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Difference between revisions of "2019 AMC 10B Problems/Problem 23"
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{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}
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{{duplicate|[[2019 AMC 10B Problems#Problem 23|2019 AMC 10B #23]] and [[2019 AMC 12B Problems#Problem 20|2019 AMC 12B #20]]}}
  
 
==Problem==
 
==Problem==

Revision as of 10:24, 5 February 2022

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 \[\sqrt{\frac{1^2}{4^2}+\frac{13^2}{4^2}}^2\pi=\frac{170\pi}{16} \implies \boxed{\textbf{(C) }\frac{85}{8}\pi}.\]

Solution 5 (power of a point)

Firstly, the point of intersection of the two tangent lines has an equal distance to points $A$ and $B$ due to power of a point theorem. This means we can easily find the point, which is $(5, 0)$. Label this point $X$. $\triangle{XAB}$ is an isosceles triangle with lengths, $\sqrt{170}$, $\sqrt{170}$, and $2\sqrt{10}$. Label the midpoint of segment $AB$ as $M$. The height of this triangle, or $\overline{XM}$, is $4\sqrt{10}$. Since $\overline{XM}$ bisects $\overline{AB}$, $\overleftrightarrow{XM}$ contains the diameter of circle $\omega$. Let the two points on circle $\omega$ where $\overleftrightarrow{XM}$ intersects be $P$ and $Q$ with $\overline{XP}$ being the shorter of the two. Now let $\overline{MP}$ be $x$ and $\overline{MQ}$ be $y$. By Power of a Point on $\overline{PQ}$ and $\overline{AB}$, $xy = (\sqrt{10})^2 = 10$. Applying Power of a Point again on $\overline{XQ}$ and $\overline{XA}$, $(4\sqrt{10}-x)(4\sqrt{10}+y)=(\sqrt{170})^2=170$. Expanding while using the fact that $xy = 10$, $y=x+\frac{\sqrt{10}}{2}$. Plugging this into $xy=10$, $2x^2+\sqrt{10}x-20=0$. Using the quadratic formula, $x = \frac{\sqrt{170}-\sqrt{10}}{4}$, and since $x+y=2x+\frac{\sqrt{10}}{2}$, $x+y=\frac{\sqrt{170}}{2}$. Since this is the diameter, the radius of circle $\omega$ is $\frac{\sqrt{170}}{4}$, and so the area of circle $\omega$ is $\frac{170}{16}\pi = \boxed{\textbf{(C) }\frac{85}{8}\pi}$.

Solution 6 (Similar to #3)

Let the tangent lines from A and B intersect at X. Let the center of $\omega$ be C. Let the intersection of AB and CX be M. Using the techniques above, we get that the coordinate of X is $(5, 0)$. 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 $AX=\sqrt{170}=\sqrt{17}\sqrt{10}$. Using the distance formula on AM, we get that $AM=\sqrt{10}$. Using the distance formula on MX, we get that $MX=4\sqrt{10}$. To get AC (the radius of $\omega$), 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 $MC=\frac{\sqrt{17}}{4}$. Thus, XC has length $XC=\frac{17\sqrt{10}}{4}$. Using the Pythagorean Theorem on CAX yields $CA=\frac{\sqrt{170}}{4}$.

Method 2: Note that CAX and AMX are similar. Thus, $\frac{AM}{MX}=\frac{AC}{AX}$. Solving for AC yields $\frac{AX \cdot AM}{MX}=\frac{\sqrt{170}}{4}$.

Using the area formula for a circle yields that the area is $\frac{85\pi}{8} \longrightarrow \boxed{(C)}$. ~Math4Life2020

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

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