Wealso notice kpq kdq and pqk pda thus pkq isosceles right-angled triangle with

Problems and Solutions: INMO-2015

1. Let ABC be a right-angled triangle with ∠B = 90◦. Let BD be the altitude from B on to AC. Let P, Q and I be the incentres of triangles ABD, CBD and ABC respectively. Show that the circumcentre of of the triangle PIQ lies on the hypotenuse AC.

that ∠XED = 90 − α. Hence E = F.

Let r1, r2 and r be the inradii of the triangles ABD, CBD and ABC respectively. Join PD and DQ. Observe that ∠PDQ = 90◦. Hence

Consider △PIQ.∠PIQ = 90+(B/2) = 135. Hence PQ Observe that

also notice ∠KPQ = ∠KDQ = 45◦and ∠PQK = ∠PDA = 45◦.

But
s1 =AB + BD + DA , 2

This gives

(s − a)(b − c) = (s − b + b − a)(b − c) = r(b − c) + (b − a)(b − c)
= r(b − c) + 2r2= r(b − c + c + a − b) = ra.

Thus KP = r. Similarly, KQ = r. This gives KP = KI = KQ = r and

therefore K is the circumcentre of △KIQ.

(infinite) decimal expansion of 1/n.

Solution: For any prime p, let νp(n) be the maximum power of p

We show that r = max(ν2(n), ν5(n)).

3. Find all real functions f from R → R satisfying the relation

f(x2+ yf(x)) = xf(x + y).

Since f This implies that f(x) = x for all x ̸= 0. Since f(0) = 0, we conclude�x2− xf(x)�= 0, we conclude that x2− xf(x) = 0 for all x ̸= 0.

that f(x) = x for all x ∈ R.

xn = 3yn−1,

and

Alternate solution: Since the ball goes back to one of the other 3 persons, we have
xn + 3yn = 3n,

since there are 3nways of passing the ball in n passes. Using xn = 3yn−1, we obtain
xn−1 + xn = 3n−1,
with x1 = 0. Thus

Thus we have to prove that a+c = b+d if and only if ars+cpq = bsp+dqr. Now we can write a + c = b + d as

a2+ c2+ 2ac = b2+ d2+ 2bd.

Similarly, by squaring ars + cpq = bsp + dqr we can show that it is equivalent to

−pq cos θ + −rs cos θ + ac = ps cos θ + qr cos θ + bd.

bers a2, b2, c2, d2, e2, f2such that

a2+ b2+ c2≡ d2+ e2+ f2 (mod 12).

whose sum is divisible by 3. This means we can find 3 pairs, say,

(x2