In the diagram, two circles and a square lie between a pair of parallel lines that are a distance of 400 apart. The square has a side length of 279 and one of its sides lies along the lower line. The circles are tangent to each other, and each circle is tangent to one of the lines. Each circle also touches the square at only one point - the lower circle touches a side of the square and the upper

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-053.jpg?height=293&width=393&top_left_y=252&top_left_x=1294)
circle touches a vertex of the square. If the upper circle has a radius of 65 , then the radius of the lower circle is closest to
(A) 151
(B) 152
(C) 153
(D) 154
(E) 155
Let $r$ be the radius of the lower circle.

We label the square as $A B C D$, the centre of the upper circle as $U$, and the centre of the lower circle as $L$. We call $E$ the point at which the upper circle touches the top line, $G$ the point at which the lower circle touches the bottom line, $F$ the point at which the two circles touch, and $H$ the point at which the lower circle touches the square.

We join $E U, U F, F L, L G, L H$, and $U A$.

We need to use two facts about circles:

- When the centre of a circle is joined to a point of tangency, the resulting line segment is perpendicular to the tangent line. Thus, $U E$ is perpendicular to the top line, $L G$ is perpendicular to the bottom line, and $L H$ is perpendicular to $A D$.
- When two circles are tangent, the line segment joining their centres passes through the point of tangency between the circles. Thus, $U F L$ is a straight line segment.

Extend $E U$ downwards to meet $L H$ at $J$. Since $E U$ and $A D$ are perpendicular to the parallel lines and $L H$ is perpendicular to $A D$, then $E J$ is perpendicular to $L H$.

Extend $B A$ to meet $E J$ at $K$. Using a similar argument, $A K$ is perpendicular to $E J$.

We focus on $\triangle U J L$, which is right-angled at $J$.

Since the radius of the upper circle is 65 , then we know that $E U=U A=U F=65$.

Since the radius of the lower circle is $r$, then we know that

$F L=L H=L G=r$.

Thus, $U L=U F+F L=65+r=r+65$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-105.jpg?height=482&width=583&top_left_y=873&top_left_x=1451)

Since the top and bottom lines are parallel, $E J$ and $L G$ are perpendicular to these lines, and $L J$ is parallel to these lines, then $E U+U J+L G$ equals the distance between the lines.

Thus, $65+U J+r=400$ and so $U J=335-r$.

Next, we note that $L J=L H-J H=r-J H=r-A K$, since $A K J H$ is a rectangle.

Now $\triangle U K A$ is right-angled and has $U A=65$.

Also, $U K=E K-E U=E K-65$.

But $E K$ equals the difference between the distance between the lines and the side length of the square, or $400-279=121$.

Thus, $U K=121-65=56$.

Using the Pythagorean Theorem in $\triangle U K A$, we obtain $A K^{2}=U A^{2}-U K^{2}=65^{2}-56^{2}$ and so $A K^{2}=1089=33^{2}$.

Since $A K>0$, then $A K=33$.

Thus, $L J=r-33$.

Finally, using the Pythagorean Theorem in $\triangle U J L$, we obtain

$$
\begin{aligned}
U J^{2}+L J^{2} & =U L^{2} \\
(335-r)^{2}+(r-33)^{2} & =(r+65)^{2} \\
r^{2}-670 r+335^{2}+r^{2}-66 r+33^{2} & =r^{2}+130 r+65^{2} \\
r^{2}-866 r+109089 & =0
\end{aligned}
$$

By the quadratic formula, $r=\frac{866 \pm \sqrt{866^{2}-4(1)(109089)}}{2}=\frac{866 \pm 560}{2}=153$ or 713 .

Since $r$ must be less than the distance between the two lines, which is 400, then $r=153$. Of the given answers, this is closest to 153 .