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

Revision as of 17:17, 6 June 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

$[asy] defaultpen(fontsize(11)+0.8); size(200); pair A,B,C,M,Ic,Ib,O,P; C=MP("C",origin,down+left); A=MP("A",8*right,down+right); B=MP("B",6*up,2*up); draw(A--B--C--A); draw(B--(B+A), gray+0.25); M=MP("M",(A+B)/2,down+left); O=MP("O",extension(B,B+A,M,M+(B-M)*dir(-90)),down); P=MP("P",extension(A,B+A,M,M+(B-M)*dir(-90)),up); draw(M--P^^A--P, gray+0.25); label("$\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); [/asy]$

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 $PA \perp AC$ .

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 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
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.

@@ Line 7: / Line 7: @@
 <math>\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</math>
+==Diagram==
+<center><asy>
+defaultpen(fontsize(11)+0.8); size(200);
+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);
+</asy></center>
 ==Solution 1 (Analytic Geometry) ==

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