## Task 34/76

In a triangle with sides $a, b$ and $c$ and the height $h_{c}$ on $c$, it is given that $2a^2 - 3ab + 2b^2 = c h_{c}$. Determine the angles $\alpha, \beta, \gamma$ opposite to the sides $a, b$ and $c$.
It is

$$
c h_{c}=2 a^{2}-3 a b+2 b^{2}=2 a^{2}-4 a b+2 b^{2}+a b=2\left(a^{2}-2 a b+b^{2}\right)+a b=2(a-b)^{2}+a b
$$

Since $\sin \gamma \leq 1$, it follows that $a b \geq a b \sin \gamma=2 A=c h_{c}$ (where $A$ denotes the area). Therefore,

$$
c h_{c}=2(a-b)^{2}+a b \geq 2(a-b)^{2}+c h_{c} \quad \text { hence } \quad 0 \geq(a-b)^{2}
$$

Since $(a-b)^{2} \geq 0$ also holds, it follows that $a=b$ and thus $\alpha=\beta$. The triangle is therefore isosceles. Substituting $a=b$ into the given equation, we get

$$
c h_{c}=2 a^{2}-3 a b+2 b^{2}=2 a^{2}-3 a^{2}+2 a^{2}=a^{2}
$$

From $c h_{c}=2 A=a^{2} \sin \gamma$, it immediately follows that $\sin \gamma=1$, hence $\gamma=\frac{\pi}{2}$. Therefore, $\alpha=\beta=\frac{\pi}{4}$, and the triangle is right-angled and isosceles.