Difference between revisions of "2019 AMC 10B Problems/Problem 23"

Latest revision as of 17:27, 3 January 2021

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) defined by the circle's center, $A$ , $B$ , and $(5, 0)$ is cyclic. Therefore, we can apply Ptolemy's Theorem to give $2\sqrt{170}x = d \sqrt{40}$ , where $x$ is the radius of the circle and $d$ is the distance between the circle's center and $(5, 0)$ . Therefore, $d = \sqrt{17}x$ . Using the Pythagorean Theorem on the triangle formed by the point $(5, 0)$ , either one of $A$ or $B$ , and the circle's center, we find that $170 + x^2 = 17x^2$ , so $x^2 = \frac{85}{8}$ , and thus the answer is $\boxed{\textbf{(C) }\frac{85}{8}\pi}$ .

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 are $13(x-6)+13=y$ and $\frac{11}{7}(x-12)+11=y$ for the tangent lines passing $A$ and $B$ respectively. Then the lines normal to them are $-\frac{1}{13}(x-6)+13=y$ and $-\frac{7}{11}(x-12)+11=y$ . Thus,

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

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

@@ Line 1: / Line 1: @@
-{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}
 ==Problem==
-Points <math>A(6,13)</math> and <math>B(12,11)</math> lie on circle <math>\omega</math> in the plane. Suppose that the tangent lines to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at a point on the <math>x</math>-axis. What is the area of <math>\omega</math>?
+Points <math>A=(6,13)</math> and <math>B=(12,11)</math> lie on circle <math>\omega</math> in the plane. Suppose that the tangent lines to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at a point on the <math>x</math>-axis. What is the area of <math>\omega</math>?
 <math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }
@@ Line 9: / Line 7: @@
 ==Solution 1==
-First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was <math>(x, 0)</math>. Using Pythagorean Theorem gives <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
+<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>.
-Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, <math>A</math>, <math>B</math>, and <math>(5, 0)</math> form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:
+==Solution 2 (coordinate bash)==
+We firstly obtain <math>x=5</math> as in Solution 1. Label the point <math>(5,0)</math> as <math>C</math>. The midpoint <math>M</math> of segment <math>AB</math> is <math>(9, 12)</math>. Notice that the center of the circle must lie on the line passing through the points <math>C</math> and <math>M</math>. Thus, the center of the circle lies on the line <math>y=3x-15</math>.
-<math>2\sqrt{170}x = d * \sqrt{40}</math>, where <math>d</math> represents the distance between circle center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}x</math>. Using Pythagorean Theorem on <math>(5, 0)</math>, either one of <math>A</math> or <math>B</math>, and the circle center, we realize that <math>170 + x^2 = 17x^2</math>, at which point <math>x^2 = \frac{85}{8}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.
+Line <math>AC</math> is <math>y=13x-65</math>. Therefore, the slope of the line perpendicular to <math>AC</math> is <math>-\frac{1}{13}</math>, so its equation is <math>y=-\frac{x}{13}+\frac{175}{13}</math>.
-==Solution 2==
+But notice that this line must pass through <math>A(6, 13)</math> and <math>(x, 3x-15)</math>. Hence <math>3x-15=-\frac{x}{13}+\frac{175}{13} \Rightarrow x=\frac{37}{4}</math>. So the center of the circle is <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>.
-First, follow solution 1 and obtain <math>x=5</math>. Label the point <math>(5,0)</math> as point <math>C</math>. The midpoint <math>M</math> of segment <math>AB</math> is <math>(9, 12)</math>. Notice that the center of the circle must lie on the line that goes through the points <math>C</math> and <math>M</math>. Thus, the center of the circle lies on the line <math>y=3x-15</math>.
-Line <math>AC</math> is <math>y=13x-65</math>. The perpendicular line must pass through <math>A(6, 13)</math> and <math>(x, 3x-15)</math>. The slope of the perpendicular line is <math>-\frac{1}{13}</math>. The line is hence <math>y=-\frac{x}{13}+\frac{175}{13}</math>. The point <math>(x, 3x-15)</math> lies on this line. Therefore, <math>3x-15=-\frac{x}{13}+\frac{175}{13}</math>. Solving this equation tells us that <math>x=\frac{37}{4}</math>. So the center of the circle is <math>(\frac{37}{4}, \frac{51}{4})</math>. The distance between the center, <math>(\frac{37}{4}, \frac{51}{4})</math>, and point A is <math>\frac{\sqrt{170}}{4}</math>. Hence, the area is <math>\frac{85}{8}\pi</math>. The answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.
+Finally, the distance between the center, <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>, and point <math>A</math> is <math>\frac{\sqrt{170}}{4}</math>. Thus the area of the circle is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.
 ==Solution 3==
-The mid point of AB is D(9,12), suppose the tanget lines at A and B intersect at C(a,0)on X axis, CD would be the perpendicular bisector of AB. Suppose the center of circle is O, then triangle AOC is similiar to DAC, that is OA/AC=AD/DC.
+The midpoint of <math>AB</math> is <math>D(9,12)</math>. Let the tangent lines at <math>A</math> and <math>B</math> intersect at <math>C(a,0)</math> on the <math>x</math>-axis. Then <math>CD</math> is the perpendicular bisector of <math>AB</math>. Let the center of the circle be <math>O</math>. Then <math>\triangle AOC</math> is similar to <math>\triangle DAC</math>, so <math>\frac{OA}{AC} = \frac{AD}{DC}</math>.
-The slope of AB is (13-11)/(6-12)=-1/3, therefore the slope of CD will be 3. the equation of CD is y-12=3*(x-9), that is y=3x-15, let y=0, we have x=5, which is the x coordiante of C(5,0)
+The slope of <math>AB</math> is <math>\frac{13-11}{6-12}=\frac{-1}{3}</math>, so the slope of <math>CD</math> is <math>3</math>. Hence, the equation of <math>CD</math> is <math>y-12=3(x-9) \Rightarrow y=3x-15</math>. Letting <math>y=0</math>, we have <math>x=5</math>, so <math>C = (5,0)</math>.
+Now, we compute <math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math>,
+<math>AD=\sqrt{(6-9)^2+(13-12)^2}=\sqrt{10}</math>, and
+<math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math>.
+Therefore <math>OA = \frac{AC\cdot AD}{DC}=\sqrt{\frac{85}{8}}</math>,
+and consequently, the area of the circle is <math>\pi\cdot OA^2 = \boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.
+==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,
+<cmath>-\frac{7}{11}(x-12)+11=-\frac{1}{13}(x-6)+13</cmath>
+<cmath>\frac{13\cdot7x-11x}{13\cdot11}=\frac{84\cdot13-6\cdot11-2\cdot11\cdot13}{11\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>
+==Solution 5 (power of a point)==
+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>.
+==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
-AC=sqrt((6-5)^2+(13-0)^2)=sqrt(170)
+==Video Solution==
-AD=sqrt((6-9)^2)+(13-12)^2)=sqrt(10)
+For those who want a video solution: (Is similar to Solution 1)
-DC=sqrt((9-5)^2+(12-0)^2)=aqrt(160)
+https://youtu.be/WI2NVuIp1Ik
-Therefore OA=AC*AD/DC=sqrt(85/5)
-Consequently, the area of the circle is pi*OA^2=pi*85/5
-(by Zhen Qin)
-(P.S. Will someone please Latex this?)
 ==See Also==
@@ Line 36: / Line 66: @@
 {{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}
 {{MAA Notice}}
-SUB2PEWDS

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

Page

Toolbox

Search