Inequalities and Optimisation

Problems

Solutions

Solution: PRMO 2012, Q2

The sides are positive integers summing to $7$, and there are exactly four ways to write $7$ as a sum of three of them, up to order: $$(1,1,5), \qquad (1,2,4), \qquad (1,3,3), \qquad (2,2,3).$$ The first two fail the triangle inequality, since $1 + 1 < 5$ and $1 + 2 < 4$. The other two are genuine triangles.

For $(1,3,3)$, take the side of length $1$ as the base. The altitude bisects it, so Pythagoras gives the height $h$: $$h^2 = 3^2 - \left(\tfrac12\right)^2 = \tfrac{35}{4}, \qquad \text{area} = \tfrac12 \cdot 1 \cdot \tfrac{\sqrt{35}}2 = \tfrac{\sqrt{35}}4.$$ For $(2,2,3)$, take the side of length $3$ as the base. Its altitude gives $$h^2 = 2^2 - \left(\tfrac32\right)^2 = \tfrac74, \qquad \text{area} = \tfrac12 \cdot 3 \cdot \tfrac{\sqrt7}2 = \tfrac{3\sqrt7}4.$$ No decimals are needed to compare them: $\tfrac{3\sqrt7}{4} = \tfrac{\sqrt{63}}{4}$, and $63 > 35$. So the maximum area is $\tfrac{3\sqrt7}{4}$.

This is one of several answers on this paper that are not whole numbers. The Mumbai regional papers of this period did not use the two-digit answer format that the national PRMO adopted later, and the questions were written accordingly.

Answer $3\sqrt7/4$

Solution: IOQM 2020, Q5

Nested absolute values look forbidding, but they unwrap one layer at a time. The outer condition $\bigl| |x| - 2020 \bigr| < 5$ says that the quantity $|x| - 2020$ lies strictly between $-5$ and $5$, which is to say that $|x|$ itself lies strictly between $2015$ and $2025$.

Since $|x|$ is a whole number whenever $x$ is, the possibilities for $|x|$ are the integers from $2016$ up to $2024$ inclusive, and there are $2024 - 2016 + 1 = 9$ of them. It is only now, at the last step, that we return to $x$. Each admissible value $v$ of $|x|$ arises from exactly two integers, $x = v$ and $x = -v$, and these are genuinely different because $v$ is not zero.

The count is therefore $9 \times 2 = 18$.

Answer 18

Solution: IOQM 2025 Part SEP, Q7

An integer-sided rectangle of area $20$ has sides $d$ and $20/d$ for a divisor $d$ of $20$, and its perimeter is $2(d + 20/d)$. The three shapes are $$1 \times 20 \ (\text{perimeter } 42), \qquad 2 \times 10 \ (24), \qquad 4 \times 5 \ (18).$$

So the minimum is $18$. The pattern is worth naming: for a fixed product the sum $d + 20/d$ is smallest when the two factors are closest together, which is the discrete shadow of the fact that among all rectangles of a given area the square has the least perimeter.

Answer 18

Solution: PRMO 2014, Q4

Let the two proportional sides be $a$ and $3a$. All three triangle inequalities must hold. The inequality $3a+17>a$ is automatic for positive $a$. The other two give $$a+3a>17 \quad\Longrightarrow\quad a\ge5,$$ $$a+17>3a \quad\Longrightarrow\quad 2a<17 \quad\Longrightarrow\quad a\le8.$$ Thus the allowed integer values are $5,6,7,8$.

The perimeter $4a + 17$ increases with $a$, so take $a = 8$: the sides are $8$, $24$ and $17$, and indeed $8 + 17 = 25 > 24$. The perimeter is $$8 + 24 + 17 = 49.$$

Answer 49

Solution: PRMO 2017, Q7

The left side increases with $n$, so the answer is a count of an initial run. Rearranged, the condition $\sqrt n + \sqrt{n+1} < 11$ says $\sqrt{n+1} < 11 - \sqrt n$, which needs $\sqrt n < 11$ and then squares to $$n + 1 < 121 - 22\sqrt n + n, \qquad 22\sqrt n < 120, \qquad \sqrt n < \frac{60}{11}.$$ Squaring once more, $$n < \frac{3600}{121}.$$ That bound lies between $29$ and $30$, since $29 \cdot 121 = 3509 < 3600 < 3630 = 30 \cdot 121$.

So the condition holds exactly for $n = 1, 2, \ldots, 29$, which is $29$ positive integers. The two boundary cases confirm it without decimals. Squaring, $\sqrt{29} + \sqrt{30} < 11$ says $59 + 2\sqrt{870} < 121$, that is $\sqrt{870} < 31$, and $870 < 961$; while $\sqrt{30} + \sqrt{31} > 11$ says $61 + 2\sqrt{930} > 121$, that is $\sqrt{930} > 30$, and $930 > 900$.

Answer 29

Solution: IOQM 2024, Q6

From $a^{20} + b^{20} + c^{20} = 1$ and every twentieth power being non-negative, each of $a^{20}, b^{20}, c^{20}$ is at most $1$, so $$|a| \le 1, \qquad |b| \le 1, \qquad |c| \le 1.$$

Now subtract the first equation from the second: $$a^{20}(a^4 - 1) + b^{20}(b^4-1) + c^{20}(c^4-1) = 0.$$ Each bracket is at most $0$, because $|a| \le 1$ gives $a^4 \le 1$, and each factor $a^{20}$ is at least $0$. A sum of three non-positive numbers is zero only if all three are zero, so for each variable either $a^{20} = 0$ or $a^4 = 1$, that is $$a \in \{0, 1, -1\},$$ and likewise for $b$ and $c$.

Finally $a^{20} + b^{20} + c^{20} = 1$ says exactly one of the three is non-zero, and that one may be $1$ or $-1$. Choosing which variable is non-zero gives $3$ ways and choosing its sign gives $2$, so there are $$3 \cdot 2 = 6$$ triples.

Answer 6

Solution: IOQM 2025 Part SEP, Q7

Let the numbers be $x$ and $y$ with $x + y = n$ and $x^2 + y^2 = n + 1012$. Since $(x-y)^2 \ge 0$, $$(x+y)^2 \le 2(x^2+y^2), \qquad \text{that is} \qquad n^2 \le 2n + 2024.$$ So $n^2 - 2n - 2024 \le 0$. The roots of $n^2 - 2n - 2024$ are $$n = \frac{2 \pm \sqrt{4 + 8096}}{2} = 1 \pm 45,$$ using $8100 = 90^2$, so the inequality holds exactly for $-44 \le n \le 46$.

The largest possible $n$ is therefore $46$, and it is attained: equality in $(x-y)^2 \ge 0$ means $x = y = 23$, and indeed $23^2 + 23^2 = 1058 = 46 + 1012$.

Answer 46

Solution: IOQM 2025 Part SEP, Q28

A square of side $n$ fits inside a circular disc of radius $R$ precisely when its circumscribed circle is no larger, that is when its diagonal $n\sqrt2$ is at most the diameter $2R$. So $$n \le R\sqrt2 = \sqrt{2R^2} = \sqrt{\frac{2000}{\pi}},$$ using $\pi R^2 = 1000$.

No decimal is needed. From $3 < \pi < \tfrac{22}{7}$, $$625 < \frac{2000}{22/7} = \frac{14000}{22} < \frac{2000}{\pi} < \frac{2000}{3} < 676,$$ and $625 = 25^2$ while $676 = 26^2$. So the largest integer is $$n = 25.$$

Answer 25

Solution: IOQM 2026, Q3

Each $x_i$ is a non-zero integer, so $x_i^2 \ge 1$, and the sum of squares is at least $49$. That bound would need every $x_i$ to be $1$ or $-1$. But then all forty-nine are odd, and a sum of an odd number of odd numbers is odd, so it cannot be $0$.

Hence some $x_i$ has $|x_i| \ge 2$ and contributes at least $4$, while the other forty-eight contribute at least $1$ each: $$\sum_{i=1}^{49} x_i^2 \ge 4 + 48 = 52.$$

The bound is reached. Take one $2$, twenty-three $1$s and twenty-five $-1$s. That is $49$ numbers, their sum is $2 + 23 - 25 = 0$, and their squares add up to $4 + 48 = 52$.

Answer 52

Solution: PRMO 2015 Part A, Q9

A construction first, so that we know what we are aiming to prove is best possible. In a square of side $5$, put the $3 \times 4$ rectangle in a bottom corner with its side of length $4$ horizontal, so it occupies the region $0 \le x \le 4$, $0 \le y \le 3$. The band above it, $0 \le x \le 5$ and $3 \le y \le 5$, measures $5$ by $2$ and comfortably holds the $2 \times 3$ rectangle laid with its side of length $3$ horizontal. Both fit, so a square of side $5$ and area $25$ suffices.

Now suppose a square of side $s$ holds both, and let us see why $s < 5$ is impossible. Certainly $s \ge 4$, since the $3 \times 4$ rectangle has a side of length $4$ and the sides are parallel to the square’s. Assume $4 \le s < 5$, and take the $3 \times 4$ rectangle to have its side of length $4$ horizontal, which we may do by rotating the whole picture if necessary.

Consider the horizontal direction. The big rectangle occupies an interval of length $4$ out of the available $s < 5$, so the space strictly to its left and strictly to its right totals $s - 4 < 1$. The $2 \times 3$ rectangle has both sides at least $2$, so it cannot fit into either of those slivers, and therefore its horizontal extent must overlap that of the $3 \times 4$ rectangle.

Two rectangles with parallel sides that overlap horizontally and do not share interior points must be separated vertically. So the $2 \times 3$ rectangle lies entirely above or entirely below the $3 \times 4$ rectangle, and the vertical room left over is $s - 3 < 2$. That is less than either side of the $2 \times 3$ rectangle, so no placement exists. Hence $s \ge 5$ and the least possible area is $25$.

Answer 25

Solution: PRMO 2017, Q15

If the two removed numbers are $m$ and $k$, the $n - 2$ that remain have total $17(n-2)$, so $$m + k = \frac{n(n+1)}{2} - 17(n-2) = \frac{n^2 - 33n + 68}{2}.$$ Call this $S(n)$; it is always an integer, since $n$ and $n - 33$ have opposite parities.

Two constraints pin down $n$. Both removed numbers lie strictly between $1$ and $n$, so $$2 + 3 = 5 \le m + k \le (n-1) + (n-2) = 2n - 3.$$

The upper bound gives $n^2 - 33n + 68 \le 4n - 6$, that is $n^2 - 37n + 74 \le 0$. Test integers rather than take a square root: at $n = 34$ the value is $-28$ and at $n = 35$ it is $4$, so $n \le 34$. The lower bound gives $n^2 - 33n + 58 \ge 0$, where at $n = 31$ the value is $-4$ and at $n = 32$ it is $26$, so $n \ge 32$ on the branch that matters (the other branch, $n \le 1$, is far too small). Hence $n \in \{32, 33, 34\}$, and $$S(32) = 18, \qquad S(33) = 34, \qquad S(34) = 51.$$

The largest is $51$, and it is attainable: with $n = 34$ remove $18$ and $33$, both strictly between $1$ and $34$, and the remaining $32$ numbers average $\tfrac{595 - 51}{32} = \tfrac{544}{32} = 17$.

Answer 51

Solution: PRMO 2017, Q18

The coefficients $1,4,16$ are squares: scaling the variables by $1,2,4$ makes the first condition a sum of three ordinary squares. The mixed terms also acquire one common coefficient. With that in mind, set $u = x$, $v = 2y$ and $w = 4z$. The first condition becomes $$u^2 + v^2 + w^2 = 48.$$ For the second, note that $xy = \tfrac{uv}{2}$, $4yz = \tfrac{vw}{2}$ and $2zx = \tfrac{wu}{2}$, so it becomes $$\frac{uv + vw + wu}{2} = 24, \qquad uv + vw + wu = 48.$$

The two right-hand sides are equal, and that is the whole problem: $$u^2 + v^2 + w^2 - uv - vw - wu = 0,$$ which is the same as $$\tfrac12\left[(u-v)^2 + (v-w)^2 + (w-u)^2\right] = 0.$$ A sum of squares vanishes only when each term does, so $u = v = w$. Then $3u^2 = 48$ gives $u = \pm4$, and $$x = u = \pm 4, \qquad y = \frac{v}{2} = \pm 2, \qquad z = \frac{w}{4} = \pm 1,$$ with the same sign throughout. Either way $$x^2 + y^2 + z^2 = 16 + 4 + 1 = 21.$$

The coefficients $1, 4, 16$ and the pairing $xy, 4yz, 2zx$ were designed to make that substitution work; spotting $u, v, w$ is the entire difficulty.

Answer 21

Solution: PRMO 2019, Q13

Square the total. Since each $x_i^2 = 1$, $$\left(\sum_{i=1}^{101} x_i\right)^{\!2} = \sum_{i=1}^{101} x_i^2 + 2\sum_{i<j} x_i x_j = 101 + 2S,$$ so, writing $s = \sum x_i$, we have $S = (s^2 - 101)/2$.

Now $s$ is a sum of $101$ terms each $\pm1$, so $s \equiv 101 \equiv 1 \pmod 2$: it is odd. For $S$ to be positive we need $s^2 > 101$, and the smallest odd square exceeding $101$ is $11^2 = 121$. That gives $$S = \frac{121 - 101}{2} = 10.$$

It remains to check that $s = 11$ is achievable, and it is: take $p$ of the values to be $+1$ and $101 - p$ to be $-1$, so $s = 2p - 101$, and $p = 56$ gives $s = 11$. So the smallest positive value of the sum is $10$.

Answer 10

Solution: IOQM 2021 Part A, Q8

We want the largest $N$ with $\bigl\lfloor \sqrt{\lfloor \sqrt{\lfloor \sqrt N \rfloor} \rfloor} \bigr\rfloor = 4$, and the way in is to name the intermediate quantities and work outwards to inwards.

Write $k = \lfloor \sqrt N \rfloor$ and $m = \lfloor \sqrt k \rfloor$, so the condition is $\lfloor \sqrt m \rfloor = 4$. That holds exactly when $16 \leq m \leq 24$, since $4^2 = 16$ and $5^2 = 25$. We want $N$ as large as possible, so take $m$ as large as possible, $m = 24$.

Next, $\lfloor \sqrt k \rfloor = 24$ holds exactly when $576 \leq k \leq 624$, so the largest useful $k$ is $624$. Finally $\lfloor \sqrt N \rfloor = 624$ holds exactly when $624^2 \leq N \leq 625^2 - 1$, so the largest $N$ is $$N = 625^2 - 1 = 390625 - 1 = 390624.$$

It is worth checking the chain forwards, since three nested floors are easy to slip on: $\lfloor \sqrt{390624} \rfloor = 624$, then $\lfloor \sqrt{624} \rfloor = 24$, then $\lfloor \sqrt{24} \rfloor = 4$. The sum of the digits of $N$ is $3 + 9 + 0 + 6 + 2 + 4 = 24$.

Answer 24

Solution: IOQM 2023, Q1

An entry $\sqrt{4n+k}$ is an integer exactly when $4n+k$ is a perfect square, so $M_n$ counts the integers $j$ with $$\sqrt{4n+1} \le j \le \sqrt{4n+1000}.$$ The number of integers in an interval is controlled by its length, and here the length is $$\ell(n) = \sqrt{4n+1000} - \sqrt{4n+1} = \frac{999}{\sqrt{4n+1000} + \sqrt{4n+1}},$$ where the second form comes from multiplying and dividing by the conjugate. Written that way the behaviour is plain: $\ell(n)$ decreases as $n$ increases, so the longest interval is at $n = 1$ and the shortest at $n = 1000$.

For a small picture, the interval from $0.4$ to $4.8$ has length $4.4$ and contains the four integers $1,2,3,4$:

In general an interval of length $\ell$ contains at least $\lfloor\ell\rfloor$ integers and at most $\lfloor\ell\rfloor+1$. For the upper bound, $q$ integer points span at least $q-1$ units, so $q-1\le\ell$. For the lower bound, the first integer at or after the left endpoint is less than one unit from it. That integer and the next $\lfloor\ell\rfloor-1$ integers all fit in the interval. If $\ell<1$, the asserted lower bound is just zero.

For the maximum, take $n = 1$. The interval runs from $\sqrt5$ to $\sqrt{1004}$, and the integers in it are $3$ through $31$, so $M_1 = 29$. No $n$ does better. For $n \ge 2$, $$\ell(n) \le \ell(2) = \sqrt{1008} - \sqrt9 < 32 - 3 = 29,$$ using $1008 < 1024$, and an interval shorter than $29$ holds at most $29$ integers. So $a = 29$.

For the minimum, take $n = 1000$. The interval runs from $\sqrt{4001}$ to $\sqrt{5000}$, and the integers it contains are exactly $64$ through $70$, since $$63^2 = 3969 < 4001 \le 4096 = 64^2,$$ $$70^2 = 4900 \le 5000 < 5041 = 71^2.$$ So $M_{1000} = 7$. No $n$ can do worse, because $\ell(n) \ge \ell(1000)$ for every $n \le 1000$ and $\ell(1000) \ge 7$. That last claim says $\sqrt{5000} \ge 7 + \sqrt{4001}$, which on squaring becomes $950 \ge 14\sqrt{4001}$, and squaring again gives $902500 \ge 784196$. An interval of length at least $7$ always contains at least $7$ integers, so $b = 7$.

Hence $a - b = 29 - 7 = 22$.

Answer 22

Solution: IOQM 2024, Q15

The requirement is that every choice of three elements of $X$, repeats allowed, forms an acute triangle, so one bad triple is enough to disqualify a value of $n$. For sides $u\le v\le w$, the angle $\theta$ opposite $w$ is the largest. By the cosine rule, $$\cos\theta=\frac{u^2+v^2-w^2}{2uv}.$$ All three angles are acute exactly when this largest one is acute, or $\cos\theta>0$. Thus the triangle is acute exactly when $u^2+v^2>w^2$; and the condition also has to include $u + v > w$, which follows from it here. Among the numbers of $X$, the inequality is hardest to satisfy when $u$ and $v$ are as small as possible and $w$ as large as possible, and since $a, b, c$ need not be distinct that worst case is $$u = v = n, \qquad w = n + 38.$$ If that case is acute then so is every other, because increasing $u$ or $v$ only helps and decreasing $w$ only helps.

So the condition is $$2n^2 > (n+38)^2, \qquad \text{that is} \qquad n^2 - 76n - 1444 > 0.$$ The positive root of $n^2 - 76n - 1444 = 0$ is $$n = \frac{76 + \sqrt{5776 + 5776}}{2} = 38 + 38\sqrt2,$$ and this lies between $91$ and $92$ without any decimal work: $38\sqrt2 < 54$ because $2 \cdot 38^2 = 2888 < 2916 = 54^2$, and $38\sqrt2 > 53$ because $2888 > 2809 = 53^2$. So the inequality holds exactly for $n \ge 92$. (It is worth pausing on the neat form $38(1 + \sqrt2)$, which is what the number $38$ was chosen to produce.)

Hence the smallest value is $n = 92$.

Answer 92

Solution: IOQM 2025 Part SEP, Q7

If $d$ is the diagonal, the other four lengths split into two pairs, one pair forming a triangle with $d$ on each side of it. So we need to split $\{10, 20, 28, 50, 75\} \setminus \{d\}$ into pairs $\{p,q\}$ and $\{r,s\}$ with $$|p - q| < d < p+q \qquad \text{and} \qquad |r-s| < d < r+s.$$

There are only three ways to split four lengths into two pairs, so each candidate is a short check.

• $d = 75$: from $\{10,20,28,50\}$ every splitting leaves a pair of sum $30$, $38$ or $48$, all below $75$. Fails.

• $d = 50$: from $\{10,20,28,75\}$ the three splittings leave a pair of sum $30$, $38$ or $48$ again, all below $50$. Fails.

• $d = 20$: from $\{10,28,50,75\}$ each splitting contains a pair differing by at least $22$, namely one of $\{50,75\}$, $\{10,50\}$ or $\{10,75\}$. Fails.

• $d = 10$: from $\{20,28,50,75\}$ every splitting contains a pair differing by at least $22$. Fails.

• $d = 28$: pair the rest as $\{10, 20\}$ and $\{50, 75\}$. Then $10 < 28 < 30$ and $25 < 28 < 125$, both satisfied.

So the diagonal can only be $28$.

Answer 28

Solution: PRMO 2018, Q18

Divide both sides by $abc$: $$\left(1 + \frac3a\right)\left(1 + \frac3b\right)\left(1 + \frac3c\right) = 4.$$ Each factor decreases as its variable increases, which is what makes the search finite. Order the variables so that $a \le b \le c$; then the largest factor is the one with $a$, and $$\left(1 + \frac3a\right)^3 \ge 4.$$ No cube root is needed: if $a \ge 6$ then $1 + \tfrac3a \le \tfrac32$ and $\left(\tfrac32\right)^3 = \tfrac{27}{8} < 4$, which fails. With $a \ge 4$ given, that leaves $a \in \{4, 5\}$.

$a = 4$

The remaining two factors must multiply to $4 \div \tfrac74 = \tfrac{16}{7}$. Since $b$ is the smaller of the two, $\left(1 + \tfrac3b\right)^2 \ge \tfrac{16}{7}$. Again no root is needed: $b \ge 6$ would give $\left(1+\tfrac3b\right)^2 \le \tfrac94 < \tfrac{16}{7}$, since $9 \cdot 7 = 63 < 64 = 16 \cdot 4$. So $b \in \{4,5\}$.

• $b = 4$: then $1 + 3/c = \tfrac{16}{7} \div \tfrac74 = \tfrac{64}{49}$, so $c = \tfrac{49}{5}$, not an integer.

• $b = 5$: then $1 + 3/c = \tfrac{16}{7} \div \tfrac85 = \tfrac{10}{7}$, so $3/c = 3/7$ and $c = 7$.

$a = 5$

Now the other two factors multiply to $4 \div \tfrac85 = \tfrac52$, and $b \ge 5$ with $\left(1+\tfrac3b\right)^2 \ge \tfrac52$ forces $b \le 5$, since $b \ge 6$ would give at most $\tfrac94 < \tfrac52$. So $b = 5$. Then $1 + 3/c = \tfrac52 \div \tfrac85 = \tfrac{25}{16}$, giving $c = \tfrac{16}{3}$, not an integer.

So up to order the only solution is $\{a,b,c\} = \{4,5,7\}$, which is not all equal as required, and indeed $4 \cdot 4 \cdot 5 \cdot 7 = 560 = 7 \cdot 8 \cdot 10$. Hence $$a + b + c = 16.$$

Answer 16

Solution: PRMO 2018, Q23

The left-hand side is a sum of squares over positive quantities, and there is an inequality that handles exactly that shape: for positive $x, y, z$, $$\frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z} \ge \frac{(a+b+c)^2}{x+y+z}.$$

It is worth seeing where that comes from, because two terms already contain the whole idea. Clearing denominators in $$\frac{a^2}{x} + \frac{b^2}{y} \ge \frac{(a+b)^2}{x+y}$$ turns it into $\left(a^2y + b^2x\right)(x+y) \ge (a+b)^2xy$, and expanding both sides leaves $a^2y^2 + b^2x^2 \ge 2abxy$, which is $(ay - bx)^2 \ge 0$. Applying the two-term version twice, first to the last pair and then to what is left, $$\frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z} \ \ge\ \frac{a^2}{x} + \frac{(b+c)^2}{y+z} \ \ge\ \frac{(a+b+c)^2}{x+y+z},$$ which is the three-term form. It is Cauchy-Schwarz wearing a convenient disguise, and equality needs $ay = bx$ at each step, that is $a : b : c$ matching $x : y : z$.

Here the three denominators are $\tfrac{b}{29} + \tfrac{c}{31}$, $\tfrac{c}{29} + \tfrac{a}{31}$ and $\tfrac{a}{29} + \tfrac{b}{31}$, and every one of $a$, $b$, $c$ appears once with each coefficient, so they add to $$(a+b+c)\left(\frac{1}{29} + \frac{1}{31}\right) = (a+b+c)\cdot\frac{60}{899}.$$ Therefore $$\text{LHS} \ge \frac{(a+b+c)^2 \cdot 899}{60(a+b+c)} = \frac{899}{60}(a+b+c),$$ and $899/60 = 14.98\ldots$

Equality, by the condition noted above, needs $a : b : c$ to match the denominators, which happens at $a = b = c$; there the left-hand side is exactly $\tfrac{899}{60}(a+b+c)$. So no constant larger than $899/60$ can work, and the largest integer that does is $$n = 14.$$

Answer 14

Solution: PRMO 2019, Q22

We want nearby terms that cancel when added. The difference of adjacent square roots is promising: $$\sqrt{n+1}-\sqrt n = \frac1{\sqrt{n+1}+\sqrt n}.$$ Its denominator is close to $2\sqrt n$, so twice this difference should be close to $1/\sqrt n$. Using the next root gives one side of the bound and using the previous root gives the other:

$$2\left(\sqrt{n+1} - \sqrt n\right) = \frac{2}{\sqrt{n+1} + \sqrt n} < \frac{2}{2\sqrt n} = \frac{1}{\sqrt n},$$ $$2\left(\sqrt{n} - \sqrt{n-1}\right) = \frac{2}{\sqrt{n} + \sqrt{n-1}} > \frac{2}{2\sqrt n} = \frac{1}{\sqrt n} \qquad (n \ge 2).$$ Both are strict, and both telescope when summed.

Write $S$ for the sum in question. From below, using the first estimate for every term, $$S > \sum_{n=1}^{1599} 2\left(\sqrt{n+1} - \sqrt n\right) = 2\left(\sqrt{1600} - 1\right) = 2(40 - 1) = 78.$$ From above, keeping the first term separate because the second estimate needs $n \ge 2$, $$S < 1 + \sum_{n=2}^{1599} 2\left(\sqrt n - \sqrt{n-1}\right) = 1 + 2\left(\sqrt{1599} - 1\right) = 2\sqrt{1599} - 1,$$ and that is smaller than $2 \times 40 - 1 = 79$, because $1599 < 1600$.

So $78 < S < 79$, and the greatest integer not exceeding $S$ is $78$.

The number $1599$ was chosen with care, though not quite as tightly as it first looks. One term fewer breaks the argument: the lower bound would read $2(\sqrt{1599} - 1)$, which is below $78$. One term more is still fine, since the upper bound becomes exactly $79$ and the inequality is strict. Two terms more does break it, the upper bound passing $79$. So the window is narrow, and the reason $1600$ appears in the lower estimate is exactly that the sum runs to $1599$.

Answer 78

Solution: PRMO 2019, Q26

Since $z \ge 1$, the constraint gives $xy \le 159$. Substituting $z = 160 - xy$, $$x + yz = x + y(160 - xy) = 160y - x(y^2 - 1).$$ For $y = 1$ this is just $x + z = 160$. For $y \ge 2$ the coefficient $y^2 - 1$ is positive, so the expression decreases as $x$ grows, and $x$ should be as large as the constraint allows: $$x = \left\lfloor \frac{159}{y} \right\rfloor.$$

Write $159 = qy + s$ with $0 \le s < y$, so that $x = q$ and $z = 160 - qy = s + 1$. Then $$x + yz = 160y - q(y^2 - 1) = 160y - y(qy) + q,$$ and $qy = 159 - s$, so the whole thing collapses to $y(1+s) + q$. Because $x$ is forced for each $y$, every row below is a single evaluation of this formula rather than a search over $x$. Running $y$ from $1$ to $24$:

$$\small \begin{array}{r|rrrrrrrr} y & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline x + yz & 160 & 83 & 56 & 55 & 56 & \mathbf{50} & 64 & 83 \\[6pt] y & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline x + yz & 80 & 115 & 80 & 61 & 64 & 95 & 160 & 265 \\[6pt] y & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 \\ \hline x + yz & 128 & 296 & 160 & 407 & 280 & 139 & 512 & 390 \end{array}$$

Larger $y$ need no table. If $s \ge 1$ then $y(1+s) + q \ge 2y + q \ge 2 \times 25 + 1 = 51$ for $y \ge 25$. If $s = 0$ then $y$ divides $159 = 3 \times 53$, so $y$ is $53$ or $159$, giving $53 + 3 = 56$ and $159 + 1 = 160$. Every value beyond the table exceeds $50$.

The minimum is therefore $50$, at $y = 6$, $x = 26$, $z = 4$: indeed $26 \times 6 + 4 = 160$ and $26 + 6 \times 4 = 50$.

Why $y = 6$ and not some neighbour? Rearranging the constraint gives the identity $x + yz = 160 + (y-1)(z - x)$, so we want $(y-1)(x-z)$ large. Raising $y$ helps the first factor and hurts the second, and $y = 6$ divides $159$ almost exactly, which keeps $z$ small while $x$ is still $26$.

Answer 50

Solution: IOQM 2022, Q15

Set $s = x + y$ and $d = x - y$. Then $s^2 - d^2 = 4xy = 4$, so $s^2 = d^2 + 4$ and in particular $s^2 - 2 = d^2 + 2$. Substituting into the expression, $$\frac{(x+y)^2 - (x-y) - 2}{(x+y)^2 + (x-y) - 2} = \frac{d^2 - d + 2}{d^2 + d + 2}.$$ The two variables have become one, and $d$ ranges over all real numbers, since for any $d$ we may take $s = \pm\sqrt{d^2+4}$ and recover a genuine pair with $xy = 1$.

To find the range, ask which values $k$ are attained. The denominator never vanishes, since $$d^2 + d + 2 = \left(d + \tfrac12\right)^2 + \tfrac74 > 0,$$ so clearing it is safe. Setting the expression equal to $k$ and clearing the denominator, $$(1-k)d^2 - (1+k)d + (2 - 2k) = 0,$$ For $k=1$ this is the linear equation $-2d=0$, so that value is attained at $d=0$. For $k\ne1$ the equation is quadratic, and a real $d$ exists exactly when its discriminant is non-negative: $$(1+k)^2 - 4(1-k)(2-2k) = (1+k)^2 - 8(1-k)^2 \geq 0.$$ Expanding, $1 + 2k + k^2 - 8 + 16k - 8k^2 \geq 0$, that is $7k^2 - 18k + 7 \leq 0$.

A quadratic is negative between its roots, so the attained values fill the closed interval between them, and $T$ and $t$ are precisely those roots. We do not even need to compute them separately, because the sum of the roots of $7k^2 - 18k + 7$ is $\tfrac{18}{7}$ by Vieta.

So $T + t = \tfrac{18}{7}$, giving $m = 18$, $n = 7$ and $m + n = 25$.

Answer 25

Solution: IOQM 2024, Q11

Write $x = a$, $y = 2b$, $z = 3c$, all positive. The two conditions are $$\frac{x}{y+1} + \frac{y}{z+1} + \frac{z}{x+1} = 1, \qquad \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = 2.$$

Add them, pairing each fraction of the first with the fraction of the second that shares its denominator: $$\frac{x+1}{y+1} + \frac{y+1}{z+1} + \frac{z+1}{x+1} = 3.$$ The three terms are positive and their product is $1$, so by the arithmetic-geometric mean

inequality their sum is at least $3$, with equality only when all three equal $1$. Equality is exactly what we have, so $$x + 1 = y + 1 = z + 1, \qquad \text{that is} \qquad x = y = z.$$

Substituting into the first equation, $\dfrac{3x}{x+1} = 1$, so $x = \tfrac12$. Unwinding, $$a = \tfrac12, \qquad 2b = \tfrac12 \Rightarrow b = \tfrac14, \qquad 3c = \tfrac12 \Rightarrow c = \tfrac16,$$ and therefore $$\frac1a + \frac1b + \frac1c = 2 + 4 + 6 = 12.$$

Answer 12

Solution: IOQM 2025 Part SEP, Q7

Write $p = MT$ and $q = MI$, so that $pq = 1$, and let $\theta$ be the angle between those two sides at $M$. The area of a parallelogram is the product of two adjacent sides times the sine of the angle between them, so $$pq \sin\theta = \sin\theta = \frac{40}{41}, \qquad \cos\theta = \pm\frac{9}{41}.$$

The diagonal $MA$ is the sum of the two side vectors, so by the cosine rule $$MA^2 = p^2 + q^2 + 2pq\cos\theta = p^2 + \frac{1}{p^2} + 2\cos\theta.$$ Two independent choices now make this as small as possible. First, $p^2 + 1/p^2 \ge 2$ by the arithmetic-geometric mean inequality, with equality when $p = 1$. Second, $\cos\theta$ should be the negative value $-\tfrac{9}{41}$, which is the obtuse choice.

Hence $$d^2 = 2 - \frac{18}{41} = \frac{82 - 18}{41} = \frac{64}{41},$$ and since $\gcd(64,41) = 1$ we read off $a = 64$, $b = 41$ and $|a - b| = 23$.

Answer 23