Difference between revisions of "2021 Fall AMC 10A Problems/Problem 15"
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label("$r$",(-0.25,1.5),E); | label("$r$",(-0.25,1.5),E); | ||
label("$r$",(0.5,2.125),S); | label("$r$",(0.5,2.125),S); | ||
| − | add(pathticks(A--F,1,0.5,0,2); | + | add(pathticks(A--F,1,0.5,0,2)); |
| − | add(pathticks(F--B,1,0.5,0,2); | + | add(pathticks(F--B,1,0.5,0,2)); |
| − | add(pathticks(A--G,1,0.5,0,2); | + | add(pathticks(A--G,1,0.5,0,2)); |
| − | add(pathticks(G--C,1,0.5,0,2); | + | add(pathticks(G--C,1,0.5,0,2)); |
Revision as of 00:44, 24 November 2021
Isosceles triangle 
has 
, and a circle with radius 
is tangent to line 
at 
and to line 
at 
. What is the area of the circle that passes through vertices 
, , and 
Solution 1
Let the center of the first circle be 
By Pythagorean Theorem,
![\[AO=\sqrt{(3\sqrt{6})^2+(5\sqrt{2})^2}=2 \sqrt{26}\]](http://latex.artofproblemsolving.com/f/a/a/faab0344e40b6da42c07a30d261447e99f9aa063.png)
Now, notice that since 
is 
degrees, so arc 
is 
degrees and is the diameter. Thus, the radius is 
so the area is 
- 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,W); label("$A$",A,S); label("$B$",B,N); label("$C$",C,W); label("$D$",F,S); label("$E$",G,W); label("$3\sqrt{6}$",(1.5,1.5),S); label("$3\sqrt{6}$",(-1.5,1.5),S); label("$5\sqrt{2}$",(1,3.625),N); label("$5\sqrt{2}$",(-1,3.625),N); label("$I$",I,N); label("$r$",(-0.25,1.5),E); label("$r$",(0.5,2.125),S); add(pathticks(A--F,1,0.5,0,2)); add(pathticks(F--B,1,0.5,0,2)); add(pathticks(A--G,1,0.5,0,2)); add(pathticks(G--C,1,0.5,0,2)); [/asy]](http://latex.artofproblemsolving.com/f/3/2/f329402a97b8cd0cbf5ee045bd1830bfb0540b77.png)
Because circle 
is tangent to 
at 
. Because O is the circumcenter of 
is the perpendicular bisector of .
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. 
