Geometry problem about Euler line

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Geometry problem about Euler line

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#1 h Apr 9, 2025, 7:36 AM

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If the Euler line of a triangle is parallel to one side of the triangle, what is the relationship between the sides of this triangle?

The relationship between the angles of this triangle

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#2 h Apr 12, 2025, 10:57 AM

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Given $\triangle ABC\ :\ A(0,a),b(-b,0),C(c,0)$ , then the centroid $Z(\frac{c-b}{3},\frac{a}{3})$ and the orthocenter $H(0,\frac{bc}{a})$ .

Slope of the line $HZ\ :\ m_{HZ}=\frac{\frac{a}{3}-\frac{bc}{a}}{\frac{c-b}{2}}$ .
$HZ // BC \Rightarrow a^{2}-3bc=0 \Rightarrow \frac{a}{b} \cdot \frac{a}{c}=3 \Rightarrow \tan B \cdot \tan C =3$ .

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#3 h Apr 12, 2025, 12:14 PM

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Simpler solution:

The Euler line passes through $H$ and $O$ . Consider three cases:

If the triangle is obtuse (e.g. $\angle A > 90^\circ$ ), then $H$ and $O$ lie on opposite sides of $BC$ , therefore $OH$ (the Euler line) cannot be parallel to $BC$ .

If the triangle is right (e.g. $\angle A = 90^\circ$ ), then the Euler line is $OA$ and it isn't parallel to $BC$ .

So the triangle must be acute, from which $O$ and $H$ lie in the triangle.

Let $M$ be the midpoint of $BC$ , and $D$ be the foot of the perpendicular from $A$ to $BC$ .

Then $AH = 2OM$ and $HD = OM$ , so $AH = 3HD$ (because $OHDM$ is a rectangle).

Since $\tan B = \frac{AD}{BD}$ and $\tan C = \tan \angle BHD = \frac{BD}{HD}$ , we have
$\tan B \tan C = \frac{AD}{BD} \cdot \frac{BD}{HD} = \frac{AD}{HD} = 3.$

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