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

(Solution 2 (Similar Triangles))
(Solution 2 (Similar Triangles))
Line 17: Line 17:
 
unitsize(50);
 
unitsize(50);
 
pair A,B,C,D,E,I,F,G,O;
 
pair A,B,C,D,E,I,F,G,O;
A=origin; B=(2,3); C=(-2,3); D=(4.25,6.25); E=(-4.25,6.25); F=(1,1.5); G=(-1,1.5);
+
A=origin; B=(2,3); C=(-2,3); D=(4.3,6.3); E=(-4.3,6.3); F=(1,1.5); G=(-1,1.5);
 
O=circumcenter(A,B,C); // olympiad - circumcenter
 
O=circumcenter(A,B,C); // olympiad - circumcenter
 
I=incenter(A,D,E);
 
I=incenter(A,D,E);
Line 45: Line 45:
 
label("$3\sqrt{6}$",(-1.25,1.5),S);
 
label("$3\sqrt{6}$",(-1.25,1.5),S);
 
label("$I$",I,N);
 
label("$I$",I,N);
 
+
label("$r$",(-1,1.25),E);
 +
label("$r$",(1,1.25),W);
  
  

Revision as of 00:01, 24 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,D,E,I,F,G,O; A=origin; B=(2,3); C=(-2,3); D=(4.3,6.3); E=(-4.3,6.3); F=(1,1.5); G=(-1,1.5); O=circumcenter(A,B,C); // olympiad - circumcenter I=incenter(A,D,E); draw(A--B--C--cycle); dot(O); dot(I); dot(F); dot(G); draw(circumcircle(A,B,C)); // olympiad - circumcircle draw(incircle(A,D,E)); draw(I--B); draw(I--C); draw(I--A); draw(rightanglemark(A,C,I));  draw(rightanglemark(A,B,I)); draw(O--F); draw(O--G); draw(rightanglemark(A,F,O));  draw(rightanglemark(A,G,O));   label("$O$",O,E); label("$A$",A,S); label("$B$",B,S); label("$C$",C,W); label("$3\sqrt{6}$",(1.25,1.5),S); label("$3\sqrt{6}$",(-1.25,1.5),S); label("$I$",I,N); label("$r$",(-1,1.25),E); label("$r$",(1,1.25),W);   [/asy]

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

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