exam: INMO

problem: Let $A B C$ be a triangle, $I$ its in-centre; $A_{1}, B_{1}, C_{1}$ be the reflections of $I$ in $B C, C A, A B$ respectively. Suppose the circum-circle of triangle $A_{1} B_{1} C_{1}$ passes through $A$. Prove that $B_{1}, C_{1}$, $I, I_{1}$ are concyclic, where $I_{1}$ is the in-centre of triangle $A_{1} B_{1} C_{1}$.

solution:
![](https://cdn.mathpix.com/cropped/2024_06_05_574c8adc52c8b9c64947g-1.jpg?height=601&width=512&top_left_y=472&top_left_x=774)

Note that $I A_{1}=I B_{1}=I C_{1}=2 r$, where $r$ is the in-radius of the triangle $A B C$. Hence $I$ is the circum-centre of the triangle $A_{1} B_{1} C_{1}$.

Let $K$ be the point of intersection of $I B_{1}$ and $A C$. Then $I K=r, I A=2 r$ and $\angle I K A=90^{\circ}$. It follows that $\angle I A K=30^{\circ}$ and hence $\angle I A B_{1}=60^{\circ}$. Thus $A I B_{1}$ is an equilateral triangle. Similarly triangle $A I C_{1}$ is also equilateral. We hence obtain $A B_{1}=A C_{1}=A I=I B_{1}=I C_{1}=2 r$.

We also observe that $\angle B_{1} I C_{1}=120^{\circ}$ and $I B_{1} A C_{1}$ is a rhombus. Thus $\angle B_{1} A C_{1}=120^{\circ}$ and by concyclicity $\angle A_{1}=60^{\circ}$. Since $A B_{1}=A C_{1}, A$ is the midpoint of the arc $B_{1} A C_{1}$. It follows that $A_{1} A$ bisects $\angle A_{1}$ and $I_{1}$ lies on the line $A_{1} A$. This implies that

$$
\angle B_{1} I_{1} C_{1}=90^{\circ}+\angle A_{1} / 2=90^{\circ}+30^{\circ}=120^{\circ}
$$

Since $\angle B_{1} I C_{1}=120^{\circ}$, we conclude that $B_{1}, I, I_{1}, C_{1}$ are concyclic. (Further $A$ is the centre.)

solution_match: ## Solution:


problem: Find all triples $(p, x, y)$ such that $p^{x}=y^{4}+4$, where $p$ is a prime and $x, y$ are natural numbers.

solution:
We begin with the standard factorisation

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

Thus we have $y^{2}-2 y+2=p^{m}$ and $y^{2}+2 y+2=p^{n}$ for some positive integers $m$ and $n$ such that $m+n=x$. Since $y^{2}-2 y+2<y^{2}+2 y+2$, we have $m<n$ so that $p^{m}$ divides $p^{n}$. Thus $y^{2}-2 y+2$ divides $y^{2}+2 y+2$. Writing $y^{2}+2 y+2=y^{2}-2 y+2+4 y$, we infer that $y^{2}-2 y+2$ divides $4 y$ and hence $y^{2}-2 y+2$ divides $4 y^{2}$. But

$$
4 y^{2}=4\left(y^{2}-2 y+2\right)+8(y-1)
$$

Thus $y^{2}-2 y+2$ divides $8(y-1)$. Since $y^{2}-2 y+2$ divides both $4 y$ and $8(y-1)$, we conclude that it also divides 8 . This gives $y^{2}-2 y+2=1,2,4$ or 8 .

If $y^{2}-2 y+2=1$, then $y=1$ and $y^{4}+4=5$, giving $p=5$ and $x=1$. If $y^{2}-2 y+2=2$, then $y^{2}-2 y=0$ giving $y=2$. But then $y^{4}+4=20$ is not the power of a prime. The equations $y^{2}-2 y+2=4$ and $y^{2}-2 y+2=8$ have no integer solutions. We conclude that $(p, x, y)=(5,1,1)$ is the only solution.

Alternatively, using $y^{2}-2 y+2=p^{m}$ and $y^{2}+2 y+2=p^{n}$, we may get

$$
4 y=p^{m}\left(p^{n-m}-1\right)
$$

If $m>0$, then $p$ divides 4 or $y$. If $p$ divides 4 , then $p=2$. If $p$ divides $y$, then $y^{2}-2 y+2=p^{m}$ shows that $p$ divides 2 and hence $p=2$. But then $2^{x}=y^{4}+4$, which shows that $y$ is even. Taking $y=2 z$, we get $2^{x-2}=4 z^{4}+1$. This implies that $z=0$ and hence $y=0$, which is a contradiction. Thus $m=0$ and $y^{2}-2 y+2=1$. This gives $y=1$ and hence $p=5, x=1$.

solution_match: Solution:


problem: Let $A$ be a set of real numbers such that $A$ has at least four elements. Suppose $A$ has the property that $a^{2}+b c$ is a rational number for all distinct numbers $a, b, c$ in $A$. Prove that there exists a positive integer $M$ such that $a \sqrt{M}$ is a rational number for every $a$ in $A$.

solution:
Suppose $0 \in A$. Then $a^{2}=a^{2}+0 \times b$ is rational and $a b=0^{2}+a b$ is also rational for all $a, b$ in $A, a \neq 0, b \neq 0, a \neq b$. Hence $a=a_{1} \sqrt{M}$ for some rational $a_{1}$ and natural number $M$. For any $b \neq 0$, we have

$$
b \sqrt{M}=\frac{a b}{a_{1}}
$$

which is a rational number.

Hence we may assume 0 is not in $A$. If there is a number $a$ in $A$ such that $-a$ is also in $A$, then again we can get the conclusion as follows. Consider two other elements $c, d$ in $A$. Then $c^{2}+d a$ is rational and $c^{2}-d a$ is also rational. It follows that $c^{2}$ is rational and $d a$ is rational. Similarly, $d^{2}$ and $c a$ are also rationals. Thus $d / c=(d a) /(c a)$ is rational. Note that we can vary $d$ over $A$ with $d \neq c$ and $d \neq a$. Again $c^{2}$ is rational implies that $c=c_{1} \sqrt{M}$ for some rational $c_{1}$ and natural number $M$. We observe that $c \sqrt{M}=c_{1} M$ is rational, and

$$
a \sqrt{M}=\frac{c a}{c_{1}}
$$

so that $a \sqrt{M}$ is a rational number. Similarly is the case with $-a \sqrt{M}$. For any other element $d$,

$$
b \sqrt{M}=M c_{1} \frac{d}{c}
$$

is a rational number.

Thus we may now assume that 0 is not in $A$ and $a+b \neq 0$ for any $a, b$ in $A$. Let $a, b, c, d$ be four distinct elements of $A$. We may assume $|a|>\mid b$. Then $d^{2}+a b$ and $d^{2}+b c$ are rational numbers and so is their difference $a b-b c$. Writing $a^{2}+a b=a^{2}+b c+(a b-b c)$, and using the facts $a^{2}+b c$, $a b-b c$ are rationals, we conclude that $a^{2}+a b$ is also a rational number. Similarly, $b^{2}+a b$ is also a rational number.

Consider

$$
q=\frac{a}{b}=\frac{a^{2}+a b}{b^{2}+a b}
$$

Note that $a^{2}+a b>0$. Thus $q$ is a rational number and $a=b q$. This gives $a^{2}+a b=b^{2}\left(q^{2}+q\right)$. Let us take $b^{2}\left(q^{2}+q\right)=l$. Then

$$
|b|=\sqrt{\frac{l}{q^{2}+q}}=\sqrt{\frac{x}{y}}
$$

where $x$ and $y$ are natural numbers. Take $M=x y$. Then $|b| \sqrt{M}=x$ is a rational number. Finally, for any $c$ in $A$, we have

$$
c \sqrt{M}=b \sqrt{M} \frac{c}{b}
$$

is also a rational number.


problem: All the points with integer coordinates in the $x y$-plane are coloured using three colours, red, blue and green, each colour being used at least once. It is known that the point $(0,0)$ is coloured red and the point $(0,1)$ is coloured blue. Prove that there exist three points with integer coordinates of distinct colours which form the vertices of a right-angled triangle.

solution:
Consider the lattice points(points with integer coordinates) on the lines $y=0$ and $y=1$, other than $(0,0)$ and $(0,1)$, If one of them, say $A=(p, 1)$, is coloured green, then we have a right-angled triangle with $(0,0),(0,1)$ and $A$ as vertices, all having different colours. (See Figures 1 and 2.)
![](https://cdn.mathpix.com/cropped/2024_06_05_574c8adc52c8b9c64947g-3.jpg?height=418&width=1278&top_left_y=190&top_left_x=389)

If not, the lattice points on $y=0$ and $y=1$ are all red or blue. We consider three different cases.

Case 1. Suppose a point $B=(c, 0)$ is blue. Consider a green point $D=(p, q)$ in the plane. Suppose $p \neq 0$. If its projection $(p, 0)$ on the $x$-axis is red, then $(p, q),(p, 0)$ and $(c, 0)$ are the vertices of a required type of right-angled triangle. If $(p, 0)$ is blue, then we can consider the triangle whose vertices are $(0,0),(p, 0)$ and $(p, q)$. If $p=0$, then the points $D,(0,0)$ and $(c, 0)$ will work.(Figure 3.)

Case 2. A point $D=(c, 1)$, on the line $y=1$, is red. A similar argument works in this case.

![](https://cdn.mathpix.com/cropped/2024_06_05_574c8adc52c8b9c64947g-3.jpg?height=539&width=762&top_left_y=1075&top_left_x=649)

Fig-4

Case 3. Suppose all the lattice points on the line $y=0$ are red and all on the line $y=1$ are blue points. Consider a green point $E=(p, q)$, where $q \neq 0$ and $q \neq 1$.(See Figure 4.) Consider an isosceles right-angled triangle $E K M$ with $\angle E=90^{\circ}$ such that the hypotenuse $K M$ is a part of the $x$-axis. Let $E M$ intersect $y=$ in $L$. Then $K$ is a red point and $L$ is a blue point. Hence $E K L$ is a desired triangle.


problem: Let $A B C$ be a triangle; $\Gamma_{A}, \Gamma_{B}, \Gamma_{C}$ be three equal, disjoint circles inside $A B C$ such that $\Gamma_{A}$ touches $A B$ and $A C ; \Gamma_{B}$ touches $A B$; and $B C$, and $\Gamma_{C}$ touches $B C$ and $C A$. Let $\Gamma$ be a circle touching circles $\Gamma_{A}, \Gamma_{B}, \Gamma_{C}$ externally. Prove that the line joining the circum-centre $O$ and the in-centre $I$ of triangle $A B C$ passes through the centre of $\Gamma$.

solution:
Let $O_{1}, O_{2}, O_{3}$ be the centres of the circles $\Gamma_{A}, \Gamma_{B}, \Gamma_{C}$ respectively, and let $P$ be the circum-centre of the triangle $O_{1} O_{2} O_{3}$. Let $x$ denote the common radius of three circles $\Gamma_{A}, \Gamma_{B}$, $\Gamma_{C}$. Note that $P$ is also the centre of the circle $\Gamma$, as $O_{1} P, O_{2} P, O_{3} P$ each exceed the radius of $\Gamma$ by $x$. Let $D, X, K, L, M$ be respectively the projections of $I, P, O, O_{1}, O_{2}$ on $B C$.

![](https://cdn.mathpix.com/cropped/2024_06_05_574c8adc52c8b9c64947g-4.jpg?height=672&width=767&top_left_y=195&top_left_x=641)

From $\frac{B L}{B D}=\frac{L O_{2}}{D I}$, we get $B L=x(s-b) / r$, as $I D=r$ and $B D=(s-b)$. Similarly, $C M=$ $x(s-c) / r$. Therefore, $L M=a-\frac{x}{r}(s-b+s-c)=\frac{a}{r}(r-x)$. Since $O_{2} L M O_{3}$ is a rectangle and $P X$ is the perpendicular bisector of $\mathrm{O}_{2} \mathrm{O}_{3}$, it is perpendicular bisector of $L M$ as well. Thus

$$
\begin{aligned}
L X & =\frac{1}{2} L M=\frac{a}{2 r}(r-x) \\
B X & =B L+L X=\frac{x}{r}(s-b)+\frac{a}{2 r}(r-x)=\frac{a}{2}-\frac{x(b-c)}{2 r} \\
D K & =B K-B D=\frac{a}{2}-(s-b)=\frac{b-c}{2} \\
X K & =B K-B X=\frac{a}{2}-\frac{a}{2}+\frac{x(b-c)}{2 r}=\frac{x(b-c)}{2 r}
\end{aligned}
$$

Hence we get

$$
\frac{X K}{D K}=\frac{x}{r}
$$

We observe that the sides of triangle $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$ are

$$
O_{2} O_{3}=L M=\frac{a}{r}(r-x), \quad O_{3} O_{1}=\frac{b}{r}(r-x), \quad O_{1} O_{2}=\frac{c}{r}(r-x)
$$

Thus the sides of $O_{1} O_{2} O_{3}$ and those of $A B C$ are in the ratio $(r-x) / r$. Further, as the sides of $O_{1} O_{2} O_{3}$ are parallel to those of $A B C$, we see that $I$ is the in-centre of $O_{1} O_{2} O_{3}$ as well. This gives $I P / I O=(r-x) / r$, and hence $P O / I O=x / r$. Thus we obtain

$$
\frac{X K}{D K}=\frac{P O}{I O}
$$

It follows that $I, P, O$ are collinear.

Alternately, we also infer that $I$ is the centre of homothety which takes the figure $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$ to $A B C$. Hence it takes $P$ to $O$. It follows that $I, P, O$ are collinear


problem: Let $P(x)$ be a given polynomial with integer coefficients. Prove that there exist two polynomials $Q(x)$ and $R(x)$, again with integer coefficients, such that (i) $P(x) Q(x)$ is a polynomial in $x^{2}$; and (ii) $P(x) R(x)$ is a polynomial in $x^{3}$.

solution:
Let $P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}$ be a polynomial with integer coefficients.

Part (i) We may write

$$
P(x)=a_{0}+a_{2} x^{2}+a_{4} x^{4}+\cdots+x\left(a_{1}+a_{3} x^{2}+a_{5} x^{5}+\cdots\right)
$$

Define

$$
Q(x)=a_{0}+a_{2} x^{2}+a_{4} x^{4}+\cdots-x\left(a_{1}+a_{3} x^{2}+a_{5} x^{5}+\cdots\right)
$$

Then $Q(x)$ is also a polynomial with integer coefficients and

$$
P(x) Q(x)=\left(a_{0}+a_{2} x^{2}+a_{4} x^{4}+\cdots\right)^{2}-x^{2}\left(a_{1}+a_{3} x^{2}+a_{5} x^{5}+\cdots\right)^{2}
$$

is a polynomial in $x^{2}$.

Part (ii) We write again

$$
P(x)=A(x)+x B(x)+x^{2} C(x)
$$

where

$$
\begin{aligned}
& A(x)=a_{0}+a_{3} x^{3}+a_{6} x^{6}+\cdots \\
& B(x)=a_{1}+a_{4} x^{3}+a_{7} x^{6}+\cdots \\
& C(x)=a_{2}+a_{5} x^{3}+a_{8} x^{6}+\cdots
\end{aligned}
$$

Note that $A(x), B(x)$ and $C(x)$ are polynomials with integer coefficients and each of these is a polynomial in $x^{3}$. We may introduce

$$
\begin{aligned}
& S(x)=A(x)+\omega x B(x)+\omega^{2} x^{2} C(x) \\
& T(x)=A(x)+\omega^{2} x B(x)+\omega x^{2} C(x)
\end{aligned}
$$

where $\omega$ is an imaginary cube-root of unity. Then

$$
\begin{aligned}
S(x) T(x)=(A(x))^{2}+x^{2}(B(x))^{2}+x^{4}(C(x))^{2} & \\
& -x A(x) B(x)-x^{3} B(x) C(x)-x^{2} C(x) A(x)
\end{aligned}
$$

since $\omega^{3}=1$ and $\omega+\omega^{2}=-1$. Taking $R(x)=S(x) T(x)$, we obtain

$$
P(x) R(x)=(A(x))^{3}+x^{3}(B(x))^{3}+x^{6}(C(x))^{3}-3 x^{3} A(x) B(x) C(x)
$$

which is a polynomial in $x^{3}$. This follows from the identity

$$
(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=a^{3}+b^{3}+c^{3}-3 a b c
$$

Alternately, $R(x)$ may be directly defined by

$$
\begin{aligned}
& R(x)=(A(x))^{2}+x^{2}(B(x))^{2}+x^{4}(C(x))^{2} \\
&-x A(x) B(x)-x^{3} B(x) C(x)-x^{2} C(x) A(x)
\end{aligned}
$$