# 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

$$
\mathrm{T}^{\prime} \mathrm{M}^{\prime}=\mathrm{S}^{\prime} \mathrm{N}^{\prime}=\mathrm{PS}=\mathrm{TQ}
$$

So

$$
\begin{aligned}
\mathrm{P}^{\prime} \mathrm{Q}^{\prime}+\mathrm{M}^{\prime} \mathrm{N}^{\prime}= & \mathrm{ST}+\mathrm{M}^{\prime} \mathrm{T}^{\prime}+\mathrm{T}^{\prime} \mathrm{S}^{\prime}+\mathrm{S}^{\prime} \mathrm{N}^{\prime} \\
& =\mathrm{ST}+\mathrm{PS}+\mathrm{TQ}+\mathrm{MN} \\
& =\mathrm{PQ}+\mathrm{MN} .
\end{aligned}
$$

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

$$
\begin{aligned}
& \mathrm{AM}+\mathrm{MN}+\mathrm{NC}+\mathrm{CQ}+\mathrm{QP}+\mathrm{PA}= \\
& \mathrm{AM}^{\prime}+\mathrm{M}^{\prime} \mathrm{M}+\mathrm{MN}+\mathrm{NN}^{\prime}+\mathrm{N}^{\prime} \mathrm{C}+\mathrm{CQ}+\mathrm{QP}+\mathrm{PA}= \\
& \mathrm{AM}^{\prime}+\mathrm{PP}^{\prime}+\mathrm{MN}+\mathrm{QQ}^{\prime}+\mathrm{N}^{\prime} \mathrm{C}+\mathrm{CQ}+\mathrm{QP}+\mathrm{PA}= \\
& \mathrm{AM}^{\prime}+\mathrm{MN}+\mathrm{QP}+\mathrm{PP}^{\prime}+\mathrm{PA}+\mathrm{QQ}^{\prime}+\mathrm{CQ}+\mathrm{N}^{\prime} \mathrm{C}= \\
& \mathrm{AM}^{\prime}+\mathrm{MN}+\mathrm{QT}+\mathrm{TS}+\mathrm{SP}+\mathrm{PP}^{\prime}+\mathrm{PA}+\mathrm{QQ}^{\prime}+\mathrm{CQ}+\mathrm{N}^{\prime} \mathrm{C}= \\
& \mathrm{AM}^{\prime}+\mathrm{M}^{\prime} \mathrm{T} \text { '+ T'S' + S'N' } \mathrm{N}^{\prime} \mathrm{C}+\mathrm{CQ}+\mathrm{QQ}^{\prime}+\mathrm{Q}^{\prime} \mathrm{P}^{\prime}+\mathrm{P}^{\prime} \mathrm{P}+\mathrm{PA}= \\
& \mathrm{AM}^{\prime}+\mathrm{M}^{\prime} \mathrm{N}^{\prime}+\mathrm{N}^{\prime} \mathrm{C}+\mathrm{CQ}^{\prime}+\mathrm{Q}^{\prime} \mathrm{P}^{\prime}+\mathrm{P}^{\prime} \mathrm{A},
\end{aligned}
$$

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:

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

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)

$$
\begin{gathered}
=\left(\prod_{i=1}^{n}\left(m^{2}+m-i^{2}+i\right)\right)\left(m^{2}+m-n^{2}-n\right) \\
=\prod_{i=1}^{n+1}\left(m^{2}+m-i^{2}+i\right)
\end{gathered}
$$

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

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

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

$$
\angle \mathrm{Q}_{4} \mathrm{P}_{1} \mathrm{P}_{2}=\frac{1}{2} \angle \mathrm{Q}_{4} \mathrm{O}_{12} \mathrm{P}_{2} \text { and } \angle \mathrm{O}_{23} \mathrm{P}_{1} \mathrm{P}_{2}=\frac{1}{2} \angle \mathrm{P}_{3} \mathrm{O}_{12} \mathrm{P}_{2}
$$

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

$$
x+y=17
$$

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

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

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

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

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

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

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

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

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

$$
\mathrm{n}=\mathrm{px}+\alpha, 0 \leq \alpha<\mathrm{x} \quad \text { and } \mathrm{n}=\mathrm{qy}-\beta \quad 0 \leq \beta<\mathrm{y}
$$

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.,

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

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

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

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

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

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

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

$$
\begin{aligned}
\sqrt{c+a-b}-\sqrt{b} & \leq \frac{1}{2 \sqrt{b}}(c+a-2 b) \\
\sqrt{b+c-a}-\sqrt{c} & \leq \frac{1}{2 \sqrt{c}}(b-a)
\end{aligned}
$$

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

$$
\begin{aligned}
& \sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}-(\sqrt{a}+\sqrt{b}+\sqrt{c}) \\
& \frac{1}{2}\left(\frac{1}{\sqrt{a}}(b-c)+\frac{1}{\sqrt{b}}(c+a-2 b)+\frac{1}{\sqrt{c}}(b-a)\right) \\
& \frac{1}{2}\left((b-c)\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)+(a-b)\left(\frac{1}{\sqrt{b}}-\frac{1}{\sqrt{c}}\right)\right) \leq 0,
\end{aligned}
$$

i.e.,

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

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