# External Tangent | AMC 10A, 2018 | Problem 15

Try this beautiful Problem on Geometry based on External Tangent from AMC 10 A, 2018. You may use sequential hints to solve the problem.

## External Tangent - AMC-10A, 2018- Problem 15

Two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points $A$ and $B$, as shown in the diagram. The distance $A B$ can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n ?$

,

- $21$
- $29$
- $58$
- $69$
- $93$

**Key Concepts**

Geometry

Triangle

Pythagoras

## Suggested Book | Source | Answer

#### Suggested Reading

Pre College Mathematics

#### Source of the problem

AMC-10A, 2018 Problem-15

#### Check the answer here, but try the problem first

$69$

## Try with Hints

#### First Hint

Given that two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points $A$ and $B$. we have to find out the length \(AB\).

Now join \(A\) & \(B\) and the points \(Y\) & \(Z\). If we can show that \(\triangle XYZ \sim \triangle XAB\) then we can find out the length of \(AB\).

Now can you finish the problem?

#### Second Hint

now the length of \(YZ=5+5=10\) (as the length of the radius of smaller circle is $5$) and \(XY=XA-AY=13-5=8\). Now \(YZ|| AB\).therefore we can say that \(\triangle XYZ \sim \triangle XAB\). therefore we can write $\frac{X Y}{X A}=\frac{Y Z}{A B}$

Now Can you finish the Problem?

#### Third Hint

From the relation we can say that $\frac{X Y}{X A}=\frac{Y Z}{A B}$

\(\Rightarrow \frac{8}{13}=\frac{10}{AB}\)

\(\Rightarrow AB=\frac{13\times 10}{8}\)

\(\Rightarrow AB=\frac{65}{4}\) which is equal to \(\frac{m}{n}\)

Therefore \(m+n=65+4=69\)

## Watch the video:

## Other useful links

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=OvduZbqenWU