Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1 (Analytic Geometry))
Line 36: Line 36:
  
 
~Wilhelm Z
 
~Wilhelm Z
 +
 +
== Solution 2 ==
 +
This solution is based on this figure: [[:Image:2021_AMC_12B_(Nov)_Problem_22,_sol.png]].
 +
 +
Because the circle with center <math>O</math> passes through points <math>A</math> and <math>B</math> and is tangent to line <math>BC</math> at point <math>B</math>, <math>O</math> is on the perpendicular bisector of segment <math>AB</math> and <math>OB \perp BC</math>.
 +
 +
Because the circle with center <math>P</math> passes through points <math>A</math> and <math>B</math> and is tangent to line <math>AC</math> at point <math>A</math>, <math>P</math> is on the perpendicular bisector of segment <math>AB</math> and <math>OA \perp AC</math>.
 +
 +
Let lines <math>OB</math> and <math>AP</math> intersect at point <math>D</math>.
 +
Hence, <math>ACBD</math> is a rectangle.
 +
 +
Denote by <math>M</math> the midpoint of segment <math>AB</math>. Hence, <math>BM = \frac{AB}{2} = 5</math>.
 +
Because <math>O</math> and <math>P</math> are on the perpendicular bisector of segment <math>AB</math>, points <math>M</math>, <math>O</math>, <math>P</math> are collinear with <math>OM \perp AB</math>.
 +
 +
We have <math>\triangle MOB \sim \triangle CBA</math>.
 +
Hence, <math>\frac{BO}{AB} = \frac{BM}{AC}</math>.
 +
Hence, <math>BO = \frac{25}{4}</math>.
 +
Hence, <math>OD = BD - BO = \frac{7}{4}</math>.
 +
 +
We have <math>\triangle DOP \sim \triangle CBA</math>.
 +
Hence, <math>\frac{OP}{BA} = \frac{DO}{CB}</math>.
 +
Therefore, <math>OP = \frac{35}{12}</math>.
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(C) }\frac{35}{12}}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)
  
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=23|num-b=21}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=23|num-b=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:21, 25 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 $O$ 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 2

This solution is based on this figure: Image:2021_AMC_12B_(Nov)_Problem_22,_sol.png.

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 $OA \perp AC$.

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)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_22&oldid=166490"