Problem 1

An acute-angled triangle

ABC

has orthocentre

H

. The circle passing through

H

with centre the midpoint of

BC

intersects the line

BC

A_1

and

A_2

. Similarly, the circle passing through

H

with centre the midpoint of

CA

intersects the line

CA

B_1

and

B_2

, and the circle passing through

H

with centre the midpoint of

AB

intersects the line

AB

C_1

and

C_2

. Show that

A_1, A_2, B_1, B_2, C_1, C_2

lie on a circle.

Step 1 of 3: Express OA_1^2 and OA_2^2 via the midpoint A_0 and the circumcenter O

Previous step Next step

In plain words

Because $O$ lies on the perpendicular bisector of $BC$ at $A_0$ and the circle centered at $A_0$ passes through $H, A_1, A_2$ , the Pythagorean theorem expresses $OA_1^2$ and $OA_2^2$ in terms of $OA_0$ and $A_0H$ .

OA_0 \perp BC, \quad A_0A_1 = A_0A_2 = A_0H \implies OA_1^2 = OA_2^2 = OA_0^2 + A_0H^2

Detailed analysis

Let $O$ be the circumcenter of $\triangle ABC$ and let $A_0, B_0, C_0$ be the midpoints of $BC, CA, AB$ . Since $OA_0 \perp BC$ and $A_1, A_2 \in BC$ lie on the circle centered at $A_0$ with radius $A_0H$ , right triangles $OA_0A_1$ and $OA_0A_2$ give $OA_1^2 = OA_2^2 = OA_0^2 + A_0A_1^2 = OA_0^2 + A_0H^2$ .