exam: INMO

problem: Let $\Gamma_{1}$ and $\Gamma_{2}$ be two circles touching each other externally at $R$. Let $l_{1}$ be a line which is tangent to $\Gamma_{2}$ at $P$ and passing through the center $O_{1}$ of $\Gamma_{1}$. Similarly, let $l_{2}$ be a line which is tangent to $\Gamma_{2}$ at $Q$ and passing through the center $O_{2}$ of $\Gamma_{2}$. Suppose $l_{1}$ and $l_{2}$ are not parallel and interesct at $K$. If $K P=K Q$, prove that the triangle $P Q R$ is equilateral.

solution:
Suppose that $P$ and $Q$ lie on the opposite sides of line joining $O_{1}$ and $O_{2}$. By symmetry we may assume that the configuration is as shown in the figure below. Then we have $K P>K O_{1}>K Q$ since $K O_{1}$ is the hypotenuse of triangle $K Q O_{1}$. This is a contradiction to the given assumption, and therefore $P$ and $Q$ lie on the same side of the line joining $O_{1}$ and $O_{2}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_b0c7cb620b2c53f6f90cg-1.jpg?height=532&width=875&top_left_y=1111&top_left_x=473)

Since $K P=K Q$ it follows that $K$ lies on the radical axis of the given circles, which is the common tangent at $R$. Therefore $K P=K Q=K R$ and hence $K$ is the cirumcenter of $\triangle P Q R$.

![](https://cdn.mathpix.com/cropped/2024_06_05_b0c7cb620b2c53f6f90cg-2.jpg?height=510&width=927&top_left_y=317&top_left_x=556)

On the other hand, $\triangle K Q O_{1}$ and $\triangle K R O_{1}$ are both right-angled triangles with $K Q=K R$ and $Q O_{1}=R O_{1}$, and hence the two triangles are congruent. Therefore $\widehat{Q K O_{1}}=\widehat{R K O_{1}}$, so $K O_{1}$, and hence $P K$ is perpendicular to $Q R$. Similarly, $Q K$ is perpendicular to $P R$, so it follows that $K$ is the orthocenter of $\triangle P Q R$. Hence we have that $\triangle P Q R$ is equilateral.

Alternate solution. We again rule out the possibility that $P$ and $Q$ are on the opposite side of the line joining $O_{1} O_{2}$, and assume that they are on the same side.

Observe that $\triangle K P O_{2}$ is congruent to $\triangle K Q O_{1}$ (since $K P=K Q$ ). Therefore $O_{1} P=O_{2} Q=r$ (say). In $\triangle O_{1} O_{2} Q$, we have $\widehat{O_{1} Q O_{2}}=\pi / 2$ and $R$ is the midpoint of the hypotenuse, so $R Q=$ $R O_{1}=r$. Therefore $\triangle O_{1} R Q$ is equilateral, so $\widehat{Q R O_{1}}=\pi / 3$. Similarly, $P R=r$ and $\widehat{P R O_{2}}=\pi / 3$, hence $\widehat{P R Q}=\pi / 3$. Since $P R=Q R$ it follows that $\triangle P Q R$ is equilateral.

problem_match: Problem 1.

solution_match: Solution.


problem:
Find all positive integers $m, n$, and primes $p \geq 5$ such that

$$
m\left(4 m^{2}+m+12\right)=3\left(p^{n}-1\right)
$$

solution:
Rewriting the given equation we have

$$
4 m^{3}+m^{2}+12 m+3=3 p^{n}
$$

The left hand side equals $(4 m+1)\left(m^{2}+3\right)$.

Suppose that $\left(4 m+1, m^{2}+3\right)=1$. Then $\left(4 m+1, m^{2}+3\right)=\left(3 p^{n}, 1\right),\left(3, p^{n}\right),\left(p^{n}, 3\right)$ or $\left(1,3 p^{n}\right)$, a contradiction since $4 m+1, m^{2}+3 \geq 4$. Therefore $\left(4 m+1, m^{2}+3\right)>1$.

Since $4 m+1$ is odd we have $\left(4 m+1, m^{2}+3\right)=\left(4 m+1,16 m^{2}+48\right)=(4 m+1,49)=7$ or 49 . This proves that $p=7$, and $4 m+1=3 \cdot 7^{k}$ or $7^{k}$ for some natural number $k$. If $(4 m+1,49)=7$ then we have $k=1$ and $4 m+1=21$ which does not lead to a solution. Therefore $\left(4 m+1, m^{2}+3\right)=49$. If $7^{3}$ divides $4 m+1$ then it does not divide $m^{2}+3$, so we get $m^{2}+3 \leq 3 \cdot 7^{2}<7^{3} \leq 4 m+1$. This implies $(m-2)^{2}<2$, so $m \leq 3$, which does not lead to a solution. Therefore we have $4 m+1=49$ which implies $m=12$ and $n=4$. Thus $(m, n, p)=(12,4,7)$ is the only solution.

problem_match: Problem 2.


problem: Let $a, b, c, d$ be positive integers such that $a \geq b \geq c \geq d$. Prove that the equation $x^{4}-a x^{3}-b x^{2}-c x-d=0$ has no integer solution.

solution: Suppose that $m$ is an integer root of $x^{4}-a x^{3}-b x^{2}-c x-d=0$. As $d \neq 0$, we have $m \neq 0$. Suppose now that $m>0$. Then $m^{4}-a m^{3}=b m^{2}+c m+d>0$ and hence $m>a \geq d$. On the other hand $d=m\left(m^{3}-a m^{2}-b m-c\right)$ and hence $m$ divides $d$, so $m \leq d$, a contradiction. If $m<0$, then writing $n=-m>0$ we have $n^{4}+a n^{3}-b n^{2}+c n-d=n^{4}+n^{2}(a n-b)+(c n-d)>0$, a contradiction. This proves that the given polynomial has no integer roots.

problem_match: Problem 3.


problem: Let $n$ be a positive integer. Call a nonempty subset $S$ of $\{1,2, \ldots, n\}$ good if the arithmetic mean of the elements of $S$ is also an integer. Further let $t_{n}$ denote the number of good subsets of $\{1,2, \ldots, n\}$. Prove that $t_{n}$ and $n$ are both odd or both even.

solution:
We show that $T_{n}-n$ is even. Note that the subsets $\{1\},\{2\}, \cdots,\{n\}$ are good. Among the other good subsets, let $A$ be the collection of subsets with an integer average which belongs to the subset, and let $B$ be the collection of subsets with an integer average which is not a member of the subset. Then there is a bijection between $A$ and $B$, because removing the average takes a member of $A$ to a member of $B$; and including the average in a member of $B$ takes it to its inverse. So $T_{n}-n=|A|+|B|$ is even.

Alternate solution. Let $S=\{1,2, \ldots, n\}$. For a subset $A$ of $S$, let $\bar{A}=\{n+1-a \mid a \in A\}$. We call a subset $A$ symmetric if $\bar{A}=A$. Note that the arithmetic mean of a symmetric subset is $(n+1) / 2$. Therefore, if $n$ is even, then there are no symmetric good subsets, while if $n$ is odd then every symmetric subset is good.

If $A$ is a proper good subset of $S$, then so is $\bar{A}$. Therefore, all the good subsets that are not symmetric can be paired. If $n$ is even then this proves that $t_{n}$ is even. If $n$ is odd, we have to show that there are odd number of symmetric subsets. For this, we note that a symmetric subset contains the element $(n+1) / 2$ if and only if it has odd number of elements. Therefore, for any natural number $k$, the number of symmetric subsets of size $2 k$ equals the number of symmetric subsets of size $2 k+1$. The result now follows since there is exactly one symmetric subset with only one element.

problem_match: Problem 4.


problem: In an acute triangle $A B C, O$ is the circumcenter, $H$ is the orthocenter and $G$ is the centroid. Let $O D$ be perpendicular to $B C$ and $H E$ be perpendicular to $C A$, with $D$ on $B C$ and $E$ on $C A$. Let $F$ be the midpoint of $A B$. Suppose the areas of triangles $O D C, H E A$ and $G F B$ are equal. Find all the possible values of $\widehat{C}$.

solution:
Let $R$ be the circumradius of $\triangle A B C$ and $\Delta$ its area. We have $O D=R \cos A$ and $D C=\frac{a}{2}$, so

$$
[O D C]=\frac{1}{2} \cdot O D \cdot D C=\frac{1}{2} \cdot R \cos A \cdot R \sin A=\frac{1}{2} R^{2} \sin A \cos A
$$

Again $H E=2 R \cos C \cos A$ and $E A=c \cos A$. Hence

$$
[H E A]=\frac{1}{2} \cdot H E \cdot E A=\frac{1}{2} \cdot 2 R \cos C \cos A \cdot c \cos A=2 R^{2} \sin C \cos C \cos ^{2} A
$$

Further

$$
[G F B]=\frac{\Delta}{6}=\frac{1}{6} \cdot 2 R^{2} \sin A \sin B \sin C=\frac{1}{3} R^{2} \sin A \sin B \sin C
$$

Equating (1) and (2) we get $\tan A=4 \sin C \cos C$. And equating (1) and (3), and using this relation we get

$$
\begin{aligned}
3 \cos A & =2 \sin B \sin C=2 \sin (C+A) \sin C \\
& =2(\sin C+\cos C \tan A) \sin C \cos A \\
& =2 \sin ^{2} C\left(1+4 \cos ^{2} C\right) \cos A
\end{aligned}
$$

Since $\cos A \neq 0$ we get $3=2 t(-4 t+5)$ where $t=\sin ^{2} C$. This implies $(4 t-3)(2 t-1)=0$ and therefore, since $\sin C>0$, we get $\sin C=\sqrt{3} / 2$ or $\sin C=1 / \sqrt{2}$. Because $\triangle A B C$ is acute, it follows that $\widehat{C}=\pi / 3$ or $\pi / 4$.

We observe that the given conditions are satisfied in an equilateral triangle, so $\widehat{C}=\pi / 3$ is a possibility. Also, the conditions are satisfied in a triangle where $\widehat{C}=\pi / 4, \widehat{A}=\tan ^{-1} 2$ and $\widehat{B}=\tan ^{-1} 3$. Therefore $\widehat{C}=\pi / 4$ is also a possibility.

Thus the two possible values of $\widehat{C}$ are $\pi / 3$ and $\pi / 4$.

problem_match: Problem 5.


problem: Let $a, b, c, x, y, z$ be positive real numbers such that $a+b+c=x+y+z$ and abc $=x y z$. Further, suppose that $a \leq x<y<z \leq c$ and $a<b<c$. Prove that $a=x, b=y$ and $c=z$.

solution:
Let

$$
f(t)=(t-x)(t-y)(t-z)-(t-a)(t-b)(t-c)
$$

Then $f(t)=k t$ for some constant $k$. Note that $k a=f(a)=(a-x)(a-y)(a-z) \leq 0$ and hence $k \leq 0$. Similarly, $k c=f(c)=(c-x)(c-y)(c-z) \geq 0$ and hence $k \geq 0$. Combining the two, it follows that $k=0$ and that $f(a)=f(c)=0$. These equalities imply that $a=x$ and $c=z$, and then it also follows that $b=y$.

problem_match: Problem 6.