Difference between revisions of "2016 AIME I Problems/Problem 6"
NumberGiant (talk | contribs) m (→Solution 1) |
Burunduchok (talk | contribs) m (→Solution 2) |
||
Line 38: | Line 38: | ||
==Solution 2== | ==Solution 2== | ||
− | 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 | + | 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 \sim \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: |
<math>x^2-2x-15=0</math>, or <math>x=5, -3</math>. <math>x</math> is positive, so <math>x=5</math>. As a result, <math>CI=\tfrac{2x}{3}=\tfrac{10}{3}</math> and the answer is <math>\boxed{013}</math> | <math>x^2-2x-15=0</math>, or <math>x=5, -3</math>. <math>x</math> is positive, so <math>x=5</math>. As a result, <math>CI=\tfrac{2x}{3}=\tfrac{10}{3}</math> and the answer is <math>\boxed{013}</math> | ||
Revision as of 21:08, 30 July 2016
Contents
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 .
Solution 4
Since and , . Also, and so . Now we can call , and , . By angle bisector theorem, . So let and for some value of . Now call . By the similar triangles we found earlier, and . We can simplify this to and . So we can plug the into the first equation and get . We can now draw a line through and that intersects at . By mass points, we can assign a mass of to , to , and to . We can also assign a mass of to by angle bisector theorem. So the ratio of . So since , we can plug this back into the original equation to get . This means that which has roots -2 and which means our and our answer is .
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.