In an isosceles trapezoid, the lengths of the parallel sides are $a$ and $b$, and its height is $m$.

a) Construct the point $P$ on the axis of symmetry of the trapezoid from which the legs are seen at a right angle.

b) Calculate the distance of point $P$ from one of the parallel sides.

c) What is the condition for the existence of point $P$?

The 663rd exercise and the 1063-1064th problems are from the tasks of the 2nd International Mathematical Olympiad for Secondary School Students held in Sinaia (Román N. K.).
Each leg is only visible at a right angle from the points of the Thales' circle written above it, so the point $P$ can only be one of the common points of these two circles. If there is a common point, due to symmetry, it lies on the axis of the trapezoid, thus it also satisfies the third requirement; therefore, in the construction and calculation, we can also look for it as the intersection point of the axis of symmetry and one of the Thales' circles. Let's denote the radius of the circle (half the length of the leg of the trapezoid) by $r$.

![](https://cdn.mathpix.com/cropped/2024_05_02_f98662931bd2a7ecf10cg-1.jpg?height=442&width=512&top_left_y=321&top_left_x=796)

The line connecting the centers of the two circles connects the midpoints of the two legs, i.e., the midline of the trapezoid, so it is perpendicular to the axis of symmetry and bisects it.

If the two circles have only one common point, this point must lie on the axis of symmetry and on the line connecting the centers, so it is at a distance of $m / 2$ from the parallel sides.

If two different points satisfy the condition, then these points are symmetric with respect to the center, so one is as far from one parallel side as the other is from the other. Let the distance of one of the points from the closer parallel side be $d$. We can calculate this by subtracting half the length of the common chord of the two circles from $m / 2$. This half-chord is the leg of a right triangle whose hypotenuse is $r$, and the other leg is half the midline. Thus, the length of the half-chord is

$$
\sqrt{r^{2}-\left(\frac{a+b}{4}\right)^{2}}
$$

We can determine the radius from a right triangle that is formed by dropping a perpendicular from one endpoint of the shorter parallel side to the longer one. In this triangle, the hypotenuse is $2 r$, and the legs are $m$ and $(a-b) / 2$, so

$$
r=\frac{1}{2} \sqrt{m^{2}+\left(\frac{a-b}{2}\right)^{2}}
$$

The sought distance is therefore:

$$
d=\frac{m}{2}-\sqrt{\frac{m^{2}}{4}+\left(\frac{a-b}{4}\right)^{2}-\left(\frac{a+b}{4}\right)^{2}}=\frac{m}{2}-\sqrt{\frac{m^{2}-a b}{4}}
$$

The condition for the existence of a solution is that

$$
m^{2} \geqq a b
$$

If the equality holds, we get the value $d=m / 2$, so the formula is also correct when there is only one solution.