\section*{Problem 2 - 201042}

Given a length \(r_{1}\).

Construct a trapezoid \(A B C D\) with \(C D < A D = A B = B C\), in which a circle \(k_{1}\) with radius \(r_{1}\) and a second circle \(k_{2}\) are positioned such that they touch each other externally, and \(k_{1}\) touches the sides \(A D, A B\), and \(B C\), while \(k_{2}\) touches the sides \(B C, C D\), and \(A D\)!

Explain and describe the construction!

Investigate whether there is exactly one trapezoid with the required properties up to congruence!
I. Suppose a trapezoid \(A B C D\) with circles \(k_{i}\) (midpoints \(M_{i}\), radii \(r_{i}, i=1,2\)) has the required properties.

If \(A D \| B C\), then \(A B C D\) is a parallelogram due to \(A D = B C\), which contradicts \(A B > C D\). Therefore, \(A B \| C D\) and the trapezoid is isosceles because \(A D = B C\).

Since \(k_{1}\) and \(k_{2}\) touch the legs \(A D\) and \(B C\), whose extensions beyond \(D\) and \(C\) intersect at \(S\), \(M_{1}\) and \(M_{2}\) lie on the angle bisector \(m\) of \(\angle A S B\), which is also the perpendicular bisector of \(A B\) and \(C D\).

Furthermore, \(Q\) (the point of tangency of \(k_{1}\) and \(k_{2}\)) also lies on \(m\). The line parallel to \(A B\) through \(Q\) is thus the common inner tangent of \(k_{1}\) and \(k_{2}\), intersecting \(A D\) at \(H\) and \(B C\) at \(G\).

The trapezoid \(H G C D\), whose sides are touched by \(k_{2}\), can be transformed into the trapezoid \(A B G H\) by a dilation centered at \(S\), whose sides are touched by \(k_{1}\).

Let \(x = r_{2} : r_{1}\), \(a = A B\), \(b = H G\), and \(c = D C\). It follows that \(x = D C : H G = H G : A B\), so \(b = a x\) and \(c = b x = a x^{2}\). Let \(U\) and \(V\) be the points of tangency of \(k_{1}\) and \(k_{2}\) with \(B C\).

We have

\[
a = A B = B C = B U + U G + G V + V C = B P + Q G + Q G + C R =
\]

by the equality of tangent segments

\[
= \frac{a}{2} + b + \frac{c}{2} = \frac{a}{2} \left(1 + 2 x + x^{2}\right)
\]

This equation is only satisfied by \(x = \sqrt{2} - 1\) for \(x > 0\). Therefore, the trapezoid \(A B C D\) can only meet the conditions if

\[
r_{2} = r_{1} (\sqrt{2} - 1)
\]

II. Construction Description

(1) From \(r_{1}\), construct \(r_{2} = r_{1} (\sqrt{2} - 1)\).

(2) Construct the circles \(k_{i}\) (centers \(M_{i}\), radii \(r_{i}, i=1,2\)) that touch each other externally at \(Q\).

(3) Construct the two external tangents \(t\) and \(t'\) of \(k_{1}\) and \(k_{2}\).

(4) The line through \(M_{1}\) and \(M_{2}\) intersects the circles \(k_{1}\) and \(k_{2}\) again at \(R\) and \(P\). Construct the perpendiculars to this line at \(R\) and \(P\). These intersect \(t\) and \(t'\) at \(A, D\) and \(B, C\).

\(A B C D\) is the desired trapezoid.

III. Proof that every trapezoid constructed in this way meets the conditions of the problem:

By construction, \(A B C D\) is an isosceles trapezoid with \(A D = B C\). The circles \(k_{1}\) and \(k_{2}\) have the prescribed tangencies with the sides. Since \(r_{2} < r_{1}\), \(C D < A B\). As derived in I., with \(x = r_{2} : r_{1}\), the equation \(B C = \frac{a}{2} (1 + 2 x + x^{2})\) holds.

Since \(r_{2} = r_{1} (\sqrt{2} - 1)\) by construction step (1), \(x\) satisfies the equation \(x^{2} + 2 x + 1 = 2\), and it follows that \(B C = a = A B\).

IV. Construction step (1) is uniquely executable. Step (2) is uniquely executable up to congruence, and steps (3) and (4) are also uniquely executable subsequently. Therefore, there is exactly one trapezoid with the required properties up to congruence.