Given two circles with different radii. Construct a right-angled triangle such that the sum of the areas of the circles drawn on its legs as diameters equals the area of the larger circle, while the difference equals the area of the smaller circle.
Let the radius of the larger circle be denoted by \( R \), the radius of the smaller circle by \( r \), and the legs of the triangle to be constructed by \( a \) and \( b \). Then we can write the following equations:

\[
\frac{a^{2}}{4} \pi + \frac{b^{2}}{4} \pi = R^{2} \pi, \quad (1) \quad \frac{a^{2}}{4} \pi - \frac{b^{2}}{4} \pi = r^{2} \pi
\]

By adding the two equations, we get \(\frac{a^{2}}{2} \pi = \left(R^{2} + r^{2}\right) \pi\), which simplifies to

\[
a^{2} = (\sqrt{2} R)^{2} + (\sqrt{2} r)^{2}
\]

Based on this, using the Pythagorean theorem, \( a \) can be easily constructed (Figure 1).

From equation (1), it is clear that the hypotenuse of the right triangle with legs \( a \) and \( b \) is \( 2R \). Therefore, the desired triangle \( ABC \) can be obtained by intersecting the perpendicular line at \( C \) on the segment \( CA \) with the circle centered at \( A \) with radius \( 2R \).

The problem always has one solution.

Gábor Poronyi (Pécs, Janus Pannonius Gymnasium, 1st year)

![](https://cdn.mathpix.com/cropped/2024_05_02_ad1bc51c17c19a7ee4f1g-1.jpg?height=222&width=569&top_left_y=927&top_left_x=775)