Problem 4

Let

ABC

be an acute-angled triangle with orthocenter

H

, and let

W

be a point on the side

BC

, lying strictly between

B

and

C

. The points

M

and

N

are the feet of the altitudes from

B

and

C

, respectively. Denote by

\omega_1

the circumcircle of

BWN

, and let

X

be the point on

\omega_1

such that

WX

is a diameter of

\omega_1

. Analogously, denote by

\omega_2

the circumcircle of

CWM

, and let

Y

be the point such that

WY

is a diameter of

\omega_2

. Prove that

X, Y

and

H

are collinear.

Step 1 of 5: A third circle through the four altitude-related points

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\omega_3 := \text{circle through } B,C,N,M \quad (\angle BNC=\angle BMC=90^\circ)

Detailed analysis

Let $L$ be the foot of the altitude from $A$ , and let $Z$ be the second intersection point of $\omega_1$ and $\omega_2$ other than $W$ . Since $\angle BNC=\angle BMC=90^\circ$ , the points $B,C,N,M$ lie on a common circle $\omega_3$ .