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 2 of 5: A is the radical centre of the three circles

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A = BN \cap CM = \text{radical center of } \omega_1,\omega_2,\omega_3 \implies A \in WZ

Detailed analysis

The line $WZ$ is the radical axis of $\omega_1$ and $\omega_2$ ; likewise $BN$ is the radical axis of $\omega_1$ and $\omega_3$ , and $CM$ is the radical axis of $\omega_2$ and $\omega_3$ . Hence $A=BN\cap CM$ is the radical centre of the three circles, so $A$ lies on line $WZ$ as well.