2021 Fall AMC 10A Problems/Problem 15

Revision as of 00:28, 24 November 2021 by Kingravi (talk | contribs) (Solution 2 (Similar Triangles))

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)


import olympiad;
unitsize(50);
pair A,B,C,D,E,I,O;
A=origin; B=(2,1); C=(-2,1); D=(6,3); E=(-6,3);
O=circumcenter(A,B,C); // olympiad - circumcenter
I=incenter(A,D,E);
draw(A--B--C--cycle);
dot(O);
dot(I)
draw(circumcircle(A,B,C)); // olympiad - circumcircle
draw(incircle(A,D,E));
label("$O$",O,S);
label("$A$",A,S);
label("$B$",B,E);
label("$C$",C,W);
label("$3\sqrt{6}$",(1.25,1),S);
label("$3\sqrt{6}$",(-1.25,1),S);



 (Error making remote request. Unknown error_msg)

Solution in Progress

~KingRavi

See Also

2021 Fall AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png