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 "2019 AMC 10B Problems/Problem 23"

m
 
(19 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2019 AMC 10B Problems#Problem 23|2019 AMC 10B #23]] and [[2019 AMC 12B Problems#Problem 20|2019 AMC 12B #20]]}}
 +
 
==Problem==
 
==Problem==
  
Line 9: Line 11:
 
First, observe that the two tangent lines are of identical length. Therefore, supposing that the point of intersection is <math>(x, 0)</math>, the Pythagorean Theorem gives <math>\sqrt{(x-6)^2 + 13^2} = \sqrt{(x-12)^2 + 11^2}</math>. This simplifies to <math>x = 5</math>.
 
First, observe that the two tangent lines are of identical length. Therefore, supposing that the point of intersection is <math>(x, 0)</math>, the Pythagorean Theorem gives <math>\sqrt{(x-6)^2 + 13^2} = \sqrt{(x-12)^2 + 11^2}</math>. This simplifies to <math>x = 5</math>.
  
−
Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) defined by the circle's center, <math>A</math>, <math>B</math>, and <math>(5, 0)</math> is cyclic. Therefore, we can apply Ptolemy's Theorem to give
+
Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) <math>AOBX</math> is cyclic.  
−
<math>2\sqrt{170}x = d \sqrt{40}</math>, where <math>x</math> is the radius of the circle and <math>d</math> is the distance between the circle's center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}x</math>. Using the Pythagorean Theorem on the triangle formed by the point <math>(5, 0)</math>, either one of <math>A</math> or <math>B</math>, and the circle's center, we find that <math>170 + x^2 = 17x^2</math>, so <math>x^2 = \frac{85}{8}</math>, and thus the answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.
+
 
 +
Therefore, we can apply [[Ptolemy's Theorem]] to give:
 +
 
 +
<math>2\sqrt{170}r = d \sqrt{40}</math>, where <math>r</math> is the radius of the circle and <math>d</math> is the distance between the circle's center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}r</math>.  
 +
 
 +
Using the Pythagorean Theorem on the right triangle <math>OAX</math> (or <math>OBX</math>), we find that <math>170 + r^2 = 17r^2</math>, so <math>r^2 = \frac{85}{8}</math>, and thus the area of the circle is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.
 +
 
 +
===Diagram for Solution 1===
 +
[[File:Desmos-graph (1).png|900px|caption]]
 +
 
 +
~BakedPotato66
  
 
==Solution 2 (coordinate bash)==
 
==Solution 2 (coordinate bash)==
Line 34: Line 46:
  
 
==Solution 4 (how fast can you multiply two-digit numbers?)==
 
==Solution 4 (how fast can you multiply two-digit numbers?)==
−
Let <math>(x,0)</math> be the intersection on the x-axis. By Power of a Point Theorem, <math>(x-6)^2+13^2=(x-12)^2+11^2\implies x=5</math>. Then the equations are <math>13(x-6)+13=y</math> and <math>\frac{11}{7}(x-12)+11=y</math> for the tangent lines passing <math>A</math> and <math>B</math> respectively. Then the lines normal to them are <math>-\frac{1}{13}(x-6)+13=y</math> and <math>-\frac{7}{11}(x-12)+11=y</math>. Thus,  
+
Let <math>(x,0)</math> be the intersection on the x-axis. By Power of a Point Theorem, <math>(x-6)^2+13^2=(x-12)^2+11^2\implies x=5</math>. Then the equations for the tangent lines passing <math>A</math> and <math>B</math>, respectively, are <math>13(x-6)+13=y</math> and <math>\frac{11}{7}(x-12)+11=y</math>. Then the lines normal (perpendicular) to them are <math>-\frac{1}{13}(x-6)+13=y</math> and <math>-\frac{7}{11}(x-12)+11=y</math>. Solving for <math>x</math>, we have
  
  
Line 42: Line 54:
 
<cmath>13\cdot7x-11x=84\cdot13-6\cdot11-2\cdot11\cdot13</cmath>
 
<cmath>13\cdot7x-11x=84\cdot13-6\cdot11-2\cdot11\cdot13</cmath>
  
−
After condensing, <math>x=\frac{37}{4}</math>. Then, the center of <math>\omega</math> is <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>. Apply distance formula. WLOG, assume you use <math>A</math>. Then, the area of <math>\omega</math> is <cmath>\sqrt{\frac{1^2}{4^2}+\frac{13^2}{4^2}}^2\pi=\frac{170\pi}{16} \implies \boxed{\textbf{(C) }\frac{85}{8}\pi}.</cmath>
+
After condensing, <math>x=\frac{37}{4}</math>. Then, the center of <math>\omega</math> is <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>. Apply distance formula. WLOG, assume you use <math>A</math>. Then, the area of <math>\omega</math> is <cmath>\left(\sqrt{\frac{1^2}{4^2}+\frac{13^2}{4^2}}\right)^2\pi=\frac{170\pi}{16} \implies \boxed{\textbf{(C) }\frac{85}{8}\pi}.</cmath>
  
−
==Solution 5 (power of a point)==
+
==Solution 5 (tangent cheese)==
 +
After getting <math>x=5</math>, let <math>C=(5,0)</math>. Get the slopes of the lines <math>AC</math> and <math>BC</math>, namely <math>\frac{13}{6-5}=13</math>, <math>\frac{11}{12-5}=\frac{11}{7}</math>. Then, use tangent angle subtraction to get <math>\tan{2x}=\frac{13-\frac{11}{7}}{1+13*\frac{11}{7}}=\frac{80}{150}=\frac{8}{15}</math>. Then, apply tangent double angle to get <math>\tan{2x}=\frac{8}{15}=\frac{2\tan{x}}{1-\tan^2{x}}</math>. Solving, we obtain <math>\tan{x}=\frac{1}{4}</math>. Then, note that <math>\tan{x}=r/{BC}</math>, so <math>r=\frac{1}{4}*\sqrt{170}</math>. Finishing off, we obtain <math>A=\pi*r^2=\pi*170/16=\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.
  
−
Firstly, the point of intersection of the two tangent lines has an equal distance to points <math>A</math> and <math>B</math> due to power of a point theorem. This means we can easily find the point, which is <math>(5, 0)</math>. Label this point <math>X</math>. <math>\triangle{XAB}</math> is an isosceles triangle with lengths, <math>\sqrt{170}</math>, <math>\sqrt{170}</math>, and <math>2\sqrt{10}</math>. Label the midpoint of segment <math>AB</math> as <math>M</math>. The height of this triangle, or <math>\overline{XM}</math>, is <math>4\sqrt{10}</math>. Since <math>\overline{XM}</math> bisects <math>\overline{AB}</math>, <math>\overleftrightarrow{XM}</math> contains the diameter of circle <math>\omega</math>. Let the two points on circle <math>\omega</math> where <math>\overleftrightarrow{XM}</math> intersects be <math>P</math> and <math>Q</math> with <math>\overline{XP}</math> being the shorter of the two. Now let <math>\overline{MP}</math> be <math>x</math> and <math>\overline{MQ}</math> be <math>y</math>. By Power of a Point on <math>\overline{PQ}</math> and <math>\overline{AB}</math>, <math>xy = (\sqrt{10})^2 = 10</math>. Applying Power of a Point again on <math>\overline{XQ}</math> and <math>\overline{XA}</math>, <math>(4\sqrt{10}-x)(4\sqrt{10}+y)=(\sqrt{170})^2=170</math>. Expanding while using the fact that <math>xy = 10</math>, <math>y=x+\frac{\sqrt{10}}{2}</math>. Plugging this into <math>xy=10</math>, <math>2x^2+\sqrt{10}x-20=0</math>. Using the quadratic formula, <math>x = \frac{\sqrt{170}-\sqrt{10}}{4}</math>, and since <math>x+y=2x+\frac{\sqrt{10}}{2}</math>, <math>x+y=\frac{\sqrt{170}}{2}</math>. Since this is the diameter, the radius of circle <math>\omega</math> is <math>\frac{\sqrt{170}}{4}</math>, and so the area of circle <math>\omega</math> is <math>\frac{170}{16}\pi = \boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.
+
~SigmaPiE
−
 
 
−
==Solution 6 (Similar to #3)==
 
−
Let the tangent lines from A and B intersect at X. Let the center of <math>\omega</math> be C. Let the intersection of AB and CX be M. Using the techniques above, we get that the coordinate of X is <math>(5, 0)</math>. However, notice that CMX is the perpendicular bisector of AB. Thus, AM is the altitude from A to CX. Using the distance formula on AX, we get that the length of <math>AX=\sqrt{170}=\sqrt{17}\sqrt{10}</math>. Using the distance formula on AM, we get that <math>AM=\sqrt{10}</math>. Using the distance formula on MX, we get that <math>MX=4\sqrt{10}</math>. To get AC (the radius of <math>\omega</math>), we use either of these methods:
 
−
 
 
−
Method 1: Since CAX is a right angle, the altitude AM is the geometric mean of XM and MC. We get that <math>MC=\frac{\sqrt{17}}{4}</math>. Thus, XC has length <math>XC=\frac{17\sqrt{10}}{4}</math>. Using the Pythagorean Theorem on CAX yields <math>CA=\frac{\sqrt{170}}{4}</math>.
 
−
 
 
−
Method 2: Note that CAX and AMX are similar. Thus, <math>\frac{AM}{MX}=\frac{AC}{AX}</math>. Solving for AC yields <math>\frac{AX \cdot AM}{MX}=\frac{\sqrt{170}}{4}</math>.
 
−
 
 
−
Using the area formula for a circle yields that the area is <math>\frac{85\pi}{8} \longrightarrow \boxed{(C)}</math>.
 
−
~Math4Life2020
 
  
 
==Video Solution==
 
==Video Solution==
Line 66: Line 69:
  
 
~IceMatrix
 
~IceMatrix
 +
==Video Solution by The Power of Logic==
 +
https://www.youtube.com/watch?v=sQIWSrio_Hc
 +
 +
~The Power of Logic
  
 
==See Also==
 
==See Also==

Latest revision as of 09:25, 20 December 2023

The following problem is from both the 2019 AMC 10B #23 and 2019 AMC 12B #20, so both problems redirect to this page.

Problem

Points $A=(6,13)$ and $B=(12,11)$ lie on circle $\omega$ in the plane. Suppose that the tangent lines to $\omega$ at $A$ and $B$ intersect at a point on the $x$-axis. What is the area of $\omega$?

$\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) } \frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}$

Solution 1

First, observe that the two tangent lines are of identical length. Therefore, supposing that the point of intersection is $(x, 0)$, the Pythagorean Theorem gives $\sqrt{(x-6)^2 + 13^2} = \sqrt{(x-12)^2 + 11^2}$. This simplifies to $x = 5$.

Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) $AOBX$ is cyclic.

Therefore, we can apply Ptolemy's Theorem to give:

$2\sqrt{170}r = d \sqrt{40}$, where $r$ is the radius of the circle and $d$ is the distance between the circle's center and $(5, 0)$. Therefore, $d = \sqrt{17}r$.

Using the Pythagorean Theorem on the right triangle $OAX$ (or $OBX$), we find that $170 + r^2 = 17r^2$, so $r^2 = \frac{85}{8}$, and thus the area of the circle is $\boxed{\textbf{(C) }\frac{85}{8}\pi}$.

Diagram for Solution 1

caption

~BakedPotato66

Solution 2 (coordinate bash)

We firstly obtain $x=5$ as in Solution 1. Label the point $(5,0)$ as $C$. The midpoint $M$ of segment $AB$ is $(9, 12)$. Notice that the center of the circle must lie on the line passing through the points $C$ and $M$. Thus, the center of the circle lies on the line $y=3x-15$.

Line $AC$ is $y=13x-65$. Therefore, the slope of the line perpendicular to $AC$ is $-\frac{1}{13}$, so its equation is $y=-\frac{x}{13}+\frac{175}{13}$.

But notice that this line must pass through $A(6, 13)$ and $(x, 3x-15)$. Hence $3x-15=-\frac{x}{13}+\frac{175}{13} \Rightarrow x=\frac{37}{4}$. So the center of the circle is $\left(\frac{37}{4}, \frac{51}{4}\right)$.

Finally, the distance between the center, $\left(\frac{37}{4}, \frac{51}{4}\right)$, and point $A$ is $\frac{\sqrt{170}}{4}$. Thus the area of the circle is $\boxed{\textbf{(C) }\frac{85}{8}\pi}$.

Solution 3

The midpoint of $AB$ is $D(9,12)$. Let the tangent lines at $A$ and $B$ intersect at $C(a,0)$ on the $x$-axis. Then $CD$ is the perpendicular bisector of $AB$. Let the center of the circle be $O$. Then $\triangle AOC$ is similar to $\triangle DAC$, so $\frac{OA}{AC} = \frac{AD}{DC}$. The slope of $AB$ is $\frac{13-11}{6-12}=\frac{-1}{3}$, so the slope of $CD$ is $3$. Hence, the equation of $CD$ is $y-12=3(x-9) \Rightarrow y=3x-15$. Letting $y=0$, we have $x=5$, so $C = (5,0)$.

Now, we compute $AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}$, $AD=\sqrt{(6-9)^2+(13-12)^2}=\sqrt{10}$, and $DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}$.

Therefore $OA = \frac{AC\cdot AD}{DC}=\sqrt{\frac{85}{8}}$, and consequently, the area of the circle is $\pi\cdot OA^2 = \boxed{\textbf{(C) }\frac{85}{8}\pi}$.


Solution 4 (how fast can you multiply two-digit numbers?)

Let $(x,0)$ be the intersection on the x-axis. By Power of a Point Theorem, $(x-6)^2+13^2=(x-12)^2+11^2\implies x=5$. Then the equations for the tangent lines passing $A$ and $B$, respectively, are $13(x-6)+13=y$ and $\frac{11}{7}(x-12)+11=y$. Then the lines normal (perpendicular) to them are $-\frac{1}{13}(x-6)+13=y$ and $-\frac{7}{11}(x-12)+11=y$. Solving for $x$, we have


\[-\frac{7}{11}(x-12)+11=-\frac{1}{13}(x-6)+13\] \[\frac{13\cdot7x-11x}{13\cdot11}=\frac{84\cdot13-6\cdot11-2\cdot11\cdot13}{11\cdot13}\] \[13\cdot7x-11x=84\cdot13-6\cdot11-2\cdot11\cdot13\]

After condensing, $x=\frac{37}{4}$. Then, the center of $\omega$ is $\left(\frac{37}{4}, \frac{51}{4}\right)$. Apply distance formula. WLOG, assume you use $A$. Then, the area of $\omega$ is \[\left(\sqrt{\frac{1^2}{4^2}+\frac{13^2}{4^2}}\right)^2\pi=\frac{170\pi}{16} \implies \boxed{\textbf{(C) }\frac{85}{8}\pi}.\]

Solution 5 (tangent cheese)

After getting $x=5$, let $C=(5,0)$. Get the slopes of the lines $AC$ and $BC$, namely $\frac{13}{6-5}=13$, $\frac{11}{12-5}=\frac{11}{7}$. Then, use tangent angle subtraction to get $\tan{2x}=\frac{13-\frac{11}{7}}{1+13*\frac{11}{7}}=\frac{80}{150}=\frac{8}{15}$. Then, apply tangent double angle to get $\tan{2x}=\frac{8}{15}=\frac{2\tan{x}}{1-\tan^2{x}}$. Solving, we obtain $\tan{x}=\frac{1}{4}$. Then, note that $\tan{x}=r/{BC}$, so $r=\frac{1}{4}*\sqrt{170}$. Finishing off, we obtain $A=\pi*r^2=\pi*170/16=\boxed{\textbf{(C) }\frac{85}{8}\pi}$.

~SigmaPiE

Video Solution

For those who want a video solution: (Is similar to Solution 1) https://youtu.be/WI2NVuIp1Ik

Video Solution by TheBeautyofMath

https://youtu.be/W1zuqrTlBtU

~IceMatrix

Video Solution by The Power of Logic

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

~The Power of Logic

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 10 Problems and Solutions
2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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=2019_AMC_10B_Problems/Problem_23&oldid=207972"