Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2 (Similar Triangles))
(→‎Solution 2 (Similar Triangles))
Line 16: Line 16:
 
import olympiad;
 
import olympiad;
 
unitsize(50);
 
unitsize(50);
−
pair A,B,C,O,I;
+
pair A,B,C,O;
 
A=origin; B=2*right; C=1.5*dir(70);
 
A=origin; B=2*right; C=1.5*dir(70);
 
O=circumcenter(A,B,C); // olympiad - circumcenter
 
O=circumcenter(A,B,C); // olympiad - circumcenter
−
I=incenter(A,B,C); // olympiad - incenter
 
 
draw(A--B--C--cycle);
 
draw(A--B--C--cycle);
 
dot(O);
 
dot(O);
 
dot(I);
 
dot(I);
 
draw(circumcircle(A,B,C)); // olympiad - circumcircle
 
draw(circumcircle(A,B,C)); // olympiad - circumcircle
−
draw(incircle(A,B,C)); // olympiad - incircle
 
−
label("$I$",I,W);
 
 
label("$O$",O,S);
 
label("$O$",O,S);
  

Revision as of 22:48, 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=2*right; C=1.5*dir(70); O=circumcenter(A,B,C); // olympiad - circumcenter draw(A--B--C--cycle); dot(O); dot(I); draw(circumcircle(A,B,C)); // olympiad - circumcircle label("$O$",O,S); [/asy]

Solution in Progress

~KingRavi

See Also

2021 Fall AMC 10A (Problems • Answer Key • Resources)
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_15&oldid=165848"