2011 AIME II Problems/Problem 13
Contents
Problem
Point lies on the diagonal of square with . Let and be the circumcenters of triangles and respectively. Given that and , then , where and are positive integers. Find .
Solution 1
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Denote the midpoint of be and the midpoint of be . Because they are the circumcenters, both Os lie on the perpendicular bisectors of and and these bisectors go through and .
It is given that $\angleO_{1}PO_{2}=120^{\circ}$ (Error compiling LaTeX. Unknown error_msg). Because and are radii of the same circle, the have the same length. This is also true of and . Because , . Thus, and are isosceles right triangles. Using the given information above and symmetry, . Because ABP and ADP share one side, have one side with the same length, and one equal angle, they are congruent by SAS. This is also true for triangle CPB and CPD. Because angles APB and APD are equal and they sum to 120 degrees, they are each 60 degrees. Likewise, both angles CPB and CPD have measures of 120 degrees.
Because the interior angles of a triangle add to 180 degrees, angle ABP has measure 75 degrees and angle PDC has measure 15 degrees. Subtracting, it is found that both angles and have measures of 30 degrees. Thus, both triangles and are 30-60-90 right triangles. Because F and E are the midpoints of AB and CD respectively, both FB and DE have lengths of 6. Thus, . Because of 45-45-90 right triangles, .
Now, using Law of Cosines on and letting ,
Using quadratic formula,
Because it is given that , , so the minus version of the above equation is too small.
Thus, and a + b = 24 + 72 =
Solution 2
Let the midpoint of side be , the midpoint of diagonal be , the midpoint of side be , the midpoint of segment be , and the midpoint of be .
Consider the special case in which is collocated with , that is that is the center of the square. Let be the half the length of the diagonal of any given square . Then, for every increment of along diagonal toward vertex , . If point is shifting at increment , then and are incrementing at the same rate (one is growing at a certain rate, the other is shortening at that same rate). This also means that the perpendicular bisectors of both and are shifting at the same rate. Circumcenters are formed by the perpendicular bisectors of the legs, so and must also be shifting at the same constant rate.
For any location of , both and will lie on line because the perpendicular bisector of the side of the square is not changing. When is located at , and are located at and , respectively. As shifts towards , and shift down along line . Both circumcenters are shifting at the same constant rate, so . Therefore, triangles and are congruent because they are both isosceles and both have congruent heights and bases. If these two triangles are congruent, then both circumscribed circles are congruent. Triangle is also isosceles because two of its legs are circumradii.
Now, given angle , angle . We know that angle . Therefore, angle . In triangle $\deltaM_4O_1P$ (Error compiling LaTeX. Unknown error_msg), . Simplifying yields . The half-angle identity gives . Solving for by subsituting gives . To find the total length , we add . Hence, .
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |