exam: INMO

problem: Let $A B C$ be a triangle with circum-circle $\Gamma$. Let $M$ be a point in the interior of triangle $A B C$ which is also on the bisector of $\angle A$. Let $A M, B M, C M$ meet $\Gamma$ in $A_{1}, B_{1}, C_{1}$ respectively. Suppose $P$ is the point of intersection of $A_{1} C_{1}$ with $A B$; and $Q$ is the point of intersection of $A_{1} B_{1}$ with $A C$. Prove that $P Q$ is parallel to $B C$.

solution:
Let $A=2 \alpha$. Then $\angle A_{1} A C=\angle B A A_{1}=\alpha$. Thus

$$
\angle A_{1} B_{1} C=\alpha=\angle B B_{1} A_{1}=\angle A_{1} C_{1} C=\angle B C_{1} A_{1}
$$

We also have $\angle B_{1} C Q=\angle A A_{1} B_{1}=\beta$, say. It follows that triangles $M A_{1} B_{1}$ and $Q C B_{1}$ are similar and hence

$$
\frac{Q C}{M A_{1}}=\frac{B_{1} C}{B_{1} A_{1}}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_9e4e2d68bb8695cca2e3g-1.jpg?height=765&width=645&top_left_y=885&top_left_x=680)

Similarly, triangles $A C M$ and $C_{1} A_{1} M$ are similar and we get

$$
\frac{A C}{A M}=\frac{C_{1} A_{1}}{C_{1} M}
$$

Using the point $P$, we get similar ratios:

$$
\frac{P B}{M A_{1}}=\frac{C_{1} B}{A_{1} C_{1}}, \quad \frac{A B}{A M}=\frac{A_{1} B_{1}}{M B_{1}}
$$

Thus,

$$
\frac{Q C}{P B}=\frac{A_{1} C_{1} \cdot B_{1} C}{C_{1} B \cdot B_{1} A_{1}}
$$

and

$$
\begin{aligned}
\frac{A C}{A B} & =\frac{M B_{1} \cdot C_{1} A_{1}}{A_{1} B_{1} \cdot C_{1} M} \\
& =\frac{M B_{1}}{C_{1} M} \frac{C_{1} A_{1}}{A_{1} B_{1}}=\frac{M B_{1}}{C_{1} M} \frac{C_{1} B \cdot Q C}{P B \cdot B_{1} C}
\end{aligned}
$$

However, triangles $C_{1} B M$ and $B_{1} C M$ are similar, which gives

$$
\frac{B_{1} C}{C_{1} B}=\frac{M B_{1}}{M C_{1}}
$$

Putting this in the last expression, we get

$$
\frac{A C}{A B}=\frac{Q C}{P B}
$$

We conclude that $P Q$ is parallel to $B C$.

solution_match: Solution:


problem: Find all natural numbers $n>1$ such that $n^{2}$ does not divide $(n-2)$ !.

solution:
Suppose $n=p q r$, where $p<q$ are primes and $r>1$. Then $p \geq 2, q \geq 3$ and $r \geq 2$, not necessarily a prime. Thus we have

$$
\begin{aligned}
& n-2 \geq n-p=p q r-p \geq 5 p>p \\
& n-2 \geq n-q=q(p r-1) \geq 3 q>q \\
& n-2 \geq n-p r=p r(q-1) \geq 2 p r>p r \\
& n-2 \geq n-q r=q r(p-1) \geq q r
\end{aligned}
$$

Observe that $p, q, p r, q r$ are all distinct. Hence their product divides $(n-2)$ !. Thus $n^{2}=p^{2} q^{2} r^{2}$ divides $(n-2)$ ! in this case. We conclude that either $n=p q$ where $p, q$ are distinct primes or $n=p^{k}$ for some prime $p$.

Case 1. Suppose $n=p q$ for some primes $p, q$, where $2<p<q$. Then $p \geq 3$ and $q \geq 5$. In this case

$$
\begin{aligned}
& n-2>n-p=p(q-1) \geq 4 p \\
& n-2>n-q=q(p-1) \geq 2 q
\end{aligned}
$$

Thus $p, q, 2 p, 2 q$ are all distinct numbers in the set $\{1,2,3, \ldots, n-2\}$. We see that $n^{2}=p^{2} q^{2}$ divides $(n-2)!$. We conclude that $n=2 q$ for some prime $q \geq 3$. Note that $n-2=2 q-2<2 q$ in this case so that $n^{2}$ does not divide $(n-2)!$.

Case 2. Suppose $n=p^{k}$ for some prime $p$. We observe that $p, 2 p, 3 p, \ldots\left(p^{k-1}-1\right) p$ all lie in the set $\{1,2,3, \ldots, n-2\}$. If $p^{k-1}-1 \geq 2 k$, then there are at least $2 k$ multiples of $p$ in the set $\{1,2,3, \ldots, n-2\}$. Hence $n^{2}=p^{2 k}$ divides $(n-2)!$. Thus $p^{k-1}-1<2 k$.

If $k \geq 5$, then $p^{k-1}-1 \geq 2^{k-1}-1 \geq 2 k$, which may be proved by an easy induction. Hence $k \leq 4$. If $k=1$, we get $n=p$, a prime. If $k=2$, then $p-1<4$ so that $p=2$ or 3 ; we get $n=2^{2}=4$ or $n=3^{2}=9$. For $k=3$, we have $p^{2}-1<6$ giving $p=2$; $n=2^{3}=8$ in this case. Finally, $k=4$ gives $p^{3}-1<8$. Again $p=2$ and $n=2^{4}=16$. However $n^{2}=2^{8}$ divides 14! and hence is not a solution.

Thus $n=p, 2 p$ for some prime $p$ or $n=8,9$. It is easy to verify that these satisfy the conditions of the problem.


problem:
Find all non-zero real numbers $x, y, z$ which satisfy the system of equations:

$$
\begin{aligned}
\left(x^{2}+x y+y^{2}\right)\left(y^{2}+y z+z^{2}\right)\left(z^{2}+z x+x^{2}\right) & =x y z \\
\left(x^{4}+x^{2} y^{2}+y^{4}\right)\left(y^{4}+y^{2} z^{2}+z^{4}\right)\left(z^{4}+z^{2} x^{2}+x^{4}\right) & =x^{3} y^{3} z^{3}
\end{aligned}
$$

solution:
Since $x y z \neq 0$, We can divide the second relation by the first. Observe that

$$
x^{4}+x^{2} y^{2}+y^{4}=\left(x^{2}+x y+y^{2}\right)\left(x^{2}-x y+y^{2}\right)
$$

holds for any $x, y$. Thus we get

$$
\left(x^{2}-x y+y^{2}\right)\left(y^{2}-y z+z^{2}\right)\left(z^{2}-z x+x^{2}\right)=x^{2} y^{2} z^{2}
$$

However, for any real numbers $x, y$, we have

$$
x^{2}-x y+y^{2} \geq|x y|
$$

Since $x^{2} y^{2} z^{2}=|x y||y z||z x|$, we get

$$
|x y||y z||z x|=\left(x^{2}-x y+y^{2}\right)\left(y^{2}-y z+z^{2}\right)\left(z^{2}-z x+x^{2}\right) \geq|x y||y z||z x|
$$

This is possible only if

$$
x^{2}-x y+y^{2}=|x y|, \quad y^{2}-y z+z^{2}=|y z|, \quad z^{2}-z x+x^{2}=|z x|
$$

hold simultaneously. However $|x y|= \pm x y$. If $x^{2}-x y+y^{2}=-x y$, then $x^{2}+y^{2}=0$ giving $x=y=0$. Since we are looking for nonzero $x, y, z$, we conclude that $x^{2}-x y+y^{2}=x y$ which is same as $x=y$. Using the other two relations, we also get $y=z$ and $z=x$. The first equation now gives $27 x^{6}=x^{3}$. This gives $x^{3}=1 / 27$ (since $x \neq 0$ ), or $x=1 / 3$. We thus have $x=y=z=1 / 3$. These also satisfy the second relation, as may be verified.


problem:
How many 6-tuples $\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\right)$ are there such that each of $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ is from the set $\{1,2,3,4\}$ and the six expressions

$$
a_{j}^{2}-a_{j} a_{j+1}+a_{j+1}^{2}
$$

for $j=1,2,3,4,5,6$ (where $a_{7}$ is to be taken as $a_{1}$ ) are all equal to one another?

solution:
Without loss of generality, we may assume that $a_{1}$ is the largest among $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$. Consider the relation

$$
a_{1}^{2}-a_{1} a_{2}+a_{2}^{2}=a_{2}^{2}-a_{2} a_{3}+a_{3}^{2}
$$

This leads to

$$
\left(a_{1}-a_{3}\right)\left(a_{1}+a_{3}-a_{2}\right)=0
$$

Observe that $a_{1} \geq a_{2}$ and $a_{3}>0$ together imply that the second factor on the left side is positive. Thus $a_{1}=a_{3}=\max \left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\right\}$. Using this and the relation

$$
a_{3}^{2}-a_{3} a_{4}+a_{4}^{2}=a_{4}^{2}-a_{4} a_{5}+a_{5}^{2}
$$

we conclude that $a_{3}=a_{5}$ as above. Thus we have

$$
a_{1}=a_{3}=a_{5}=\max \left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\right\}
$$

Let us consider the other relations. Using

$$
a_{2}^{2}-a_{2} a_{3}+a_{3}^{2}=a_{3}^{2}-a_{3} a_{4}+a_{4}^{2}
$$

we get $a_{2}=a_{4}$ or $a_{2}+a_{4}=a_{3}=a_{1}$. Similarly, two more relations give either $a_{4}=a_{6}$ or $a_{4}+a_{6}=a_{5}=a_{1}$; and either $a_{6}=a_{2}$ or $a_{6}+a_{2}=a_{1}$. Let us give values to $a_{1}$ and count the number of six-tuples in each case.

(A) Suppose $a_{1}=1$. In this case all $a_{j}$ 's are equal and we get only one six-tuple $(1,1,1,1,1,1)$.

(B) If $a_{1}=2$, we have $a_{3}=a_{5}=2$. We observe that $a_{2}=a_{4}=a_{6}=1$ or $a_{2}=a_{4}=$ $a_{6}=2$. We get two more six-tuples: $(2,1,2,1,2,1),(2,2,2,2,2,2)$.

(C) Taking $a_{1}=3$, we see that $a_{3}=a_{5}=3$. In this case we get nine possibilities for $\left(a_{2}, a_{4}, a_{6}\right)$

$$
(1,1,1),(2,2,2),(3,3,3),(1,1,2),(1,2,1),(2,1,1),(1,2,2),(2,1,2),(2,2,1)
$$

(D) In the case $a_{1}=4$, we have $a_{3}=a_{5}=4$ and

$$
\begin{aligned}
\left(a_{2}, a_{4}, a_{6}\right)=(2,2,2),(4,4,4) & ,(1,1,1),(3,3,3) \\
& (1,1,3),(1,3,1),(3,1,1),(1,3,3),(3,1,3),(3,3,1)
\end{aligned}
$$

Thus we get $1+2+9+10=22$ solutions. Since $\left(a_{1}, a_{3}, a_{5}\right)$ and $\left(a_{2}, a_{4}, a_{6}\right)$ may be interchanged, we get 22 more six-tuples. However there are 4 common among these, namely, $(1,1,1,1,1,1),(2,2,2,2,2,2),(3,3,3,3,3,3)$ and $(4,4,4,4,4,4)$. Hence the total number of six-tuples is $22+22-4=40$.


problem: Let $A B C$ be an acute-angled triangle with altitude $A K$. Let $H$ be its ortho-centre and $O$ be its circum-centre. Suppose $K O H$ is an acute-angled triangle and $P$ its circum-centre. Let $Q$ be the reflection of $P$ in the line $H O$. Show that $Q$ lies on the line joining the mid-points of $A B$ and $A C$.

solution:
Let $D$ be the mid-point of $B C ; M$ that of $H K$; and $T$ that of $O H$. Then $P M$ is perpendicular to $H K$ and $P T$ is perpendicular to $O H$. Since $Q$ is the reflection of $P$ in $H O$, we observe that $P, T, Q$ are collinear, and $P T=T Q$. Let $Q L, T N$ and $O S$ be the perpendiculars drawn respectively from $Q, T$ and $O$ on to the altitude $A K$.(See the figure.)

![](https://cdn.mathpix.com/cropped/2024_06_05_9e4e2d68bb8695cca2e3g-4.jpg?height=640&width=688&top_left_y=1159&top_left_x=707)

We have $L N=N M$, since $T$ is the mid-point of $Q P ; H N=N S$, since $T$ is the mid-point of $O H$; and $H M=M K$, as $P$ is the circum-centre of $K H O$. We obtain

$$
L H+H N=L N=N M=N S+S M
$$

which gives $L H=S M$. We know that $A H=2 O D$. Thus

$$
\begin{aligned}
A L=A H & -L H=2 O D-L H=2 S K-S M=S K+(S K-S M)=S K+M K \\
& =S K+H M=S K+H S+S M=S K+H S+L H=S K+L S=L K
\end{aligned}
$$

This shows that $L$ is the mid-point of $A K$ and hence lies on the line joining the midpoints of $A B$ and $A C$. We observe that the line joining the mid-points of $A B$ and $A C$ is also perpendicular to $A K$. Since $Q L$ is perpendicular to $A K$, we conclude that $Q$ also lies on the line joining the mid-points of $A B$ and $A C$.

Remark: It may happen that $H$ is above $L$ as in the adjoining figure, but the result remains true here as well. We have $H N=N S, L N=N M$, and $H M=$ $M K$ as earlier. Thus $H N=H L+L N$ and $N S=$ $S M+N M$ give $H L=S M$. Now $A L=A H+H L=$ $2 O D+S M=2 S K+S M=S K+(S K+S M)=$ $S K+M K=S K+H M=S K+H L+L M=S K+$ $S M+L M=L K$. The conclusion that $Q$ lies on the line joining the mid-points of $A B$ and $A C$ follows as earlier.

![](https://cdn.mathpix.com/cropped/2024_06_05_9e4e2d68bb8695cca2e3g-5.jpg?height=511&width=280&top_left_y=196&top_left_x=1499)


problem:
Define a sequence $\left\langle a_{n}\right\rangle_{n \geq 0}$ by $a_{0}=0, a_{1}=1$ and

$$
a_{n}=2 a_{n-1}+a_{n-2}
$$

for $n \geq 2$.

(a) For every $m>0$ and $0 \leq j \leq m$, prove that $2 a_{m}$ divides $a_{m+j}+(-1)^{j} a_{m-j}$.

(b) Suppose $2^{k}$ divides $n$ for some natural numbers $n$ and $k$. Prove that $2^{k}$ divides $a_{n}$.

solution:
(a) Consider $f(j)=a_{m+j}+(-1)^{j} a_{m-j}, 0 \leq j \leq m$, where $m$ is a natural number. We observe that $f(0)=2 a_{m}$ is divisible by $2 a_{m}$. Similarly,

$$
f(1)=a_{m+1}-a_{m-1}=2 a_{m}
$$

is also divisible by $2 a_{m}$. Assume that $2 a_{m}$ divides $f(j)$ for all $0 \leq j<l$, where $l \leq m$. We prove that $2 a_{m}$ divides $f(l)$. Observe

$$
\begin{aligned}
& f(l-1)=a_{m+l-1}+(-1)^{l-1} a_{m-l+1} \\
& f(l-2)=a_{m+l-2}+(-1)^{l-2} a_{m-l+2}
\end{aligned}
$$

Thus we have

$$
\begin{aligned}
a_{m+l} & =2 a_{m+l-1}+a_{m+l-2} \\
& =2 f(l-1)-2(-1)^{l-1} a_{m-l+1}+f(l-2)-(-1)^{l-2} a_{m-l+2} \\
& =2 f(l-1)+f(l-2)+(-1)^{l-1}\left(a_{m-l+2}-2 a_{m-l+1}\right) \\
& =2 f(l-1)+f(l-2)+(-1)^{l-1} a_{m-l}
\end{aligned}
$$

This gives

$$
f(l)=2 f(l-1)+f(l-2)
$$

By induction hypothesis $2 a_{m}$ divides $f(l-1)$ and $f(l-2)$. Hence $2 a_{m}$ divides $f(l)$. We conclude that $2 a_{m}$ divides $f(j)$ for $0 \leq j \leq m$.

(b) We see that $f(m)=a_{2 m}$. Hence $2 a_{m}$ divides $a_{2 m}$ for all natural numbers $m$. Let $n=2^{k} l$ for some $l \geq 1$. Taking $m=2^{k-1} l$, we see that $2 a_{m}$ divides $a_{n}$. Using an easy induction, we conclude that $2^{k} a_{l}$ divides $a_{n}$. In particular $2^{k}$ divides $a_{n}$.

solution_match: ## Solution: