# INMO 2004 - Solutions 

1. Consider a convex quadrilateral $A B C D$, in which $K, L, M, N$ are the midpoints of the sides $A B$, $B C, C D, D A$ respectively. Suppose

(a) $B D$ bisects $K M$ at $Q$;

(b) $Q A=Q B=Q C=Q D$; and

(c) $L K / L M=C D / C B$.

Prove that $A B C D$ is a square.

## Solution:

![](https://cdn.mathpix.com/cropped/2024_06_05_ef678323fa24a191ae6eg-1.jpg?height=556&width=640&top_left_y=714&top_left_x=840)

Fig. 1.

Observe that $K L M N$ is a paralellogram, $Q$ is the midpoint of $M K$ and hence $N L$ also passes through $Q$. Let $T$ be the point of intersection of $A C$ and $B D$; and let $S$ be the point of intersection of $B D$ and $M N$.

Consider the triangle $M N K$. Note that $S Q$ is parallel to $N K$ and $Q$ is the midpoint of $M K$. Hence $S$ is the mid-point of $M N$. Since $M N$ is parallel to $A C$, it follows that $T$ is the mid-point of $A C$. Now $Q$ is the circumcentre of $\triangle A B C$ and the median $B T$ passes through $Q$. Here there are two possibilities:

(i) $A B C$ is a right triangle with $\angle A B C=90^{\circ}$ and $T=Q$; and

(ii) $T \neq Q$ in which case $B T$ is perpendicular to $A C$.

Suppose $\angle A B C=90^{\circ}$ and $T=Q$. Observe that $Q$ is the circumcentre of the triangle $D C B$ and hence $\angle D C B=90^{\circ}$. Similarly $\angle D A B=90^{\circ}$. It follows that $\angle A D C=90^{\circ}$. and $A B C D$ is a rectangle. This implies that $K L M N$ is a rhombus. Hence $L K / L M=1$ and this gives $C D=C B$. Thus $A B C D$ is a square.

In the second case, observe that $B D$ is perpendicular to $A C, K L$ is parallel to $A C$ and $L M$ is parallel to $B D$. Hence it follows that $M L$ is perpendicular to $L K$. Similar reasoning shows that $K L M N$ is a rectangle.

Using $L K / L M=C D / C B$, we get that $C B D$ is similar to $L M K$. In particular, $\angle L M K=$ $\angle C B D=\alpha$ say. Since $L M$ is parallel to $D B$, we also get $\angle B Q K=\alpha$. Since $K L M N$ is a cyclic quadrilateral we also get $\angle L N K=\angle L M K=\alpha$. Using the fact that $B D$ is parallel to $N K$, we get $\angle L Q B=\angle L N K=\alpha$. Since $B D$ bisects $\angle C B A$, we also have $\angle K B Q=\alpha$. Thus

$$
Q K=K B=B L=L Q
$$

and $B L$ is parallel to $Q K$. This gives $Q M$ is parallel to $L C$ and

$$
Q M=Q L=B L=L C
$$

It follows that $Q L C M$ is a parallelogram. But $\angle L C M=90^{\circ}$. Hence $\angle M Q L=90^{\circ}$. This implies that $K L M N$ is a square. Also observe that $\angle L Q K=90^{\circ}$ and hence $\angle C B A=\angle L Q K=90^{\circ}$. This gives $\angle C D A=90^{\circ}$ and hence $A B C D$ is a rectangle. Since $B A=B C$, it follows that $A B C D$ is a square.

2. Suppose $p$ is a prime greater than 3 . Find all pairs of integers $(a, b)$ satisfying the equation

$$
a^{2}+3 a b+2 p(a+b)+p^{2}=0
$$

Solution: We write the equation in the form

$$
a^{2}+2 a p+p^{2}+b(3 a+2 p)=0
$$

Hence

$$
b=\frac{-(a+p)^{2}}{3 a+2 p}
$$

is an integer. This shows that $3 a+2 p$ divides $(a+p)^{2}$ and hence also divides $(3 a+3 p)^{2}$. But, we have

$$
(3 a+3 p)^{2}=(3 a+2 p+p)^{2}=(3 a+2 p)^{2}+2 p(3 a+2 p)+p^{2}
$$

It follows that $3 a+2 p$ divides $p^{2}$. Since $p$ is a prime, the only divisors of $p^{2}$ are $\pm 1, \pm p$ and $\pm p^{2}$. Since $p>3$, we also have $p=3 k+1$ or $3 k+2$.

Case 1: Suppose $p=3 k+1$. Obviously $3 a+2 p=1$ is not possible. Infact, we get $1=3 a+2 p=$ $3 a+2(3 k+1) \Rightarrow 3 a+6 k=-1$ which is impossible. On the other hand $3 a+2 p=-1$ gives $3 a=-2 p-1=-6 k-3 \Rightarrow a=-2 k-1$ and $a+p=-2 k-1+3 k+1=k$.

Thus $b=\frac{-(a+p)^{2}}{(3 k+2 p)}=k^{2}$. Thus $(a, b)=\left(-2 k-1, k^{2}\right)$ when $p=3 k+1$. Similarly, $3 a+2 p=p \Rightarrow$ $3 a=-p$ which is not possible. Considering $3 a+2 p=-p$, we get $3 a=-3 p$ or $a=-p \Rightarrow b=0$. Hence $(a, b)=(-3 k-1,0)$ where $p=3 k+1$.

Let us consider $3 a+2 p=p^{2}$. Hence $3 a=p^{2}-2 p=p(p-2)$ and neither $p$ nor $p-2$ is divisible by 3 . If $3 a+2 p=-p^{2}$, then $3 a=-p(p+2) \Rightarrow a=-(3 k+1)(k+1)$.

Hence $a+p=(3 k+1)(-k-1+1)=-(3 k+1) k$. This gives $b=k^{2}$. Again $(a, b)=(-(k+$ 1) $\left.(3 k+1), k^{2}\right)$ when $p=3 k+1$.

Case 2: Suppose $p=3 k-1$. If $3 a+2 p=1$, then $3 a=-6 k+3$ or $a=-2 k+1$. We also get

$$
b=\frac{-(a+p)^{2}}{1}=\frac{-(-2 k+1+3 k-1)^{2}}{1}=-k^{2}
$$

and we get the solution $(a, b)=\left(-2 k+1, k^{2}\right)$. On the other hand $3 a+2 p=-1$ does not have any solution integral solution for $a$. Similarly, there is no solution in the case $3 a+2 p=p$. Taking $3 a+2 p=-p$, we get $a=-p$ and hence $b=0$. We get the solution $(a, b)=(-3 k+1,0)$. If $3 a+2 p=p^{2}$, then $3 a=p(p-2)=(3 k-1)(3 k-3)$ giving $a=(3 k-1)(k-1)$ and hence $a+p=(3 k-1)(1+k-1)=k(3 k-1)$. This gives $b=-k^{2}$ and hence $(a, b)=\left(3 k-1,-k^{2}\right)$. Finally $3 a+2 p=-p^{2}$ does not have any solution.

3. If $\alpha$ is a real root of the equation $x^{5}-x^{3}+x-2=0$, prove that $\left[\alpha^{6}\right]=3$. (For any real number $a$, we denote by $[a]$ the greatest integer not exceeding $a$.)

Solution: Suppose $\alpha$ is a real root of the given equation. Then

$$
\alpha^{5}-\alpha^{3}+\alpha-2=0
$$

This gives $\alpha^{5}-\alpha^{3}+\alpha-1=1$ and hence $(\alpha-1)\left(\alpha^{4}+\alpha^{3}+1\right)=1$. Observe that $\alpha^{4}+\alpha^{3}+1 \geq$ $2 \alpha^{2}+\alpha^{3}=\alpha^{2}(\alpha+2)$. If $-1 \leq \alpha<0$, then $\alpha+2>0$, giving $\alpha^{2}(\alpha+2)>0$ and hence $(\alpha-1)\left(\alpha^{4}+\alpha^{3}+1\right)<0$. If $\alpha<-1$, then $\alpha^{4}+\alpha^{3}=\alpha^{3}(\alpha+1)>0$ and hence $\alpha^{4}+\alpha^{3}+1>0$. This again gives $(\alpha-1)\left(\alpha^{4}+\alpha^{3}+1\right)<0$.

The above resoning shows that for $\alpha<0$, we have $\alpha^{5}-\alpha^{3}+\alpha-1<0$ and hence cannot be equal to 1 . We conclude that a real root $\alpha$ of $x^{5}-x^{3}+x-2=0$ is positive (obviously $\alpha \neq 0$ ).

Now using $\alpha^{5}-\alpha^{3}+\alpha-2=0$, we get

$$
\alpha^{6}=\alpha^{4}-\alpha^{2}+2 \alpha
$$

The statement $\left[\alpha^{6}\right]=3$ is equivalent to $3 \leq \alpha^{6}<4$.

Consider $\alpha^{4}-\alpha^{2}+2 \alpha<4$. Since $\alpha>0$, this is equivalent to $\alpha^{5}-\alpha^{3}+2 \alpha^{2}<4 \alpha$. Using the relation (1), we can write $2 \alpha^{2}-\alpha+2<4 \alpha$ or $2 \alpha^{2}-5 \alpha+2<0$. Treating this as a quadratic, we get this is equivalent to $\frac{1}{2}<\alpha<2$. Now observe that if $\alpha \geq 2$ then $1=(\alpha-1)\left(\alpha^{4}+\alpha^{3}+1\right) \geq 25$ which is impossible. If $0<\alpha \leq \frac{1}{2}$, then $1=(\alpha-1)\left(\alpha^{4}+\alpha^{3}+1\right)<0$ which again is impossible. We conlude that $\frac{1}{2}<\alpha<2$. Similarly $\alpha^{4}-\alpha^{2}+2 \alpha \geq 3$ is equivalent to $\alpha^{5}-\alpha^{3}+2 \alpha^{2}-3 \alpha \geq 0$ which is equivalent to $2 \alpha^{2}-4 \alpha+2 \geq 0$. But this is $2(\alpha-1)^{2} \geq 0$ which is valid. Hence $3 \leq \alpha^{6}<4$ and we get $\left[\alpha^{6}\right]=3$.

4. Let $R$ denote the circumradius of a triangle $A B C ; a, b, c$ its sides $B C, C A, A B$; and $r_{a}, r_{b}, r_{c}$ its exradii opposite $A, B, C$. If $2 R \leq r_{a}$, prove that

(i) $a>b$ and $a>c$;

(ii) $2 R>r_{b}$ and $2 R>r_{c}$.

Solution: We know that $2 R=\frac{a b c}{2 \triangle}$ and $r_{a}=\frac{\triangle}{s-a}$, where $a, b, c$ are the sides of the triangle $A B C$, $s=\frac{a+b+c}{2}$ and $\triangle$ is the area of $A B C$. Thus the given condition $2 R \leq r_{a}$ translates to

$$
a b c \leq \frac{2 \triangle^{2}}{s-a}
$$

Putting $s-a=p, s-b=q, s-c=r$, we get $a=q+r, b=r+p, c=p+q$ and the condition now is

$$
p(p+q)(q+r)(r+p) \leq 2 \triangle^{2}
$$

But Heron's formula gives, $\triangle^{2}=s(s-a)(s-b)(s-c)=p q r(p+q+r)$. We obtain $(p+q)(q+$ $r)(r+p) \leq 2 q r(p+q+r)$. Expanding and effecting some cancellations, we get

$$
p^{2}(q+r)+p\left(q^{2}+r^{2}\right) \leq q r(q+r)
$$

Suppose $a \leq b$. This implies that $q+r \leq r+p$ and hence $q \leq p$. This implies that $q^{2} r \leq p^{2} r$ and $q r^{2} \leq p r^{2}$ giving $q r(q+r) \leq p^{2} r+p r^{2}<p^{2} r+p r^{2}+p^{2} q+p q^{2}=p^{2}(q+r)+p\left(q^{2}+r^{2}\right)$ which contradicts $(\star)$. Similarly, $a \leq c$ is also not possible. This proves (i).

Suppose $2 R \leq r_{b}$. As above this takes the form

$$
q^{2}(r+p)+q\left(r^{2}+p^{2}\right) \leq p r(p+r)
$$

Since $a>b$ and $a>c$, we have $q>p, r>p$. Thus $q^{2} r>p^{2} r$ and $q r^{2}>p r^{2}$. Hence

$$
q^{2}(r+p)+q\left(r^{2}+p^{2}\right)>q^{2} r+q r^{2}>p^{2} r+p r^{2}=p r(p+r)
$$

which contradicts ( $\star \star$ ). Hence $2 R>r_{b}$. Similarly, we can prove that $2 R>r_{c}$. This proves (ii)

5. Let $S$ denote the set of all 6-tuples $(a, b, c, d, e, f)$ of positive integers such that $a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=$ $f^{2}$. Consider the set

$$
T=\{a b c d e f:(a, b, c, d, e, f) \in S\}
$$

Find the greatest common divisor of all the members of $T$.

Solution: We show that the required gcd is 24 . Consider an element $(a, d, c, d, e, f) \in S$. We have

$$
a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=f^{2}
$$

We first observe that not all $a, b, c, d, e$ can be odd. Otherwise, we have $a^{2} \equiv b^{2} \equiv c^{2} \equiv d^{2} \equiv e^{2} \equiv 1$ $(\bmod 8)$ and hence $f^{2} \equiv 5(\bmod 8)$, which is impossible because no square can be congruent to 5 modulo 8. Thus at least one of $a, b, c, d, e$ is even.

Similarly if none of $a, b, c, d, e$ is divisible by 3 , then $a^{2} \equiv b^{2} \equiv c^{2} \equiv d^{2} \equiv e^{2} \equiv 1(\bmod 3)$ and hence $f^{2} \equiv 2(\bmod 3)$ which again is impossible because no square is congruent to 2 modulo 3 . Thus 3 divides $a b c d e f$.

There are several possibilities for $a, b, c, d, e$.

Case 1: Suppose one of them is even and the other four are odd; say $a$ is even, $b, c, d, e$ are odd. Then $b^{2}+c^{2}+d^{2}+e^{2} \equiv 4(\bmod 8)$. If $a^{2} \equiv 4(\bmod 8)$, then $f^{2} \equiv 0(\bmod 8)$ and hence $2|a, 4| f$ giving $8 \mid a f$. If $a^{2} \equiv 0(\bmod 8)$, then $f^{2} \equiv 4(\bmod 8)$ which again gives that $4 \mid a$ and $2 \mid f$ so that $8 \mid a f$. It follows that $8 \mid a b c d e f$ and hence $24 \mid a b c d e f$.

Case 2: Suppose $a, b$ are even and $c, d, e$ are odd. Then $c^{2}+d^{2}+e^{2} \equiv 3(\bmod 8)$. Since $a^{2}+b^{2} \equiv 0$ or 4 modulo 8 , it follows that $f^{2} \equiv 3$ or $7(\bmod 8)$ which is impossible. Hence this case does not arise.

Case 3: If three of $a, b, c, d, e$ are even and two odd, then $8 \mid a b c d e f$ and hence 24|abcdef.

Case 4: If four of $a, b, c, d, e$ are even, then again $8 \mid a b c d e f$ and 24|abcdef. Here again for any six tuple $(a, b, c, d, e, f)$ in $S$, we observe that $24 \mid a b c d e f$. Since

$$
1^{2}+1^{2}+1^{2}+2^{2}+3^{2}=4^{2}
$$

We see that $(1,1,1,2,3,4) \in S$ and hence $24 \in T$. Thus 24 is the gcd of $T$.

6. Prove that the number of 5 -tuples of positive integers $(a, b, c, d, e)$ satisfying the equation

$$
a b c d e=5(b c d e+a c d e+a b d e+a b c e+a b c d)
$$

is an odd integer.

Solution: We write the equation in the form:

$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}=\frac{1}{5}
$$

The number of five tuple ( $a, b, c, d, e)$ which satisfy the given relation and for which $a \neq b$ is even, because for if $(a, b, c, d, e)$ is a solution, then so is $(b, a, c, d, e)$ which is distinct from $(a, b, c, d, e)$. Similarly the number of five tuples which satisfy the equation and for which $c \neq d$ is also even. Hence it suffices to count only those five tuples $(a, b, c, d, e)$ for which $a=b, c=d$. Thus the equation reduces to

$$
\frac{2}{a}+\frac{2}{c}+\frac{1}{e}=\frac{1}{5}
$$

Here again the tuple ( $a, a, c, c, e)$ for which $a \neq c$ is even because we can associate different solution $(c, c, a, a, e)$ to this five tuple. Thus it suffices to consider the equation

$$
\frac{4}{a}+\frac{1}{e}=\frac{1}{5}
$$

and show that the number of pairs $(a, e)$ satisfying this equation is odd.

This reduces to

$$
a e=20 e+5 a
$$

or

$$
(a-20)(e-5)=100
$$

But observe that

$$
\begin{aligned}
& 100=1 \times 100=2 \times 50=4 \times 25=5 \times 20 \\
& \quad=10 \times 10=20 \times 5=25 \times 4=50 \times 2=100 \times 1
\end{aligned}
$$

Note that no factorisation of 100 as product of two negative numbers yield a positive tuple $(a, e)$. Hence we get these 9 solutions. This proves that the total number of five tuples $(a, b, c, d, e)$ satisfying the given equation is odd.