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Revision as of 17:32, 16 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, 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 $AB$ 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 $y=-\frac{x}{13}+\frac{175}{13}$. The point $(x, 3x-15)$ lies on this line. Therefore, $3x-15=-\frac{x}{13}+\frac{175}{13}$. 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 $\boxed{\textbf{(C) }\frac{85}{8}\pi}$.

Solution 3

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

$AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}$, $AD=\sqrt{(6-9)^2)+(13-12)^2}=\sqrt{10}$, $DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}$, Therefore $OA = \frac{AC\cdot AD}{DC}=\sqrt{\frac{85}{8}}$, Consequently, the area of the circle is $\pi\cdot OA^2 = \pi\cdot\frac{85}{8}$. (by Zhen Qin) (P.S. Will someone please Latex this?) ($\LaTeX$ed by a pewdiepie subscriber)

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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