Circles, Coordinates and Trigonometry

Problems

Solutions

Solution: IOQM 2025 Part SEP, Q28

Since $AB$ lies along the horizontal axis and $CD$ is parallel to it, the point $D$ has the same $y$-coordinate as $C$, so $b = 5$.

The condition $AD = BC$ now fixes $a^2$: $$BC^2 = (10-8)^2 + 5^2 = 29, \qquad AD^2 = a^2 + 25,$$ so $a^2 + 25 = 29$ and $a^2 = 4$.

Both signs of $a$ give a legitimate quadrilateral, $a = 2$ giving the isosceles trapezium and $a = -2$ the parallelogram, and the question is untroubled by the ambiguity because it asks only for $$a^2 b = 4 \cdot 5 = 20.$$

Answer 20

Solution: IOQM 2026, Q5

Every point on the perpendicular bisector of $BC$ is the same distance from $B$ as from $C$, so $NB = NC$ and triangle $NBC$ is isosceles.

The points $A$ and $N$ lie on the circumcircle on the same side of the chord $BC$, so they are in the same segment, and angles in the same segment are equal: $$\angle BNC = \angle BAC = 80^{\circ}.$$ The two base angles of the isosceles triangle $NBC$ share what is left, so $$\angle NBC = \tfrac12\,(180^{\circ} - 80^{\circ}) = 50^{\circ}.$$

$N$ is the midpoint of the arc $BAC$, and the question does not care where on that arc $A$ sits, which is why only the angle at $A$ was given.

Answer 50

Solution: PRMO 2012, Q7

Let $M$ be the midpoint of $AB$. Since $CA = CB$, the point $C$ lies directly above $M$, and $$AM = MB = 1, \qquad CM = \sqrt{7^2 - 1^2} = \sqrt{48}.$$

Now put $D$ on the far side of $B$, at distance $BD$ from it, so $MD = 1 + BD$. In the right triangle $CMD$, $$8^2 = CM^2 + MD^2 = 48 + (1 + BD)^2, \qquad (1 + BD)^2 = 16,$$ so $BD = 3$.

Answer 3

Solution: PRMO 2015 Part A, Q8

Let $O$ be the centre of the plate and $R$ its radius. Because $C$ is the midpoint of the chord $AB$, the line $OC$ is perpendicular to $AB$, and because $D$ is the midpoint of arc $AB$, the same line $OC$ extended passes through $D$. So $O$, $C$ and $D$ are collinear, and $$OC = OD - CD = R - 6.$$

Now apply Pythagoras in the right triangle $OCA$, whose legs are $OC$ and $CA = \tfrac{24}{2} = 12$ and whose hypotenuse is the radius $OA = R$: $$R^2 = 12^2 + (R-6)^2 = 144 + R^2 - 12R + 36.$$ The $R^2$ terms cancel, which is the whole reason this configuration is set up the way it is, and what remains is linear: $$12R = 180, \qquad R = 15.$$

The same relation is worth remembering in the form it usually appears: if a chord of length $2c$ sits at distance $h$ below the arc’s midpoint, then $c^2 = h(2R - h)$. Here $12^2 = 6(2R-6)$, which again gives $R = 15$.

Answer 15

Solution: IOQM 2024, Q7

The two points $(1,2,0)$ and $(3,3,2)$ differ by $(2,1,2)$, whose length is $$\sqrt{4+1+4} = 3.$$

In a cube of edge $s$, two distinct vertices are joined by an edge, a face diagonal or a space diagonal, and those have lengths $s$, $s\sqrt2$ and $s\sqrt3$. Each case is realisable, since a cube may be placed in space with any given segment as an edge, as a face diagonal or as a space diagonal. So: $$\begin{array}{c|c|c} \text{the segment is} & s^2 & \text{surface area } 6s^2 \\ \hline \text{an edge} & 9 & 54 \\ \text{a face diagonal} & 9/2 & 27 \\ \text{a space diagonal} & 3 & 18 \end{array}$$ The sum of all possible surface areas is $54 + 27 + 18 = 99$.

Answer 99

Solution: PRMO 2018, Q5

Since $AD \perp AB$ and $AB \parallel CD$, the side $AD$ is perpendicular to both parallel sides. So $AD$ is a common perpendicular between two tangent lines of the incircle, and its length is the diameter $2r$. The touch points $Q$ on $AB$ and $P$ on $CD$ sit directly opposite each other on the circle, so $$PQ = 2r$$ as well, and the whole problem is to find $r$.

Set up coordinates with $A$ at the origin, $AB$ along the positive $x$-axis and $AD$ down the negative $y$-axis. The incircle has centre $(r, -r)$ and radius $r$, so it touches $AB$ at $Q = (r, 0)$ and $CD$ at $P = (r, -2r)$. Then $QB = 49$ places $B$ at $(r + 49, 0)$, and $PC = 36$ places $C$ at $(r + 36, -2r)$.

Now use the fourth side. Equal tangents from $B$ and from $C$ give $$BC = BQ + CP = 49 + 36 = 85,$$ while the coordinates give $BC^2 = (49 - 36)^2 + (2r)^2$. Hence $$4r^2 = 85^2 - 13^2 = 7225 - 169 = 7056, \qquad r^2 = 1764, \qquad r = 42,$$ and $PQ = 2r = 84$.

The numbers were chosen so that $(13, 84, 85)$ is a Pythagorean triple, which is what makes the final square root come out whole.

Answer 84

Solution: PRMO 2018, Q7

Let $O$ be the centre and $r$ the circumradius, and let the three consecutive vertices be $A$, $B$, $C$ with $PA = PB = 8$ and $PC = 16$. Place the hexagon with $$A = (r, 0), \qquad B = \left(\tfrac{r}{2}, \tfrac{r\sqrt3}{2}\right), \qquad C = \left(-\tfrac{r}{2}, \tfrac{r\sqrt3}{2}\right).$$

Because $PA = PB$, the point $P$ lies on the perpendicular bisector of $AB$, which passes through $O$ since $OA = OB$. That line has direction $\left(\tfrac{\sqrt3}{2}, \tfrac12\right)$, so $P = s\left(\tfrac{\sqrt3}{2}, \tfrac12\right)$ for some $s$, and $|OP| = |s|$.

Now compute the two given distances. Expanding, $$PA^2 = \left(\tfrac{s\sqrt3}{2} - r\right)^2 + \tfrac{s^2}{4} = s^2 - \sqrt3\,rs + r^2 = 64,$$ $$PC^2 = \left(\tfrac{s\sqrt3}{2} + \tfrac{r}{2}\right)^2 + \left(\tfrac{s}{2} - \tfrac{r\sqrt3}{2}\right)^2 = s^2 + r^2 = 256.$$ The second is the cleaner of the two: the cross terms cancel because $OP$ and $OC$ happen to be perpendicular. Substituting $s^2 + r^2 = 256$ into the first gives $\sqrt3\,rs = 192$, so $$s^2 + r^2 = 256, \qquad rs = 64\sqrt3.$$

The numbers $r^2$ and $s^2$ have sum $256$ and product $(64\sqrt3)^2=12288$, so they are the roots of $$z^2-256z+12288=0, \qquad (z-64)(z-192)=0.$$ Thus $\{r^2,s^2\}=\{64,192\}$, and since $r>0$ and $rs=64\sqrt3>0$ also gives $s>0$, $$\{r,s\}=\{8,8\sqrt3\}.$$

Which of the two is the radius? If $r$ were $8$ then $|OP| = 8\sqrt3 > r$ and $P$ would lie outside the hexagon, against the hypothesis. So $$r = 8\sqrt3,$$ and the nearest integer is $14$: squaring the three quantities, $13.5^2 = 182.25$ and $14.5^2 = 210.25$, while $\left(8\sqrt3\right)^2 = 192$ sits between them. As a check, $P$ is then $8$ from the centre while the hexagon’s inradius is $r\sqrt3/2 = 12$, so $P$ is comfortably inside.

Answer 14

Solution: PRMO 2018, Q8

The angles at the base

Since $OB = OC$, triangle $OBC$ is isosceles, and $\angle OCB = 20^{\circ}$ gives $\angle OBC = 20^{\circ}$. Hence $$\angle ABO = \angle ABC - \angle OBC = 30^{\circ} - 20^{\circ} = 10^{\circ},$$ and since $OA = OB$ as well, $\angle OAB = 10^{\circ}$ too. The apex angle of the isosceles triangle $OAB$ is therefore $\angle AOB = 160^{\circ}$, and the inscribed angle on the same arc is half of it: $$\angle ACB = 80^{\circ}.$$

Two isosceles triangles, and no trigonometry

The angle $\angle OCA$ is whatever is left of $\angle ACB$ once $\angle OCB$ is taken off: $$\angle OCA = 80^{\circ} - 20^{\circ} = 60^{\circ}.$$ But $OA = OC$, so triangle $OAC$ is isosceles with a base angle of $60^{\circ}$, and that forces the apex to be $60^{\circ}$ as well. It is equilateral, so $$AC = R.$$

Now take triangle $ACD$. Its angle at $A$ is $\angle CAB$, and since $O$ lies inside the triangle the ray $AO$ falls between $AB$ and $AC$, so $$\angle CAB = \angle OAB + \angle OAC = 10^{\circ} + 60^{\circ} = 70^{\circ}.$$ Its angle at $D$ comes from triangle $BCD$, where $\angle DBC = 30^{\circ}$ and $\angle BCD = \angle DCO + \angle OCB = 40^{\circ}$, so $\angle BDC = 110^{\circ}$ and therefore $\angle ADC = 70^{\circ}$. The angles at $A$ and at $D$ are equal, so triangle $ACD$ is isosceles and $$CD = AC = R.$$

Finishing

So $CD = R = OC$, and triangle $OCD$ is isosceles with apex angle $\angle OCD = 20^{\circ}$. Its base angles are equal, so $$\angle CDO = \frac{180^{\circ} - 20^{\circ}}{2} = 80^{\circ}.$$

The equality $CD = R$ is what the problem is built around, and it is not a coincidence of lengths. The $20^{\circ}$ at $C$ leaves exactly $60^{\circ}$ of $\angle ACB$ behind, which is what an equilateral triangle needs, and after that every length in the picture is $R$.

Answer 80

Solution: PRMO 2019, Q19

Let $O$ be the centre, so $O$ is the midpoint of $AB$ and also of the diameter $DE$. Write $AC = 6k$ and $CB = 7k$, so that $AB = 13k$ and the radius is $\tfrac{13}{2}k$. The offset of $C$ from the centre is $$OC = AO - AC = \frac{13}{2}k - 6k = \frac{k}{2}.$$

For the first triangle, $CD$ is the height from $D$ onto the base $AB$, since $DC \perp AB$: $$[ABD] = \tfrac12 \cdot AB \cdot CD = \tfrac12 \cdot 13k \cdot CD.$$

For the second, split it at the centre. Since $O$ is the midpoint of $DE$, the triangles $CDO$ and $COE$ have equal areas, so $[CDE] = 2\,[CDO]$. In triangle $CDO$ take $OC$ as the base, which lies along $AB$; the height from $D$ is again $CD$. Hence $$[CDE] = 2 \cdot \tfrac12 \cdot OC \cdot CD = \frac{k}{2}\, CD.$$

Dividing, the length $CD$ cancels and never has to be computed: $$\frac{[ABD]}{[CDE]} = \frac{\tfrac12 \cdot 13k \cdot CD}{\tfrac{k}{2} \cdot CD} = 13.$$

The ratio is exactly $13$, so the instruction to round is a courtesy rather than a necessity. Notice also what the answer is: $AB / (2 \cdot OC)$, which is $13$ because $6 : 7$ splits the diameter into thirteen parts and leaves the centre half a part away from $C$.

Answer 13

Solution: IOQM 2020, Q12

By symmetry, every chord of the outer circle that touches the inner circle subtends the same angle at the common centre, which means the figure is really one rigid step repeated over and over. Identifying that step is the whole problem.

Look at the vertex $B$, where the chords $AB$ and $BC$ meet at $75^{\circ}$. Passing from one chord to the next rotates the configuration about the centre through the supplement of that angle, because the chord turns by the exterior angle rather than the interior one, so each step is a rotation through $$180^{\circ} - 75^{\circ} = 105^{\circ}.$$

After $k$ chords the accumulated rotation is $105k$ degrees, and the chain closes up precisely when this amounts to a whole number of complete turns, that is when $$105k \equiv 0 \pmod{360}.$$ Dividing through by $\gcd(105, 360) = 15$ turns this into $7k \equiv 0 \pmod{24}$, and since $7$ and $24$ share no factor the smallest positive $k$ is $24$. The path closes after twenty-four chords, having wound seven times around the centre on the way.

Answer 24

Solution: IOQM 2023, Q14

Treat the points as position vectors with $B$ at the origin, so that $B = \mathbf{0}$. Each construction is a statement about a vector.

Since $BC : CD = 1 : 1$, the point $C$ is the midpoint of $BD$, so $D = 2C$. Since $CA : AE = 1 : 2$, the point $E$ lies on the extension of $CA$ beyond $A$ with $AE$ twice $CA$, so $E = A + 2(A - C) = 3A - 2C$. Since $AB : BF = 1 : 3$, the point $F$ lies beyond $B$ with $BF$ three times $AB$, so $F = B + 3(B-A) = -3A$, using $B = \mathbf{0}$.

Now add: $$D + E + F = 2C + (3A - 2C) - 3A = \mathbf{0}.$$ The centroid of $DEF$ is $K = \tfrac13(D+E+F) = \mathbf{0}$, which is the point $B$. The three constructions were built to cancel, and no coordinates for $A$ and $C$ were ever needed.

Therefore $GK$ is the distance from $G(32,24)$ to the origin: $$GK = \sqrt{32^2 + 24^2} = \sqrt{1024 + 576} = \sqrt{1600} = 40.$$

Answer 40

Solution: IOQM 2024, Q17

Since $AB = AC$, the foot $D$ is the midpoint of $BC$, so $BD = DC = 15$ and $$AD = \sqrt{20^2 - 15^2} = \sqrt{175} = 5\sqrt7 .$$ Put $D$ at the origin with $BC$ along the horizontal axis, so $$B = (-15, 0), \quad C = (15,0), \quad A = (0, 5\sqrt7), \quad M = \left(0, \tfrac{5\sqrt7}{2}\right).$$

The circumcentre lies on the perpendicular bisector of $BC$, which is the vertical axis, so write it as $O = (0,k)$. Equating $OB^2$ and $OA^2$, $$225 + k^2 = \left(5\sqrt7 - k\right)^2 = 175 - 10\sqrt7\,k + k^2,$$ so $10\sqrt7\, k = -50$ and $k = -\dfrac{5}{\sqrt7}$. The centre sits just below $BC$, which is what one expects for a triangle this wide. The circumradius follows: $$R^2 = 225 + \frac{25}{7} = \frac{1600}{7}.$$

The chord $PQ$ is horizontal at height $\tfrac{5\sqrt7}{2}$, so its distance from $O$ is $$\frac{5\sqrt7}{2} + \frac{5}{\sqrt7} = \frac{35 + 10}{2\sqrt7} = \frac{45}{2\sqrt7}.$$ Half the chord is then $$\sqrt{R^2 - \frac{2025}{28}} = \sqrt{\frac{6400 - 2025}{28}} = \sqrt{\frac{4375}{28}} = \frac{25\sqrt7}{2\sqrt7} = \frac{25}{2},$$ using $4375 = 625 \cdot 7$. Hence $PQ = 25$.

Answer 25

Solution: IOQM 2025 Part SEP, Q28

Place $A = (0,0)$, $B = (1,0)$, $C = (1,1)$ and $D = (0,1)$. Since $AB$ is a chord of $S$, the centre lies on the perpendicular bisector $x = \tfrac12$, say at $\left(\tfrac12, h\right)$. There is no need to guess whether $h$ is positive or negative: leave it free and let the arithmetic decide. Because $A$ lies on $S$, the radius satisfies $$r^2 = \tfrac14 + h^2.$$

Let $O$ be the centre. The radius $OK$ is perpendicular to tangent $CK$, so Pythagoras gives $CK^2=CO^2-r^2$, also the power of $C$. Since $CK=2$, computing this difference gives $$CK^2 = \left(1 - \tfrac12\right)^2 + (1-h)^2 - r^2 = \tfrac14 + 1 - 2h + h^2 - \tfrac14 - h^2 = 1 - 2h.$$ So $1 - 2h = 4$ and $h = -\tfrac32$. The negative sign is the answer to the question we declined to prejudge: the centre does lie below $AB$, which is consistent with $CD$ missing the circle, but that was a conclusion rather than an assumption.

Therefore $r^2 = \tfrac14 + \tfrac94 = \tfrac{10}{4}$, and the diameter satisfies $$d^2 = 4r^2 = 10.$$

Answer 10

Solution: IOQM 2026, Q19

Since $\angle AEC = 45^{\circ} + 45^{\circ} = 90^{\circ}$, the point $E$ lies on the circle with diameter $AC$, whose centre is $F$. So $FE = FC$. In the same way $\angle BED = 90^{\circ}$, and $GE = GB$.

In triangle $EBC$, write $\beta = \angle EBC$ and $\gamma = \angle ECB$ for the angles at $B$ and $C$. The angle at $E$ is $45^{\circ}$, so $\beta + \gamma = 135^{\circ}$.

The point $F$ lies on the segment $AC$, on the same side of $C$ as $B$, so triangle $FEC$ has $\angle FCE = \gamma$, and being isosceles it has $\angle FEC = \gamma$ too. Likewise $G$ lies on $BD$ on the same side of $B$ as $C$, and $\angle GEB = \angle GBE = \beta$.

Now measure every angle at $E$ from the ray $EC$, counting towards $A$ as positive. The ray $EB$ is at $45^{\circ}$. The ray $EF$ lies between $EC$ and $EA$, at $\gamma$. The ray $EG$ is $\beta$ past $EB$ in the direction of $D$, so it is at $45^{\circ} - \beta$, which is negative when it has gone past $EC$. The angle between $EF$ and $EG$ is $$\gamma - (45^{\circ} - \beta) = \beta + \gamma - 45^{\circ} = 135^{\circ} - 45^{\circ} = 90^{\circ}.$$

Answer 90

Solution: PRMO 2012, Q20

Put $O$ at the origin, so $P = (-2,0)$, $S = (2,0)$ and $X = (0,2)$, and let $Q = (-q, h)$, $R = (q, h)$ with $q^2 + h^2 = 4$, since $Q$ and $R$ lie on the big arc.

The semicircle on $QR$ has centre $(0,h)$ and radius $q$, and it is drawn downwards. It touches the line $PS$ exactly when its centre is at height equal to its radius: $$h = q, \qquad \text{so} \qquad 2q^2 = 4, \qquad q = h = \sqrt2.$$ In particular that semicircle passes through $O$, which is why the region in the question is named $QXROQ$.

Now the area, in two pieces.

Above the chord $QR$

The points $Q$ and $R$ sit at $135^{\circ}$ and $45^{\circ}$ on the big circle, so the arc $QXR$ spans a right angle. The segment between the chord and that arc is a quarter disc minus the triangle $OQR$: $$\frac{\pi \cdot 2^2}{4} - \frac12 \cdot 2 \cdot 2 = \pi - 2.$$

Below the chord $QR$

That is exactly the small semicircle, of radius $\sqrt2$: $$\frac{\pi\left(\sqrt2\right)^2}{2} = \pi.$$

Adding, $$(\pi - 2) + \pi = 2\pi - 2.$$

Answer $2\pi-2$

Solution: PRMO 2017, Q26

The chord $AB = 6$ lies at distance $\sqrt{25 - 9} = 4$ from the centre, and $CD = 8$ at distance $\sqrt{25 - 16} = 3$. Since the centre lies between them, the region asked for is the disc with the two outer segments removed.

Take the segment cut off by $AB$. Its chord subtends a central angle $2\theta$ with $\sin\theta = \tfrac35$ and $\cos\theta = \tfrac45$, with $\theta$ measured in radians. Subtracting the triangle from the sector gives the segment area $$\tfrac12 r^2\left(2\theta - \sin 2\theta\right) = \tfrac12 \cdot 25\left(2\theta - \tfrac{24}{25}\right) = 25\theta - 12,$$ using $\sin 2\theta = 2 \cdot \tfrac35 \cdot \tfrac45 = \tfrac{24}{25}$.

For $CD$ the half-angle $\phi$ has $\sin\phi = \tfrac45$ and $\cos\phi = \tfrac35$, and $\sin 2\phi$ is again $\tfrac{24}{25}$, so that segment has area $25\phi - 12$.

Now the pleasant part: $\sin\theta = \cos\phi$, so $\theta + \phi = \tfrac{\pi}{2}$ and the two segments together have area $$25(\theta + \phi) - 24 = \frac{25\pi}{2} - 24.$$ Subtracting from the whole disc, $$25\pi - \left(\frac{25\pi}{2} - 24\right) = \frac{25\pi}{2} + 24 = \frac{25\pi + 48}{2}.$$

So $m = 25$, $n = 48$, $k = 2$, whose greatest common divisor is $1$, and $$m + n + k = 75.$$

Answer 75

Solution: PRMO 2018, Q14

Both products are products of sines

Complementary angles convert cosines to sines: $\cos k^{\circ} = \sin(90-k)^{\circ}$. As $k$ runs from $1$ to $89$ so does $90-k$, so the first product is unchanged by the swap: $$x = \prod_{k=1}^{89}\cos k^{\circ} = \prod_{k=1}^{89}\sin k^{\circ}.$$ The angles in $y$ are $2, 6, 10, \ldots, 86$, whose complements are $88, 84, \ldots, 4$, which are the multiples of $4$ up to $88$. Hence $$y = \prod_{j=1}^{22}\sin (4j)^{\circ}.$$ So both products run over evenly spaced angles, and $y$ steps twice as fast as the even part of $x$. That is the whole shape of the problem: doubling is the move.

Doubling, twice

The first factor shows the folding explicitly: $\sin2^\circ=2\sin1^\circ\cos1^\circ=2\sin1^\circ\sin89^\circ$. The next gives $\sin4^\circ=2\sin2^\circ\sin88^\circ$. Each doubled angle supplies a pair at opposite ends of the full list. Let $E$ be the product over the even angles below $90^{\circ}$, $$E = \prod_{j=1}^{44}\sin (2j)^{\circ},$$ and expand each factor with $\sin 2\theta = 2\sin\theta\cos\theta$: $$E = 2^{44}\prod_{j=1}^{44}\sin j^{\circ}\prod_{j=1}^{44}\cos j^{\circ}.$$ The second product is $\prod_{j=1}^{44}\sin(90-j)^{\circ} = \prod_{m=46}^{89}\sin m^{\circ}$, so between them the two products supply every factor of $x$ except the middle one, $\sin 45^{\circ} = 1/\sqrt2$. Hence $$E = 2^{44}\cdot\frac{x}{\sin 45^{\circ}} = 2^{44}\sqrt2\,x = 2^{89/2}\,x.$$

Now do the same to $y$, whose angles are the even angles doubled again: $$y = \prod_{m=1}^{22}\sin (4m)^{\circ} = 2^{22}\prod_{m=1}^{22}\sin (2m)^{\circ}\prod_{m=1}^{22}\cos (2m)^{\circ}.$$ This time $\cos (2m)^{\circ} = \sin(90-2m)^{\circ}$, and $90-2m$ runs through $88, 86, \ldots, 46$, the even angles from $46$ to $88$. The two products between them are the even angles from $2$ to $88$, once each, with no middle factor to divide away this time: $$y = 2^{22}\prod_{j=1}^{44}\sin (2j)^{\circ} = 2^{22}E.$$

The ratio

The two doublings compose, and $E$ cancels: $$\frac{y}{x} = \frac{2^{22}E}{x} = 2^{22}\cdot 2^{89/2} = 2^{133/2}.$$ Neither $x$ nor $y$ was ever evaluated, which is the point: only their ratio is asked for, and the ratio is a power of $2$. So $\log_2(y/x) = 133/2$ and $$\frac{2}{7}\log_2\frac{y}{x} = \frac{2}{7}\cdot\frac{133}{2} = \frac{133}{7} = 19$$ exactly, so the nearest integer is $19$.

Answer 19

Solution: PRMO 2019, Q11

In any triangle all three angles lie strictly between $0^{\circ}$ and $180^{\circ}$, so $\sin A > 0$ and $\sin B > 0$. Using $\sin kC \le 1$, $$\cos A \cos B + \sin A \sin B \sin kC \le \cos A \cos B + \sin A \sin B,$$ and the right-hand side is exactly $\cos(A-B)$, which is at most $1$. For the given equation to hold, both inequalities must be equalities. The first needs $\sin kC = 1$, since it is multiplied by the strictly positive $\sin A \sin B$; the second needs $A = B$.

Now $\sin kC = 1$ means $kC = 90^{\circ} + 360^{\circ}m$ for a non-negative integer $m$, and the constraint $kC \le 360^{\circ}$ leaves only $$kC = 90^{\circ}.$$ So $C$ is a positive integer dividing $90$. Since $A = B$ we also have $C = 180^{\circ} - 2A$, which is even, and conversely any even $C$ with $0 < C < 180$ gives the integer angle $A = B = (180 - C)/2$.

So we need the even divisors of $90 = 2 \cdot 3^2 \cdot 5$. Each is twice a divisor of $45$, and $45$ has $(2+1)(1+1) = 6$ divisors, namely $1, 3, 5, 9, 15, 45$. The even divisors of $90$ are therefore $$2,\ 6,\ 10,\ 18,\ 30,\ 90,$$ giving the six triangles $(A,B,C) = (89,89,2)$, $(87,87,6)$, $(85,85,10)$, $(81,81,18)$, $(75,75,30)$ and $(45,45,90)$, all with integer angles and all genuinely different in shape. Hence there are $6$ triangles up to similarity.

Answer 6

Solution: PRMO 2019, Q23

Where $P$ must be

For any point $X$ in the plane, the triangle inequality gives $XA + XC \ge AC$ with equality exactly when $X$ lies on the segment $AC$, and likewise $XB + XD \ge BD$ with equality exactly on $BD$. Adding, $$XA + XB + XC + XD \ge AC + BD,$$ and both equalities hold together at the point where the diagonals cross, which exists and is interior because $ABCD$ is convex. So $P$ is the intersection of the diagonals, and $$AC = PA + PC, \qquad BD = PB + PD.$$

Which lengths pair up

Because $ABCD$ is cyclic and $P$ lies inside, the intersecting chords give $$PA \cdot PC = PB \cdot PD.$$ Among the products available from $\{3,4,6,8\}$ the only equal pair is $3 \times 8 = 4 \times 6 = 24$, so one diagonal is split into $3$ and $8$ and the other into $4$ and $6$. Hence $$AC = 11, \qquad BD = 10.$$

The area

Let $\theta$ be the angle between the diagonals at $P$. Splitting $ABCD$ into the four triangles meeting at $P$, each of which has two sides along the diagonals and the included angle $\theta$ or $180^{\circ} - \theta$, $$[ABCD] = \tfrac12 \sin\theta\,\bigl(PA \cdot PB + PB \cdot PC + PC \cdot PD + PD \cdot PA\bigr).$$ The bracket factorises as $(PA + PC)(PB + PD)$, which is $AC \cdot BD$, so $$[ABCD] = \tfrac12 \cdot 11 \cdot 10 \cdot \sin\theta = 55 \sin\theta \le 55,$$ with equality exactly when the diagonals are perpendicular.

That case really occurs. Put $P$ at the origin with $A = (3,0)$, $C = (-8,0)$, $B = (0,4)$ and $D = (0,-6)$. This is convex, and since $PA \cdot PC = PB \cdot PD = 24$ the triangles $APB$ and $DPC$ are similar, so $\angle PAB = \angle PDC$ and the four points are concyclic. Hence the maximum area is $$55.$$

Answer 55

Solution: PRMO 2019, Q25

Let the wall have centre $O$ and radius $r$, with $N$ the north gate and $S$ the south gate, so $NS = 2r$ is a diameter. Put the tree at $T$, with $TN = 16$, and the viewing point at $P$, with $SP = 48$. Write $$m = TS = TN + NS = 2r + 16.$$

Three facts pin the picture down.

First, $SP$ is tangent to the circle at $S$. The gates point east and west as well as north and south, so $SP$ runs east from $S$ and is perpendicular to the north-south diameter $NS$ there, which is precisely the tangency condition. In particular the tangent length from $P$ is $48$.

Second, the sight line $TP$ touches the wall, at $Q$ say, because the tree is just seen: any nearer to the village and the wall would hide it. Since $P$ is outside the circle, both tangents from $P$ have length $48$, so $PQ = 48$. And the power of the point $T$, computed along the line $TNS$ which meets the circle at $N$ and $S$, gives $$TQ^2 = TN \cdot TS = 16m.$$

Third, $\angle TSP = 90^{\circ}$, so $TP^2 = m^2 + 48^2$.

Since $Q$ lies between $T$ and $P$, we have $TQ + QP = TP$, that is $$\sqrt{16m} + 48 = \sqrt{m^2 + 2304}.$$ Squaring, the $2304$ cancels and $m^2 = 16m + 384\sqrt m$. Writing $s = \sqrt m > 0$ and dividing by $s$, $$s^3 = 16s + 384, \qquad \text{that is} \qquad s^3 - 16s - 384 = 0.$$ Trying small values, $s = 8$ works, since $512 - 128 - 384 = 0$, and factoring it out leaves $$s^3 - 16s - 384 = (s - 8)(s^2 + 8s + 48),$$ whose quadratic factor has discriminant $64 - 192 < 0$ and so no real root. Hence $s = 8$ is the only possibility, giving $m = 64$, then $2r + 16 = 64$, so $r = 24$ and the diameter is $$2r = 48 \text{ metres}.$$

Answer 48

Solution: IOQM 2020, Q24

A point of the ground is in shadow exactly when the segment joining it to the light source meets the disc, so the shadow runs between the two points where the tangent lines from the source touch down on the $x$-axis. Every line through the source $(0,16)$ has the form $y = mx + 16$, and it is tangent to the disc when its distance from the centre $(6,10)$ equals the radius $2$. To compute that distance, a normal vector to the line is $(m,-1)$, with length $\sqrt{m^2+1}$. The vector from the source $(0,16)$ to the centre is $(6,-6)$. Its component in the normal direction is therefore $|(6,-6)\mathbin{\cdot}(m,-1)|/\sqrt{m^2+1}=|6m+6|/\sqrt{m^2+1}$. So the tangency condition is $$\frac{|6m - 10 + 16|}{\sqrt{m^2+1}} = 2.$$ Squaring and tidying, $36(m+1)^2 = 4(m^2+1)$, which reduces to $9(m+1)^2 = m^2 + 1$ and then to $$4m^2 + 9m + 4 = 0.$$

Rather than solve this, notice what the question needs. Setting $y = 0$ in $y = mx+16$ gives the intercept $x = -16/m$, so the shadow has length $$\left| \frac{-16}{m_1} - \frac{-16}{m_2} \right| = 16 \cdot \frac{|m_1 - m_2|}{|m_1 m_2|},$$ and both of those quantities come straight from the coefficients. The product of the roots is $m_1 m_2 = 4/4 = 1$, and the difference satisfies $(m_1-m_2)^2 = (m_1+m_2)^2 - 4m_1m_2 = \tfrac{81}{16} - 4 = \tfrac{17}{16}$, so $|m_1 - m_2| = \tfrac{\sqrt{17}}{4}$.

The shadow therefore has length $16 \cdot \tfrac{\sqrt{17}}{4} = 4\sqrt{17}$. Since $17$ is prime and so certainly square-free, $m = 4$ and $n = 17$, giving $m + n = 21$.

Answer 21

Solution: IOQM 2021 Part A, Q9

Start with the first four points

The large index suggests looking for something that stays unchanged. First compute a small window so that we know what kind of quadrilateral the rule produces: $$P_0=(3,1), \qquad P_1=(-4,-2), \qquad P_2=(5,3), \qquad P_3=(-6,-4).$$ Read round its boundary as $P_3,P_0,P_2,P_1$, as in the figure. Its opposite sides have the same vectors, $(9,5)$ and $(2,2)$, so it is a parallelogram. Taking $(2,2)$ as the base gives length $2\sqrt2$. A unit vector perpendicular to that base is $(1,-1)/\sqrt2$, so the perpendicular height is the component of $(9,5)$ in that direction: $$h=\left|(9,5)\mathbin{\cdot}\frac{(1,-1)}{\sqrt2}\right|=\frac4{\sqrt2}=2\sqrt2.$$ Hence the first window has area $$(2\sqrt2)(2\sqrt2)=8.$$

Why the next window has the same area

A vertical shear moves each vertical line up or down while keeping every point on that line together. For example, $(x,y)\mapsto(x,y+x)$ moves the line at $x$ upward by $x$. To see why this keeps a triangle’s area, draw the vertical line through its middle vertex $U$, meeting the opposite side at $V$. The two pieces have the same vertical base $UV$. A vertical shear keeps that base’s length and keeps each horizontal height to it, so both triangle areas stay the same. If two vertices share a vertical line, that line is already a base and no split is needed. Splitting a polygon into triangles gives the same conclusion for its area.

A horizontal shear preserves area by the same argument with horizontal bases and vertical heights. Stretching horizontally by a factor of $2$ doubles area; stretching vertically by $1/2$ halves it. Reflecting in an axis does not change area.

Now split our rule into those operations. Starting from $(x,y)$, apply $$(x,y)\longmapsto (x,x+y)=(X,Y),$$ then stretch and reflect to get $$(X,Y)\longmapsto(-2X,-Y/2)=(U,V),$$ and finally apply the horizontal shear $$(U,V)\longmapsto(U-V,V).$$ Their combined effect is $$(x,y)\longmapsto\left(-2x+\frac{x+y}{2},-\frac{x+y}{2}\right) =\left(-\frac{3x-y}{2},-\frac{x+y}{2}\right),$$ exactly the given recurrence. The shears preserve area, and the two stretch factors multiply to $2\cdot(1/2)=1$, so the whole rule preserves area.

It sends the quadrilateral on $P_n,P_{n+1},P_{n+2},P_{n+3}$ to the one on $P_{n+1},P_{n+2},P_{n+3},P_{n+4}$. Repeating the rule ninety-six times therefore carries the first window to the requested window without changing its area. The answer is $8$.

Answer 8

Solution: IOQM 2022, Q11

Because $AB$ is a diameter and $CE$ is perpendicular to it, $AB$ is the perpendicular bisector of $CE$, so $E$ is the mirror image of $C$ and $CE = 2\,CD$. The circle centred at $C$ has radius $CD$, so its intersections $P$ and $Q$ with $\omega$ satisfy $$CP = CQ = CD = \tfrac12\,CE.$$

All four of $P$, $C$, $Q$, $E$ lie on $\omega$, and since $CP = CQ$ the point $C$ is the midpoint of the arc $PQ$ not containing $E$, so the four points lie in the order $P$, $C$, $Q$, $E$ around the circle. What we now need is a relation among the six distances between four points on a circle, and there is exactly one: the product of the diagonals equals the sum of the products of the two pairs of opposite sides. For our four points that reads $$PC \cdot QE + CQ \cdot PE = PQ \cdot CE.$$ The next paragraph proves it and nothing else, so a reader who knows it already may skip to the finish.

The four-point relation, proved where it is used

For a cyclic quadrilateral $WXYZ$ in that order, the claim is $WY \cdot XZ = WX \cdot YZ + XY \cdot WZ$. Take the point $K$ on the diagonal $WY$ with $\angle WXK = \angle YXZ$; since that angle is smaller than $\angle WXY$, the ray $XK$ runs inside the angle at $X$ and so does meet the segment $WY$.

Triangles $WXK$ and $ZXY$ now have two equal angles, the one just arranged and $\angle XWK = \angle XWY = \angle XZY$, both standing on the arc $XY$. So they are similar, and comparing the sides about the equal angles, $$\frac{WK}{ZY} = \frac{WX}{ZX}, \qquad \text{that is} \qquad WK \cdot XZ = WX \cdot YZ.$$ Subtracting the arranged angle from the whole angle at $X$ gives $\angle KXY = \angle WXZ$, and $\angle XYK = \angle XYW = \angle XZW$ stand on the arc $WX$, so triangles $KXY$ and $WXZ$ are similar too, giving $$\frac{KY}{WZ} = \frac{XY}{XZ}, \qquad \text{that is} \qquad KY \cdot XZ = XY \cdot WZ.$$ Adding the two, and using $WK + KY = WY$ because $K$ lies on the segment, $$WY \cdot XZ = WX \cdot YZ + XY \cdot WZ.$$ That is Ptolemy’s theorem, and applying it to $W, X, Y, Z = P, C, Q, E$ gives the identity displayed above.

The finish

Writing $t$ for the common length $PC = CQ$ and $CE = 2t$, $$t\,(QE + PE) = PQ \cdot 2t, \qquad \text{so} \qquad PE + QE = 2\,PQ.$$

The perimeter of triangle $PEQ$ is therefore $PE + QE + QP = 2\,PQ + PQ = 3\,PQ$. Given that it equals $24$, we get $PQ = 8$.

The argument never needed the position of $C$, and it is worth being precise about what that does and does not mean. The relation $PE + QE = 2\,PQ$ holds wherever $C$ sits, so the perimeter is always $3\,PQ$; but the triangle $PEQ$ does change shape. It is equilateral only when $PE = QE$, which puts $E$ on the perpendicular bisector of $PQ$. That bisector is the line joining the centre of $\omega$ to $C$, so the condition is that $C$, the centre and $E$ be collinear, which happens exactly when $C$ is the midpoint of one of the arcs $AB$.

Answer 8

Solution: IOQM 2022, Q23

The right angle at $B$ in triangle $ABE$ does the first job. With $BE = 3$ and $BA = 4$, the angle $\alpha = \angle BAE$ satisfies $\tan\alpha = \tfrac34$, so $\cos\alpha = \tfrac45$. Since $E$ lies on the ray $AD$, this $\alpha$ is also $\angle BAD$, and the given ratio $1:2$ makes $\angle DAC = 2\alpha$.

Now use the median. Because $D$ is the midpoint of $BC$, the triangles $ABD$ and $ADC$ have equal areas. Writing each with the common side $AD$ and the angle at $A$, $$\tfrac12 \cdot AB \cdot AD \sin\alpha = \tfrac12 \cdot AC \cdot AD \sin 2\alpha,$$ and cancelling $AD$ and $\sin\alpha$, which are non-zero, together with $\sin 2\alpha = 2\sin\alpha\cos\alpha$, $$AB = 2\,AC\cos\alpha \qquad\Longrightarrow\qquad AC = \frac{4}{2 \cdot \frac45} = \frac52.$$

Finally apply the cosine rule in triangle $ABC$, whose angle at $A$ is $3\alpha$. Using the triple-angle formula, $\cos 3\alpha = 4\cos^3\alpha - 3\cos\alpha = 4 \cdot \tfrac{64}{125} - \tfrac{12}{5} = -\tfrac{44}{125}$, so $$BC^2 = 4^2 + \left(\tfrac52\right)^2 - 2 \cdot 4 \cdot \tfrac52 \cdot \left(-\tfrac{44}{125}\right),$$ which collects over a denominator of $100$ as $$BC^2 = 16 + \tfrac{25}{4} + \tfrac{176}{25} = \frac{1600 + 625 + 704}{100} = \frac{2929}{100}.$$

No decimal is needed to round this. Since $2900 < 2929 < 2950$, the value lies strictly between $29$ and $29.5$, so the nearest integer is $29$.

Answer 29

Solution: IOQM 2025 Part SEP, Q7

The paper does not state convexity; its keyed answer assumes the convex configuration drawn here. Name the quadrilateral $PQRS$ with $PQ = a = 4\sqrt3$, $QR = b = 9$ and $RS = c = \sqrt3$, so that the angle at $Q$ is $30^{\circ}$ and the angle at $R$ is $90^{\circ}$.

Put $Q$ at the origin and $R$ at $(9,0)$. The side $QP$ leaves $Q$ at $30^{\circ}$ to $QR$, so $$P = 4\sqrt3\left(\cos 30^{\circ}, \sin 30^{\circ}\right) = 4\sqrt3 \left(\tfrac{\sqrt3}{2}, \tfrac12\right) = \left(6,\ 2\sqrt3\right).$$ The side $RS$ leaves $R$ at right angles to $RQ$, on the same side as $P$, so $$S = \left(9,\ \sqrt3\right).$$

Now the diagonals. They are $PR$ and $QS$, with direction vectors $$R - P = \left(3,\ -2\sqrt3\right), \qquad S - Q = \left(9,\ \sqrt3\right).$$ Their dot product is $27 - 2\sqrt3 \cdot \sqrt3 = 21$, while their lengths are $\sqrt{9+12} = \sqrt{21}$ and $\sqrt{81+3} = \sqrt{84}$. Hence $$\cos x^{\circ} = \frac{21}{\sqrt{21}\sqrt{84}} = \frac{21}{\sqrt{1764}} = \frac{21}{42} = \frac12,$$ so the acute angle is $60^{\circ}$ and $x = 60$.

Answer 60

Solution: IOQM 2025 Part SEP, Q7

Place $B = (0,0)$, $A = (0,1)$ and $C = (2,0)$, so that $BC$ lies along the horizontal axis and the circle through $B$ centred at $A$ is $x^2 + (y-1)^2 = 1$.

Let the square have side $s$, with $D = (d, 0)$ and $E = (e, 0)$ on $BC$ and the other two vertices directly above them, $$F = (e, s), \qquad G = (d, s), \qquad s = e - d.$$

Now use the two conditions. The line $AC$ is $y = 1 - \tfrac x2$, so $F$ on it gives $$s = 1 - \frac e2, \qquad e = 2(1-s).$$ And $G$ on the circle gives $$d^2 + (s-1)^2 = 1.$$ Together with $d = e - s = 2 - 3s$, the second condition becomes $$(2-3s)^2 = 1 - (1-s)^2 = 2s - s^2,$$ that is $10s^2 - 14s + 4 = 0$, or $5s^2 - 7s + 2 = 0$. The roots are $$s = 1 \qquad \text{and} \qquad s = \tfrac25.$$

The root $s = 1$ puts $d = 2 - 3 = -1$, which is outside the segment $BC$, so it is rejected. With $s = \tfrac25$ we get $d = \tfrac45$ and $e = \tfrac65$, both between $0$ and $2$ and in the right order, and the area is $$s^2 = \frac{4}{25}.$$

So $m = 4$, $n = 25$ and $m + n = 29$.

Answer 29

Solution: PRMO 2013, Q17

The size of everything

Let $O$ be the centre of $S$ and $R$ its radius. The circle $S_2$ touches $S$ internally, so its centre $O_2$ satisfies $OO_2 = R - 100$; and it passes through $O$, so $OO_2 = 100$. Hence $$R = 200.$$

Set up coordinates with $O$ at the origin and $AB$ the horizontal line $y = d$, where $d > 0$ is its distance from the centre, so that $R_1$ is the region above it. Since $S_2$ touches $AB$ from below, its centre is $100$ lower: $$O_2 = (u, d - 100), \qquad u^2 + (d-100)^2 = 100^2,$$ the second equation being $OO_2 = 100$.

Where the $30^{\circ}$ goes

The point $Y$ is directly above $O_2$, so $XY$ lies along $AB$, which is horizontal. The point $X$ lies on the line through $O$ and $O_2$, so the segment $XO$ has slope $(d-100)/u$ in absolute value. The angle between $XO$ and the horizontal $XY$ is therefore $$\tan 30^{\circ} = \left|\frac{d - 100}{u}\right|, \qquad |u| = \sqrt3\,|d - 100|.$$ Combined with $u^2 + (d-100)^2 = 10^4$ this gives $(d-100)^2 = 2500$, so $d = 150$ or $d = 50$.

The first fails, and squares settle it without decimals. With $d = 150$ we have $|u| = 50\sqrt3$, so the point $X$, which sits at $x$-coordinate $ud/(d-100)$, has $$|X_x|^2 = \left(50\sqrt3 \cdot \tfrac{150}{50}\right)^2 = \left(150\sqrt3\right)^2 = 67500,$$ while the chord at that height reaches only to $200^2 - 150^2 = 17500$ in the same units. As $67500 > 17500$, the point misses the chord. So $$d = 50, \qquad |u| = 50\sqrt3, \qquad X = (\mp 50\sqrt3,\ 50),$$ and this one does lie on the chord, since $\left(50\sqrt3\right)^2 = 7500$ is comfortably below $200^2 - 50^2 = 37500$.

The circle $S_1$

Its centre is directly above $X$ at height $50 + r_1$, and touching $S$ internally puts it at distance $200 - r_1$ from $O$: $$7500 + \left(50 + r_1\right)^2 = \left(200 - r_1\right)^2.$$ Expanding, the $r_1^2$ terms cancel and $$10000 + 100r_1 = 40000 - 400r_1, \qquad 500 r_1 = 30000, \qquad r_1 = 60.$$

Answer 60

Solution: PRMO 2015 Part A, Q19

Let $T$ be the point where $\omega$ touches the diameter $XY$, and $r$ and $R$ the radii of $\omega$ and $\Omega$.

The first step is the standard lemma about a circle inscribed in a segment. Because the two circles are internally tangent at $P$, they share a tangent line there, and their centres lie with $P$ on one line. The scaling centred at $P$ that multiplies all distances from $P$ by $R/r$ therefore carries the centre of $\omega$ to the centre of $\Omega$ and the radius $r$ to the radius $R$, so it carries $\omega$ to $\Omega$. Under it the tangent line $XY$ at $T$ goes to a parallel tangent line of $\Omega$, and the image of $T$ is the point $M$ of $\Omega$ where that parallel tangent touches. Since the image lies on the ray from $P$ through $T$ and beyond, $M$ is the endpoint of the diameter perpendicular to $XY$ on the opposite side from $\omega$, in other words the midpoint of the arc $XY$ not containing $P$.

Equal arcs $XM$ and $MY$ subtend equal angles at $P$, so $PM$, and hence the line $PT$, bisects $\angle XPY$. But $XY$ is a diameter, so $\angle XPY = 90^{\circ}$, and therefore $$\angle TPY = 45^{\circ}.$$

Now bring in the given ratio. Let $\theta = \angle PYX$. The line $YP$ meets $\omega$ at $Z$ and at $P$, and $YT$ is tangent to $\omega$, so the power of the point $Y$ gives $$YT^2 = YZ \cdot YP.$$ With $YZ = 2\,PZ$ we have $YP = YZ + ZP = 3\,PZ$, so $YT^2 = 2PZ \cdot 3PZ = 6\,PZ^2$, that is $$YT = \sqrt{6}\; PZ, \qquad YP = 3\,PZ = \frac{3}{\sqrt6}\,YT = \frac{\sqrt6}{2}\,YT.$$

Finally look at triangle $TPY$. Its angle at $P$ is $45^{\circ}$, its angle at $Y$ is $\theta$, and so its angle at $T$ is $135^{\circ} - \theta$. The sine rule gives $$\frac{YT}{\sin 45^{\circ}} = \frac{YP}{\sin(135^{\circ} - \theta)},$$ and substituting $YP = \tfrac{\sqrt6}{2}YT$ and cancelling $YT$, $$\sin(135^{\circ} - \theta) = \frac{\sqrt6}{2}\sin 45^{\circ} = \frac{\sqrt6}{2}\cdot\frac{\sqrt2}{2} = \frac{\sqrt3}{2}.$$ So $135^{\circ} - \theta$ is $60^{\circ}$ or $120^{\circ}$, giving $\theta = 75^{\circ}$ or $\theta = 15^{\circ}$.

The condition $PY > PX$ decides between them. In the right triangle $XPY$ with hypotenuse $XY = 2R$ we have $PY = 2R\cos\theta$ and $PX = 2R\sin\theta$, so $PY > PX$ means $\theta < 45^{\circ}$. Hence $$\angle PYX = 15^{\circ}.$$

The condition that $O$ lies outside $\omega$ is what makes $\omega$ small enough to sit inside one of the two half-discs and be tangent to $XY$ at a single interior point, which is the picture the argument assumes throughout.

Answer 15

Solution: PRMO 2017, Q27

Let $R$ be the radius of $\Omega_1$ and $r$ that of $\Omega_2$. Since $AB$ is a diameter, the angle it subtends at any other point of $\Omega_1$ is a right angle, so triangles $ABA_1$ and $ABB_1$ are right-angled at $A_1$ and $B_1$.

The first tangent

In the right triangle $ABA_1$, $$\sin\angle A_1AB = \frac{A_1B}{AB} = \frac{5}{2R}.$$ The line $AA_1$ is tangent to $\Omega_2$, so the distance from $P$ to it equals $r$. That distance is the leg of a right triangle with hypotenuse $AP$ and the angle $\angle A_1AB$ at $A$: $$r = AP \sin\angle A_1AB = (R + 10)\cdot\frac{5}{2R},$$ since $P$ lies on $OB$ with $OP = 10$, so $AP = R + 10$.

The second tangent

Identically, in the right triangle $ABB_1$ we have $\sin\angle B_1BA = \tfrac{AB_1}{AB} = \tfrac{15}{2R}$, and $BP = R - 10$, so $$r = (R - 10)\cdot\frac{15}{2R}.$$

Solving

Equating the two expressions and cancelling $\tfrac{1}{2R}$, $$5(R+10) = 15(R-10), \qquad 5R + 50 = 15R - 150, \qquad R = 20.$$

As a check, $r$ then comes out as $\tfrac{5 \times 30}{40} = 3.75$ from both formulas, and the circle of radius $3.75$ about a point $10$ from the centre does sit inside a circle of radius $20$.

Answer 20

Solution: IOQM 2021 Part B, Q1

Write $A$, $B$, $C$ for the angles of the triangle, $a$, $b$, $c$ for its sides, and $\Omega$ for its circumcircle.

This is a long problem, so here is the route before we set off. There are two incentres to place on one circle, and they are placed one at a time and by quite different means. We first show that $I_1$ lies on the circle through $A$, $E$ and $F$, which turns out to need no angles at all, only lengths; then we show separately that $I$ lies on that same circle, which is where the angles come in. Four claims do it:

• $CD_1 = CE_1$ and $BD_1 = BF_1$.

• $I_1$ is the reflection of $D_1$ in the line $BC$.

• $A$, $E$, $F$, $I_1$ lie on one circle.

• $A$, $E$, $F$, $I$ lie on that same circle.

The last line of the solution collects them.

Setting up: two powers of a point

Read the power of $C$ with respect to the circle $ADBE$ along its two lines through $C$: $$CE \cdot CA = CD \cdot CB, \qquad \text{so} \qquad CE = \frac{CD \cdot a}{b},$$ and $E$ lies on the ray $CA$. The same reading at $B$ gives $$BF \cdot BA = BD \cdot BC,$$ with $F$ on the ray $BA$. Both identities are used again at the end.

One consequence is worth recording now: $E$ and $F$ cannot both fall outside their sides. For $E$ falls beyond $A$ exactly when $CD \cdot a \ge b^2$, and $F$ beyond $A$ exactly when $BD \cdot a \ge c^2$; adding those and using $CD + BD = a$ would give $a^2 \ge b^2 + c^2$, which an acute triangle forbids. Here is a concrete reason to keep both configurations in view. Take $$a = 5, \qquad b = c = 4, \qquad BD = \tfrac{1}{10},$$ which is an acute triangle with $D$ properly inside $BC$. Then $CD = \tfrac{49}{10}$ and the power of $C$ gives $$CE = \frac{CD \cdot a}{b} = \frac{49}{10}\cdot\frac54 = \frac{49}{8} = 6.125 > 4 = b,$$ so $E$ lies beyond $A$ on the ray $CA$. That is why nothing below assumes both $E$ and $F$ fall inside their sides.

The tool: a chord and its inscribed angle

In a circle of radius $\rho$, a chord $XY$ and any point $Z$ of the circle satisfy $XY = 2\rho \sin \angle XZY$. Two points of the circle on opposite arcs give angles that are supplementary rather than equal, but the sine does not notice, which is exactly why this survives every configuration. Every step below uses it and nothing sharper.

Claim 1: $CD_1 = CE_1$ and $BD_1 = BF_1$

Take $CD_1$ and $CE_1$, both chords of $\Omega$. The first subtends $\angle CAD_1$ at $A$, and the line $AD_1$ is the line $AD$, so $CD_1 = 2R \sin \angle CAD$. The second subtends $\angle CBE_1$ at $B$, and the line $BE_1$ is the line $BE$, so $CE_1 = 2R \sin \angle DBE$. Now use the circle $ADBE$: the chord $DE$ subtends $\angle DBE$ at $B$ and $\angle DAE$ at $A$, so those two sines agree; and the line $AE$ is the line $AC$, so $\sin \angle DAE = \sin \angle DAC$. Hence $$CE_1 = 2R \sin \angle DAC = CD_1.$$ The mirror argument through the circle $ADCF$ gives $BF_1 = BD_1$.

Claim 2: $I_1$ is the reflection of $D_1$ in $BC$

First, sides. The point $E$ lies on the ray $CA$, so it is on the same side of $BC$ as $A$; the ray $BE$ therefore leaves $B$ into that side, and $E_1$ lies on the arc $BC$ containing $A$. The same holds for $F_1$. Meanwhile $D$ is interior to $BC$, so $D_1$ lies on the other arc.

Next, angles. Equal chords cut equal arcs, so $CD_1 = CE_1$ gives $\angle D_1BC = \angle E_1BC$, and $BD_1 = BF_1$ gives $\angle D_1CB = \angle F_1CB$. Each pair is measured on opposite sides of $BC$, so the line $BE_1$ is the mirror image of the line $BD_1$ in $BC$, and $CF_1$ is the mirror image of $CD_1$.

Let $J$ be the reflection of $D_1$ in $BC$. Reflection fixes $B$ and carries the line $BD_1$ to the line $BE_1$, so $J$ lies on $BE_1$; likewise on $CF_1$. Hence $$J = BE_1 \cap CF_1, \qquad BJ = BD_1, \qquad CJ = CD_1.$$

It remains to say that $J$ is the incentre and not something else. Since $BD_1 = BF_1$, the point $B$ is a midpoint of an arc $D_1F_1$, so the equal arcs give $\angle D_1E_1B = \angle BE_1F_1$ and the line $BE_1$ is one of the two bisectors of the angle at $E_1$; likewise $CF_1$ at $F_1$. A bisector at $E_1$ meets a bisector at $F_1$ in the incentre or in one of the three excentres. Each excentre is outside $\Omega$: the excentre opposite a vertex of angle $\phi$ lies across the opposite side and sees it under $90^\circ-\phi/2$. Points of the circle on that side see it under $180^\circ-\phi$, a larger angle, so the excentre lies beyond the circle. Meanwhile the incentre is inside the triangle and so inside $\Omega$. And $J$ is inside $\Omega$: it sees $BC$ under $\angle BJC = \angle BD_1C = 180^{\circ} - A$, whereas the points of $\Omega$ on that side of $BC$ see it under $A$, and $180^{\circ} - A > A$ because the triangle is acute. A larger angle means nearer, so $J$ is inside. Therefore $$J = I_1.$$

Claim 3: $A$, $E$, $F$, $I_1$ lie on one circle

The whole claim rests on one length identity, $BE \cdot BD_1 = BD \cdot BC$, so take that first. In the circle $ADBE$, of radius $\rho$, the chords $BE$ and $BD$ subtend $\angle BAE$ and $\angle BAD$ at $A$, so $$BE = 2\rho \sin \angle BAE, \qquad BD = 2\rho \sin \angle BAD,$$ and the line $AE$ is the line $AC$, so $\sin \angle BAE = \sin A$. In $\Omega$, $$BD_1 = 2R \sin \angle BAD, \qquad BC = 2R \sin A.$$ Multiplying the first pair against the second, $$BE \cdot BD_1 = \frac{BD \sin A}{\sin \angle BAD} \cdot 2R \sin \angle BAD = BD \cdot 2R \sin A = BD \cdot BC.$$

Now $I_1$ lies on the line $BE$ with $BI_1 = BD_1$, and $I_1$ and $E$ are on the same side of $B$, both being on the $A$ side of $BC$. So $BE \cdot BI_1 = BD \cdot BC$, and the power of $B$ at the start of the solution turns that into $$BE \cdot BI_1 = BF \cdot BA.$$ That is the converse of the power of a point at $B$, applied to the lines $BEI_1$ and $BFA$: $$A, \quad E, \quad F, \quad I_1 \quad \text{lie on one circle.}$$ Not one angle was chased, so no configuration can spoil it.

Claim 4: the incentre $I$ lies on that circle too

The angle between the line $DE$ and the line $BC$ equals the angle between $AC$ and $AB$: the chord $DE$ of the circle $ADBE$ gives $\sin \angle EDB = \sin \angle EAB$, and the line $AE$ is the line $AC$. As $E$ is on the $A$ side and $D$ is between $B$ and $C$, this reads $\angle EDC = A$, and in the same way $\angle FDB = A$. The three angles at $D$ above $BC$ add to a straight angle, so $$\angle EDF = 180^{\circ} - 2A, \qquad \angle EIF = 90^{\circ} + \tfrac12 \angle EDF = 180^{\circ} - A,$$ the second because the incentre lies on the bisectors from $E$ and from $F$, which halves those two angles of triangle $DEF$. Also $I$ is inside triangle $DEF$, so $I$ and $D$ are on the same side of the line $EF$.

Two positions for $A$ remain, and they pair up exactly.

If $E$ and $F$ both lie inside their sides, the line $EF$ cuts triangle $AEF$ off the corner at $A$, so $A$ is on the far side of $EF$ from $B$, $C$ and hence from $D$ and $I$. Here $\angle EAF = A$, which is supplementary to $\angle EIF$, and opposite angles of $AEIF$ summing to $180^{\circ}$ make it cyclic.

If instead one of them lies outside, say $E$ beyond $A$, then the ray $AE$ is opposite to $AC$ and $\angle EAF = 180^{\circ} - A$, equal to $\angle EIF$; so this time $A$ must be on the same side of $EF$ as $I$, and it is. Since $E$ is not between $A$ and $C$ but $F$ is between $A$ and $B$, the line $EF$ leaves $A$ and $C$ together and puts $B$ opposite, so it crosses $BC$ at an interior point $G$, and what is needed is $BG < BD$. Writing $u = BD$, $v = DC$ and using the transversal ratio on triangle $ABC$ cut by the line $EFG$, $$\frac{BG}{GC} = \frac{FB}{AF} \cdot \frac{EA}{CE} = \frac{ua}{c^2 - ua} \cdot \frac{va - b^2}{va} = \frac{u\,(va - b^2)}{v\,(c^2 - ua)},$$ using $BF = ua/c$ and $CE = va/b$ from the two powers. Then $BG < BD$ says $(va - b^2) < (c^2 - ua)$, that is $a(u+v) < b^2 + c^2$, that is $$a^2 < b^2 + c^2,$$ which is exactly what an acute angle at $A$ provides. So $D$, and with it $I$, lies on the same side of $EF$ as $A$, and equal angles on one side of $EF$ make $AEIF$ cyclic again.

Finishing

Claim 3 put $I_1$ on the circle through $A$, $E$ and $F$; Claim 4 put $I$ on it. In particular $$E, \quad F, \quad I, \quad I_1$$ are concyclic, which is what was asked, and rather more is true: $A$ lies on that circle as well.

The remaining tool

The transversal ratio used once above is the companion of Ceva proved in Chapter 7, and it follows the same way: drop perpendiculars from $A$, $B$ and $C$ to the line $EFG$, read each of the three ratios as a ratio of two of those distances, and multiply, whereupon everything cancels.

Answer proof

Solution: IOQM 2023, Q15

Put $A$ at the origin with $B = (1,0)$, $C = (1,1)$ and $D = (0,1)$, and write $M = (1,m)$ on $BC$ and $N = (n,1)$ on $CD$. Set $u = 1-m$ and $v = 1-n$, the two legs $MC$ and $NC$ of the corner triangle. The perimeter condition is $$u + v + \sqrt{u^2+v^2} = 2.$$ Squaring $\sqrt{u^2+v^2} = 2 - u - v$ and cancelling $u^2+v^2$ from both sides leaves $$2uv - 4u - 4v + 4 = 0, \qquad \text{that is} \qquad uv - 2u - 2v + 2 = 0.$$ Rewriting this in terms of $m = 1-u$ and $n = 1-v$ turns it into $m + n + mn = 1$, and since $\tan \angle BAM = m$ and $\tan \angle DAN = n$, the addition formula gives $$\tan\bigl(\angle BAM + \angle DAN\bigr) = \frac{m+n}{1-mn} = 1.$$ Both angles are acute and their sum is less than $90^{\circ}$, so $\angle BAM + \angle DAN = 45^{\circ}$ and hence $$\angle MAN = 90^{\circ} - 45^{\circ} = 45^{\circ}.$$

Everything now follows from that one angle. Let $O$ be the circumcentre of $AMN$ and $R = OA = OM = ON$ its circumradius. The central angle standing on the chord $MN$ is twice the inscribed angle at $A$, so $$\angle MON = 2 \cdot 45^{\circ} = 90^{\circ}.$$ Triangle $MON$ is therefore right-angled at $O$ with the two equal legs $OM = ON = R$. The circumcentre of a right triangle is the midpoint of its hypotenuse, so $P$ is the midpoint of $MN$, and in an isosceles right triangle the distance from the right-angle vertex to that midpoint is half the hypotenuse: $$OP = \tfrac12 MN = \tfrac12 \cdot R\sqrt2 = \frac{R}{\sqrt2}.$$

Hence $\left(\dfrac{OP}{OA}\right)^2 = \dfrac{R^2/2}{R^2} = \dfrac12$, so $m = 1$, $n = 2$ and $m + n = 3$.

Note that the positions of $M$ and $N$ never had to be found. The perimeter condition was worth exactly one angle, and that angle was worth the whole problem.

Answer 3

Solution: IOQM 2023, Q23

Since the angle at $A$ is right and the incircle touches both legs, the incentre lies on the bisector of that angle at distance $r\sqrt2$ from $A$, where $r$ is the inradius. The incentre is the origin, so $$r\sqrt2 = OA = \sqrt{12^2 + 84^2} = \sqrt{7200} = 60\sqrt2, \qquad r = 60.$$

Now find the directions of the legs, since $B$ and $C$ have to be lattice points and it is the directions that decide which lattice points are even available. The unit vector from $A$ towards $O$ is $$\frac{(-12,-84)}{60\sqrt2} = \frac{(-1,-7)}{5\sqrt2},$$ and the two legs make angles of $45^{\circ}$ with it on either side, since $AO$ bisects the right angle. Turning a vector $(x,y)$ through $45^{\circ}$ sends it to $\left(\tfrac{x-y}{\sqrt2},\ \tfrac{x+y}{\sqrt2}\right)$, and through $-45^{\circ}$ to $\left(\tfrac{x+y}{\sqrt2},\ \tfrac{y-x}{\sqrt2}\right)$. Applying both to $\tfrac{(-1,-7)}{5\sqrt2}$, every $\sqrt2$ meets another and disappears: $$\frac{1}{5\sqrt2}\left(\frac{-1+7}{\sqrt2},\ \frac{-1-7}{\sqrt2}\right) = \frac{(3,-4)}{5},$$ $$\frac{1}{5\sqrt2}\left(\frac{-1-7}{\sqrt2},\ \frac{-7+1}{\sqrt2}\right) = \frac{(-4,-3)}{5}.$$ That is the pleasant surprise of this problem: both directions are rational, and $(3,-4)$ and $(-4,-3)$ are lattice vectors of length $5$ with no common factor to strip out, so the lattice points along each leg are spaced exactly one such vector apart. They are therefore $$B = A + t(3,-4), \qquad C = A + w(-4,-3),$$ with $t$ and $w$ positive integers, and every such choice puts $B$ and $C$ at lattice points with $AB = 5t$ and $AC = 5w$.

One condition is left, that the incircle also touches $BC$. For a right triangle the inradius is half of (sum of legs minus hypotenuse), so with $BC = 5\sqrt{t^2+w^2}$, $$60 = \tfrac12\left(5t + 5w - 5\sqrt{t^2+w^2}\right),$$ that is, $$t + w - 24 = \sqrt{t^2+w^2}.$$ Squaring and cancelling $t^2 + w^2$, $$0 = 576 + 2tw - 48t - 48w, \qquad \text{so} \qquad (t-24)(w-24) = 288.$$

Both factors must be positive. If both were negative, writing $t = 24-a$ and $w = 24-b$ with $a, b > 0$ and $ab = 288$ would need $a+b<24$, because the unsquared equation requires $t+w-24=24-a-b=\sqrt{t^2+w^2}>0$. But $(a-b)^2 \ge 0$ gives $$(a+b)^2 \ge 4ab = 1152 > 576 = 24^2,$$ so $a + b > 24$, a contradiction.

So the solutions correspond exactly to the ways of writing $288$ as an ordered product of two positive integers, and $288 = 2^5 \cdot 3^2$ has $$(5+1)(2+1) = 18$$ divisors. Each gives a different pair $(t,w)$, hence a different triangle, so the answer is $18$.

Answer 18

Solution: IOQM 2023, Q25

For a square there is one perpendicular crossing pair, its two diagonals. The same four vertices can be listed starting at any of the four corners, so counting a starting corner counts that one pair four times. This is the repetition to keep track of in a larger polygon.

Any two diagonals that cross inside the polygon are determined by the four vertices they use: for vertices $P, Q, R, T$ in cyclic order the crossing pair is $PR$ and $QT$. So we may count four-element subsets instead of pairs of diagonals.

Take such a subset and let $a, b, c, d$ be the four arcs it cuts, measured in steps around the polygon, so that $a+b+c+d = n$ with each part at least $1$. Place the vertices at $P, Q = P + a$, $R = P + a + b$, $T = P+a+b+c$. A chord joining vertices $i$ and $j$ is perpendicular to the radius bisecting it, so its direction, as an angle taken modulo $180^{\circ}$, is determined by $i + j$ modulo $n$; two chords are perpendicular exactly when their index sums differ by $n/2$ modulo $n$. In particular $n$ must be even, say $n = 2m$, or no two diagonals are ever perpendicular.

For our two chords the index sums are $P + R$ and $Q + T$, and $$(P+R) - (Q+T) = (P - Q) + (R - T) = -a - c.$$ So the perpendicularity condition is $a + c \equiv m \pmod{n}$, and since $2 \le a + c \le n-2$ this says $$a + c = m, \qquad \text{and hence also} \qquad b + d = m.$$

Counting is now easy. Choosing the starting vertex $P$ in $n$ ways and then the arcs, the pairs $(a,c)$ with $a, c \ge 1$ and $a + c = m$ number $m-1$, and likewise for $(b,d)$. Each four-element subset is produced four times, once for each of its vertices playing the role of $P$, so the number of perpendicular crossing pairs is $$N(n) = \frac{n(m-1)^2}{4} = \frac{m(m-1)^2}{2}, \qquad n = 2m.$$

This increases with $m$, so we need the least $m$ with $m(m-1)^2 \ge 2000$. Since $$13 \cdot 12^2 = 1872 < 2000 \le 2366 = 14 \cdot 13^2,$$ that is $m = 14$, giving $N = 1183 \ge 1000$ while $m = 13$ gives only $936$. Hence $$n = 2m = 28.$$

Answer 28

Solution: IOQM 2025 Part SEP, Q7

The four points $C,D,M,N$ lie on one circle. Equal chords subtend equal angles, so $MD=MN$ gives $$\angle MCD=\angle MCN.$$ The rays $CD$ and $CN$ follow the perpendicular sides of the rectangle, and $CM$ lies between them. The two equal angles therefore add to $90^\circ$, so each is $45^\circ$.

Triangle $MBC$ is right-angled at $B$, with $\angle MCB=\angle MCN=45^\circ$. Hence $$BC=MC\cos45^\circ=\frac{MC}{\sqrt2}.$$ The given equality $MC=CD$, and the rectangle’s $CD=AB$, now give $$BC=\frac{AB}{\sqrt2}, \qquad \left(\frac{AB}{BC}\right)^2=2.$$ Thus $m=2$, $n=1$ and $m+n=3$.

Answer 3

Solution: IOQM 2025 Part SEP, Q7

The two small circles meet at $A$ and $B$, so $AB$ is a chord of both, and the whole picture is symmetric about the line through the two centres, which is the perpendicular bisector of $AB$. That symmetry is what to put on the axes.

Let $AB$ lie along the vertical line $x = 0$, with $A = (0,k)$ and $B = (0,-k)$. The centres $O_1$ and $O_2$ lie on the perpendicular bisector of $AB$, so write them as $(p,0)$ and $(q,0)$, and then $$r_1^2 = p^2+k^2, \qquad r_2^2 = q^2+k^2.$$ The condition $\angle OAB = 90^{\circ}$ says $OA$ is perpendicular to the vertical line $AB$, so $O$ is at the same height as $A$: write $O = (o, k)$ with $R = 10$.

Internal tangency of $S_1$ with $S$ means $OO_1 = R - r_1$, that is $$(o-p)^2 + k^2 = (R - r_1)^2 = R^2 - 2Rr_1 + r_1^2 = R^2 - 2Rr_1 + p^2 + k^2.$$ Cancelling and rearranging, $$2Rr_1 = R^2 - o^2 + 2op. \tag{1}$$ Square $r_1 = \dfrac{R^2 - o^2 + 2op}{2R}$ and use $r_1^2 = p^2+k^2$. Writing $D = R^2 - o^2$ and keeping everything for the moment, $$4R^2\left(p^2 + k^2\right) = \left(D + 2op\right)^2,$$ which expands and rearranges to $$4D\,p^2 - 4Dop + 4R^2k^2 - D^2 = 0, \tag{2}$$ using $4R^2 - 4o^2 = 4D$ on the $p^2$ term.

Before dividing by $D$, check that it is not zero. If $D = 0$ then $(2)$ collapses to $4R^2k^2 = 0$, and $R = 10$, so $k = 0$. But $A = (0,k)$ and $B = (0,-k)$, so $k = 0$ would make $A$ and $B$ the same point, and the two circles are given as meeting at two distinct points. Hence $D \neq 0$, and dividing $(2)$ by $4D$ is legitimate: $$p^2 - op + \frac{R^2k^2}{D} - \frac D4 = 0.$$ The identical computation for the second circle produces the same equation with $q$ in place of $p$. So $p$ and $q$ are the two roots of one quadratic $$t^2 - ot + \left(\frac{R^2k^2}{D} - \frac D4\right) = 0,$$ and they are distinct because the circles are. By Vieta, $$p + q = o.$$

Now add the two copies of $(1)$: $$2R(r_1+r_2) = 2\left(R^2 - o^2\right) + 2o(p+q) = 2R^2 - 2o^2 + 2o^2 = 2R^2,$$ so $r_1 + r_2 = R = 10$, whatever the two circles happen to be.

Answer 10

Solution: IOQM 2026, Q27

Call the quadrilateral $ABCD$, with $AB = \sqrt a$, $BC = \sqrt{a + 3}$, $CD = \sqrt{a + 2}$, $DA = \sqrt{2a + 5}$, and both diagonals $\sqrt{2a + 5}$.

Two things can be read off. First, $BC^2 + CD^2 = 2a + 5 = BD^2$, so $\angle BCD = 90^{\circ}$. Second, $AC = AD$, so triangle $ACD$ is isosceles and $\angle ADC = \angle ACD$.

Finding $a$. The diagonal $CA$ splits the right angle at $C$ into $\angle BCA$ and $\angle ACD$, so the cosine of one is the sine of the other, and $\cos^2 \angle BCA + \cos^2 \angle ACD = 1$. The cosine rule in triangle $ABC$ gives $$\cos \angle BCA = \frac{(a + 3) + (2a + 5) - a}{2\sqrt{(a + 3)(2a + 5)}} = \frac{a + 4}{\sqrt{(a + 3)(2a + 5)}},$$ and in the isosceles triangle $ACD$ the perpendicular from $A$ meets $CD$ at its midpoint, so $\cos \angle ACD = \dfrac{\sqrt{a + 2}}{2\sqrt{2a + 5}}$. Squaring and adding, $$\frac{(a + 4)^2}{(a + 3)(2a + 5)} + \frac{a + 2}{4(2a + 5)} = 1.$$ Multiplying by $4(a+3)(2a+5)$ gives $$4(a+4)^2+(a+2)(a+3)=4(a+3)(2a+5).$$ Expanding the two sides, $$5a^2+37a+70=8a^2+44a+60, \qquad 3a^2+7a-10=0.$$ Thus $(a-1)(3a+10)=0$. Since $a>0$ is the square of a side length, $a=1$.

The angles. Now $AB = 1$, $BC = 2$, $CD = \sqrt3$ and $AC = \sqrt7$, and the cosine rule in triangle $ABC$ gives $$\cos \angle ABC = \frac{1 + 4 - 7}{2 \cdot 1 \cdot 2} = -\frac12, \qquad \angle ABC = 120^{\circ}.$$ The angle at $C$ is $90^{\circ}$. The angle at $D$ equals $\angle ACD$, a part of the right angle, so it is less than $90^{\circ}$, and its cosine is $\sqrt3/(2\sqrt7)$, less than $\tfrac12$, so it is more than $60^{\circ}$. The angle at $A$ is $360^{\circ} - 120^{\circ} - 90^{\circ}$ minus the angle at $D$, so it is less than $90^{\circ}$ too.

The largest angle is $120^{\circ}$, the second largest $90^{\circ}$, and $\theta = 30$.

Answer 30