In the diagram, the two smaller circles have equal radii. Each of the three circles is tangent to the other two circles, and each is also tangent to sides of the rectangle. If the width of the rectangle is 4 , then its length is
(A) $2+\sqrt{8}$
(B) $3+\sqrt{8}$
(C) $3+\sqrt{10}$
(D) $\sqrt{32}$
(E) $4+\sqrt{3}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-135.jpg?height=250&width=391&top_left_y=344&top_left_x=1344)
In the diagram, the two smaller circles have equal radii. Each of the three circles is tangent to the other two circles, and each is also tangent to sides of the rectangle. If the width of the rectangle is 4 , then its length is
(A) $2+\sqrt{8}$
(D) $\sqrt{32}$
(B) $3+\sqrt{8}$
(E) $4+\sqrt{3}$
(C) $3+\sqrt{10}$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-264.jpg?height=279&width=415&top_left_y=1519&top_left_x=1408)

## Solution

Let the radius of the larger circle be $R$.

Let the radii of the smaller circle be $r$.

From the diagram, $2 R=4$ so $R=2$ and $4 r=4$ so $r=1$, as the radii are perpendicular to the sides of the rectangle since the circles are tangent to the sides of the rectangle.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-264.jpg?height=263&width=412&top_left_y=1945&top_left_x=1407)

Join $C_{1}$ to $P$ and extend until it hits both vertical sides of the rectangle.

Therefore the length of the rectangle is $R+C_{1} P+r=3+C_{1} P$.

Now $C_{2} P \perp C_{1} P$ and $C_{2} C_{1}$ goes through the point of tangency between the larger circle and the upper circle, so $C_{1} C_{2}=R+r=3$.

By Pythagoras $3^{2}=\left(C_{1} P\right)^{2}+1^{2} \Rightarrow\left(C_{1} P\right)^{2}=8 \Rightarrow C_{1} P=\sqrt{8}$.

Thus the length is $3+\sqrt{8}$.

ANSWER: (B)