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
Solution 8 (Cyclic Quadrilaterals)
Connect to and to to form quadrilateral . Since quadrilateral is cyclic, we can apply Ptolemy's Theorem on the quadrilateral.
Denote the length of and as (they must be congruent, as and are both inscribed in arcs that have the same degree measure due to the angle bisector intersecting the circumcircle at ), and the lengths of , , , and as , respectively.
After applying Ptolemy's, one will get that:
Next, since is cyclic, triangles and are similar, yielding the following equation once simplifications are made to the equation , with the length of written in terms of using the angle bisector theorem on triangle :
Next, drawing in the bisector of to the incenter , and applying the angle bisector theorem, we have that:
Now, solving for in the second equation, and in the third equation and plugging them both back into the first equation, and making the substitution , we get the quadratic equation:
Solving, we get , which gives and , when we rewrite the above equations in terms of . Thus, our answer is and we're done.
Let , and . Then, notice that , so . Also, by the incenter-excenter lemma, . Therefore, by Ptolemy's Theorem on cyclic quadrilateral , , so , so . Solving, we get that , so and the answer is .
Perform a Inversion followed by a reflection along the angle bisector of .
It's well known that where is the excenter.
Also by Fact 5, .
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