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

Revision as of 23:51, 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

Solution 2 (Similar Triangles)

$[asy] import olympiad; unitsize(50); pair A,B,C,O; A=origin; B=(5.196,5.196); C=(-5.196,5.196); O=circumcenter(A,B,C); // olympiad - circumcenter draw(A--B--C--cycle); dot(O); draw(circumcircle(A,B,C)); // olympiad - circumcircle label("$O$",O,S); [/asy]$

Solution in Progress

~KingRavi

2021 Fall AMC 10A (Problems • Answer Key • Resources)
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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 15"

Revision as of 23:51, 23 November 2021

Solution 1

Solution 2 (Similar Triangles)

See Also

@@ Line 17: / Line 17: @@
 unitsize(50);
 pair A,B,C,O;
-A=origin; B=(27,27); C=(-27,27);
+A=origin; B=(5.196,5.196); C=(-5.196,5.196);
 O=circumcenter(A,B,C); // olympiad - circumcenter
 draw(A--B--C--cycle);