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Isotomic point of the height foot

Source: All Russian MO 2015, grade 10, problem 7

August 7, 2015

geometrycircumcircle

Problem Statement

In an acute-angled and not isosceles triangle

ABC,

we draw the median

AM

and the height

PM \perp AB

Points

Q

and

P

are marked on the lines

AB

and

AC

, respectively, so that the

QM \perp AC

and

PM \perp AB

. The circumcircle of

PMQ

intersects the line

BC

for second time at point

X.

Prove that

BH = CX.

M. Didin

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