Selection tests for the International Mathematical Olympiad 2013

Question

Let ABC be an acute triangle, M be the midpoint of BC and P be a point on line segment AM. Lines BP and CP meet the circumcircle of ABC again at X and Y, respectively, and sides AC at D and AB at E, respectively. Prove that the circumcircles of AXD and AYE have a common point T A on line AM. FIGURE_0

Accepted Answer

By applying Ceva to the concurrent cevians

A M

,

B D

and

C E

, we obtain

\frac{A E}{E B} = \frac{A D}{D C} \cdot \frac{C M}{M B} = \frac{A D}{D C} .

We deduce from Thales' theorem that segments

E D

and

B C

are parallel.

Since quadrilateral

B C X Y

is cyclic, we have

∠ B X Y = ∠ B C Y = ∠ C E D

. We deduce that quadrilateral

E D X Y

is cyclic and therefore

P D \cdot P X = P E \cdot P Y

Hence, point

P

lies on the radical axis of circumcircles of triangles

A X D

and

A Y E

which passes through

A

.

If these two circles are tangent to

A P

at

A

then

∠ M A C = ∠ A X B = ∠ A C B, and ∠ B A M = ∠ C Y A = ∠ C B A,

which implies that triangle

A B C

is a right triangle at

A

and this is a contradiction.

We conclude that the two circles intersect in a second point

T \neq = A

on line

A M

.