74th Romanian Mathematical Olympiad

Question

Let ABC be a triangle inscribed in the circle C with center O and radius 1. For any point M C 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 C A, B, C. b) Prove that if there exist three distinct points M_1, M_2, M_3 C A, B, C such that s(M_1) = s(M_2) = s(M_3), then triangle ABC is equilateral.

Accepted Answer

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

A B C

. Therefore, we obtain:

s (M) = 6 + ∣ h ∣^{2} + 2 m \overline{h} + 2 \overline{m} h .

If triangle

A B C

is equilateral, then

h = a + b + c = 0

, hence

s (M) = 6

.

b. Assume, by contradiction, that triangle

A B C

is not equilateral, which is equivalent to

h \neq = 0

. Then, since

s (M_{1}) = s (M_{2})

, we have:

∣ h ∣^{2} + 2 m_{1} \overline{h} + 2 \overline{m}_{1} h = ∣ h ∣^{2} + 2 m_{2} \overline{h} + 2 \overline{m}_{2} h \Leftrightarrow \overline{h} (m_{1} - m_{2}) + h \frac{m _{2} - m _{1}}{m _{1} m _{2}} = 0 \Leftrightarrow m_{1} m_{2} = \frac{h}{h}

.

Similarly, we obtain

m_{1} m_{3} = \frac{h}{h}

. Since

m_{1} \neq = 0

, we get

m_{2} = m_{3}

, which is a contradiction with

M_{2} \neq = M_{3}

. Therefore, triangle

A B C

is equilateral.