INMO 2021

Put $x=0$ and we get $f(y f(0))=0$. If $f(0) \neq 0$, then $y f(0)$ takes all real values when $y$ varies over real line. We get $f(x) \equiv 0$. Suppose $f(0)=0$. Taking $y=-x$, we get $f\left(x^{2}-x f(x)\right)=0$ for all real $x$.

Suppose there exists $x_{0} \neq 0$ in $\mathbb{R}$ such that $f\left(x_{0}\right)=0$. Putting $x=x_{0}$ in the given relation we get

$$f\left(x_{0}^{2}\right)=x_{0} f\left(x_{0}+y\right)$$

for all $y \in \mathbb{R}$. Now the left side is a constant and hence it follows that $f$ is a constant function. But the only constant function which satisfies the equation is identically zero function, which is already obtained. Hence we may consider the case where $f(x) \neq 0$ for all $x \neq 0$.

Since $f\left(x^{2}-x f(x)\right)=0$, we conclude that $x^{2}-x f(x)=0$ for all $x \neq 0$. This implies that $f(x)=x$ for all $x \neq 0$. Since $f(0)=0$, we conclude that $f(x)=x$ for all $x \in \mathbb{R}$. Thus we have two functions: $f(x) \equiv 0$ and $f(x)=x$ for all $x \in \mathbb{R}$.

Let $x_{n}$ be the number of ways in which $A$ can get back the ball after $n$ passes. Let $y_{n}$ be the number of ways in which the ball goes back to a fixed person other than $A$ after $n$ passes. Then

$$x_{n}=3 y_{n-1},$$

and

$$y_{n}=x_{n-1}+2 y_{n-1}$$

We also have $x_{1}=0$, $x_{2}=3$, $y_{1}=1$ and $y_{2}=2$.

Eliminating $y_{n}$ and $y_{n-1}$, we get $x_{n+1}=3 x_{n-1}+2 x_{n}$. Thus

$$\begin{aligned} & x_{3}=3 x_{1}+2 x_{2}=2 \times 3=6 \\ & x_{4}=3 x_{2}+2 x_{3}=(3 \times 3)+(2 \times 6)=9+12=21 \\ & x_{5}=3 x_{3}+2 x_{4}=(3 \times 6)+(2 \times 21)=18+42=60 \\ & x_{6}=3 x_{4}+2 x_{5}=(3 \times 21)+(2 \times 60)=63+120=183 \\ & x_{7}=3 x_{5}+2 x_{6}=(3 \times 60)+(2 \times 183)=180+366=546 \end{aligned}$$

Alternate solution: Since the ball goes back to one of the other 3 persons, we have

$$x_{n}+3 y_{n}=3^{n}$$

since there are $3^{n}$ ways of passing the ball in $n$ passes. Using $x_{n}=3 y_{n-1}$, we obtain

$$x_{n-1}+x_{n}=3^{n-1}$$

with $x_{1}=0$. Thus

$$\begin{array}{r} x_{7}=3^{6}-x_{6}=3^{6}-3^{5}+x_{5}=3^{6}-3^{5}+3^{4}-x_{4}=3^{6}-3^{5}+3^{4}-3^{3}+x_{3} \\ =3^{6}-3^{5}+3^{4}-3^{3}+3^{2}-x_{2}=3^{6}-3^{5}+3^{4}-3^{3}+3^{2}-3 \\ =\left(2 \times 3^{5}\right)+\left(2 \times 3^{3}\right)+(2 \times 3)=486+54+6=546 \end{array}$$

Let $AP = p$, $BP = q$, $CP = r$, $DP = s$; $AB = a$, $BC = b$, $CD = c$ and $DA = d$. Let $\angle APB = \angle CPD = \theta$. Then $\angle BPC = \angle DPA = \pi - \theta$. Let us also write $PE = h_1$, $PF = h_2$, $PG = h_3$ and $PH = h_4$.

Observe that

$$h_1 a = p q \sin \theta, \quad h_2 b = q r \sin \theta, \quad h_3 c = r s \sin \theta, \quad h_4 d = s p \sin \theta$$

Hence

$$\frac{1}{h_1} + \frac{1}{h_3} = \frac{1}{h_2} + \frac{1}{h_4}$$

is equivalent to

$$\frac{a}{p q} + \frac{c}{r s} = \frac{b}{q r} + \frac{d}{s p}$$

This is the same as

$$a r s + c p q = b s p + d q r$$

Thus we have to prove that $a + c = b + d$ if and only if $a r s + c p q = b s p + d q r$. Now we can write $a + c = b + d$ as

$$a^2 + c^2 + 2 a c = b^2 + d^2 + 2 b d$$

But we know that

$$\begin{aligned} & a^2 = p^2 + q^2 - 2 p q \cos \theta, \quad c^2 = r^2 + s^2 - 2 r s \cos \theta \\ & b^2 = q^2 + r^2 + 2 q r \cos \theta, \quad d^2 = p^2 + s^2 + 2 p s \cos \theta \end{aligned}$$

Hence $a + c = b + d$ is equivalent to

$$- p q \cos \theta - r s \cos \theta + a c = p s \cos \theta + q r \cos \theta + b d$$

Similarly, by squaring $a r s + c p q = b s p + d q r$ we can show that it is equivalent to

$$- p q \cos \theta - r s \cos \theta + a c = p s \cos \theta + q r \cos \theta + b d$$

We conclude that $a + c = b + d$ is equivalent to $c p q + a r s = b p s + d q r$. Hence $ABCD$ has an incircle if and only if

$$\frac{1}{h_1} + \frac{1}{h_3} = \frac{1}{h_2} + \frac{1}{h_4}$$

The first observation is that we can find 5 pairs of squares such that the two numbers in a pair have the same parity. We can see this as follows:

Odd numbers	Even numbers	Odd pairs	Even pairs	Total pairs
0	11	0	5	5
1	10	0	5	5
2	9	1	4	5
3	8	1	4	5
4	7	2	3	5
5	6	2	3	5
6	5	3	2	5
7	4	3	2	5
8	3	4	1	5
9	2	4	1	5
10	1	5	0	5
11	0	5	0	5

Let us take such 5 pairs: say $\left(x_{1}^{2}, y_{1}^{2}\right),\left(x_{2}^{2}, y_{2}^{2}\right), \ldots,\left(x_{5}^{2}, y_{5}^{2}\right)$. Then $x_{j}^{2}-y_{j}^{2}$ is divisible by 4 for $1 \leq j \leq 5$. Let $r_{j}$ be the remainder when $x_{j}^{2}-y_{j}^{2}$ is divisible by $3, 1 \leq j \leq 5$. We have 5 remainders $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$. But these can be 0, 1 or 2. Hence either one of the remainders occurs 3 times or each of the remainders occurs once. If, for example, $r_{1}=r_{2}=r_{3}$, then 3 divides $r_{1}+r_{2}+r_{3}$; if $r_{1}=0, r_{2}=1$ and $r_{3}=2$, then again 3 divides $r_{1}+r_{2}+r_{3}$. Thus we can always find three remainders whose sum is divisible by 3. This means we can find 3 pairs, say, $\left(x_{1}^{2}, y_{1}^{2}\right),\left(x_{2}^{2}, y_{2}^{2}\right),\left(x_{3}^{2}, y_{3}^{2}\right)$ such that 3 divides $\left(x_{1}^{2}-y_{1}^{2}\right)+\left(x_{2}^{2}-y_{2}^{2}\right)+\left(x_{3}^{2}-y_{3}^{2}\right)$. Since each difference is divisible by 4, we conclude that we can find 6 numbers $a^{2}, b^{2}, c^{2}, d^{2}, e^{2}, f^{2}$ such that

$$a^{2}+b^{2}+c^{2} \equiv d^{2}+e^{2}+f^{2} \quad(\bmod 12)$$

Let $\angle C B A=\alpha$ and $\angle D B A=\beta$. Then $\angle B D A=\alpha$ and $\angle B C A=\beta$. We also observe that $\angle A O_{1} O_{2}=\left(\angle A O_{1} B / 2\right)=\alpha$ and, simiarly, $\angle A O_{2} O_{1}=\beta$. Hence

$$\angle O_{1} A O_{2}=180^{\circ}-(\alpha+\beta)$$

We also have

$$\angle P O_{1} A=\frac{\angle D O_{1} A}{2}=\frac{2 \angle D B A}{2}=\angle D B A=\beta$$

Hence $\angle P O_{1} O_{2}=\angle P O_{1} A+\angle A O_{1} O_{2}=\beta+\alpha$. Similarly, we can get $\angle P O_{2} O_{1}=\alpha+\beta$. It follows that $P$ lies on the perpendicular bisector of $O_{1} O_{2}$.

Now we observe that

$$\angle X Q Y=360^{\circ}-2 \angle X A Y=360^{\circ}-2\left(180^{\circ}-\alpha-\beta\right)=2(\alpha+\beta)$$

This gives

$$\angle O_{1} Q O_{2}=\frac{1}{2}(\angle X Q A+\angle Y Q A)=\frac{\angle X Q Y}{2}=\alpha+\beta$$

This shows that $A, O_{1}, O_{2}, Q$ are concyclic. We also have

$$\begin{aligned} & \angle A B X=\angle A B D+\angle D B X=\beta+\angle D A X=\beta+\frac{\angle D A B}{2} \\ & \angle A B Y=\angle A B C+\angle C B Y=\alpha+\angle C A Y=\alpha+\frac{\angle B A C}{2} \end{aligned}$$

Adding we obtain

$$\angle A B X+\angle A B Y=\alpha+\beta+\frac{1}{2}(\angle D A B+\angle B A C)=\alpha+\beta+\left(180^{\circ}-\alpha-\beta\right)=180^{\circ}$$

Hence $X, B, Y$ are collinear. Now

$$\begin{gathered} \angle Q A X=\frac{1}{2}\left(180^{\circ}-\angle A Q X\right)=90^{\circ}-\beta \\ \angle X A O_{1}=\frac{1}{2}\left(180^{\circ}-\angle X O_{1} A\right)=90^{\circ}-\frac{1}{2}\left(360^{\circ}-2 \angle A B X\right)=\angle A B X-90^{\circ} \end{gathered}$$

Hence

$$\angle Q A O_{1}=90^{\circ}-\beta+\angle A B X-90^{\circ}=\angle A B X-\beta=\frac{\angle D A B}{2}=\frac{\angle O_{1} A O_{2}}{2}$$

This shows that $A Q$ bisects $\angle O_{1} A O_{2}$ and therefore the chords $Q O_{1}$ and $Q O_{2}$ subtend equal angles on the circumference of the circle passing through $Q O_{2} A O_{1}$. Hence $Q O_{2}=Q O_{1}$. This means $Q$ lies on the perpendicular bisector of $O_{1} O_{2}$.

Combining, we get that $P Q$ is the perpendicular bisector of $O_{1} O_{2}$.

Changing $\theta$ to $\theta - \pi / 2$, we see that

$$P(\sin \theta + \cos \theta) = P(\sin \theta - \cos \theta)$$

This shows that $P(x) = P(-x)$ for all $x \in [-\sqrt{2}, \sqrt{2}]$ and as $P$ is a polynomial, in fact,

$$P(x) = P(-x)$$

for all $x \in \mathbb{R}$. Hence $P(x)$ is an even polynomial; $P(x) = Q(x^{2})$ for some polynomial $Q(x)$. This gives

$$Q(1 + \sin(2\theta)) = P(\cos \theta + \sin \theta) = P(\cos \theta - \sin \theta) = Q(1 - \sin(2\theta))$$

Taking $t = \sin(2\theta)$, we see that $Q(1 + t) = Q(1 - t)$. Hence $Q(0) = Q(2)$.

Consider $Q(t) - Q(0)$. This vanishes both at $t = 0$ and $t = 2$. Hence $t(2 - t)$ is a factor of $Q(t) - Q(0)$. We obtain

$$Q(t) - Q(0) = t(2 - t) h(t)$$

for some polynomial $h(t)$. Using $Q(1 + t) = Q(1 - t)$, it follows that $h(1 + t) = h(1 - t)$. Hence by induction we get

$$Q(t) = \sum_{k=0}^{n} b_{k} t^{k} (2 - t)^{k}$$

Hence

$$P(x) = Q(x^{2}) = \sum_{k=0}^{n} b_{k} (x^{2}(2 - x^{2}))^{k} = \sum_{k=0}^{n} b_{k} (1 - (1 - x^{2})^{2})^{k}$$

Using binomial theorem, we can write this as

$$P(x) = \sum_{k=0}^{n} a_{k} (1 - x^{2})^{2k}$$

for some coefficients $a_{k}$, $0 \leq k \leq n$.