Difference between revisions of "2016 AIME I Problems/Problem 6"
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Mathgeek2006 (talk | contribs) (→Solution 1) |
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=Solution= | =Solution= | ||
==Solution 1== | ==Solution 1== | ||
− | + | Suppose we label the angles as shown below. | |
+ | <asy> | ||
+ | size(150); | ||
+ | import olympiad; | ||
+ | real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; | ||
+ | pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2)); | ||
+ | pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; | ||
+ | pair I=incenter(A,B,C); | ||
+ | pair L=extension(C,D,A,B); | ||
+ | dot(I^^A^^B^^C^^D); | ||
+ | draw(C--D); | ||
+ | path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} | ||
+ | draw(A--B--D--cycle); | ||
+ | draw(circumcircle(A,B,D)); | ||
+ | draw(A--C--B); | ||
+ | draw(A--I--B^^C--I); | ||
+ | draw(incircle(A,B,C)); | ||
+ | label("$A$",A,SW,fontsize(8)); | ||
+ | label("$B$",B,SE,fontsize(8)); | ||
+ | label("$C$",C,N,fontsize(8)); | ||
+ | label("$D$",D,S,fontsize(8)); | ||
+ | label("$I$",I,NE,fontsize(8)); | ||
+ | label("$L$",L,SW,fontsize(8)); | ||
+ | label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8)); | ||
+ | label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8)); | ||
+ | label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8)); | ||
+ | label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8)); | ||
+ | label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8)); | ||
+ | label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8)); | ||
+ | </asy> | ||
+ | As <math>\angle BCD</math> and <math>\angle BAD</math> intercept the same arc, we know that <math>\angle BAD=\gamma</math>. Similarly, <math>\angle ABD=\gamma</math>. Also, using <math>\triangle ICA</math>, we find <math>\angle CIA=180-\alpha-\gamma</math>. Therefore, <math>\angle AID=\alpha+\gamma</math>. Therefore, <math>\angle DAI=\angle AID=\alpha+\gamma</math>, so <math>\triangle AID</math> must be isosceles with <math>AD=ID=5</math>. Similarly, <math>BD=ID=5</math>. Then <math>\triangle DLB \sim \triangle ALC</math>, hence <math>\frac{AL}{AC} = \frac{3}{5}</math>. Also, <math>AI</math> bisects <math>\angle LAI</math>, so by the Angle Bisector Theorem <math>\frac{CI}{IL} =\frac{AC}{AL}= \frac{5}{3}</math>. Thus <math>CI = \frac{10}{3}</math>, and the answer is <math>\boxed{013}</math>. | ||
==Solution 2== | ==Solution 2== |
Revision as of 00:15, 7 March 2016
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
This is a cheap solution.
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 .
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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