\section*{Problem 6 - 081226}

A trapezoid \(A B C D\), whose bases have lengths \(a\) and \(b(a>b)\) and whose two other (non-parallel) sides, when sufficiently extended, form an angle of size \(\alpha\), has an incircle.

Calculate the diameter \(d\) of this incircle from the quantities \(a, b\) and \(\alpha\)!
}

Let \(|A B|=a,|C D|=b, A B \| C D\) and \(S\) be the intersection of \(A D\) and \(B C\). Furthermore, let \(F_{T}\) be the area of the trapezoid, \(F_{S A B}\) the area of the triangle \(\triangle S A B\) and \(F_{S C D}\) the area of the triangle \(\triangle S C D\). Then it holds that \(F_{S A B}=F_{T}+F_{S C D}\).

By the intercept theorem, \(F_{S C D}=F_{S A B} \cdot\left(\frac{b}{a}\right)^{2}\), so \(F_{T}=F_{S A B} \cdot\left(1-\frac{b^{2}}{a^{2}}\right)=F_{S A B} \cdot \frac{a^{2}-b^{2}}{a^{2}}\).

It is \(F_{T}=d \cdot \frac{a+b}{2}\), since \(\frac{a+b}{2}\) is the length of the midline of the trapezoid and \(d\) is the distance between the parallel sides. Therefore,

\[
\frac{d}{2}=F_{S A B} \cdot \frac{a^{2}-b^{2}}{a^{2} \cdot(a+b)}=F_{S A B} \cdot \frac{a-b}{a^{2}}
\]

On the other hand, \(F_{S A B}=\frac{1}{2} \cdot(|A S|+|B S|+a) \cdot \frac{d}{2}\). This follows from the fact that the incircle of the trapezoid is also an incircle of the triangle \(\triangle S A B\). Drawing the connections from the incenter to the vertices of the triangle and the tangents to the three sides, the triangle \(\triangle S A B\) is divided into six right triangles, each of which has one leg as a radius of the incircle, and the other legs together form the perimeter of the triangle.

Substituting the previously obtained term for \(\frac{d}{2}\) and canceling \(F_{S A B}\), we get

\[
1=\frac{1}{2} \cdot(|A S|+|B S|+a) \cdot \frac{a-b}{a^{2}} \quad \text { or } \quad \frac{2 a^{2}}{a-b}=a+|A S|+|B S|
\]
thus

\[
|A S|+|B S|=\frac{2 a^{2}}{a-b}-a=\frac{a^{2}+a b}{a-b}=a \cdot \frac{a+b}{a-b}
\]

and

\[
(|A S|+|B S|)^{2}=|A S|^{2}+|B S|^{2}+2 \cdot|A S| \cdot|B S|=a^{2} \cdot \frac{(a+b)^{2}}{(a-b)^{2}}
\]

By the cosine rule in the triangle \(\triangle S A B\), \(a^{2}=|A S|^{2}+|B S|^{2}-2 \cdot|A S| \cdot|B S| \cdot \cos \alpha\). The difference of the last two identities yields

\[
2 \cdot|A S| \cdot|B S| \cdot(1+\cos \alpha)=a^{2} \cdot\left(\frac{(a+b)^{2}}{(a-b)^{2}}-1\right)=a^{2} \cdot \frac{4 a b}{(a-b)^{2}}
\]

thus

\[
F_{S A B}=\frac{1}{2} \cdot|A S| \cdot|B S| \cdot \sin \alpha=a^{2} \cdot \frac{a b}{(a-b)^{2}} \cdot \frac{\sin \alpha}{1+\cos \alpha}
\]

Therefore, we obtain

\[
d=2 F_{S A B} \cdot \frac{a-b}{a^{2}}=2 a^{2} \cdot \frac{a b}{(a-b)^{2}} \cdot \frac{\sin \alpha}{1+\cos \alpha} \cdot \frac{a-b}{a^{2}}=2 \frac{a b}{a-b} \cdot \frac{\sin \alpha}{1+\cos \alpha}
\]

Considering that

\[
\frac{\sin \alpha}{1+\cos \alpha}=\frac{\sin \alpha \cdot(1-\cos \alpha)}{(1+\cos \alpha) \cdot(1-\cos \alpha)}=\frac{\sin \alpha \cdot(1-\cos \alpha)}{1-\cos ^{2} \alpha}=\frac{\sin \alpha \cdot(1-\cos \alpha)}{\sin ^{2} \alpha}=\frac{1-\cos \alpha}{\sin \alpha}
\]

the result can also be written as

\[
d=2 \frac{a b}{a-b} \cdot \frac{1-\cos \alpha}{\sin \alpha}
\]