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
This takes a slightly different route than Solution 1.
Solution 1 proves that $\angleDPB=120^{\circ}$ (Error compiling LaTeX. Unknown error_msg) and that .
Construct diagonal
and using the two statements above it quickly becomes clear that $\angleBDP = \angleDBP = 30^{\circ}$ (Error compiling LaTeX. Unknown error_msg) by isosceles triangle base angles.
Let the midpoint of diagonal
be
, and since the diagonals are perpendicular, both triangle
and triangle
are 30-60-90 right triangles.
Since
,
and
.
30-60-90 triangles' sides are in the ratio
, so
.
$\overlineAP = \overline{MP} + \overline{BM} = 6\sqrt{2} + 2\sqrt{6} = \sqrt{72} + \sqrt{24}$ (Error compiling LaTeX. Unknown error_msg).
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 |