Dot product with equilateral triangle

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Source: Own, syllabus for 10th Grade Geometry at HSGS 2024

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#1 h Dec 2, 2024, 12:40 PM

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Let $H$ be the orthocenter of triangle $ABC$ . Let $R$ be the circumradius of $ABC$ . Prove that triangle $ABC$ is equilateral iff
$\overrightarrow{HA}\cdot\overrightarrow{HB}+\overrightarrow{HB}\cdot\overrightarrow{HC}+\overrightarrow{HC}\cdot\overrightarrow{HA}=-\frac{3R^2}{2}.$

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#2 h Apr 9, 2025, 8:31 PM

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We have $\sum \overrightarrow{HA} \cdot \overrightarrow{HB} = \sum \frac{HA^2 + HB^2 - AB^2}{2} = -\frac{3R^2}{2} \iff 2 \sum HA^2 - \sum AB^2 = -3R^2.$ Since $HA = 2R \cdot |\cos A|$ and $AB = 2R \cdot \sin C$ , the above equation becomes
$8 \sum \cos^2 A - 4 \sum \sin^2 A = -3 \iff 8 \sum \cos^2 A - 4 \sum (1 - \cos^2 A) = -3 \iff \sum \cos^2 A = \frac{3}{4}.$ Now using the fact that $\cos^2 A = \dfrac{\cos 2A + 1}{2}$ , we have $\sum \cos 2A = -\frac{3}{2}$ .
This can be rewritten as
$\begin{align*} 0 = \sum \cos 2A + \frac{3}{2} = \cos 2A + \cos 2B + \cos 2C + \frac{3}{2} &= 2 \cos (A + B) \cos (A - B) + 2 \cos^2 C - 1 + \frac{3}{2} \\ &= \frac{1}{2}\left( 4 \cos^2 C - 4 \cos C \cos (A - B) + \cos^2 (A - B) + \sin^2 (A - B) \right) \\ &= \frac{1}{2} \left((2\cos C - \cos (A - B))^2 + \sin^2 (A - B) \right) \leqslant 0, \end{align*}$ and since we have equality, $\sin (A - B) = 0 \implies A = B$ . Similarly $B = C$ , so $\triangle ABC$ is equilateral.
Conversely, it is easy to verify that, if $\triangle ABC$ is equilateral, there holds the equation from the problem statement.

This post has been edited 3 times. Last edited by ericdimc, Apr 19, 2025, 6:02 PM

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#3 h Apr 9, 2025, 8:42 PM

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Another solution. Let $O$ be the circumcircle of $\triangle ABC$ . Then using Sylvester's Theorem,
$\overrightarrow{OH} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC},$ we have the following:
$\begin{align*} -\dfrac{3R^2}{2} = \sum \overrightarrow{HA} \cdot \overrightarrow{HB} &= \sum (\overrightarrow{HO} + \overrightarrow{OA}) \cdot (\overrightarrow{HO} + \overrightarrow{OB}) \\ &= \sum (HO^2 + \overrightarrow{HO} \cdot \overrightarrow{OB} + \overrightarrow{HO} \cdot \overrightarrow{OA} + \overrightarrow{OA} \cdot \overrightarrow{OB}) \\ &= 3OH^2 + 2 \overrightarrow{HO} \cdot (\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}) + \sum \frac{R^2 + R^2 - c^2}{2} \\ &= OH^2 + 3R^2 - \sum \dfrac{c^2}{2} = -\dfrac{3R^2}{2} \iff OH^2 = 0, \end{align*}$ so $O = H$ , meaning that $\triangle ABC$ is equilateral, since the altitudes are simultaneously medians.

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