# 38th Canadian Mathematical Olympiad 

Wednesday, March 29, 2006

Solutions to the 2006 CMO paper

1. Let $f(n, k)$ be the number of ways of distributing $k$ candies to $n$ children so that each child receives at most 2 candies. For example, if $n=3$, then $f(3,7)=0, f(3,6)=1$ and $f(3,4)=6$.

Determine the value of

$$
f(2006,1)+f(2006,4)+f(2006,7)+\cdots+f(2006,1000)+f(2006,1003) .
$$

Comment. Unfortunately, there was an error in the statement of this problem. It was intended that the sum should continue to $f(2006,4012)$.

Solution 1. The number of ways of distributing $k$ candies to 2006 children is equal to the number of ways of distributing 0 to a particular child and $k$ to the rest, plus the number of ways of distributing 1 to the particular child and $k-1$ to the rest, plus the number of ways of distributing 2 to the particular child and $k-2$ to the rest. Thus $f(2006, k)=$ $f(2005, k)+f(2005, k-1)+f(2005, k-2)$, so that the required sum is

$$
1+\sum_{k=1}^{1003} f(2005, k)
$$

In evaluating $f(n, k)$, suppose that there are $r$ children who receive 2 candies; these $r$ children can be chosen in $\left(\begin{array}{l}n \\ r\end{array}\right)$ ways. Then there are $k-2 r$ candies from which at most one is given to each of $n-r$ children. Hence

$$
f(n, k)=\sum_{r=0}^{\lfloor k / 2\rfloor}\left(\begin{array}{l}
n \\
r
\end{array}\right)\left(\begin{array}{c}
n-r \\
k-2 r
\end{array}\right)=\sum_{r=0}^{\infty}\left(\begin{array}{l}
n \\
r
\end{array}\right)\left(\begin{array}{c}
n-r \\
k-2 r
\end{array}\right)
$$

with $\left(\begin{array}{l}x \\ y\end{array}\right)=0$ when $x<y$ and when $y<0$. The answer is

$$
\sum_{k=0}^{1003} \sum_{r=0}^{\infty}\left(\begin{array}{c}
2005 \\
r
\end{array}\right)\left(\begin{array}{c}
2005-r \\
k-2 r
\end{array}\right)=\sum_{r=0}^{\infty}\left(\begin{array}{c}
2005 \\
r
\end{array}\right) \sum_{k=0}^{1003}\left(\begin{array}{c}
2005-r \\
k-2 r
\end{array}\right)
$$

Solution 2. The desired number is the sum of the coefficients of the terms of degree not exceeding 1003 in the expansion of $\left(1+x+x^{2}\right)^{2005}$, which is equal to the coefficient of $x^{1003}$ in the expansion of

$$
\begin{aligned}
\left(1+x+x^{2}\right)^{2005}\left(1+x+\cdots+x^{1003}\right) & =\left[\left(1-x^{3}\right)^{2005}(1-x)^{-2005}\right]\left(1-x^{1004}\right)(1-x)^{-1} \\
& =\left(1-x^{3}\right)^{2005}(1-x)^{-2006}-\left(1-x^{3}\right)^{2005}(1-x)^{-2006} x^{1004}
\end{aligned}
$$

Since the degree of every term in the expansion of the second member on the right exceeds 1003, we are looking for the coefficient of $x^{1003}$ in the expansion of the first member:

$$
\left(1-x^{3}\right)^{2005}(1-x)^{-2006}=\sum_{i=0}^{2005}(-1)^{i}\left(\begin{array}{c}
2005 \\
i
\end{array}\right) x^{3 i} \sum_{j=0}^{\infty}(-1)^{j}\left(\begin{array}{c}
-2006 \\
j
\end{array}\right) x^{j}
$$

$$
\begin{aligned}
& =\sum_{i=0}^{2005} \sum_{j=0}^{\infty}(-1)^{i}\left(\begin{array}{c}
2005 \\
i
\end{array}\right)\left(\begin{array}{c}
2005+j \\
j
\end{array}\right) x^{3 i+j} \\
& =\sum_{k=0}^{\infty}\left(\sum_{i=1}^{2005}(-1)^{i}\left(\begin{array}{c}
2005 \\
i
\end{array}\right)\left(\begin{array}{c}
2005+k-3 i \\
2005
\end{array}\right)\right) x^{k} .
\end{aligned}
$$

The desired number is

$$
\sum_{i=1}^{334}(-1)^{i}\left(\begin{array}{c}
2005 \\
i
\end{array}\right)\left(\begin{array}{c}
3008-3 i \\
2005
\end{array}\right)=\sum_{i=1}^{334}(-1)^{i} \frac{(3008-3 i) !}{i !(2005-i) !(1003-3 i) !} .
$$

(Note that $\left(\begin{array}{c}3008-3 i \\ 2005\end{array}\right)=0$ when $i \geq 335$.)

2. Let $A B C$ be an acute-angled triangle. Inscribe a rectangle $D E F G$ in this triangle so that $D$ is on $A B, E$ is on $A C$ and both $F$ and $G$ are on $B C$. Describe the locus of (i.e., the curve occupied by) the intersections of the diagonals of all possible rectangles $D E F G$.

Solution. The locus is the line segment joining the midpoint $M$ of $B C$ to the midpoint $K$ of the altitude $A H$. Note that a segment $D E$ with $D$ on $A B$ and $E$ on $A C$ determines an inscribed rectangle; the midpoint $F$ of $D E$ lies on the median $A M$, while the midpoint of the perpendicular from $F$ to $B C$ is the centre of the rectangle. This lies on the median $M K$ of the triangle $A M H$.

Conversely, any point $P$ on $M K$ is the centre of a rectangle with base along $B C$ whose height is double the distance from $K$ to $B C$.

3. In a rectangular array of nonnegative real numbers with $m$ rows and $n$ columns, each row and each column contains at least one positive element. Moreover, if a row and a column intersect in a positive element, then the sums of their elements are the same. Prove that $m=n$.

Solution 1. Consider first the case where all the rows have the same positive sum $s$; this covers the particular situation in which $m=1$. Then each column, sharing a positive element with some row, must also have the sum $s$. Then the sum of all the entries in the matrix is $m s=n s$, whence $m=n$.

We prove the general case by induction on $m$. The case $m=1$ is already covered. Suppose that we have an $m \times n$ array not all of whose rows have the same sum. Let $r<m$ of the rows have the sum $s$, and each of the of the other rows have a different sum. Then every column sharing a positive entry with one of these rows must also have sum $s$, and these are the only columns with the sum $s$. Suppose there are $c$ columns with sum $s$. The situation is essentially unchanged if we permute the rows and then the column so that the first $r$ rows have the sum $s$ and the first $c$ columns have the sum $s$. Since all the entries of the first $r$ rows not in the first $c$ columns and in the first $c$ columns not in the first $r$ rows must be 0 , we can partition the array into a $r \times c$ array in which all rows and columns have sum $s$ and which satisfies the hypothesis of the problem, two rectangular arrays of zeros in the upper right and lower left and a rectangular $(m-r) \times(n-c)$ array in the lower right that satisfies the conditions of the problem. By the induction hypothesis, we see that $r=c$ and so $m=n$.

Solution 2. [Y. Zhao] Let the term in the $i$ th row and the $j$ th column of the array be denoted by $a_{i j}$, and let $S=\{(i, j)$ : $\left.a_{i j}>0\right\}$. Suppose that $r_{i}$ is the sum of the $i$ th row and $c_{j}$ the sum of the $j$ th column. Then $r_{i}=c_{j}$ whenever $(i, j) \in S$. Then we have that

$$
\sum\left\{\frac{a_{i j}}{r_{i}}:(i, j) \in S\right\}=\sum\left\{\frac{a_{i j}}{c_{j}}:(i, j) \in S\right\}
$$

We evaluate the sums on either side independently.

$$
\begin{aligned}
& \sum\left\{\frac{a_{i j}}{r_{i}}:(i, j) \in S\right\}=\sum\left\{\frac{a_{i j}}{r_{i}}: 1 \leq i \leq m, 1 \leq j \leq n\right\}=\sum_{i=1}^{m} \frac{1}{r_{i}} \sum_{j=1}^{n} a_{i j}=\sum_{i=1}^{m}\left(\frac{1}{r_{i}}\right) r_{i}=\sum_{i=1}^{m} 1=m . \\
& \sum\left\{\frac{a_{i j}}{c_{j}}:(i, j) \in S\right\}=\sum\left\{\frac{a_{i j}}{c_{j}}: 1 \leq i \leq m, 1 \leq j \leq n\right\}=\sum_{j=1}^{n} \frac{1}{c_{j}} \sum_{i=1}^{m} a_{i j}=\sum_{j=1}^{n}\left(\frac{1}{c_{j}}\right) c_{j}=\sum_{j=1}^{n} 1=n .
\end{aligned}
$$

Hence $m=n$.

Comment. The second solution can be made cleaner and more elegant by defining $u_{i j}=a_{i j} / r_{i}$ for all $(i, j)$. When $a_{i j}=0$, then $u_{i j}=0$. When $a_{i j}>0$, then, by hypothesis, $u_{i j}=a_{i j} / c_{j}$, a relation that in fact holds for all $(i, j)$. We find that

$$
\sum_{j=1}^{n} u_{i j}=1 \quad \text { and } \quad \sum_{i=1}^{n} u_{i j}=1
$$

for $1 \leq i \leq m$ and $1 \leq j \leq n$, so that $\left(u_{i j}\right)$ is an $m \times n$ array whose row sums and column sums are all equal to 1 . Hence

$$
m=\sum_{i=1}^{m}\left(\sum_{j=1}^{n} u_{i j}\right)=\sum\left\{u_{i j}: 1 \leq i \leq m, 1 \leq j \leq n\right\}=\sum_{j=1}^{n}\left(\sum_{i=1}^{m} u_{i j}\right)=n
$$

(being the sum of all the entries in the array).

4. Consider a round-robin tournament with $2 n+1$ teams, where each team plays each other team exactly once. We say that three teams $X, Y$ and $Z$, form a cycle triplet if $X$ beats $Y, Y$ beats $Z$, and $Z$ beats $X$. There are no ties.

(a) Determine the minimum number of cycle triplets possible.

(b) Determine the maximum number of cycle triplets possible.

Solution 1. (a) The minimum is 0 , which is achieved by a tournament in which team $T_{i}$ beats $T_{j}$ if and only if $i>j$.

(b) Any set of three teams constitutes either a cycle triplet or a "dominated triplet" in which one team beats the other two; let there be $c$ of the former and $d$ of the latter. Then $c+d=\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)$. Suppose that team $T_{i}$ beats $x_{i}$ other teams; then it is the winning team in exactly $\left(\begin{array}{c}x_{i} \\ 2\end{array}\right)$ dominated triples. Observe that $\sum_{i=1}^{2 n+1} x_{i}=\left(\begin{array}{c}2 n+1 \\ 2\end{array}\right)$, the total number of games. Hence

$$
d=\sum_{i=1}^{2 n+1}\left(\begin{array}{c}
x_{i} \\
2
\end{array}\right)=\frac{1}{2} \sum_{i=1}^{2 n+1} x_{i}^{2}-\frac{1}{2}\left(\begin{array}{c}
2 n+1 \\
2
\end{array}\right)
$$

By the Cauchy-Schwarz Inequality, $(2 n+1) \sum_{i=1}^{2 n+1} x_{i}^{2} \geq\left(\sum_{i=1}^{2 n+1} x_{i}\right)^{2}=n^{2}(2 n+1)^{2}$, whence

$$
c=\left(\begin{array}{c}
2 n+1 \\
3
\end{array}\right)-\sum_{i=1}^{2 n+1}\left(\begin{array}{c}
x_{i} \\
2
\end{array}\right) \leq\left(\begin{array}{c}
2 n+1 \\
3
\end{array}\right)-\frac{n^{2}(2 n+1)}{2}+\frac{1}{2}\left(\begin{array}{c}
2 n+1 \\
2
\end{array}\right)=\frac{n(n+1)(2 n+1)}{6} .
$$

To realize the upper bound, let the teams be $T_{1}=T_{2 n+2}, T_{2}=T_{2 n+3}$. $\cdots, T_{i}=T_{2 n+1+i}, \cdots, T_{2 n+1}=T_{4 n+2}$. For each $i$, let team $T_{i}$ beat $T_{i+1}, T_{i+2}, \cdots, T_{i+n}$ and lose to $T_{i+n+1}, \cdots, T_{i+2 n}$. We need to check that this is a consistent assignment of wins and losses, since the result for each pair of teams is defined twice. This can be seen by noting that $(2 n+1+i)-(i+j)=2 n+1-j \geq n+1$ for $1 \leq j \leq n$. The cycle triplets are $\left(T_{i}, T_{i+j}, T_{i+j+k}\right)$ where $1 \leq j \leq n$ and $(2 n+1+i)-(i+j+k) \leq n$, i.e., when $1 \leq j \leq n$ and $n+1-j \leq k \leq n$. For each $i$, this counts $1+2+\cdots+n=\frac{1}{2} n(n+1)$ cycle triplets. When we range over all $i$, each cycle triplet gets counted three times, so the number of cycle triplets is

$$
\frac{2 n+1}{3}\left(\frac{n(n+1)}{2}\right)=\frac{n(n+1)(2 n+1)}{6} .
$$

Solution 2. [S. Eastwood] (b) Let $t$ be the number of cycle triplets and $u$ be the number of ordered triplets of teams $(X, Y, Z)$ where $X$ beats $Y$ and $Y$ beats $Z$. Each cycle triplet generates three ordered triplets while other triplets generate exactly one. The total number of triplets is

$$
\left(\begin{array}{c}
2 n+1 \\
3
\end{array}\right)=\frac{n\left(4 n^{2}-1\right)}{3} .
$$

The number of triples that are not cycle is

$$
\frac{n\left(4 n^{2}-1\right)}{3}-t
$$

Hence

$$
u=3 t+\left(\frac{n\left(4 n^{2}-1\right)}{3}-t\right) \Longrightarrow
$$

$$
t=\frac{3 u-n\left(4 n^{2}-1\right)}{6}=\frac{u-(2 n+1) n^{2}}{2}+\frac{n(n+1)(2 n+1)}{6}
$$

If team $Y$ beats $a$ teams and loses to $b$ teams, then the number of ordered triples with $Y$ as the central element is $a b$. Since $a+b=2 n$, by the Arithmetic-Geometric Means Inequality, we have that $a b \leq n 2$. Hence $u \leq(2 n+1) n 2$, so that

$$
t \leq \frac{n(n+1)(2 n+1)}{6}
$$

The maximum is attainable when $u=(2 n+1) n 2$, which can occur when we arrange all the teams in a circle with each team beating exactly the $n$ teams in the clockwise direction.

Comment. Interestingly enough, the maximum is $\sum_{i=1}^{n} i^{2}$; is there a nice argument that gives the answer in this form?

5. The vertices of a right triangle $A B C$ inscribed in a circle divide the circumference into three arcs. The right angle is at $A$, so that the opposite $\operatorname{arc} B C$ is a semicircle while $\operatorname{arc} A B$ and $\operatorname{arc} A C$ are supplementary. To each of the three arcs, we draw a tangent such that its point of tangency is the midpoint of that portion of the tangent intercepted by the extended lines $A B$ and $A C$. More precisely, the point $D$ on $\operatorname{arc} B C$ is the midpoint of the segment joining the points $D^{\prime}$ and $D^{\prime \prime}$ where the tangent at $D$ intersects the extended lines $A B$ and $A C$. Similarly for $E$ on arc $A C$ and $F$ on arc $A B$.

Prove that triangle $D E F$ is equilateral.

![](https://cdn.mathpix.com/cropped/2024_04_17_87adde2ca6b3faf10a84g-4.jpg?height=509&width=783&top_left_y=1042&top_left_x=662)

Solution 1. A prime indicates where a tangent meets $A B$ and a double prime where it meets $A C$. It is given that $D D^{\prime}=D D^{\prime \prime}, E E^{\prime}=E E^{\prime \prime}$ and $F F^{\prime}=F F^{\prime \prime}$. It is required to show that arc $E F$ is a third of the circumference as is arc $D B F$.

$A F$ is the median to the hypotenuse of right triangle $A F^{\prime} F^{\prime \prime}$, so that $F F^{\prime}=F A$ and therefore

$$
\operatorname{arc} A F=2 \angle F^{\prime \prime} F A=2\left(\angle F F^{\prime} A+\angle F A F^{\prime}\right)=4 \angle F A F^{\prime}=4 \angle F A B=2 \text { arc } B F \text {, }
$$

whence $\operatorname{arc} F A=(2 / 3)$ arc $B F A$. Similarly, arc $A E=(2 / 3)$ arc $A E C$. Therefore, $\operatorname{arc} F E$ is $2 / 3$ of the semicircle, or $1 / 3$ of the circumference as desired.

As for arc $D B F$, arc $B D=2 \angle B A D=\angle B A D+\angle B D^{\prime} D=\angle A D D^{\prime \prime}=(1 / 2)$ arc $A C D$. But, arc $B F=(1 / 2)$ arc $A F$, so arc $D B F=(1 / 2)$ arc $F A E D$. That is, arc $D B F$ is $1 / 3$ the circumference and the proof is complete.

Solution 2. Since $A E^{\prime} E^{\prime \prime}$ is a right triangle, $A E=E E^{\prime}=E E^{\prime \prime}$ so that $\angle C A E=\angle C E^{\prime \prime} E$. Also $A D=D^{\prime} D=D D^{\prime \prime}$, so that $\angle C D D^{\prime \prime}=\angle C A D=\angle C D^{\prime \prime} D$. As $E A D C$ is a concyclic quadrilateral,

$$
\begin{aligned}
180^{\circ} & =\angle E A D+\angle E C D \\
& =\angle D A C+\angle C A E+\angle E C A+\angle A C D \\
& =\angle D A C+\angle C A E+\angle C E E^{\prime \prime}+\angle C E^{\prime \prime} E+\angle C D D^{\prime \prime}+\angle C D^{\prime \prime} D \\
& =\angle D A C+\angle C A E+\angle C A E+\angle C A E+\angle C A D+\angle C A D \\
& =3(\angle D A C+\angle D A E)=3(\angle D A E)
\end{aligned}
$$

Hence $\angle D F E=\angle D A E=60^{\circ}$. Similarly, $\angle D E F=60^{\circ}$. It follows that triangle $D E F$ is equilateral.