u$ the horizontal drop from $Q$ to $R$ is $\sqrt{u^2 - \tfrac34 u^2} = \tfrac{u}{2}$ , placing $R$ at $1-u$ in the isosceles trapezium. The other sign would put $R$ at $1-2u$ and make $BR=PQ$ , a parallelogram, rather than this isosceles trapezium.

So the trapezium has parallel sides $BR = 1 - u$ and $PQ = 1 - 2u$ , and height $\tfrac{\sqrt3}{2}u$ , giving $[BPQR] = \tfrac12\left((1-u) + (1-2u)\right)\cdot \tfrac{\sqrt3}{2}u = \frac{\sqrt3}{4}\,u\,(2 - 3u).$

The ratio we must minimise is therefore $\frac{2[ABC]}{[BPQR]} = \frac{2 \cdot \frac{\sqrt3}{8}}{\frac{\sqrt3}{4}u(2-3u)} = \frac{1}{u(2-3u)},$ so minimising it means maximising the quadratic. Completing the square, $u(2-3u)=\frac13-3\left(u-\frac13\right)^2\le\frac13.$ Equality holds at $u=1/3$ , which lies in $0<u<1/2$ .

The minimum of the ratio is thus $1 \big/ \tfrac13 = 3$ .

Answer 3

Solution: IOQM 2024, Q27

Key idea

The three angles at $P$ add to $360^{\circ}$ and the three angles of the triangle to $180^{\circ}$ , so the common difference is $60^{\circ}$ ; and the angles of the triangle formed by the three perpendicular feet are exactly those differences, so $DEF$ is equilateral.

Call the common value $k$ , so $\angle BPC = \angle BAC + k, \qquad \angle CPA = \angle CBA + k,$ $\angle APB = \angle ACB + k.$ Adding all three and using that the angles at an interior point total $360^{\circ}$ while the angles of the triangle total $180^{\circ}$ , $360^{\circ} = 180^{\circ} + 3k, \qquad k = 60^{\circ}.$

The feet are on the sides, as the angle data ensure. Put $\alpha=\angle PAB$ , so $0<\alpha<30^\circ$ . The three equations above give $\angle PBC=30^\circ+\alpha$ , $\angle PCB=60^\circ-\alpha$ , $\angle ABP=B-30^\circ-\alpha$ and $\angle ACP=60^\circ-\alpha$ . Also $\angle PCA=C-60^\circ+\alpha>0$ , so $C>60^\circ-\alpha$ and $\angle ABP=120^\circ-C-\alpha<60^\circ$ . Thus the angles made by $P$ with both ends of each side are acute, and each perpendicular foot lies on its side.

Now consider the triangle $DEF$ formed by the perpendicular feet; this is called the pedal triangle. Since $\angle PFB = \angle PDB = 90^{\circ}$ , the points $B, D, P, F$ lie on the circle with diameter $BP$ , so $\angle FDP = \angle FBP$ . Similarly $C, D, P, E$ lie on the circle with diameter $CP$ , so $\angle PDE = \angle PCE$ . Adding, $\angle FDE = \angle ABP + \angle ACP.$ On the other hand, splitting the angles of the triangle at $B$ and at $C$ , $\angle BPC = 180^{\circ} - \angle PBC - \angle PCB = \angle BAC + \angle ABP + \angle ACP,$ because $\angle BAC = 180^{\circ} - (\angle ABP + \angle PBC) - (\angle ACP + \angle PCB)$ . Comparing the two displays, $\angle FDE = \angle BPC - \angle BAC,$ and by symmetry each angle of the pedal triangle is the corresponding difference. The hypothesis therefore says all three angles of $DEF$ are equal, so $DEF$ is equilateral.

Its side is now a single computation. Since $\angle AFP = \angle AEP = 90^{\circ}$ , the points $A, F, P, E$ lie on the circle with diameter $AP$ , and in that circle the chord $EF$ subtends the inscribed angle $\angle FAE = \angle BAC$ . By the extended sine rule in that circle, $EF = AP \sin \angle BAC = 12 \sin 30^{\circ} = 6.$

An equilateral triangle of side $6$ has area $\frac{\sqrt3}{4} \cdot 36 = 9\sqrt3,$ so $m = 9$ , $n = 3$ and $mn = 27$ .

Answer 27

Solution: IOQM 2025 Part SEP, Q28

Key idea

The step from the circumcentre to the midpoint of $BC$ is exactly half the step from $A$ to the orthocentre, and in the same direction. So $AH = 2 \cdot OD = 12$ , and the whole triangle can be placed on coordinates.

G is the centroid, so GA = 2\,GD and GH = 2\,GO : the shaded triangles are similar with ratio 2 — $G$ is the centroid, so $GA = 2\,GD$ and $GH = 2\,GO$ :
the shaded triangles are similar with ratio $2$

The one fact, proved

Let $D$ be the midpoint of $BC$ . The claim is that the journey from $A$ to $H$ is the journey from $O$ to $D$ , doubled: same direction, twice the length. Nothing about it is standard school material, so here is the proof, and it needs only the centroid.

Let $G$ be the centroid, which lies on the median $AD$ with $AG : GD = 2 : 1$ . Take the point $H'$ on the line $OG$ , on the far side of $G$ from $O$ , with $GH' = 2\,GO$ . Now compare the triangles $GOD$ and $GH'A$ . The angles at $G$ are vertically opposite, and the sides about them are in the ratio $\frac{GH'}{GO} = 2 = \frac{GA}{GD},$ so the two triangles are similar with ratio $2$ . Hence $AH'$ is parallel to $OD$ and twice as long, and since $O$ and $D$ sit on the opposite side of $G$ from $H'$ and $A$ , the step from $A$ to $H'$ runs the same way as the step from $O$ to $D$ , not the reverse.

That parallelism is what we want. $O$ is the circumcentre, so $OD$ is perpendicular to the chord $BC$ ; hence $AH'$ is perpendicular to $BC$ , which puts $H'$ on the altitude from $A$ . Repeating the argument with the midpoints of $CA$ and $AB$ puts $H'$ on the other two altitudes, using the same point $H'$ each time because $G$ and $O$ do not move. So $H'$ is the orthocentre $H$ , and along the way we have shown $AH = 2 \cdot OD, \qquad AH \parallel OD.$ In the language of vectors that reads $\vec{AH} = 2\,\vec{OD}$ , and the same computation packaged differently gives the familiar $\vec{OH} = \vec{OA} + \vec{OB} + \vec{OC}$ .

Placing the picture

Put $O$ at the origin and $D = (6,0)$ , which is legitimate because $OD = 6$ . Since $D$ is the midpoint of the chord $BC$ , the line $BC$ is perpendicular to $OD$ , so it is the vertical line $x = 6$ . The triangle $ODH$ is equilateral of side $6$ , so $H = \left(3,\ 3\sqrt3\right),$ choosing one of the two possible positions; the other is its mirror image and gives the same area. By the fact above, $A = H - 2\,\vec{OD} = (3-12,\ 3\sqrt3) = \left(-9,\ 3\sqrt3\right)$ .

The circumradius is then $R = OA = \sqrt{81 + 27} = \sqrt{108} = 6\sqrt3,$ and $B$ , $C$ are the points of the line $x = 6$ at distance $R$ from $O$ : $36 + y^2 = 108, \qquad y = \pm 6\sqrt2,$ so $BC = 12\sqrt2$ . The distance from $A$ to that line is $|-9-6| = 15$ , so $[ABC] = \tfrac12 \cdot 12\sqrt2 \cdot 15 = 90\sqrt2.$

Hence $a = 90$ , $b = 2$ and $a + b = 92$ .

Answer 92

Solution: IOQM 2025 Part SEP, Q28

Key idea

Each tangent parallel to a side cuts off a small triangle similar to $ABC$ , and the similarity ratio is $\dfrac{h - 2r}{h}$ for the corresponding height $h$ , so the hexagon is what remains after removing three known fractions of the area.

For the triangle with sides $13, 13, 10$ the semiperimeter is $18$ , so by Heron’s formula $[ABC] = \sqrt{18 \cdot 5 \cdot 5 \cdot 8} = 60, \qquad r = \frac{60}{18} = \frac{10}{3}.$ The heights are $h_a = \frac{2 \cdot 60}{10} = 12 \quad \text{(to the side } 10), \qquad h_b = h_c = \frac{2 \cdot 60}{13} = \frac{120}{13}.$

Consider the tangent to the incircle parallel to the side of length $10$ . It and that side are the two tangents to the incircle perpendicular to $h_a$ , so they are $2r$ apart, and the little triangle cut off at the opposite vertex is similar to $ABC$ with ratio $k_a = \frac{h_a - 2r}{h_a} = \frac{12 - \tfrac{20}{3}}{12} = \frac49.$ The same computation on each side of length $13$ gives $k_b = k_c = \frac{\tfrac{120}{13} - \tfrac{20}{3}}{\tfrac{120}{13}} = \frac{5}{18}.$

The three small triangles meet the hexagon only along its sides, so $[DEFGHK] = [ABC]\left(1 - k_a^2 - k_b^2 - k_c^2\right),$ which here is $[DEFGHK] = 60\left(1 - \frac{16}{81} - \frac{2 \cdot 25}{324}\right).$ Over the common denominator $324$ the removed part is $\dfrac{64 + 50}{324} = \dfrac{114}{324} = \dfrac{19}{54}$ , so $[DEFGHK] = 60 \cdot \frac{35}{54} = \frac{350}{9} = 38 + \frac89.$

Hence $m = 38$ , $n = 8$ , $l = 9$ and $m+n+l = 55$ .

Answer 55

Problems

Solutions

Solution: PRMO 2014, Q10

Solution: PRMO 2019, Q1

Solution: PRMO 2012, Q8

Solution: PRMO 2013, Q8

Solution: PRMO 2014, Q3

Solution: PRMO 2014, Q15

Solution: PRMO 2018, Q2

Solution: IOQM 2020, Q1

Solution: IOQM 2020, Q4

Solution: IOQM 2023, Q5

Solution: IOQM 2024, Q4

Solution: IOQM 2026, Q8

Solution: PRMO 2012, Q10

Solution: PRMO 2013, Q9

Both cases give the same angle

Solution: PRMO 2013, Q12

Solution: PRMO 2013, Q19

Solution: PRMO 2015 Part A, Q11

Solution: PRMO 2017, Q13

Solution: PRMO 2017, Q17

Solution: PRMO 2017, Q25

Solution: PRMO 2018, Q10

Solution: PRMO 2018, Q17

Solution: PRMO 2018, Q21

Solution: PRMO 2019, Q10

Solution: IOQM 2020, Q7

Solution: IOQM 2020, Q9

Solution: IOQM 2020, Q19

Solution: IOQM 2021 Part A, Q1

Solution: IOQM 2021 Part A, Q5

Solution: IOQM 2022, Q1

Solution: IOQM 2022, Q2

Solution: IOQM 2022, Q9

Solution: IOQM 2024, Q12

Solution: IOQM 2025 Part SEP, Q28

Solution: IOQM 2026, Q13

Solution: IOQM 2026, Q24

Solution: PRMO 2012, Q14

Solution: PRMO 2013, Q15

Solution: PRMO 2014, Q12

The first triangle

The second triangle

Putting them together

Solution: PRMO 2014, Q16

Solution: PRMO 2015 Part A, Q16

Solution: PRMO 2017, Q24

Solution: PRMO 2018, Q13

Solution: PRMO 2019, Q28

Solution: IOQM 2020, Q16

Solution: IOQM 2021 Part A, Q6

Solution: IOQM 2022, Q3

Solution: IOQM 2022, Q13

Solution: IOQM 2023, Q13

Solution: IOQM 2024, Q22

Solution: IOQM 2025 Part SEP, Q28

Solution: IOQM 2026, Q30

Solution: PRMO 2017, Q30

Solution: PRMO 2018, Q29

Solution: PRMO 2019, Q29

The isosceles triangle

How far along BEBEBE the foot lies

The area

Solution: IOQM 2020, Q22

Solution: IOQM 2020, Q23

Solution: IOQM 2022, Q12

Solution: IOQM 2024, Q27

Solution: IOQM 2025 Part SEP, Q28

The one fact, proved

Placing the picture

Solution: IOQM 2025 Part SEP, Q28

How far along $BE$ the foot lies