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"

(→‎Solution: Added solution.)
(→‎Solution)
Line 14: Line 14:
  
 
<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>.
 
<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>.
 +
 +
==Solution 2==
 +
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\overline</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{1}{13}+\frac{175}{13}</math>. The point <math>(x, 3x-15)</math> lies on this line. Therefore, <math>3x-15=-\frac{1}{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>.
  
 
==See Also==
 
==See Also==

Revision as of 20:00, 14 February 2019

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

First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was $(x, 0)$. Using Pythagorean Theorem gives $x=5$.

Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, $A$, $B$, and $(5, 0)$ form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:

$2\sqrt{170}x = d * \sqrt{40}$, where $d$ represents the distance between circle center and $(5, 0)$. Therefore, $d = \sqrt{17}x$. Using Pythagorean Theorem on $(5, 0)$, either one of $A$ or $B$, and the circle center, we realize that $170 + x^2 = 17x^2$, at which point $x^2 = \frac{85}{8}$, so the answer is $\boxed{\textbf{(C) }\frac{85}{8}\pi}$.

Solution 2

First, follow solution 1 and obtain $x=5$. Label the point $(5,0)$ as point $C$. The midpoint $M$ of segment $AB\overline$ (Error compiling LaTeX. Unknown error_msg) is $(9, 12)$. Notice that the center of the circle must lie on the line that goes through the points $C$ and $M$. Thus, the center of the circle lies on the line $y=3x-15$.

Line $AC$ is $y=13x-65$. The perpendicular line must pass through $A(6, 13)$ and $(x, 3x-15)$. The slope of the perpendicular line is $-\frac{1}{13}$. The line is hence $y=-\frac{1}{13}+\frac{175}{13}$. The point $(x, 3x-15)$ lies on this line. Therefore, $3x-15=-\frac{1}{13}+\frac{175}{13}$. Solving this equation tells us that $x=\frac{37}{4}$. So the center of the circle is $(\frac{37}{4}, \frac{51}{4})$. The distance between the center, $(\frac{37}{4}, \frac{51}{4})$, and point A is $\frac{\sqrt{170}}{4}$. Hence, the area is $\frac{85}{8}\pi$. The answer is $\boxed{\textbf{(C) }\frac{85}{8}\pi}$.

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=102689"