exam: INMO

problem:
Let $A B C D$ be a quadrilateral inscribed in a circle. Suppose $A B=$ $\sqrt{2+\sqrt{2}}$ and $A B$ subtends $135^{\circ}$ at the centre of the circle. Find the maximum possible area of $A B C D$.
![](https://cdn.mathpix.com/cropped/2024_06_05_177a0e3dd76b8909557ag-1.jpg?height=360&width=1090&top_left_y=478&top_left_x=515)

solution:
Let $O$ be the centre of the circle in which $A B C D$ is inscribed and let $R$ be its radius. Using cosine rule in triangle $A O B$, we have

$$
2+\sqrt{2}=2 R^{2}\left(1-\cos 135^{\circ}\right)=R^{2}(2+\sqrt{2})
$$

Hence $R=1$.

Consider quadrilateral $A B C D$ as in the second figure above. Join $A C$. For $[A D C]$ to be maximum, it is clear that $D$ should be the mid-point of the arc $A C$ so that its distance from the segment $A C$ is maximum. Hence $A D=D C$ for $[A B C D]$ to be maximum. Similarly, we conclude that $B C=C D$. Thus $B C=C D=D A$ which fixes the quadrilateral $A B C D$. Therefore each of the sides $B C, C D, D A$ subtends equal angles at the centre $O$.

Let $\angle B O C=\alpha, \angle C O D=\beta$ and $\angle D O A=\gamma$. Observe that

$$
[A B C D]=[A O B]+[B O C]+[C O D]+[D O A]=\frac{1}{2} \sin 135^{\circ}+\frac{1}{2}(\sin \alpha+\sin \beta+\sin \gamma)
$$

Now $[A B C D]$ has maximum area if and only if $\alpha=\beta=\gamma=\left(360^{\circ}-\right.$ $\left.135^{\circ}\right) / 3=75^{\circ}$. Thus

$$
[A B C D]=\frac{1}{2} \sin 135^{\circ}+\frac{3}{2} \sin 75^{\circ}=\frac{1}{2}\left(\frac{1}{\sqrt{2}}+3 \frac{\sqrt{3}+1}{2 \sqrt{2}}\right)=\frac{5+3 \sqrt{3}}{4 \sqrt{2}}
$$

Alternatively, we can use Jensen's inequality. Observe that $\alpha, \beta, \gamma$ are all less than $180^{\circ}$. Since $\sin x$ is concave on $(0, \pi)$, Jensen's inequality gives

$$
\frac{\sin \alpha+\sin \beta+\sin \gamma}{3} \leq \sin \left(\frac{\alpha+\beta+\gamma}{3}\right)=\sin 75^{\circ}
$$

Hence

$$
[A B C D] \leq \frac{1}{2 \sqrt{2}}+\frac{3}{2} \sin 75^{\circ}=\frac{5+3 \sqrt{3}}{4 \sqrt{2}}
$$

with equality if and only if $\alpha=\beta=\gamma=75^{\circ}$.

solution_match: Solution:


solution:
Since $p_{4}-p_{1}=8$, and no prime is even, we observe that $\left\{p_{1}, p_{2}, p_{3}, p_{4}\right\}$ is a subset of $\left\{p_{1}, p_{1}+2, p_{1}+4, p_{1}+6, p_{1}+8\right\}$. Moreover $p_{1}$ is larger than 3 . If $p_{1} \equiv 1(\bmod 3)$, then $p_{1}+2$ and $p_{1}+8$ are divisible by 3 . Hence we do not get 4 primes in the set $\left\{p_{1}, p_{1}+2, p_{1}+4, p_{1}+6, p_{1}+8\right\}$. Thus $p_{1} \equiv 2(\bmod 3)$ and $p_{1}+4$ is not a prime. We get $p_{2}=p_{1}+2, p_{3}=$ $p_{1}+6, p_{4}=p_{1}+8$.

Consider the remainders of $p_{1}, p_{1}+2, p_{1}+6, p_{1}+8$ when divided by 5 . If $p_{1} \equiv 2(\bmod 5)$, then $p_{1}+8$ is divisible by 5 and hence is not a prime. If $p_{1} \equiv 3(\bmod 5)$, then $p_{1}+2$ is divisible by 5. If $p_{1} \equiv 4(\bmod 5)$, then $p_{1}+6$ is divisible by 5 . Hence the only possibility is $p_{1} \equiv 1(\bmod 5)$.

Thus we see that $p_{1} \equiv 1(\bmod 2), p_{1} \equiv 2(\bmod 3)$ and $p_{1} \equiv 1(\bmod 5)$. We conclude that $p_{1} \equiv 11(\bmod 30)$.

Similarly $q_{1} \equiv 11(\bmod 30)$. It follows that 30 divides $p_{1}-q_{1}$.


problem:
Define a sequence $\left\langle f_{0}(x), f_{1}(x), f_{2}(x), \ldots\right\rangle$ of functions by

$$
f_{0}(x)=1, \quad f_{1}(x)=x, \quad\left(f_{n}(x)\right)^{2}-1=f_{n+1}(x) f_{n-1}(x), \text { for } n \geq 1
$$

Prove that each $f_{n}(x)$ is a polynomial with integer coefficients.

solution:
Observe that

$$
f_{n}^{2}(x)-f_{n-1}(x) f_{n+1}(x)=1=f_{n-1}^{2}(x)-f_{n-2}(x) f_{n}(x)
$$

This gives

$$
f_{n}(x)\left(f_{n}(x)+f_{n-2}(x)\right)=f_{n-1}\left(f_{n-1}(x)+f_{n+1}(x)\right)
$$

We write this as

$$
\frac{f_{n-1}(x)+f_{n+1}(x)}{f_{n}(x)}=\frac{f_{n-2}(x)+f_{n}(x)}{f_{n-1}(x)}
$$

Using induction, we get

$$
\frac{f_{n-1}(x)+f_{n+1}(x)}{f_{n}(x)}=\frac{f_{0}(x)+f_{2}(x)}{f_{1}(x)}
$$

Observe that

$$
f_{2}(x)=\frac{f_{1}^{2}(x)-1}{f_{0}(x)}=x^{2}-1
$$

Hence

$$
\frac{f_{n-1}(x)+f_{n+1}(x)}{f_{n}(x)}=\frac{1+\left(x^{2}-1\right)}{x}=x
$$

Thus we obtain

$$
f_{n+1}(x)=x f_{n}(x)-f_{n-1}(x)
$$

Since $f_{0}(x), f_{1}(x)$ and $f_{2}(x)$ are polynomials with integer coefficients, induction again shows that $f_{n}(x)$ is a polynomial with integer coefficients.

Note: We can get $f_{n}(x)$ explicitly:

$$
f_{n}(x)=x^{n}-\binom{n-1}{1} x^{n-2}+\binom{n-2}{2} x^{n-4}-\binom{n-3}{3} x^{n-6}+\cdots
$$


problem: Let $A B C$ be a triangle. An interior point $P$ of $A B C$ is said to be good if we can find exactly 27 rays emanating from $P$ intersecting the sides of the triangle $A B C$ such that the triangle is divided by these rays into 27 smaller triangles of equal area. Determine the number of good points for a given triangle $A B C$.

solution:
Let $P$ be a good point. Let $l, m, n$ be respetively the number of parts the sides $B C, C A, A B$ are divided by the rays starting from $P$. Note that a ray must pass through each of the vertices the triangle $A B C$; otherwise we get some quadrilaterals.

Let $h_{1}$ be the distance of $P$ from $B C$. Then $h_{1}$ is the height for all the triangles with their bases on $B C$. Equality of areas implies that all these bases have equal length. If we denote this by $x$, we get $l x=a$. Similarly, taking $y$ and $z$ as the lengths of the bases of triangles on $C A$ and $A B$ respectively, we get $m y=b$ and $n z=c$. Let $h_{2}$ and $h_{3}$ be the distances of $P$ from $C A$ and $A B$ respectively. Then

$$
h_{1} x=h_{2} y=h_{3} z=\frac{2 \Delta}{27}
$$

where $\Delta$ denotes the area of the triangle $A B C$. These lead to

$$
h_{1}=\frac{2 \Delta}{27} \frac{l}{a}, \quad h_{1}=\frac{2 \Delta}{27} \frac{m}{b}, \quad h_{1}=\frac{2 \Delta}{27} \frac{n}{c}
$$

But

$$
\frac{2 \Delta}{a}=h_{a}, \quad \frac{2 \Delta}{b}=h_{b}, \quad \frac{2 \Delta}{c}=h_{c}
$$

Thus we get

$$
\frac{h_{1}}{h_{a}}=\frac{l}{27}, \quad \frac{h_{2}}{h_{b}}=\frac{m}{27}, \quad \frac{h_{3}}{h_{c}}=\frac{n}{27}
$$

However, we also have

$$
\frac{h_{1}}{h_{a}}=\frac{[P B C]}{\Delta}, \quad \frac{h_{2}}{h_{b}}=\frac{[P C A]}{\Delta}, \quad \frac{h_{3}}{h_{c}}=\frac{[P A B]}{\Delta}
$$

Adding these three relations,

$$
\frac{h_{1}}{h_{a}}+\frac{h_{2}}{h_{b}}+\frac{h_{3}}{h_{c}}=1
$$

Thus

$$
\frac{l}{27}+\frac{m}{27}+\frac{n}{27}=\frac{h_{1}}{h_{a}}+\frac{h_{2}}{h_{b}}+\frac{h_{3}}{h_{c}}=1
$$

We conclude that $l+m+n=27$. Thus every good point $P$ determines a partition $(l, m, n)$ of 27 such that there are $l, m, n$ equal segments respectively on $B C, C A, A B$.

Conversely, take any partition $(l, m, n)$ of 27 . Divide $B C, C A, A B$ respectively in to $l, m, n$ equal parts. Define

$$
h_{1}=\frac{2 l \Delta}{27 a}, \quad h_{2}=\frac{2 m \Delta}{27 b}
$$

Draw a line parallel to $B C$ at a distance $h_{1}$ from $B C$; draw another line parallel to $C A$ at a distance $h_{2}$ from $C A$. Both lines are drawn such that they intersect at a point $P$ inside the triangle $A B C$. Then

$$
[P B C]=\frac{1}{2} a h_{1}=\frac{l \Delta}{27}, \quad[P C A]=\frac{m \Delta}{27}
$$

Hence

$$
[P A B]=\frac{n \Delta}{27}
$$

This shows that the distance of $P$ from $A B$ is

$$
h_{3}=\frac{2 n \Delta}{27 c}
$$

Therefore each traingle with base on $C A$ has area $\frac{\Delta}{27}$. We conclude that all the triangles which partitions $A B C$ have equal areas. Hence $P$ is a good point.

Thus the number of good points is equal to the number of positive integral solutions of the equation $l+m+n=27$. This is equal to

$$
\binom{26}{2}=325
$$


problem: Let $A B C$ be an acute-angled triangle, and let $D, E, F$ be points on $B C$, $C A, A B$ respectively such that $A D$ is the median, $B E$ is the internal angle bisector and $C F$ is the altitude. Suppose $\angle F D E=\angle C, \angle D E F=$ $\angle A$ and $\angle E F D=\angle B$. Prove that $A B C$ is equilateral.

solution:
Since $\triangle B F C$ is right-angled at $F$, we have

![](https://cdn.mathpix.com/cropped/2024_06_05_177a0e3dd76b8909557ag-4.jpg?height=371&width=393&top_left_y=2037&top_left_x=519)
$F D=B D=C D=a / 2$. Hence $\angle B F D=\angle B$. Since $\angle E F D=$ $\angle B$, we have $\angle A F E=\pi-2 \angle B$. Since $\angle D E F=\angle A$, we also get $\angle C E D=\pi-2 \angle B$. Applying sine rule in $\triangle D E F$, we have

$$
\frac{D F}{\sin A}=\frac{F E}{\sin C}=\frac{D E}{\sin B}
$$

Thus we get $F E=c / 2$ and $D E=b / 2$. Sine rule in $\triangle C E D$ gives

$$
\frac{D E}{\sin C}=\frac{C D}{\sin (\pi-2 B)}
$$

Thus $(b / \sin C)=(a / 2 \sin B \cos B)$. Solving for $\cos B$, we have

$$
\cos B=\frac{a \sin c}{2 b \sin B}=\frac{a c}{2 b^{2}}
$$

Similarly, sine rule in $\triangle A E F$ gives

$$
\frac{E F}{\sin A}=\frac{A E}{\sin (\pi-2 B)}
$$

This gives (since $A E=b c /(a+c)$ ), as earlier,

$$
\cos B=\frac{a}{a+c}
$$

Comparing the two values of $\cos B$, we get $2 b^{2}=c(a+c)$. We also have

$$
c^{2}+a^{2}-b^{2}=2 c a \cos B=\frac{2 a^{2} c}{a+c}
$$

Thus

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

This reduces to $2 a^{3}-3 a^{2} c+c^{3}=0$. Thus $(a-c)^{2}(2 a+c)=0$. We conclude that $a=c$. Finally

$$
2 b^{2}=c(a+c)=2 c^{2}
$$

We thus get $b=c$ and hence $a=c=b$. This shows that $\triangle A B C$ is equilateral.


problem:
Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be a function satisfying $f(0) \neq 0, f(1)=0$ and

(i) $f(x y)+f(x) f(y)=f(x)+f(y)$;

(ii) $(f(x-y)-f(0)) f(x) f(y)=0$,

for all $x, y \in \mathbb{Z}$, simultaneously.

(a) Find the set of all possible values of the function $f$.

(b) If $f(10) \neq 0$ and $f(2)=0$, find the set of all integers $n$ such that $f(n) \neq 0$.

solution:
Setting $y=0$ in the condition (ii), we get

$$
(f(x)-f(0)) f(x)=0
$$

for all $x$ (since $f(0) \neq 0$ ). Thus either $f(x)=0$ or $f(x)=f(0)$, for all $x \in \mathbb{Z}$. Now taking $x=y=0$ in (i), we see that $f(0)+f(0)^{2}=2 f(0)$. This shows
that $f(0)=0$ or $f(0)=1$. Since $f(0) \neq 0$, we must have $f(0)=1$. We conclude that

$$
\text { either } f(x)=0 \text { or } f(x)=1 \text { for each } x \in \mathbb{Z}
$$

This shows that the set of all possible value of $f(x)$ is $\{0,1\}$. This completes (a).

Let $S=\{n \in \mathbb{Z} \mid f(n) \neq 0\}$. Hence we must have $S=\{n \in \mathbb{Z} \mid f(n)=1\}$ by (a). Since $f(1)=0,1$ is not in $S$. And $f(0)=1$ implies that $0 \in S$. Take any $x \in \mathbb{Z}$ and $y \in S$. Using (ii), we get

$$
f(x y)+f(x)=f(x)+1
$$

This shows that $x y \in S$. If $x \in \mathbb{Z}$ and $y \in \mathbb{Z}$ are such that $x y \in S$, then (ii) gives

$$
1+f(x) f(y)=f(x)+f(y)
$$

Thus $(f(x)-1)(f(y)-1)=0$. It follows that $f(x)=1$ or $f(y)=1$; i.e., either $x \in S$ or $y \in S$. We also observe from (ii) that $x \in S$ and $y \in S$ implies that $f(x-y)=1$ so that $x-y \in S$. Thus $S$ has the properties:

(A) $x \in \mathbb{Z}$ and $y \in S$ implies $x y \in S$;

(B) $x, y \in \mathbb{Z}$ and $x y \in S$ implies $x \in S$ or $y \in S$;

(C) $x, y \in S$ implies $x-y \in S$.

Now we know that $f(10) \neq 0$ and $f(2)=0$. Hence $f(10)=1$ and $10 \in S$; and $2 \notin S$. Writing $10=2 \times 5$ and using (B), we conclude that $5 \in S$ and $f(5)=1$. Hence $f(5 k)=1$ for all $k \in \mathbb{Z}$ by (A).

Suppose $f(5 k+l)=1$ for some $l, 1 \leq l \leq 4$. Then $5 k+l \in S$. Choose $u \in \mathbb{Z}$ such that $l u \equiv 1(\bmod 5)$. We have $(5 k+l) u \in S$ by (A). Moreover, $l u=1+5 m$ for some $m \in \mathbb{Z}$ and

$$
(5 k+l) u=5 k u+l u=5 k u+5 m+1=5(k u+m)+1
$$

This shows that $5(k u+m)+1 \in S$. However, we know that $5(k u+m) \in S$. By (C), $1 \in S$ which is a contradiction. We conclude that $5 k+l \notin S$ for any $l, 1 \leq l \leq 4$. Thus

$$
S=\{5 k \mid k \in \mathbb{Z}\}
$$

$\qquad$