4. Find all right triangles $s$ with side lengths $a, b, c$ and medians $t_{a}$, $t_{b}, t_{c}$, for which $a+t_{a}=b+t_{b}$. Consider both cases, when $AB$ is a) the hypotenuse, b) a leg.
Solution. a) Let $a$ and $b$ be the legs (Fig. 3). Then, according to the Pythagorean theorem,

$$
t_{a}=\sqrt{b^{2}+\left(\frac{a}{2}\right)^{2}}, \quad t_{b}=\sqrt{a^{2}+\left(\frac{b}{2}\right)^{2}}
$$

so the condition $a+t_{a}=b+t_{b}$ has the form

$$
a+\sqrt{b^{2}+\left(\frac{a}{2}\right)^{2}}=b+\sqrt{a^{2}+\left(\frac{b}{2}\right)^{2}}.
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_8eb7f58dfd33c03bd7dfg-4.jpg?height=466&width=623&top_left_y=795&top_left_x=728)

Fig. 3

Since the inequality $a>b$ implies (see introductory problem 1) $t_{b}>t_{a}$, the following transformations are equivalent:

$$
\begin{aligned}
2 a-2 b & =\sqrt{4 a^{2}+b^{2}}-\sqrt{4 b^{2}+a^{2}}, \\
4 a^{2}-8 a b+4 b^{2} & =5 a^{2}+5 b^{2}-2 \sqrt{\left(4 a^{2}+b^{2}\right)\left(4 b^{2}+a^{2}\right)}, \\
2 \sqrt{4 a^{4}+17 a^{2} b^{2}+4 b^{4}} & =a^{2}+8 a b+b^{2}, \\
16 a^{4}+68 a^{2} b^{2}+16 b^{4} & =a^{4}+16 a^{3} b+66 a^{2} b^{2}+16 a b^{3}+b^{4}, \\
15 a^{4}-16 a^{3} b+2 a^{2} b^{2}-16 a b^{3}+15 b^{4} & =0 .
\end{aligned}
$$

The polynomial on the left side of the last equation is clearly divisible by the binomial $a-b$ (since it is zero for $a=b$). By division, we find that the resulting polynomial of the third degree has the same property, so after repeated division, we convert the equation under consideration into a product form

$$
(a-b)^{2}\left(15 a^{2}+14 a b+15 b^{2}\right)=0 .
$$

The last equality holds if and only if $a=b$, because $15 a^{2}+14 a b+15 b^{2}>0$ for any pair of real numbers $a, b$.

In case a), we can also proceed as follows: By subtracting the equations

$$
t_{a}^{2}=b^{2}+\left(\frac{a}{2}\right)^{2}, \quad t_{b}^{2}=a^{2}+\left(\frac{b}{2}\right)^{2}
$$

we get

$$
t_{a}^{2}-t_{b}^{2}=\frac{3}{4}\left(b^{2}-a^{2}\right)
$$

On both sides of the last equation, there are differences of squares. We convert them into products and then use the given equation $a+t_{a}=b+t_{b}$ rewritten as $t_{a}-t_{b}=b-a$:

$$
\begin{aligned}
\left(t_{a}-t_{b}\right)\left(t_{a}+t_{b}\right) & =\frac{3}{4}(b-a)(b+a) \\
(b-a)\left(t_{a}+t_{b}\right) & =\frac{3}{4}(b-a)(a+b) .
\end{aligned}
$$

If $a \neq b$, we get $t_{a}+t_{b}=\frac{3}{4}(a+b)$; this together with the equation $t_{a}-t_{b}=b-a$ gives $t_{a}=\frac{7}{8} b-\frac{1}{8} a$, thus $t_{a}<b$, which contradicts the assumption $t_{a}>b$. Therefore, $a=b$.

For the case a), the following transformations are equivalent:

$$
\begin{aligned}
3 a-2 b & =\sqrt{4 a^{2}-3 b^{2}}, \\
9 a^{2}-12 a b+4 b^{2} & =4 a^{2}-3 b^{2}, \\
5 a^{2}-12 a b+7 b^{2} & =0, \\
(a-b)(5 a-7 b) & =0, \\
5 a-7 b & =0 .
\end{aligned}
$$

Conclusion: The equality $a+t_{a}=b+t_{b}$ holds for right-angled isosceles triangles with legs $a=b$ and for right-angled triangles with sides in the ratio $5: \sqrt{24}: 7$, where the shortest side is (the third) side $c$.

GUIDING AND SIMILAR PROBLEMS: