In rectangle $A B C D$, where $A B=6, A D=3\left(1+\frac{\sqrt{2}}{2}\right)$, two circles are placed. A circle with radius 2

has its center at point $K$ and touches sides $A B$ and $A D$. A circle with radius 1 has its center at point $L$ and touches side $C D$ and the first circle. Find the area of triangle $C L M$, where $M$ is the foot of the perpendicular dropped from vertex $B$ to the line passing through points $K$ and $L$.
Let $N$ be the point of tangency of the first circle with side $AB$, and $G$ be the projection of point $L$ onto line $NK$. Then $\angle LMBN = \angle LKG = 45^{\circ}$.

## Solution

Let $N$ and $H$ be the points of tangency of the first circle with sides $AB$ and $AD$ respectively, $F$ be the point of tangency of the second circle with side $CD$, $E$ be the projection of point $K$ onto $CD$, $G$ be the projection of point $L$ onto $KE$, and $P$ be the intersection of lines $KL$ and $CD$.

Consider the right triangle $KGL$. In it,

$$
\begin{gathered}
K L=2+1=3, \quad K G=N E-N K-G E=N E-N K-L F= \\
=\left(1+\frac{\sqrt{2}}{2}\right)-2-1=\frac{3 \sqrt{2}}{2} \\
\cos \angle G K L=\frac{K G}{K L}=\frac{\sqrt{2}}{2}, \quad \angle G K L=45^{\circ}
\end{gathered}
$$

Since $\angle HAK = 45^{\circ}$, points $A$, $K$, and $L$ lie on the same line. From the isosceles right triangle $ABM$, we find that

$$
A M=A B \sin 45^{\circ}=3 \sqrt{2}
$$

Therefore,

$$
L M=A K+K L-A M=2 \sqrt{2}+3-3 \sqrt{2}=3-\sqrt{2} .
$$

Since $\angle LFP = \angle KGL = 45^{\circ}$, then $FP = LF = 1$, so

$$
C P=C D-D E-E F-F P=6-2-\frac{3 \sqrt{2}}{2}-1=3-\frac{3 \sqrt{2}}{2}
$$

Let $CQ$ be the height of triangle $CLM$. Then

$$
C Q=C P \sin \angle C P Q=C P \sin 45^{\circ}=\left(3-\frac{3 \sqrt{2}}{2}\right) \cdot \frac{\sqrt{2}}{2}=3(\sqrt{2}-1)
$$

Therefore,

$$
S_{\triangle \mathrm{CLM}}=\frac{1}{2} L M \cdot C Q=\frac{1}{2}(3-\sqrt{2})(3 \sqrt{2}-1)=\frac{3(4 \sqrt{2}-5)}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-16.jpg?height=535&width=598&top_left_y=693&top_left_x=730)

## Answer

$\frac{3(4 \sqrt{2}-5)}{4}$.