APMO/md/en-apmo1996_sol.md

Solutions

Problem 1. Let M'N' and $\mathrm{P}^{\prime} \mathrm{Q}^{\prime}$ be two segments perpendicular to BD at a distance d . We need to prove that the perimeters $A M N C Q P$ and $A M^{\prime} N^{\prime} C Q^{\prime} P^{\prime}$ are equal. Denote by $\mathrm{S}^{\prime}$ the projection of N into M'N' and by S the projection of $\mathrm{P}^{\prime}$ into PQ . The triangles $\mathrm{NS}^{\prime} \mathrm{N}^{\prime}$ and $\mathrm{P}^{\prime} \mathrm{SP}$ are equal. (right triangles such that NS' = P'S and equal angles $\angle \mathrm{P}^{\prime} \mathrm{PS}=\angle \mathrm{NN}^{\prime} \mathrm{S}^{\prime}$.) So $\mathrm{PP}^{\prime}=\mathrm{NN}^{\prime}$. Since it is clear that $\mathrm{MM}^{\prime}=\mathrm{NN}^{\prime}$, we have $\mathrm{MM}^{\prime}=\mathrm{NN}^{\prime}=\mathrm{PP}^{\prime}=\mathrm{QQ} \mathrm{Q}^{\prime}$. On the other hand, since $\mathrm{SP}=\mathrm{S}^{\prime} \mathrm{N}^{\prime}$, using $\mathrm{T}^{\prime}$ and T the projections of M and $\mathrm{Q}^{\prime}$ into $\mathrm{M}^{\prime} \mathrm{N}^{\prime}$ and PQ , respectively, we have

$${\mathrm{T}}^{\mathrm{\prime }}{\mathrm{M}}^{\mathrm{\prime }}={\mathrm{S}}^{\mathrm{\prime }}{\mathrm{N}}^{\mathrm{\prime }}=\mathrm{P}\mathrm{S}=\mathrm{T}\mathrm{Q}\backslash mathrm\{T\}^\{\backslash prime\}\; \backslash mathrm\{M\}^\{\backslash prime\}=\backslash mathrm\{S\}^\{\backslash prime\}\; \backslash mathrm\{N\}^\{\backslash prime\}=\backslash mathrm\{PS\}=\backslash mathrm\{TQ\}$$

So

(Without loss of generality $\mathrm{M}^{\prime}$ lies between M and A ) So the perimeter of $A M N C Q P$ is:

which is the perimeter of $\mathrm{AM}^{\prime} \mathrm{N}^{\prime} \mathrm{CQ}^{\prime} \mathrm{P}$ '.

Points to be given for showing that: $\mathrm{MM}^{\prime}=\mathrm{NN}^{\prime}=\mathrm{PP}^{\prime}=\mathrm{QQ}^{\prime}$ 1 point $\mathrm{T}^{\prime} \mathrm{M}^{\prime}=\mathrm{S}^{\prime} \mathrm{N}^{\prime}=\mathrm{PS}=\mathrm{TQ} . \quad 1$ point $P^{\prime} Q^{\prime}+M^{\prime} N^{\prime}=P Q+M N$. 2 points

The perimeter of $A M N C Q P$ isthe same as the perimeter of $A M^{\prime} N^{\prime} \mathrm{CQ}^{\prime} \mathrm{P}^{\prime}$

3 points Problem 2. We first prove by induction on n that:

$$\frac{(m+n)!}{(m-n)!}=\prod _{i=1}^n(m^2+m-i^2+i)\backslash frac\{(m+n)!\}\{(m-n)!\}=\backslash prod\_\{i=1\}^\{n\}\backslash left(m^\{2\}+m-i^\{2\}+i\backslash right)$$

1. $\frac{(m+1)!}{(m-1)!}=m(m+1)=m^{2}+m$.
2. $\frac{(m+n+1)!}{(m-n-1)!}=\left(\prod_{i=1}^{n}\left(m^{2}+m-i^{2}+i\right)\right)(m+n+1)(m-n)$ (by induction)

$$\begin{array}{c}{\displaystyle =\left(\prod _{i=1}^n(m^2+m-i^2+i)\right)(m^2+m-n^2-n)}\\ {\displaystyle =\prod _{i=1}^{n+1}(m^2+m-i^2+i)}\end{array}\backslash begin\{gathered\}\; =\backslash left(\backslash prod\_\{i=1\}^\{n\}\backslash left(m^\{2\}+m-i^\{2\}+i\backslash right)\backslash right)\backslash left(m^\{2\}+m-n^\{2\}-n\backslash right)\; \backslash \backslash \; =\backslash prod\_\{i=1\}^\{n+1\}\backslash left(m^\{2\}+m-i^\{2\}+i\backslash right)\; \backslash end\{gathered\}$$

But $m^{2}+m \geq m^{2}+m-i^{2}+i \geq i^{2}+i-i^{2}+i=2 i$, for $i \geq m$. Therefore

$$2^{n}n!\le \frac{(m+n)!}{(m-n)!}\le {(m^2+m)}^{n}2^\{n\}\; n!\backslash leq\; \backslash frac\{(m+n)!\}\{(m-n)!\}\; \backslash leq\backslash left(m^\{2\}+m\backslash right)^\{n\}$$

Points to be given for showing that: the innequlities hold for special values up to 1 point $\frac{(m+n)!}{(m-n)!}=\prod_{i=1}^{n}\left(m^{2}+m-i^{2}+i\right)$ for all $n$ 4 points $m^{2}+m \geq m^{2}+m-i^{2}+i \geq i^{2}+i-i^{2}+i=2 i$, for $i \geq m$, and therefore the innequalities hold

2 points

Problem 3. Let C be the given circle. Draw four circles $\mathrm{C}{12}, \mathrm{C}{23}, \mathrm{C}{34}, \mathrm{C}{41}$ with centers $\mathrm{O}{12}, \mathrm{O}{23}$, $\mathrm{O}{34}, \mathrm{O}{41}$, respectively, on the circle C such that $\mathrm{C}{12}$ passes through $\mathrm{P}{1}$ and $\mathrm{P}{2}, \mathrm{C}{23}$ passes through $\mathrm{P}{2}$ and $P{3}, C_{34}$ passes through $P_{3}$ and $P_{4}, C_{41}$ passes through $P_{4}$ and $P_{1}$. Let the other point of intersection of $\mathrm{C}{12}$ and $\mathrm{C}{23}$ be $\mathrm{Q}{4}$, the other point of intersection of $\mathrm{C}{23}$ and $\mathrm{C}{34}$ be $\mathrm{Q}{11}$, the other point of intersection of $C_{34}$ and $C_{41}$ be $Q_{2}$, and the other point of intersection of $C_{41}$ and $C_{12}$ be $Q_{3}$. Then

$$\mathrm{\angle }{\mathrm{Q}}_4{\mathrm{P}}_1{\mathrm{P}}_2=\frac{1}{2}\mathrm{\angle }{\mathrm{Q}}_4{\mathrm{O}}_{12}{\mathrm{P}}_2\text{ and }\mathrm{\angle }{\mathrm{O}}_{23}{\mathrm{P}}_1{\mathrm{P}}_2=\frac{1}{2}\mathrm{\angle }{\mathrm{P}}_3{\mathrm{O}}_{12}{\mathrm{P}}_2\backslash angle\; \backslash mathrm\{Q\}\_\{4\}\; \backslash mathrm\{P\}\_\{1\}\; \backslash mathrm\{P\}\_\{2\}=\backslash frac\{1\}\{2\}\; \backslash angle\; \backslash mathrm\{Q\}\_\{4\}\; \backslash mathrm\{O\}\_\{12\}\; \backslash mathrm\{P\}\_\{2\}\; \backslash text\; \{\; and\; \}\; \backslash angle\; \backslash mathrm\{O\}\_\{23\}\; \backslash mathrm\{P\}\_\{1\}\; \backslash mathrm\{P\}\_\{2\}=\backslash frac\{1\}\{2\}\; \backslash angle\; \backslash mathrm\{P\}\_\{3\}\; \backslash mathrm\{O\}\_\{12\}\; \backslash mathrm\{P\}\_\{2\}$$

Clearly, $\mathrm{O}{23}, \mathrm{Q}{4}$ and $\mathrm{P}{1}$ are collinear. It follows that $\angle \mathrm{Q}{4} \mathrm{Q}{3} \mathrm{P}{2}=\angle \mathrm{Q}{4} \mathrm{P}{1} \mathrm{P}{2}=\angle \mathrm{O}{23} \mathrm{P}{1} \mathrm{P}{2} \angle \mathrm{O}{23} \mathrm{O}{41} \mathrm{P}{2}$. Since also $\mathrm{O}{41}, \mathrm{Q}{3}$ and $\mathrm{P}{2}$ are collinear, it follows that $\mathrm{Q}{3} \mathrm{Q}{4}$ and $\mathrm{O}{41} \mathrm{O}{23}$ are parallel. Since $\mathrm{O}{\mathrm{ij}}$ bisects The arcs $\mathrm{P}{\mathrm{i}} \mathrm{P}{\mathrm{j}}$, for $(\mathrm{i}, \mathrm{j})=(1,2),(2,3),(3,4),(4,1)$ we conclude that $\mathrm{O}{41} \mathrm{O}{23}$ and $\mathrm{O}{12} \mathrm{O}{34}$ are perpendicular, and hence $\mathrm{Q}{3} \mathrm{Q}{4}$ and $\mathrm{O}{12} \mathrm{O}{34}$ are perpendicular. Since both $\left(\mathrm{O}{12}, \mathrm{Q}{3}, \mathrm{P}{4}\right)$ and $\left(\mathrm{O}{12}, \mathrm{Q}{4}, \mathrm{P}{3}\right)$ are collinear triples of points, we have $\angle \mathrm{P}{4} \mathrm{O}{12} \mathrm{P}{3}=\angle$ $\mathrm{Q}{3} \mathrm{O}{12} \mathrm{Q}{4}$, and this angle is bisected by $\mathrm{O}{12} \mathrm{O}{34}$. Thus $Q{3}$ and $Q_{4}$ are reflections through the axis $\mathrm{O}{12} \mathrm{O}{34}$, and so are, by a similar argument $\mathrm{Q}{1}$ and $\mathrm{Q}{2}$. We have thus shown: $Q_{1}, Q_{2}, Q_{3}, Q_{4}$ form a rectangle. But as $Q_{4}$ lies on both the angle bisector $\mathrm{O}{12} \mathrm{P}{3}$ and the angle bisector $\mathrm{O}{23} \mathrm{P}{1}$ of the triangle $\mathrm{P}{1} \mathrm{P}{2} \mathrm{P}{3}$, the point $\mathrm{Q}{4}$ must coincide with the incenter $\mathrm{I}{4}$ of the triangle $\mathrm{P}{1} \mathrm{P}{2} \mathrm{P}{3}$, and by a similar argument, $\mathrm{Q}{1}=\mathrm{I}{1}, \mathrm{Q}{2}=\mathrm{I}{2}$ and $\mathrm{Q}{3}=\mathrm{I}{3}$.

Points to be given for showing that $\mathrm{Q}{3} \mathrm{Q}{4}$ and $\mathrm{O}{41} \mathrm{O}{23}$ are parallel $\mathrm{Q}{3} \mathrm{Q}{4}$ and $\mathrm{O}{12} \mathrm{O}{34}$ are perpendicular 1 points $\mathrm{Q}{1}, \mathrm{Q}{2}, \mathrm{Q}{3}, \mathrm{Q}{4}$ form a rectangle 2 points $\mathrm{Q}{4}=\mathrm{I}{4}, \mathrm{Q}{1}=\mathrm{I}{1}, \mathrm{Q}{2}=\mathrm{I}{2}$ and $\mathrm{Q}{3}=\mathrm{I}{3} \quad 1$ point

Problem 4. . We may assume that the n couples will form x male groups and y female groups. Without loss of generality, let $x \geq y$, and

$$x+y=17x+y=17$$

Then, by the pigeonhole theorem, there exists a male group of size $\leq\left[\frac{n}{x}\right]$ and a female group of size $\geq\left[\frac{n}{y}\right]$. By condition (2), we have

$$\left[\frac{n}{y}\right]-\left[\frac{n}{x}\right]\le 1\backslash left[\backslash frac\{n\}\{y\}\backslash right]-\backslash left[\backslash frac\{n\}\{x\}\backslash right]\; \backslash leq\; 1$$

From the conditions $\mathrm{x}+\mathrm{y}=17$ and $\mathrm{x} \geq \mathrm{y}$ follows that $\mathrm{x} \geq 9$, and $\mathrm{y} \leq 8$, which in turn implies

$$\left[\frac{n}{y}\right]-\left[\frac{n}{x}\right]>\left[\frac{n}{8}\right]-\left[\frac{n}{9}\right]\backslash left[\backslash frac\{n\}\{y\}\backslash right]-\backslash left[\backslash frac\{n\}\{x\}\backslash right]>\backslash left[\backslash frac\{n\}\{8\}\backslash right]-\backslash left[\backslash frac\{n\}\{9\}\backslash right]$$

Therefore, we only need to exclude those n such that $\left[\frac{n}{8}\right]-\left[\frac{n}{9}\right]>1$. Let $\mathrm{n}=9 \mathrm{u}+\mathrm{s}, 0 \leq \mathrm{s}<9$. Then

$$\left[\frac{n}{8}\right]-\left[\frac{n}{9}\right]>1\iff \left[\frac{u+s}{8}\right]>1\backslash left[\backslash frac\{n\}\{8\}\backslash right]-\backslash left[\backslash frac\{n\}\{9\}\backslash right]>1\; \backslash Leftrightarrow\backslash left[\backslash frac\{u+s\}\{8\}\backslash right]>1$$

By analyzing this condition it is clear that the only values of n that are allowed are

$$\begin{array}{c}{\displaystyle \mathrm{n}=9,10,11,12,13,14,15,16,18,19,20,21,22,23,24,27,28,29,30,}\\ {\displaystyle 31,32,36,37,38,39,40,45,46,46,47,48,54,55,56,63,63,72.}\end{array}\backslash begin\{gathered\}\; \backslash mathrm\{n\}=9,10,11,12,13,14,15,16,18,19,20,21,22,23,24,27,28,29,30,\; \backslash \backslash \; \backslash quad\; 31,32,36,37,38,39,40,45,46,46,47,48,54,55,56,63,63,72\; .\; \backslash end\{gathered\}$$

Conversely, conditions $\left(^{}\right)$ and $\left({ }^{ *}\right)$ give rise to a set of discussing groups according to the following description:

Let $\left[\frac{n}{x}\right]=p$ and $\left[\frac{n}{y}\right]=q$, then $p \leq q \leq p+1$ We have

So we can arrange the males in $\alpha$ discussing groups of size $p+1$ and $x-\alpha$ groups of $p$ elements. The females are distributed in $\beta$ discussing groups of size $q-1$, and $(y-\beta)$ discussing groups of size q.

Points to be given for showing that $x \geq 9$, and $y \leq 8 \quad 1$ points one only needs to exclude those n such that $\left[\frac{n}{8}\right]-\left[\frac{n}{9}\right]>1 \quad 2$ points analyzing the condition ian giving the values of $n$ that are allowed 2 points given p and q as described in the solution one can arrange the males in $\alpha$ discussion groups of size $p+1$ and $\mathrm{x}-\alpha$ groups of p elements and the females in $\beta$ discussion groups of size $q-1$, and $(y-\beta)$ groups of size $q$. 2 points

Problem 5. Without loss of generality we can assume that $a \geq b \geq c$. Note that if $x \geq y>0$, then $\sqrt{y} \leq 1 / 2(\sqrt{x}+\sqrt{y})$, i.e.,

$$\frac{1}{2\sqrt{y}}(\sqrt{x}+\sqrt{y})\ge 1\backslash frac\{1\}\{2\; \backslash sqrt\{y\}\}(\backslash sqrt\{x\}+\backslash sqrt\{y\})\; \backslash geq\; 1$$

Similarly, if $y \geq x>0$, then

$$\frac{1}{2\sqrt{y}}(\sqrt{x}+\sqrt{y})\le 1\backslash frac\{1\}\{2\; \backslash sqrt\{y\}\}(\backslash sqrt\{x\}+\backslash sqrt\{y\})\; \backslash leq\; 1$$

Multiplying both inequalities by $\sqrt{x}-\sqrt{y}$ we obtain

$$\sqrt{x}-\sqrt{y}\le \frac{1}{2\sqrt{y}}(x-y)\backslash sqrt\{x\}-\backslash sqrt\{y\}\; \backslash leq\; \backslash frac\{1\}\{2\; \backslash sqrt\{y\}\}(x-y)$$

for every $\mathrm{x}, \mathrm{y}>0$. Moreover, it is easily seen that equality occurs if and only if $\mathrm{x}=\mathrm{y}$. By applying this last inequality we obtain

$$\sqrt{a+b-c}-\sqrt{a}\le \frac{1}{2\sqrt{a}}(b-c)\backslash sqrt\{a+b-c\}-\backslash sqrt\{a\}\; \backslash leq\; \backslash frac\{1\}\{2\; \backslash sqrt\{a\}\}(b-c)$$

and by adding up the left hand and the right hand sides of these inequalities we have

i.e.,

$$\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}\le \sqrt{a}+\sqrt{b}+\sqrt{c},\backslash sqrt\{a+b-c\}+\backslash sqrt\{b+c-a\}+\backslash sqrt\{c+a-b\}\; \backslash leq\; \backslash sqrt\{a\}+\backslash sqrt\{b\}+\backslash sqrt\{c\},$$

and equality occurs if and only if the three relations in $\left({ }^{* *}\right)$ are equalities, i.e., if and only if $a=b=c$.

Points to be given for showing that $\sqrt{x}-\sqrt{y} \leq \frac{1}{2 \sqrt{y}}(x-y)$ 2 points the relations in $\left({ }^{* }\right)$ hold true 1 point each $\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}-(\sqrt{a}+\sqrt{b}+\sqrt{c}) \leq$ 1 point equality occurs if and only if the three relations in $\left({ }^{()}\right.$ are equalities, i.e., if and only $a=b=c$ 1 point