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

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<cmath>OP=MP-MO=AM\cot\theta - BM\tan\theta = 5(\tfrac 43 - \tfrac 34) = \boxed{\textbf{(C)}\ \tfrac{35}{12}}.</cmath>
 
<cmath>OP=MP-MO=AM\cot\theta - BM\tan\theta = 5(\tfrac 43 - \tfrac 34) = \boxed{\textbf{(C)}\ \tfrac{35}{12}}.</cmath>
  
−
==Solution 2 (Analytic Geometry) ==
+
== Solution 2==
 +
 
 +
This one uses the same diagram as Solution 1. After doing angle chasing we find <math>\triangle BPM \sim \triangle BAC</math> and <math>\frac{BM}{BC} = \frac{PM}{AC}</math>, resulting in <math>PM = \frac{20}{3}</math>.
 +
 
 +
We also find that <math>\triangle BOM \sim \triangle ABC</math> and <math>\frac{BM}{AC} = \frac{OM}{BC}</math>, resulting in <math>OM = \frac{15}{4}</math>. <math>OP = PM - OM = \frac{35}{12}</math>.
 +
 
 +
==Solution 3 (Analytic Geometry) ==
  
 
In a Cartesian plane, let <math>C, B,</math> and <math>A</math> be <math>(0,0),(0,6),(8,0)</math> respectively.
 
In a Cartesian plane, let <math>C, B,</math> and <math>A</math> be <math>(0,0),(0,6),(8,0)</math> respectively.

Revision as of 09:17, 10 November 2022

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. 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 3

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)


Video Solution

https://youtu.be/0TeJ-9XUkAA

~MathProblemSolvingSkills.com


Video Solution by Mathematical Dexterity

https://www.youtube.com/watch?v=ctx67nltpE0

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

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