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))
(→‎Solution 4)
 
(17 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
Right triangle <math>ABC</math> has side lengths <math>BC=6</math>, <math>AC=8</math>, and <math>AB=10</math>.  
+
Right triangle <math>ABC</math> has side lengths <math>BC=6</math>, <math>AC=8</math>, and <math>AB=10</math>. A circle centered at <math>O</math> is tangent to line <math>BC</math> at <math>B</math> and passes through <math>A</math>. A circle centered at <math>P</math> is tangent to line <math>AC</math> at <math>A</math> and passes through <math>B</math>. What is <math>OP</math>?
−
 
 
−
A circle centered at <math>O</math> is tangent to line <math>BC</math> at <math>B</math> and passes through <math>A</math>. A circle centered at <math>P</math> is tangent to line <math>AC</math> at <math>A</math> and passes through <math>B</math>. What is <math>OP</math>?
 
  
 
<math>\textbf{(A)}\ \frac{23}{8} \qquad\textbf{(B)}\  \frac{29}{10} \qquad\textbf{(C)}\  \frac{35}{12} \qquad\textbf{(D)}\
 
<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>
 
\frac{73}{25} \qquad\textbf{(E)}\ 3</math>
  
−
==Solution 1 (Analytic Geometry) ==
+
==Diagram==
 +
<center><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></center>
 +
 
 +
==Solution 1==
 +
 
 +
Let <math>M</math> be the midpoint of <math>AB</math>; so <math>BM=AM=5</math>. Let <math>D</math> be the point such that <math>ABCD</math> is a rectangle. Then <math>MO\perp AB</math> and <math>MP\perp AB</math>. Let <math>\theta = \angle BAC</math>; so <math>\tan\theta = \tfrac 68 = \tfrac 34</math>. Then
 +
<cmath>OP=MP-MO=AM\cot\theta - BM\tan\theta = 5(\tfrac 43 - \tfrac 34) = \boxed{\textbf{(C)}\ \tfrac{35}{12}}.</cmath>
 +
 
 +
== Solution 2==
 +
 
 +
This one uses the same diagram as Solution 1, except we draw <math>BP</math>. 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>.
 +
 
 +
-ThisUsernameIsTaken
 +
 
 +
==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.
Line 36: Line 54:
  
 
~Wilhelm Z
 
~Wilhelm Z
 +
 +
== Solution 4 ==
 +
 +
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>PA \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)
 +
 +
==Solution 5==
 +
 +
Let <math>C</math> be the origin, making <math>B=(0,6)</math> and <math>A=(8,0)</math>. Let <math>D</math> be the midpoint of <math>AB</math>; <math>D=(4,3)</math>.
 +
 +
Notice that both <math>O</math> and <math>P</math> must be on the perpendicular bisector <math>l</math> of <math>AB</math>. The slope of <math>AB</math> is <math>-\dfrac{3}{4}</math>, making the <math>l</math>'s slope be <math>\dfrac{4}{3}</math>. Since <math>l</math> passes through <math>D</math>, the equation for <math>l</math> becomes
 +
 +
<cmath>y-3=\dfrac{4}{3} (x-4),</cmath>
 +
 +
using the slope intersect form. Since <math>OB</math> is perpendicular to <math>AC</math> and <math>AP</math> is perpendicular to <math>AC</math> (cause of tangencies), the <math>y</math>-coordinate for <math>O</math> is <math>6</math> and the <math>x</math>-coordinate for <math>P</math> is <math>8</math>. Plugging these numbers in the equation for <math>l</math> gives <math>O=\left( \dfrac{25}{4}, 6 \right)</math> and <math>P=\left( 8, \dfrac{25}{3} \right)</math>. Thus,
 +
 +
<cmath>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)}}</cmath>
 +
 +
~ sml1809
 +
 +
==Video Solution==
 +
https://youtu.be/0TeJ-9XUkAA
 +
 +
~MathProblemSolvingSkills.com
 +
 +
 +
 +
==Video Solution by Mathematical Dexterity==
 +
https://www.youtube.com/watch?v=ctx67nltpE0
  
 
{{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}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:41, 5 August 2023

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 $OM = \frac{15}{4}$. $OP = PM - OM = \frac{35}{12}$.

-ThisUsernameIsTaken

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

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
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=196481"