Difference between revisions of "2013 Mock AIME I Problems/Problem 1"

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== Solution ==
 
== Solution ==
  
Let <math>A_3</math> be the center of circle <math>C_3</math> and <math>Q</math> be the point of tangency between <math>C_3</math> and <math>C_2</math>. Note that triangles <math>PQA_2</math> and <math>A_3PA_2</math> are similar, so <math>\frac{A_3A_2}{PA_2}=\frac{PA_2}{A_2Q}=3</math> and <math>A_3A_2=r+1=9</math>. Thus the radius of <math>C_3</math> is <math>\boxed{8}</math>.
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Let <math>A_3</math> be the center of circle <math>C_3</math> and <math>Q</math> be the point of tangency between <math>C_3</math> and <math>C_2</math>. Note that triangles <math>PQA_2</math> and <math>A_3PA_2</math> are similar, so <math>\frac{A_3A_2}{PA_2}=\frac{PA_2}{A_2Q}=3</math> and <math>A_3A_2=r+1=9</math>. Thus the radius of <math>C_3</math> is <math>\boxed{008}</math>.

Revision as of 18:24, 29 July 2024

Problem 1

Two circles $C_1$ and $C_2$, each of unit radius, have centers $A_1$ and $A_2$ such that $A_1A_2=6$. Let $P$ be the midpoint of $A_1A_2$ and let $C_3$ be a circle externally tangent to both $C_1$ and $C_2$. $C_1$ and $C_3$ have a common tangent that passes through $P$. If this tangent is also a common tangent to $C_2$ and $C_1$, find the radius of circle $C_3$.

Solution

Let $A_3$ be the center of circle $C_3$ and $Q$ be the point of tangency between $C_3$ and $C_2$. Note that triangles $PQA_2$ and $A_3PA_2$ are similar, so $\frac{A_3A_2}{PA_2}=\frac{PA_2}{A_2Q}=3$ and $A_3A_2=r+1=9$. Thus the radius of $C_3$ is $\boxed{008}$.