, 2010

The equality given in the problem is equivalent to
$$(a^2 + b^2) \sin \varphi \cos \varphi + ab(\sin^2 \varphi + \cos^2 \varphi) = 2ab$$
and hence also to
$$(a^2 + b^2) \sin \varphi \cos \varphi = ab. \qquad (1)$$

Let $\alpha$ and $\beta$ be the angles opposite to legs with length $a$ and $b$, respectively. Then $\sin \alpha = a/\sqrt{a^2 + b^2}$, $\sin \beta = \cos \alpha = b/\sqrt{a^2 + b^2}$, implying
$$(a^2 + b^2) \sin \alpha \cos \alpha = ab.$$

Comparing this to (1) shows the equivalence of the equality of the problem and the equality $\sin \varphi \cos \varphi = \sin \alpha \cos \alpha$, i.e., equality $\sin 2\varphi = \sin 2\alpha$. As $0 < \alpha, \beta < 90^\circ$, this implies $2\varphi = 2\alpha$ or $2\varphi = 180^\circ - 2\alpha$, whence $\varphi = \alpha$ or $\varphi = 90^\circ - \alpha = \beta$. Hence, $\varphi$ satisfies the equality if and only if it equals one of the acute angles of the right triangle.

Let $ABC$ be the given triangle with right angle at vertex $C$. Let $c$ be the length of its hypothenuse and $h$ be the height corresponding to the hypothenuse. Let $C'$ be a point on the circumcircle of $ABC$ such that one acute angle of triangle $ABC'$ is $\varphi$ (Fig. 8). Let $a'$ and $b'$ be the lengths of the legs of triangle $ABC'$ and $h'$ be the height of the triangle $ABC'$ corresponding to its hypothenuse. Then $(a^2 + b^2) \sin \varphi \cos \varphi = c^2 \sin \varphi \cos \varphi = (c \sin \varphi)(c \cos \varphi) = a'b' = ch'$. Since the equality in the problem is equivalent to the equality (1) from the Solution 1, it is also equivalent to $ch' = ab$. But $ab = ch$, hence it is also equivalent to $h = h'$. This condition holds if and only if $ABC \sim ABC'$ or $ABC \sim BAC'$, i.e., $\varphi$ equals one of the acute angles of triangle $ABC$.

Fig. 8