2016 AIME I Problems/Problem 6
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 .
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 .
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
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 .
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 .
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 .
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)
We can just say that quadrilateral is a right kite with right angles at and . Let us construct another similar right kite with the points of tangency on and called and respectively, point , and point . Note that we only have to look at one half of the circle since the diagram is symmetrical. Let us call for simplicity's sake. Based on the fact that is similar to we can use triangle proportionality to say that is . Using geometric mean theorem we can show that must be . With Pythagorean Theorem we can say that . Multiplying both sides by and moving everything to LHS will give you Since must be in the form we can assume that is most likely a positive fraction in the form where is a factor of . Testing the factors in synthetic division would lead , giving us our desired answer . ~Lopkiloinm
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