Two non-congruent triangles have the same area. Each triangle has sides of length $8$ and $9$, and the third side of each triangle has integer length. What is the sum of the lengths of the third sides? $\textbf{(A) } 20 \qquad$ $ \textbf{(B) } 22 \qquad$ $ \textbf{(C) } 24 \qquad$ $ \textbf{(D) } 26 \qquad$ $ \textbf{(E) } 28$
Solution
Let the angle between the sides $8$ and $9$ be $\gamma$ and the third side be $c$. The area is $\frac12\cdot8\cdot9\sin\gamma=36\sin\gamma$. Two non-congruent triangles with equal area have $\sin\gamma=\sin\gamma'$ with $\gamma'\ne\gamma$, so $\gamma'=180^\circ-\gamma$.
By the law of cosines, $c^2=8^2+9^2-144\cos\gamma=145-144\cos\gamma$ and $c'^2=145+144\cos\gamma$. Adding, $c^2+c'^2=290$.
The third side satisfies $1<c<17$, so $2\le c\le16$ and $c^2\le256$. Then $c'^2=290-c^2\le256$ forces $c^2\ge34$, i.e. $6\le c\le16$. Checking $290-c^2$ for $c=6,\dots,16$, only $c=11$ gives a square: $290-121=169=13^2$. So the third sides are $11$ and $13$, with sum $24$.
Two non-congruent triangles have the same area. Each triangle has sides of length $7$ and $9$, and the third side of each triangle has integer length. What is the sum of the lengths of the third sides?
