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

Try this beautiful Problem on geometry based on circle from AMC 10A, 2018. Problem-15. You may use sequential hints to solve the problem.

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 $m+n ?$

,

• $21$
• $29$
• $58$
• $69$
• $93$

Geometry

Triangle

Pythagoras

## Suggested Book | Source | Answer

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 $m+n=65+4=69$

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