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 2 of 3: Set up position vectors from the circumcenter O

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In plain words

Taking the circumcenter $O$ as origin makes the three vertex vectors have equal length $R$ and gives the classical formula $\overrightarrow{OH} = \mathbf{a}+\mathbf{b}+\mathbf{c}$ for the orthocenter.

\overrightarrow{OA}=\mathbf{a},\ \overrightarrow{OB}=\mathbf{b},\ \overrightarrow{OC}=\mathbf{c}, \quad |\mathbf{a}|^2=|\mathbf{b}|^2=|\mathbf{c}|^2=R^2, \quad \overrightarrow{OH}=\mathbf{a}+\mathbf{b}+\mathbf{c}

Detailed analysis

Let $R$ be the circumradius of $\triangle ABC$ and set $\overrightarrow{OA}=\mathbf{a}$ , $\overrightarrow{OB}=\mathbf{b}$ , $\overrightarrow{OC}=\mathbf{c}$ , so $|\mathbf{a}|^2=|\mathbf{b}|^2=|\mathbf{c}|^2=R^2$ and the orthocenter satisfies $\overrightarrow{OH}=\mathbf{a}+\mathbf{b}+\mathbf{c}$ . Then $\overrightarrow{OA_0} = \dfrac{\mathbf{b}+\mathbf{c}}{2}$ and $\overrightarrow{A_0H} = \overrightarrow{OH} - \overrightarrow{OA_0} = \mathbf{a} + \dfrac{\mathbf{b}+\mathbf{c}}{2} = \dfrac{2\mathbf{a}+\mathbf{b}+\mathbf{c}}{2}$ .