## [

In an isosceles trapezoid with bases 1 and 4, two circles are placed, each touching the other circle, the two lateral sides, and one of the bases. Find the area of the trapezoid.

#
The segment of the common tangent of the given circles, passing through their point of tangency, divides the given trapezoid into two similar trapezoids. Therefore, the square of the length of this segment is equal to the product of the bases of the given trapezoid.

## Solution

Let the common tangent of the circles, passing through their point of tangency $K$, intersect the lateral sides $A B$ and $C D$ of the trapezoid $A B C D$ at points $Q$ and $P$ respectively; the circle with center $O_{1}$ and radius $r$, inscribed in the isosceles trapezoid $Q B C P$, touches its sides $B C=1$ and $C P$ at points $M$ and $E$ respectively, and the circle with center $O_{2}$ and radius $R$, inscribed in the isosceles trapezoid $A Q P D$, touches its sides $A D=4$ and $D P$ at points $N$ and $F$ respectively. Then

$$
C E=C M=\frac{1}{2} B C=\frac{1}{2}, D F=D N=\frac{1}{2} A D=2
$$

The trapezoids $Q B C P$ and $A Q P D$ are similar with respect to the point of intersection of the lines $A B$ and $C D$, with a coefficient equal to the ratio of the radii of the circles inscribed in them. Therefore, these trapezoids are similar. Hence, $B C: Q P = Q P: A D$. From this, we find that

$$
Q P=\sqrt{B C \cdot A D}=\sqrt{1 \cdot 4}=2
$$

Therefore,

$$
P K=\frac{1}{2} Q P=1, E P=P K=P F=1
$$

Since $\mathrm{CO}_{1}$ and $P O_{1}$ are the bisectors of angles $B C P$ and $C P Q$, the sum of which is $180^{\circ}$, then $\angle C O_{1} P=90^{\circ}$. From the right triangle $C O_{1} P$, we find that

$$
r=O_{1} E=\sqrt{C E \cdot E P}=\sqrt{\frac{1}{2} \cdot 1}=\frac{\sqrt{2}}{2}
$$

Similarly, we find that

$$
R=O_{2} F=\sqrt{D F \cdot F P}=\sqrt{2 \cdot 1}=\sqrt{2} .
$$

Therefore,

$$
S_{\mathrm{ABCD}}=\frac{1}{2}(A D+B C) M N=\frac{1}{2}(1+4)(2 r+2 R)=\frac{5}{2} \cdot 3 \sqrt{2}=\frac{15 \sqrt{2}}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-59.jpg?height=548&width=395&top_left_y=2127&top_left_x=840)

## Answer

The project is supported by the Department of Education of Moscow and the Federal Targeted Program "Personnel".