Difference between revisions of "2021 Fall AMC 10A Problems/Problem 15"

Revision as of 16:40, 23 November 2021

Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6$ , and a circle with radius $5\sqrt2$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$ . What is the area of the circle that passes through vertices $A$ , $B$ , and $C?$

$\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi$

Solution 1

Let the center of the first circle be $O.$ By Pythagorean Theorem, $AO=\sqrt{(3\sqrt{6})^2+(5\sqrt{2})^2}=2 \sqrt{26}$ Now, notice that since $\angle ABO$ is $90$ degrees, so arc $AO$ is $180$ degrees and $AO$ is the diameter. Thus, the radius is $\sqrt{26},$ so the area is $\boxed{26\pi}.$

- kante314

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 10 Problems and Solutions

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 - kante314
+==See Also==
+{{AMC10 box|year=2021 Fall|ab=A|num-b=14|num-a=16}}
+{{MAA Notice}}

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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 15"

Revision as of 16:40, 23 November 2021

Solution 1

See Also