2021 AIME II Problems/Problem 14
Contents
Problem
Let be an acute triangle with circumcenter and centroid . Let be the intersection of the line tangent to the circumcircle of at and the line perpendicular to at . Let be the intersection of lines and . Given that the measures of and are in the ratio the degree measure of can be written as where and are relatively prime positive integers. Find .
Diagram
~MRENTHUSIASM
Solution 1
In this solution, all angle measures are in degrees.
Let be the midpoint of so that and are collinear. Let and
Note that:
- Since quadrilateral is cyclic by the Converse of the Inscribed Angle Theorem.
It follows that as they share the same intercepted arc
- Since quadrilateral is cyclic by the supplementary opposite angles.
It follows that as they share the same intercepted arc
Together, we conclude that by AA, so
Next, we express in terms of By angle addition, we have Substituting back gives from which
For the sum of the interior angles of we get Finally, we obtain from which the answer is
~Constance-variance ~MRENTHUSIASM
Solution 2
Let be the midpoint of . Because , and are cyclic, so is the center of the spiral similarity sending to , and . Because , it's easy to get from here.
~Lcz
Solution 3 (Easy and Simple)
Firstly, let be the midpoint of . Then, . Now, note that since , quadrilateral is cyclic. Also, because , is also cyclic. Now, we define some variables: let be the constant such that and . Also, let and (due to the fact that and are cyclic). Then, Now, because is tangent to the circumcircle at , , and . Finally, notice that . Then, Thus, and However, from before, , so . To finish the problem, we simply compute so our final answer is .
~advanture
Solution 4 (Why Isosceles)
quadrilateral is cyclic
as they share the same intersept
quadrilateral is cyclic
as they share the same intercept
In triangles and two pairs of angles are equal, which means that the third angles are also equal.
so
According to the Claim, is isosceles,
Claim
Let be an acute triangle with circumcenter
Let be the midpoint of so
If then
We define as the sum of this angle can be greater than
Proof
as they share the same intercept (an inscribed angle and half of central angle).
as they share the same intercept
If then
vladimir.shelomovskii@gmail.com, vvsss
Solution 5
Extend and meet line at . Extend to meet at . Since is the median from to , are collinear. Furthermore, is perpendicular to
Draw the circumcircle of , as , are collinear, lies on as is the Simson line of with respect to . Thus, , the answer is which is .
~bluesoul
Video Solution 1
https://www.youtube.com/watch?v=zFH1Z7Ydq1s
~Mathematical Dexterity
Video Solution 2
https://www.youtube.com/watch?v=7Bxr2h4btWo
~Osman Nal
Video Solution by Interstigation
https://www.youtube.com/watch?v=yIWe7ME6fpA
~Interstigation
See Also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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