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 3 of 3: Expand OA_0^2 + A_0H^2 and observe complete symmetry

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In plain words

Expanding the two squared norms combines into an expression symmetric in $\mathbf{a}, \mathbf{b}, \mathbf{c}$ , so cyclic permutation leaves the distance from $O$ unchanged.

OA_1^2 = \left|\frac{\mathbf{b}+\mathbf{c}}{2}\right|^2 + \left|\frac{2\mathbf{a}+\mathbf{b}+\mathbf{c}}{2}\right|^2 = 2R^2 + \mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{c} + \mathbf{c}\cdot\mathbf{a}

Detailed analysis

Expanding the dot products gives $OA_1^2 = OA_2^2 = \dfrac{|\mathbf{b}|^2 + 2\mathbf{b}\cdot\mathbf{c} + |\mathbf{c}|^2}{4} + \dfrac{4|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 + 4\mathbf{a}\cdot\mathbf{b} + 4\mathbf{a}\cdot\mathbf{c} + 2\mathbf{b}\cdot\mathbf{c}}{4} = 2R^2 + \mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{c} + \mathbf{c}\cdot\mathbf{a}$ . Because the right-hand side is symmetric under permutations of $\mathbf{a}, \mathbf{b}, \mathbf{c}$ , the exact same value equals $OB_1^2 = OB_2^2 = OC_1^2 = OC_2^2$ , so all six points $A_1, A_2, B_1, B_2, C_1, C_2$ lie on a single circle centered at $O$ .