MMO2025 Round 4

Question

Let ABC be a triangle with AB AC, and let H be its orthocenter. Let T be a point on the arc BC of the circumcircle of triangle ABC that does not contain point A. Let l be the line through H parallel to BC, and let l intersect lines TB and TC at points P and Q, respectively. Let the circumcircles of triangles PAB and QAC intersect again at point S. Suppose that PAQ = 2 BAC. Prove the following. (1) The point S lies on the circumcircle of triangle BHC. (2) PAH = HAQ or PAB = BAH.

Accepted Answer

(1) By the cyclic quadrilateral property,

∠ P S A = ∠ P B A = 18 0^{\circ} - ∠ A B T = ∠ A C T = 18 0^{\circ} - ∠ A C Q = 18 0^{\circ} - ∠ A S Q,

so point

S

lies on line

P Q

.

From the given condition, we have

∠ P A B + ∠ C A Q = ∠ B A C

. Hence,

∠ P S B + ∠ C S Q = ∠ B A C

. Therefore,

∠ B S C = 18 0^{\circ} - ∠ B A C = ∠ B H C

, and thus

S

lies on the circum-circle of triangle

B H C

.

Hence,

B T C H

is a parallelogram, and under a homothety centered at

T

with ratio 2, segment

B C

maps to

P Q

, so

T B = B P

,

T C = C Q

. Since

A B ⊥ T P

and

A C ⊥ T Q

, point

A

is the center of the circle through

T P Q

. As

A H ⊥ P Q

, point

H

is the midpoint of

P Q

, giving

∠ P A H = ∠ H A Q .

In this case, point

P

lies on the circle through triangle

A B S

. From angle chasing:

∠ P A B = ∠ P S B = ∠ S B C = ∠ H C B = ∠ B A H,

so

∠ P A B = ∠ B A H .