2021 Fall AMC 10A Problems/Problem 15

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

Solution 2 (Similar Triangles)

$[asy] import olympiad; unitsize(50); pair A,B,C,O; A=origin; B=(2,2); C=(-2,2); O=circumcenter(A,B,C); // olympiad - circumcenter draw(A--B--C--cycle); dot(O); draw(circumcircle(A,B,C)); // olympiad - circumcircle label("$O$",O,S); label("$A$",A,S); label("$B$",B,E); label("$C$",C,W); label("$3\sqrt{6}$",(1,1),S); label("$3\sqrt{6}$",(-1,1),S); [/asy]$

Solution in Progress

~KingRavi

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2021 Fall AMC 10A Problems/Problem 15

Solution 1

Solution 2 (Similar Triangles)

See Also