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

Revision as of 22:00, 19 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 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 $x=5$ .

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 $d$ 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

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 $\pi\cdot OA^2 = \boxed{\textbf{(C) }\frac{85}{8}\pi}$ .

@@ Line 28: / Line 28: @@
 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>AD=\sqrt{(6-9)^2+(13-12)^2}=\sqrt{10}</math>, and
 <math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math>.

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