What is the NO-SHORTCUT approach for learning great Mathematics?

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 $m+n=65+4=69$

Watch the video:

This site uses Akismet to reduce spam. Learn how your comment data is processed.