3. What are the sizes of the angles of a scalene triangle if the middle angle by size is the arithmetic mean of the remaining two angles and if the length of the longest side of the triangle is twice the length of the shortest side?
3. What are the sizes of the angles of a scalene triangle if the middle angle by size is the arithmetic mean of the remaining two angles and if the length of the longest side of the triangle is twice the length of the shortest side?

## Solution.

Let the lengths of the sides of the scalene triangle $ABC$ be $|AB|=c, |BC|=a, |AC|=b$, where $a < b < c$.

Let point $P$ be the midpoint of the longest side $|AB|=c$ of triangle $ABC$.

Then $|AP|=|BP|=\frac{c}{2}$, and since $a=\frac{c}{2}$, it follows that $|BP|=|BC|=\frac{c}{2}$.

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This means that triangle $BCP$ is isosceles with legs $\overline{BP}$ and $\overline{BC}$

and angles at the base $|\angle BPC|=|\angle PCB|=60^{\circ}$.

It follows that triangle $BCP$ is equilateral.

Therefore, $|PB|=|PC|$, and since $|PA|=|PB|$, it follows that $|PA|=|PC|$. Thus, triangle $PCA$ is isosceles with angles at the base $|\angle PAC|=|\angle ACP|=\alpha$.

Since $|\angle BPC|=60^{\circ}$ is an exterior angle of triangle $PCA$, the equality $\alpha+\alpha=60^{\circ}$ holds, i.e., $\alpha=30^{\circ}$.