Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.3 National olympiad, round 2 Prove it Romania

Let $ABC$ be a triangle inscribed in the circle $C$ with center $O$ and radius $1$ . For any point $M \in C \setminus \{A, B, C\}$ , we denote $s(M) = OH_1^2 + OH_2^2 + OH_3^2$ , where $H_1$ , $H_2$ , and $H_3$ are the orthocenters of triangles $MAB$ , $MBC$ , and $MCA$ , respectively.

a) Prove that if triangle $ABC$ is equilateral, then $s(M) = 6$ , for any $M \in C \setminus \{A, B, C\}$ .

b) Prove that if there exist three distinct points $M_1, M_2, M_3 \in C \setminus \{A, B, C\}$ such that $s(M_1) = s(M_2) = s(M_3)$ , then triangle $ABC$ is equilateral.

Solution

Consider an orthonormal coordinate system with the origin at $O$ . For any point $Z$ in the plane, we denote its complex coordinate by $z$ .

a.
From Sylvester's relation, we have $h_1 = m + a + b$ , $h_2 = m + b + c$ , and $h_3 = m + c + a$ . Also, let $h = a + b + c$ be the complex coordinate of the orthocenter of triangle $ABC$ . Therefore, we obtain:
$s(M) = 6 + |h|^2 + 2m\overline{h} + 2\overline{m}h.$
If triangle $ABC$ is equilateral, then $h = a + b + c = 0$ , hence $s(M) = 6$ .

b.
Assume, by contradiction, that triangle $ABC$ is not equilateral, which is equivalent to $h \neq 0$ . Then, since $s(M_1) = s(M_2)$ , we have: $|h|^2 + 2m_1\overline{h} + 2\overline{m}_1h = |h|^2 + 2m_2\overline{h} + 2\overline{m}_2h \Leftrightarrow \overline{h}(m_1 - m_2) + h\frac{m_2 - m_1}{m_1m_2} = 0 \Leftrightarrow m_1m_2 = \frac{h}{\overline{h}}$ .

Similarly, we obtain $m_1m_3 = \frac{h}{\overline{h}}$ . Since $m_1 \neq 0$ , we get $m_2 = m_3$ , which is a contradiction with $M_2 \neq M_3$ . Therefore, triangle $ABC$ is equilateral.

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