# 2021 Fall AMC 12B Problems/Problem 22

## 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 $OP$?

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

## Diagram

$[asy] defaultpen(fontsize(10)+0.8); size(150); pair A,B,C,M,Ic,Ib,O,P; C=MP("C",origin,down+left); A=MP("A",8*right,down+right); B=MP("B",6*up,2*up); draw(A--B--C--A); draw(B--(B+A), gray+0.25); M=MP("M",(A+B)/2,down+left); O=MP("O",extension(B,B+A,M,M+(B-M)*dir(-90)),down); P=MP("P",extension(A,B+A,M,M+(B-M)*dir(-90)),up); draw(M--P^^A--P, gray+0.25); label("\theta", A, 7*dir(162)); label("\theta", B, 7*dir(-20)); label("\theta", P, 7*dir(-110)); label("6", B--C, left); label("8", A--C, down); label("D", A+B, right); [/asy]$

## Solution 1

Let $M$ be the midpoint of $AB$; so $BM=AM=5$. Let $D$ be the point such that $ABCD$ is a rectangle. Then $MO\perp AB$ and $MP\perp AB$. Let $\theta = \angle BAC$; so $\tan\theta = \tfrac 68 = \tfrac 34$. Then $$OP=MP-MO=AM\cot\theta - BM\tan\theta = 5(\tfrac 43 - \tfrac 34) = \boxed{\textbf{(C)}\ \tfrac{35}{12}}.$$

## Solution 2

This one uses the same diagram as Solution 1, except we draw $BP$. After doing angle chasing we find $\triangle BPM \sim \triangle BAC$ and $\frac{BM}{BC} = \frac{PM}{AC}$, resulting in $PM = \frac{20}{3}$.

We also find that $\triangle BOM \sim \triangle ABC$ and $\frac{BM}{AC} = \frac{OM}{BC}$, resulting in $OM = \frac{15}{4}$. $OP = PM - OM = \frac{35}{12}$.

## Solution 3 (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 $O$ to 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,

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

~Wilhelm Z

## Solution 4

Because the circle with center $O$ passes through points $A$ and $B$ and is tangent to line $BC$ at point $B$, $O$ is on the perpendicular bisector of segment $AB$ and $OB \perp BC$.

Because the circle with center $P$ passes through points $A$ and $B$ and is tangent to line $AC$ at point $A$, $P$ is on the perpendicular bisector of segment $AB$ and $PA \perp AC$.

Let lines $OB$ and $AP$ intersect at point $D$. Hence, $ACBD$ is a rectangle.

Denote by $M$ the midpoint of segment $AB$. Hence, $BM = \frac{AB}{2} = 5$. Because $O$ and $P$ are on the perpendicular bisector of segment $AB$, points $M$, $O$, $P$ are collinear with $OM \perp AB$.

We have $\triangle MOB \sim \triangle CBA$. Hence, $\frac{BO}{AB} = \frac{BM}{AC}$. Hence, $BO = \frac{25}{4}$. Hence, $OD = BD - BO = \frac{7}{4}$.

We have $\triangle DOP \sim \triangle CBA$. Hence, $\frac{OP}{BA} = \frac{DO}{CB}$. Therefore, $OP = \frac{35}{12}$.

Therefore, the answer is $\boxed{\textbf{(C) }\frac{35}{12}}$.

~Steven Chen (www.professorchenedu.com)

## Solution 5

Let $C$ be the origin, making $B=(0,6)$ and $A=(8,0)$. Let $D$ be the midpoint of $AB$; $D=(4,3)$.

Notice that both $O$ and $P$ must be on the perpendicular bisector $l$ of $AB$. The slope of $AB$ is $-\dfrac{3}{4}$, making the $l$'s slope be $\dfrac{4}{3}$. Since $l$ passes through $D$, the equation for $l$ becomes

$$y-3=\dfrac{4}{3} (x-4),$$

using the slope intersect form. Since $OB$ is perpendicular to $AC$ and $AP$ is perpendicular to $AC$ (cause of tangencies), the $y$-coordinate for $O$ is $6$ and the $x$-coordinate for $P$ is $8$. Plugging these numbers in the equation for $l$ gives $O=\left( \dfrac{25}{4}, 6 \right)$ and $P=\left( 8, \dfrac{25}{3} \right)$. Thus,

$$OP=\sqrt{\left(\dfrac{7}{4}\right)^2 + \left(\dfrac{7}{3}\right)^2} = 7\sqrt{\dfrac{3^2+4^2}{(3^2)(4^2)}} = 7\cdot \dfrac{5}{12} = \boxed{\dfrac{35}{12} \textbf{ (C)}}$$

~ sml1809

## Video Solution

~MathProblemSolvingSkills.com

## Video Solution by Mathematical Dexterity

 2021 Fall AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 23 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions