74th NMO Selection Tests for JBMO

Question

In the exterior of the acute-angled triangle

A B C

, we construct the isosceles triangles

D A B

and

E A C

with bases

A B

and

A C

, respectively, such that

∠ D B C = ∠ E C B = 9 0^{\circ}

. Let

M

and

N

be the reflections of

A

with respect to

D

and

E

, respectively. Prove that the line

M N

passes through the orthocenter of the triangle

A B C

.

Accepted Answer

FIGURE_0 Since AH BC and CP BC, we deduce that AH CP. Therefore AHCP is a parallelogram, and CP = AH. Similarly, we deduce that AHFC' is a parallelogram, hence C'F = AH = CP. Since C'E = CE, we obtain that EP = EF. Consequently, D and E are the midpoints of the bases of the trapezoid BB'PF, and Q is the intersection point of the lines B'P and BF, therefore D, E, Q are collinear. Since D and I are the midpoints of the bases of the trapezoid AHBB', and Q is the intersection point of the lines B'A and BH, it follows that D, I, Q are also collinear, consequently I lies on the line DE. DI and QI are midlines in the triangles AMH and ANH, respectively, hence DI MH and IE NH, therefore H MN.