11.4. In a right-angled trapezoid $A B C D$, with right angles at vertices $A$ and $B$, there are two circles. One of them is tangent to the lateral sides and to the larger base $A D$, while the other is tangent to the lateral sides, to the smaller base $B C$, and to the first circle.

1) The line passing through the centers of the circles intersects $A D$ at point $P$. Prove that $\frac{A P}{P D}=\sin (\angle D)$.
2) Determine the area of the trapezoid, if the radii of the circles are $\frac{4}{3}$ and $\frac{1}{3}$.
## Solution.

1) Since the circles are tangent to each other, we construct their common tangent $L Z$. Let: $m(\angle D)=\alpha, R$ - the radius of the larger circle with center at $O_{1}, r$ - the radius of the smaller circle with center at $O_{2}, M$ - the foot of the perpendicular from $O_{1}$ to $A D, N$ - the foot of the perpendicular from $O_{1}$ to $A B, K$ - the foot of the perpendicular from $O_{1}$ to $C D, E$ - the foot of the perpendicular from $O_{2}$ to $A B, F$ - the foot of the perpendicular from $O_{2}$ to $C D, T$ - the foot of the perpendicular from $O_{2}$ to $B C, S$ - the intersection point of the circles, $G=\left(O_{1} O_{2}\right) \cap(B C), P=\left(O_{1} O_{2}\right) \cap(A D)$ and $m\left(\angle M O_{1} P\right)=\varphi$.

![](https://cdn.mathpix.com/cropped/2024_06_06_3aea3bae54655b1bda95g-3.jpg?height=911&width=688&top_left_y=138&top_left_x=44)

According to the theorem of equal segments determined by tangents drawn from an external point to a circle, we have

$$
\begin{gathered}
A M=A N=R, B E=B T=r, L E=L S=L N \\
Z F=Z S=Z K, D M=D K, C T=C F
\end{gathered}
$$

From the equality of angles determined by two parallel lines and a transversal, the theorem of vertical angles, and the perpendicularity relations $O_{1} M \perp A D$ and $O_{2} T \perp B C$, we obtain

$$
\begin{gathered}
m\left(\angle G O_{2} T\right)=\varphi, m\left(\angle N O_{1} S\right)=90^{\circ}-\varphi, m(\angle N L S)=90^{\circ}+\varphi \\
m\left(\angle E O_{2} G\right)=90^{\circ}-\varphi, m\left(\angle E O_{2} S\right)=90^{\circ}+\varphi, m(\angle E L S)=90^{\circ}-\varphi \\
m\left(\angle M O_{1} K\right)=180^{\circ}-\alpha, m\left(\angle T O_{2} F\right)=\alpha, m(\angle B C D)=180^{\circ}-\alpha \\
m\left(\angle S O_{1} K\right)=\alpha+\varphi, m(\angle S Z K)=180^{\circ}-(\alpha+\varphi) \\
m(\angle S Z F)=\alpha+\varphi, m\left(\angle S O_{2} F\right)=180^{\circ}-(\alpha+\varphi)
\end{gathered}
$$

Then $M P=R \operatorname{tg} \varphi$ and $A P=A M+M P=R(1+\operatorname{tg} \varphi)$ (1).

According to the cosine theorem for $\mathrm{VNSO}_{1}$ and VNSL, we have $2 R^{2}(1-\sin \varphi)=N S^{2}=2 L N^{2}(1+\sin \varphi)$, i.e., $L N=R \sqrt{\frac{1-\sin \varphi}{1+\sin \varphi}}$ (2). Similarly, from $\mathrm{VEO}_{2} S$ and VELS we have $2 r^{2}(1+\sin \varphi)=E S^{2}=2 L E^{2}(1-\sin \varphi)$, i.e., $L E=r \sqrt{\frac{1+\sin \varphi}{1-\sin \varphi}}$ (3). Since $L N=L E$, from (2) and (3) we obtain $r=R \cdot \frac{1-\sin \varphi}{1+\sin \varphi}$ (4). Performing similar reasoning for the pairs of triangles ( $\mathrm{VSO}_{1} \mathrm{~K}, \mathrm{VSZK}$ ) and ( $\mathrm{VSO}_{2} F, \mathrm{VSZF}$ ), we obtain $r=R \cdot \frac{1-\cos (\alpha+\varphi)}{1+\cos (\alpha+\varphi)}$

From (4) and (5) we have $\frac{1-\cos (\alpha+\varphi)}{1+\cos (\alpha+\varphi)}=\frac{1-\sin \varphi}{1+\sin \varphi}$, which implies $\sin \varphi=\cos (\alpha+\varphi)$, i.e., $\cos \left(90^{\circ}-\varphi\right)=\cos (\alpha+\varphi)$. Since $0^{\circ}<\varphi<90^{\circ}$ and $0^{\circ}<\alpha+\varphi<180^{\circ}$, then $90^{\circ}-\varphi=\alpha+\varphi$, i.e., $\alpha=90^{\circ}-2 \varphi$ (6).

Since $M D=K D$, then in a similar manner for $\mathrm{VO}_{1} K M$ and VDKM we obtain $M D=R \sqrt{\frac{1+\cos \alpha}{1-\cos \alpha}}$. We have $P D=M D-M P=R \sqrt{\frac{1+\cos \alpha}{1-\cos \alpha}}-R \operatorname{tg} \varphi$ (7). From relations (1), (6), and (7) we obtain

$$
\frac{A P}{P D}=\frac{R(1+\operatorname{tg} \varphi)}{R\left(\sqrt{\frac{1+\cos \alpha}{1-\cos \alpha}}-\operatorname{tg} \varphi\right)}=\frac{1+\operatorname{tg} \varphi}{\sqrt{\frac{1+\cos \alpha}{1-\cos \alpha}}-\operatorname{tg} \varphi}=\frac{1+\operatorname{tg} \varphi}{\sqrt{\frac{1+\sin 2 \varphi}{1-\sin 2 \varphi}}-\operatorname{tg} \varphi}=\frac{1+\operatorname{tg} \varphi}{\frac{\cos \varphi+\sin \varphi}{\cos \varphi-\sin \varphi}-\operatorname{tg} \varphi}=\cos 2 \varphi=\sin \alpha
$$

2) If $R=\frac{4}{3}$ and $r=\frac{1}{3}$, then from (4) and (6) we obtain $\frac{1}{3}=\frac{4}{3} \cdot \frac{1-\sin \varphi}{1+\sin \varphi}$, which implies $\sin \varphi=\frac{3}{5}$ and $\cos \varphi=\frac{4}{5}$, i.e., $\sin \alpha=\cos 2 \varphi=\frac{7}{25}$ and $\cos \alpha=\frac{24}{25}$

(8). Then, from (8) we have $A D=A M+M D=R+R \sqrt{\frac{1+\cos \alpha}{1-\cos \alpha}}=\frac{32}{3}$

(9). From (2) and

(3) we obtain $A B=A N+N L+L E+E B=R+r+2 R \sqrt{\frac{1-\sin \varphi}{1+\sin \varphi}}=3$ (10). Finally, $B C=B T+T C=r+T C$. Similarly with

(3), from $\mathrm{VTO}_{2} F$ and VTCF we have $T C=r \sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}}$, which implies $B C=r+r \sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}}=\frac{8}{21}$ (11), which implies $A_{A B C D}=\frac{A D+B C}{2} \cdot A B=\frac{1}{2}\left(\frac{32}{3}+\frac{8}{21}\right) \cdot 3=\frac{116}{7}$.

## Grading Rubric