Difference between revisions of "2018 AMC 12B Problems/Problem 21"
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− | Let the triangle have coordinates <math>(0,0),( | + | Let the triangle have coordinates <math>(0,0),(12,0),(0,5).</math> Then the coordinates of the incenter and circumcenter are <math>(2,2)</math> and <math>(6,2.5),</math> respectively. If we let <math>M=(x,x),</math> then <math>x</math> satisfies |
<cmath>\sqrt{(2.5-x)^2+(6-x)^2}+x=6.5</cmath> | <cmath>\sqrt{(2.5-x)^2+(6-x)^2}+x=6.5</cmath> | ||
<cmath>2.5^2-5x+x^2+6^2-12x+x^2=6.5^2-13x+x^2</cmath> | <cmath>2.5^2-5x+x^2+6^2-12x+x^2=6.5^2-13x+x^2</cmath> | ||
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{{AMC12 box|year=2018|ab=B|num-b=20|num-a=22}} | {{AMC12 box|year=2018|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
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+ | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 20:24, 25 January 2019
Contents
Problem
In with side lengths , , and , let and denote the circumcenter and incenter, respectively. A circle with center is tangent to the legs and and to the circumcircle of . What is the area of ?
Solution 1
Let the triangle have coordinates Then the coordinates of the incenter and circumcenter are and respectively. If we let then satisfies Now the area of our triangle can be calculated with the Shoelace Theorem. The answer turns out to be
Solution 2
Notice that we can let . If , then and . Using shoelace formula, we get .
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.