Difference between revisions of "2016 AIME I Problems/Problem 6"
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==Solution 6== | ==Solution 6== | ||
− | Let <math>I_C</math> be the excenter opposite to <math>C</math> in <math>ABC</math>.By | + | Let <math>I_C</math> be the excenter opposite to <math>C</math> in <math>ABC</math>. By the incenter-excenter lemma <math>DI=DC \therefore</math> <math>LI_C=8,LI=2,II_C=10</math>. Its well known that <math>(I_C,I,L,C)=-1 \implies \dfrac{LI_C}{LI}.\dfrac{CI}{CI_C}=-1 \implies \dfrac{CI}{CI+10}=\dfrac{1}{4} \implies \boxed{CI=\dfrac{10}{3}}</math>.<math>\blacksquare</math> |
~Pluto1708 | ~Pluto1708 | ||
+ | |||
+ | Alternate solution: "We can use the angle bisector theorem on <math>\triangle CBL</math> and bisector <math>BI</math> to get that <math>\tfrac{CI}{IL}=\tfrac{CI}{2}=\tfrac{BC}{BL}</math>. Since <math>\triangle CBL \sim \triangle ADL</math>, we get <math>\tfrac{BC}{BL}=\tfrac{AD}{DL}=\tfrac{5}{3}</math>. Thus, <math>CI=\tfrac{10}{3}</math> and <math>p+q=\boxed{13}</math>." | ||
+ | (https://artofproblemsolving.com/community/c759169h1918283_geometry_problem) | ||
== See also == | == See also == | ||
{{AIME box|year=2016|n=I|num-b=5|num-a=7}} | {{AIME box|year=2016|n=I|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:17, 25 September 2019
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)
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.