exam: EGMO

problem:
Determine all real constants $t$ such that whenever $a, b, c$ are the lengths of the sides of a triangle, then so are $a^{2}+b c t, b^{2}+c a t, c^{2}+a b t$.

Proposed by S. Khan, UNK
The answer is the interval $[2 / 3,2]$.

solution:
If $t<2 / 3$, take a triangle with sides $c=b=1$ and $a=2-\epsilon$. Then $b^{2}+c a t+c^{2}+$ $a b t-a^{2}-b c t=3 t-2+\epsilon(4-2 t-\epsilon) \leq 0$ for small positive $\epsilon$; for instance, for any $0<\epsilon<(2-3 t) /(4-2 t)$.

On the other hand, if $t>2$, then take a triangle with sides $b=c=1$ and $a=\epsilon$. Then $b^{2}+c a t+c^{2}+a b t-a^{2}-b c t=2-t+\epsilon(2 t-\epsilon) \leq 0$ for small positive $\epsilon$; for instance, for any $0<\epsilon<(t-2) /(2 t)$.
Now assume that $2 / 3 \leq t \leq 2$ and $b+c>a$. Then using $(b+c)^{2} \geq 4 b c$ we obtain

$$
\begin{aligned}
b^{2}+c a t+c^{2}+a b t-a^{2}-b c t & =(b+c)^{2}+a t(b+c)-(2+t) b c-a^{2} \\
& \geq(b+c)^{2}+a t(b+c)-\frac{1}{4}(2+t)(b+c)^{2}-a^{2} \\
& \geq \frac{1}{4}(2-t)(b+c)^{2}+a t(b+c)-a^{2}
\end{aligned}
$$

As $2-t \geq 0$ and $t>0$, this last expression is an increasing function of $b+c$, and hence using $b+c>a$ we obtain

$$
b^{2}+c a t+c^{2}+a b t-a^{2}-b c t>\frac{1}{4}(2-t) a^{2}+t a^{2}-a^{2}=\frac{3}{4}\left(t-\frac{2}{3}\right) a^{2} \geq 0
$$

as $t \geq 2 / 3$. The other two inequalities follow by symmetry.

solution_match: # Solution 1.


solution:
After showing that $t$ must be in the interval $[2 / 3,2]$ as in Solution 1, we let $x=$ $(c+a-b) / 2, y=(a+b-c) / 2$ and $z=(b+c-a) / 2$ so that $a=x+y, b=y+z$, $c=z+x$. Then we have:
$b^{2}+c a t+c^{2}+a b t-a^{2}-b c t=\left(x^{2}+y^{2}-z^{2}+x y+x z+y z\right) t+2\left(z^{2}+x z+y z-x y\right)$
Since this linear function of $t$ is positive both at $t=2 / 3$ where
$\frac{2}{3}\left(x^{2}+y^{2}-z^{2}+x y+x z+y z\right)+2\left(z^{2}+x z+y z-x y\right)=\frac{2}{3}\left((x-y)^{2}+4(x+y) z+2 z^{2}\right)>0$
and at $t=2$ where
$2\left(x^{2}+y^{2}-z^{2}+x y+x z-y z\right)+2\left(z^{2}+x z+y z+x y\right)=2\left(x^{2}+y^{2}\right)+4(x+y) z>0$,
it is positive on the entire interval $[2 / 3,2]$.

solution_match: # Solution 2.


solution:
After the point in Solution 2 where we obtain
$b^{2}+c a t+c^{2}+a b t-a^{2}-b c t=\left(x^{2}+y^{2}-z^{2}+x y+x z+y z\right) t+2\left(z^{2}+x z+y z-x y\right)$
we observe that the right hand side can be rewritten as

$$
(2-t) z^{2}+(x-y)^{2} t+(3 t-2) x y+z(x+y)(2+t)
$$

As the first three terms are non-negative and the last term is positive, the result follows.

solution_match: # Solution 3.


solution:
First we show that $t$ must be in the interval $[2 / 3,2]$ as in Solution 1. Then:
Case 1: If $a \geq b, c$, then $a b+a c-b c>0,2\left(b^{2}+c^{2}\right) \geq(b+c)^{2}>a^{2}$ and $t \geq 2 / 3$ implies:

$$
\begin{aligned}
b^{2}+c a t+c^{2}+a b t-a^{2}-b c t & =b^{2}+c^{2}-a^{2}+(a b+a c-b c) t \\
& \geq\left(b^{2}+c^{2}-a^{2}\right)+\frac{2}{3}(a b+a c-b c) \\
& \geq \frac{1}{3}\left(3 b^{2}+3 c^{2}-3 a^{2}+2 a b+2 a c-2 b c\right) \\
& \geq \frac{1}{3}\left[\left(2 b^{2}+2 c^{2}-a^{2}\right)+(b-c)^{2}+2 a(b+c-a)\right] \\
& >0
\end{aligned}
$$

Case 2: If $b \geq a, c$, then $b^{2}+c^{2}-a^{2}>0$. If also $a b+a c-b c \geq 0$, then $b^{2}+c a t+$ $c^{2}+a b t-a^{2}-b c t>0$. If, on the other hand, $a b+a c-b c \leq 0$, then since $t \leq 2$, we have:

$$
\begin{aligned}
b^{2}+c a t+c^{2}+a b t-a^{2}-b c t & \geq b^{2}+c^{2}-a^{2}+2(a b+a c-b c) \\
& \geq(b-c)^{2}+a(b+c-a)+a(b+c) \\
& >0
\end{aligned}
$$

By symmetry, we are done.

solution_match: # Solution 4.


problem:
Let $D$ and $E$ be two points on the sides $A B$ and $A C$, respectively, of a triangle $A B C$, such that $D B=B C=C E$, and let $F$ be the point of intersection of the lines $C D$ and $B E$. Prove that the incenter $I$ of the triangle $A B C$, the orthocenter $H$ of the triangle $D E F$ and the midpoint $M$ of the $\operatorname{arc} B A C$ of the circumcircle of the triangle $A B C$ are collinear.

Proposed by Danylo Khilko, UKR

solution:
As $D B=B C=C E$ we have $B I \perp C D$ and $C I \perp B E$. Hence $I$ is orthocenter of triangle $B F C$. Let $K$ be the point of intersection of the lines $B I$ and $C D$, and let $L$ be the point of intersection of the lines $C I$ and $B E$. Then we have the power relation $I B \cdot I K=I C \cdot I L$. Let $U$ and $V$ be the feet of the perpendiculars from $D$ to $E F$ and $E$ to $D F$, respectively. Now we have the power relation $D H \cdot H U=E H \cdot H V$.
![](https://cdn.mathpix.com/cropped/2024_11_22_d396c315567a32ce5b15g-5.jpg?height=1058&width=1025&top_left_y=908&top_left_x=496)

Let $\omega_{1}$ and $\omega_{2}$ be the circles with diameters $B D$ and $C E$, respectively. From the power relations above we conclude that $I H$ is the radical axis of the circles $\omega_{1}$ and $\omega_{2}$.
Let $O_{1}$ and $O_{2}$ be centers of $\omega_{1}$ and $\omega_{2}$, respectively. Then $M B=M C, B O_{1}=C O_{2}$ and $\angle M B O_{1}=\angle M C O_{2}$, and the triangles $M B O_{1}$ and $M C O_{2}$ are congruent. Hence $M O_{1}=M O_{2}$. Since radii of $\omega_{1}$ and $\omega_{2}$ are equal, this implies that $M$ lies on the radical axis of $\omega_{1}$ and $\omega_{2}$ and $M, I, H$ are collinear.


solution:
Let the points $K, L, U, V$ be as in Solution 1. Le $P$ be the point of intersection of $D U$ and $E I$, and let $Q$ be the point of intersection of $E V$ and $D I$.

Since $D B=B C=C E$, the points $C I$ and $B I$ are perpendicular to $B E$ and $C D$, respectively. Hence the lines $B I$ and $E V$ are parallel and $\angle I E B=\angle I B E=$ $\angle U E H$. Similarly, the lines $C I$ and $D U$ are parallel and $\angle I D C=\angle I C D=\angle V D H$. Since $\angle U E H=\angle V D H$, the points $D, Q, F, P, E$ are concyclic. Hence $I P \cdot I E=$ $I Q \cdot I D$.

Let $R$ be the second point intersection of the circumcircle of triangle $H E P$ and the line $H I$. As $I H \cdot I R=I P \cdot I E=I Q \cdot I D$, the points $D, Q, H, R$ are also concyclic. We have $\angle D Q H=\angle E P H=\angle D F E=\angle B F C=180^{\circ}-\angle B I C=90^{\circ}-\angle B A C / 2$. Now using the concylicity of $D, Q, H, R$, and $E, P, H, R$ we obtain $\angle D R H=$ $\angle E R H=\angle 180^{\circ}-\left(90^{\circ}-\angle B A C / 2\right)=90^{\circ}+\angle B A C / 2$. Hence $R$ is inside the triangle $D E H$ and $\angle D R E=360^{\circ}-\angle D R H-\angle E R H=180^{\circ}-\angle B A C$ and it follows that the points $A, D, R, E$ are concyclic.
![](https://cdn.mathpix.com/cropped/2024_11_22_d396c315567a32ce5b15g-6.jpg?height=1063&width=1014&top_left_y=1038&top_left_x=512)

As $M B=M C, B D=C E, \angle M B D=\angle M C E$, the triangles $M B D$ and $M C E$ are congruent and $\angle M D A=\angle M E A$. Hence the points $M, D, E, A$ are concylic. Therefore the points $M, D, R, E, A$ are concylic. Now we have $\angle M R E=180^{\circ}-$ $\angle M A E=180^{\circ}-\left(90^{\circ}+\angle B A C / 2\right)=90^{\circ}-\angle B A C / 2$ and since $\angle E R H=90^{\circ}+$ $\angle B A C / 2$, we conclude that the points $I, H, R, M$ are collinear.


solution:
Suppose that we have a coordinate system and $\left(b_{x}, b_{y}\right),\left(c_{x}, c_{y}\right),\left(d_{x}, d_{y}\right),\left(e_{x}, e_{y}\right)$ are the coordinates of the points $B, C, D, E$, respectively. From $\overrightarrow{B I} \cdot \overrightarrow{C D}=0, \overrightarrow{C I} \cdot \overrightarrow{B E}=$ $0, \overrightarrow{E H} \cdot \overrightarrow{C D}=0, \overrightarrow{D H} \cdot \overrightarrow{B E}=0$ we obtain $\overrightarrow{I H} \cdot(\vec{B}-\vec{C}-\vec{E}+\vec{D})=0$. Hence the slope of the line $I H$ is $\left(c_{x}+e_{x}-b_{x}-d_{x}\right) /\left(b_{y}+d_{y}-c_{y}-e_{y}\right)$.
Assume that the $x$-axis lies along the line $B C$, and let $\alpha=\angle B A C, \beta=\angle A B C$, $\theta=\angle A C B$. Since $D B=B C=C E$, we have $c_{x}-b_{x}=B C, e_{x}-d_{x}=B C-$ $B C \cos \beta-B C \cos \theta, b_{y}=c_{y}=0, d_{y}-e_{y}=B C \sin \beta-B C \sin \theta$. Therefore the slope of $I H$ is $(2-\cos \beta-\cos \theta) /(\sin \beta-\sin \theta)$.

Now we will show that the slope of the line $M I$ is the same. Let $r$ and $R$ be the inradius and circumradius of the triangle $A B C$, respectively. As $\angle B M C=$ $\angle B A C=\alpha$ and $B M=M C$, we have

$$
m_{y}-i_{y}=\frac{B C}{2} \cot \left(\frac{\alpha}{2}\right)-r \text { and } m_{x}-i_{x}=\frac{A C-A B}{2}
$$

where $\left(m_{x}, m_{y}\right)$ and $\left(i_{x}, i_{y}\right)$ are the coordinates of $M$ and $I$, respectively. Therefore the slope of $M I$ is $(B C \cot (\alpha / 2)-2 r) /(A C-A B)$.

Now the equality of these slopes follows using

$$
\frac{B C}{\sin \alpha}=\frac{A C}{\sin \beta}=\frac{A B}{\sin \theta}=2 R
$$

hence

$$
B C \cot \left(\frac{\alpha}{2}\right)=4 R \cos ^{2}\left(\frac{\alpha}{2}\right)=2 R(1+\cos \alpha)
$$

and

$$
\frac{r}{R}=\cos \alpha+\cos \beta+\cos \theta-1
$$

as

$$
\frac{B C \cot (\alpha / 2)-2 r}{A C-A B}=\frac{2 R(1+\cos \alpha)-2 r}{2 R(\sin \beta-\sin \theta)}=\frac{2-\cos \beta-\cos \theta}{\sin \beta-\sin \theta}
$$

giving the collinearity of the points $I, H, M$.


solution:
Let the bisectors $B I$ and $C I$ meet the circumcircle of $A B C$ again at $P$ and $Q$, respectively. Let the altitude of $D E F$ belonging to $D$ meet $B I$ at $R$ and the one belonging to $E$ meet $C I$ at $S$.

Since $B I$ is angle bisector of the iscosceles triangle $C B D, B I$ and $C D$ are perpendicular. Since $E H$ and $D F$ are also perpendicular, $H S$ and $R I$ are parallel. Similarly, $H R$ and $S I$ are parallel, and hence $H S I R$ is a parallelogram.

On the other hand, as $M$ is the midpoint of the $\operatorname{arc} B A C$, we have $\angle M P I=$ $\angle M P B=\angle M Q C=\angle M Q I$, and $\angle P I Q=(\widehat{P A}+\widehat{C B}+\widehat{A Q}) / 2=(\widehat{P C}+\widehat{C B}+$ $\widehat{B Q}) / 2=\angle P M Q$. Therefore $M P I Q$ is a parallelogram.

Since $C I$ is angle bisector of the iscosceles triangle $B C E$, the triangle $B S E$ is also isosceles. Hence $\angle F B S=\angle E B S=\angle S E B=\angle H E F=\angle H D F=\angle R D F=$ $\angle F C S$ and $B, S, F, C$ are concyclic. Similarly, $B, F, R, C$ are concyclic. Therefore $B, S, R, C$ are concyclic. As $B, Q, P, C$ are also concyclic, $S R$ an $Q P$ are parallel.

Now it follows that HSIR and MQIP are homothetic parallelograms, and therefore $M, H, I$ are collinear.
![](https://cdn.mathpix.com/cropped/2024_11_22_d396c315567a32ce5b15g-8.jpg?height=1196&width=1075&top_left_y=1142&top_left_x=563)


problem:
We denote the number of positive divisors of a positive integer $m$ by $d(m)$ and the number of distinct prime divisors of $m$ by $\omega(m)$. Let $k$ be a positive integer. Prove that there exist infinitely many positive integers $n$ such that $\omega(n)=k$ and $d(n)$ does not divide $d\left(a^{2}+b^{2}\right)$ for any positive integers $a, b$ satisfying $a+b=n$.

Proposed by JPN

solution:
We will show that any number of the form $n=2^{p-1} m$ where $m$ is a positive integer that has exactly $k-1$ prime factors all of which are greater than 3 and $p$ is a prime number such that $(5 / 4)^{(p-1) / 2}>m$ satisfies the given condition.

Suppose that $a$ and $b$ are positive integers such that $a+b=n$ and $d(n) \mid d\left(a^{2}+b^{2}\right)$. Then $p \mid d\left(a^{2}+b^{2}\right)$. Hence $a^{2}+b^{2}=q^{c p-1} r$ where $q$ is a prime, $c$ is a positive integer and $r$ is a positive integer not divisible by $q$. If $q \geq 5$, then

$$
2^{2 p-2} m^{2}=n^{2}=(a+b)^{2}>a^{2}+b^{2}=q^{c p-1} r \geq q^{p-1} \geq 5^{p-1}
$$

gives a contradiction. So $q$ is 2 or 3 .
If $q=3$, then $a^{2}+b^{2}$ is divisible by 3 and this implies that both $a$ and $b$ are divisible by 3 . This means $n=a+b$ is divisible by 3 , a contradiction. Hence $q=2$.

Now we have $a+b=2^{p-1} m$ and $a^{2}+b^{2}=2^{c p-1} r$. If the highest powers of 2 dividing $a$ and $b$ are different, then $a+b=2^{p-1} m$ implies that the smaller one must be $2^{p-1}$ and this makes $2^{2 p-2}$ the highest power of 2 dividing $a^{2}+b^{2}=2^{c p-1} r$, or equivalently, $c p-1=2 p-2$, which is not possible. Therefore $a=2^{t} a_{0}$ and $b=2^{t} b_{0}$ for some positive integer $t<p-1$ and odd integers $a_{0}$ and $b_{0}$. Then $a_{0}^{2}+b_{0}^{2}=2^{c p-1-2 t} r$. The left side of this equality is congruent to 2 modulo 4 , therefore $c p-1-2 t$ must be 1. But then $t<p-1$ gives $(c / 2) p=t+1<p$, which is not possible either.

solution_match: # Solution.