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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 $m+n ?$

,

• $21$
• $29$
• $58$
• $69$
• $93$

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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