## Task B-3.6.

Two isosceles triangles have the same perimeter and area, but they are not congruent. The side lengths of one triangle are 29, 29, 40. The side lengths of the other triangle are integers. Determine what these numbers are.
## Solution.

First, let's calculate the perimeter and area of the given triangles

$$
\begin{aligned}
& O=a+2 b=98 \Longrightarrow s=49 \\
& P=\sqrt{s(s-a)(s-b)^{2}}=\sqrt{49 \cdot 9 \cdot 20^{2}}=420
\end{aligned}
$$

Let $x$ be the length of the base and $y$ be the length of the legs of the new triangle. Then we have

$$
\begin{array}{ll}
x+2 y=98 \Longrightarrow y=49-\frac{x}{2} & 1 \text{ point} \\
420=\sqrt{49(49-x)(49-y)^{2}} & 1 \text{ point} \\
420=\sqrt{49(49-x)\left(\frac{x}{2}\right)^{2}} & 1 \text{ point} \\
120=x \sqrt{49-x} . &
\end{array}
$$

Thus, $x$ and $\sqrt{49-x}$ are natural numbers and divisors of the number $120 \Longrightarrow \sqrt{49-x}<7$.

By checking, we find that the equality is valid only for $\sqrt{49-x}=3$ and $\sqrt{49-x}=5 . \quad 1$ point From $\sqrt{49-x}=3$ we get the given triangle, and from $\sqrt{49-x}=5$ we get $x=24$ and $y=37 . \quad 1$ point

Thus, the desired triangle has sides $24,37,37$.