What could be the vertex angle of an isosceles triangle whose altitudes can form a triangle when used as sides?
Solution. Let the base of the isosceles triangle $ABC$ ($AB = AC$) be $a$, the legs be $b$, the vertex angle be $\alpha$, the altitude to the base be $m_a$, and the altitudes to the legs be $m_b$.

The necessary and sufficient condition for the constructibility of the triangle with sides $m_a, m_b, m_b$ is that $2 m_b > m_a$ (the condition $m_a + m_b > m_b$ is always satisfied).

![](https://cdn.mathpix.com/cropped/2024_05_02_5650736a4988120a1bccg-1.jpg?height=352&width=290&top_left_y=320&top_left_x=904)

The area of triangle $ABC$ is: $T = \frac{a \cdot m_a}{2} = \frac{b \cdot m_b}{2}$, hence $m_a = \frac{2T}{a}$ and $m_b = \frac{2T}{b}$. Substituting these into the inequality $2 m_b > m_a$ and simplifying, we get

$$
\frac{a}{b} > \frac{1}{2}
$$

In triangle $ABC$, let the foot of the angle bisector of $\alpha$ be $D$ ($AD \perp BC$). In the right triangle $ADC$, $\sin \frac{\alpha}{2} = \frac{a}{2b}$. Substituting this into (1), we get $\sin \frac{\alpha}{2} > \frac{1}{4}$, and thus $\alpha > 28.95^\circ$.

Therefore, the altitudes of isosceles triangles can form a triangle if the vertex angle is greater than $28.95^\circ$.

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Gergely Tassy (Budapest, Veres Péter Gymnasium, 10th grade)