Difference between revisions of "2021 Fall AMC 10A Problems/Problem 15"
(→Solution 2 (Similar Triangles)) |
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| Line 17: | Line 17: | ||
unitsize(50); | unitsize(50); | ||
pair A,B,C,O; | pair A,B,C,O; | ||
| − | A=origin; B=( | + | A=origin; B=(2,2); C=(-2,2); |
O=circumcenter(A,B,C); // olympiad - circumcenter | O=circumcenter(A,B,C); // olympiad - circumcenter | ||
draw(A--B--C--cycle); | draw(A--B--C--cycle); | ||
| Line 23: | Line 23: | ||
draw(circumcircle(A,B,C)); // olympiad - circumcircle | draw(circumcircle(A,B,C)); // olympiad - circumcircle | ||
label("$O$",O,S); | label("$O$",O,S); | ||
| − | label("$A$",A, | + | label("$A$",A,S); |
| − | label("$B$",B, | + | label("$B$",B,E); |
| − | label("$C$",C, | + | label("$C$",C,W); |
| + | label("$3\sqrt{6}$",(1,1),S); | ||
| + | label("$3\sqrt{6}$",(-1,1),S); | ||
| + | |||
Revision as of 23:16, 23 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,O; A=origin; B=(2,2); C=(-2,2); 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); label("$A$",A,S); label("$B$",B,E); label("$C$",C,W); label("$3\sqrt{6}$",(1,1),S); label("$3\sqrt{6}$",(-1,1),S); [/asy]](http://latex.artofproblemsolving.com/d/1/2/d12aaa5e04f0a38aa7d9156f47ebb046df5a4279.png)
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. 
