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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 22"

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Considering the coordinates of A and <math>B</math> for the two equations respectively, we get:
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Considering the coordinates of <math>A</math> and <math>B</math> for the two equations respectively, we get:
  
 
<math>(8-a)^2+(0-6)^2=a^2</math>
 
<math>(8-a)^2+(0-6)^2=a^2</math>

Revision as of 02:41, 24 November 2021

Problem

Right triangle $ABC$ has side lengths $BC=6$, $AC=8$, and $AB=10$.

A circle centered at $O$ is tangent to line $BC$ at $B$ and passes through $A$. A circle centered at $P$ is tangent to line $AC$ at $A$ and passes through $B$. What is $OP$?

$\textbf{(A)}\ \frac{23}{8} \qquad\textbf{(B)}\ \frac{29}{10} \qquad\textbf{(C)}\ \frac{35}{12} \qquad\textbf{(D)}\ \frac{73}{25} \qquad\textbf{(E)}\ 3$

Solution 1 (Analytic Geometry)

In a Cartesian plane, let $C, B,$ and $A$ be $(0,0),(0,6),(8,0)$ respectively.

By analyzing the behaviors of the two circles, we set $O$ be $(a,6)$ and $P$ be $(8,b)$.

Hence derive the two equations:

$(x-a)^2+(y-6)^2=a^2$

$(x-8)^2+(y-b)^2=b^2$


Considering the coordinates of $A$ and $B$ for the two equations respectively, we get:

$(8-a)^2+(0-6)^2=a^2$

$(0-8)^2+(6-b)^2=b^2$

Solve to get $a=\frac{25}{4}$ and $b=\frac{25}{3}$


Through using the distance formula,

$OP=\sqrt{(8-\frac{25}{4})^2+(\frac{25}{3}-6)^2}=\frac{35}{12} \Rightarrow \boxed{\textbf{(C)}}$.

~Wilhelm Z

2021 Fall AMC 12B (Problems • Answer Key • Resources)
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Problem 21 		Followed by
Problem 23
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All AMC 12 Problems and Solutions

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