Difference between revisions of "2016 AIME I Problems/Problem 6"
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WLOG assume <math>\triangle ABC</math> is isosceles. Then, <math>L</math> is the midpoint of <math>AB</math>, and <math>\angle CLB=\angle CLA=90^\circ</math>. Draw the perpendicular from <math>I</math> to <math>CB</math>, and let it meet <math>CB</math> at <math>E</math>. Since <math>IL=2</math>, <math>IE</math> is also <math>2</math> (they are both inradii). Set <math>BD</math> as <math>x</math>. Then, triangles <math>BLD</math> and <math>CEI</math> are similar, and <math>\tfrac{2}{3}=\tfrac{CI}{x}</math>. Thus, <math>CI=\tfrac{2x}{3}</math>. <math>\triangle CBD~\triangle CEI</math>, so <math>\tfrac{IE}{DB}=\tfrac{CI}{CD}</math>. Thus <math>\tfrac{2}{x}=\tfrac{(2x/3)}{(2x/3+5)}</math>. Solving for <math>x</math>, we have: | WLOG assume <math>\triangle ABC</math> is isosceles. Then, <math>L</math> is the midpoint of <math>AB</math>, and <math>\angle CLB=\angle CLA=90^\circ</math>. Draw the perpendicular from <math>I</math> to <math>CB</math>, and let it meet <math>CB</math> at <math>E</math>. Since <math>IL=2</math>, <math>IE</math> is also <math>2</math> (they are both inradii). Set <math>BD</math> as <math>x</math>. Then, triangles <math>BLD</math> and <math>CEI</math> are similar, and <math>\tfrac{2}{3}=\tfrac{CI}{x}</math>. Thus, <math>CI=\tfrac{2x}{3}</math>. <math>\triangle CBD~\triangle CEI</math>, so <math>\tfrac{IE}{DB}=\tfrac{CI}{CD}</math>. Thus <math>\tfrac{2}{x}=\tfrac{(2x/3)}{(2x/3+5)}</math>. Solving for <math>x</math>, we have: |
Revision as of 10:57, 13 March 2016
Problem
In let be the center of the inscribed circle, and let the bisector of intersect at . The line through and intersects the circumscribed circle of at the two points and . If and , then , where and are relatively prime positive integers. Find .
Solution
Solution 1
Suppose we label the angles as shown below. As and intercept the same arc, we know that . Similarly, . Also, using , we find . Therefore, . Therefore, , so must be isosceles with . Similarly, . Then , hence . Also, bisects , so by the Angle Bisector Theorem . Thus , and the answer is .
Solution 2
WLOG assume is isosceles. Then, is the midpoint of , and . Draw the perpendicular from to , and let it meet at . Since , is also (they are both inradii). Set as . Then, triangles and are similar, and . Thus, . , so . Thus . Solving for , we have: , or . is positive, so . As a result, and the answer is
Solution 3
WLOG assume is isosceles (with vertex ). Let be the center of the circumcircle, the circumradius, and the inradius. A simple sketch will reveal that must be obtuse (as an acute triangle will result in being greater than ) and that and are collinear. Next, if , and . Euler gives us that , and in this case, . Thus, . Solving for , we have , then , yielding . Next, so . Finally, gives us , and . Our answer is then .
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.