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
, and the midpoint of side
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
,
is equivalent to
. From this, we know that both the midpoint of
and the midpoint of
shift
for every shift
of point
. Since
is a diagonal of a square, we know that angle
is
. So, angle
is also
. From this, the triangle formed by the midpoint of
,
, and
is and isosceles right triangle. This is also true for the triangle formed by the midpoint of
,
, and
. Using this idea, Both
and
are shifting down along line
at the rate of
per
shift of
. So,
, and since side triangles
and
share that height, congruent bases, and the fact that they are both isosceles (the sets of two congruent legs are circumradii of the same circle), they are congruent. This means that triangle
is also isosceles since the segments from
to each of the circumcenters are both radii of the congruent circles (the two triangles are congruent, so the radii of each circle is congruent, so the circles are congruent). Given any angle
that represents angle
, by equal base angles of an isosceles triagnle, angle
. From earlier, angle
, and so angle
. It follows that angle
. In the triangle formed by the midpoint of
,
, and
,
. Simplifying yields
. Substituting
and
, we get
. Using the half angle trig identity,
, so solving for
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 |