Difference between revisions of "2021 Fall AMC 12B Problems/Problem 22"
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==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. | ||
− | By analyzing the behaviors of the two circles, we set <math>O</math> be <math>(a,6)</math> and <math>P</math> be <math>(8,b)</math>. | + | By analyzing the behaviors of the two circles, we set <math>O</math> to be <math>(a,6)</math> and <math>P</math> be <math>(8,b)</math>. |
Hence derive the two equations: | Hence derive the two equations: | ||
Line 21: | Line 39: | ||
− | Considering the coordinates of A and <math>B</math> for the two equations respectively, we get: | + | Considering the coordinates of <math>A</math> and <math>B</math> for the two equations respectively, we get: |
<math>(8-a)^2+(0-6)^2=a^2</math> | <math>(8-a)^2+(0-6)^2=a^2</math> | ||
Line 32: | Line 50: | ||
Through using the distance formula, | Through using the distance formula, | ||
− | <math>OP=\sqrt{(8-\frac{25}{4})^2+(\frac{25}{3}-6)^2}= | + | <math>OP=\sqrt{(8-\frac{25}{4})^2+(\frac{25}{3}-6)^2}= \boxed{\textbf{(C)}\ \frac{35}{12}}</math>. |
+ | |||
~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 00:41, 6 August 2023
Contents
Problem
Right triangle has side lengths
,
, and
. A circle centered at
is tangent to line
at
and passes through
. A circle centered at
is tangent to line
at
and passes through
. What is
?
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]](http://latex.artofproblemsolving.com/4/f/c/4fce5013af1499cc86ddd3841b213fd1049e6b70.png)
Solution 1
Let be the midpoint of
; so
. Let
be the point such that
is a rectangle. Then
and
. Let
; so
. Then
Solution 2
This one uses the same diagram as Solution 1, except we draw . After doing angle chasing we find
and
, resulting in
.
We also find that and
, resulting in
.
.
-ThisUsernameIsTaken
Solution 3 (Analytic Geometry)
In a Cartesian plane, let and
be
respectively.
By analyzing the behaviors of the two circles, we set to be
and
be
.
Hence derive the two equations:
Considering the coordinates of and
for the two equations respectively, we get:
Solve to get and
Through using the distance formula,
.
~Wilhelm Z
Solution 4
Because the circle with center passes through points
and
and is tangent to line
at point
,
is on the perpendicular bisector of segment
and
.
Because the circle with center passes through points
and
and is tangent to line
at point
,
is on the perpendicular bisector of segment
and
.
Let lines and
intersect at point
.
Hence,
is a rectangle.
Denote by the midpoint of segment
. Hence,
.
Because
and
are on the perpendicular bisector of segment
, points
,
,
are collinear with
.
We have .
Hence,
.
Hence,
.
Hence,
.
We have .
Hence,
.
Therefore,
.
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 5
Let be the origin, making
and
. Let
be the midpoint of
;
.
Notice that both and
must be on the perpendicular bisector
of
. The slope of
is
, making the
's slope be
. Since
passes through
, the equation for
becomes
using the slope intersect form. Since is perpendicular to
and
is perpendicular to
(cause of tangencies), the
-coordinate for
is
and the
-coordinate for
is
. Plugging these numbers in the equation for
gives
and
. Thus,
~ sml1809
Video Solution
~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.