2016 AIME I Problems/Problem 6
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 .
Solution 5
Since and both intercept arc , it follows that . Note that by the external angle theorem. It follows that , so we must have that is isosceles, yielding . Note that , so . This yields . It follows that , giving a final answer of .
Solution 6
Let be the excenter opposite to in . By the incenter-excenter lemma . Its well known that . ~Pluto1708
Alternate solution: "We can use the angle bisector theorem on and bisector to get that . Since , we get . Thus, and ." (https://artofproblemsolving.com/community/c759169h1918283_geometry_problem)
Solution 7
First, we know that by the incenter-excenter lemma (Fact 5). If you are not familiar with this lemma, it can be pretty useful in some problems so it might be a good chance to get acquainted with it. :)
Now because we are dealing with circumcircles and angle bisectors, we try to solve the problem using similar triangles and the angle bisector theorem.
(Another cool fact is because bisects angle , is the midpoint of arc .)
Therefore, we know that , and by similar reasoning, . (Note: I let ).
So we try now to exploit that we have 2 pairs of equal triangles:
(First, let ) Triangle is similar to triangle : Triangle is similar to triangle :
But they don’t really help us. Hm....
Well, usually when we have an in center of a triangle, it is usually good to connect all the vertices pod the triangle to the incenter, so let’s try that. Maybe then we can apply the angle bisected theorem then!
First connect to and to . Then by the angle bisector theorem, . Wait! Since triangle and are similar, we have . And therefore, , so , therefore, , and so
Key Concepts:
- 6 usually isn’t too hard a problem on the AIME’s. The method usually is around 3 lines. Maybe not all the work, but the method shouldn’t be too bad.
When dealing with intersecting circles, or circles in general, always remember similar triangles! Try angle chasing!
On top of that, when dealing with angle bisectors, think about using the angle bisector theorem and similar triangles. Circles + angles bisectors usually means similar triangles!
(Professor-Mom)
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 |
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