Difference between revisions of "2011 AIME II Problems/Problem 13"
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<math>0=x^{2}-12x\sqrt{2}+48</math> | <math>0=x^{2}-12x\sqrt{2}+48</math> | ||
− | Using quadratic formula, | + | Using quadratic formula, we arrive at |
− | < | + | <cmath>x = \sqrt{72} \pm \sqrt{24}</cmath> |
− | + | Taking the positive root, <math>AP=\sqrt{72}+ \sqrt{24}</math> and a + b = 24 + 72 = <math>\framebox[1.5\width]{096.}</math> | |
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==Solution 2== | ==Solution 2== |
Revision as of 20:31, 27 February 2018
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
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 . 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, we arrive at
Taking the positive root, and a + b = 24 + 72 =
Solution 2
This takes a slightly different route than Solution 1.
Solution 1 proves that and that
.
Construct diagonal
and using the two statements above it quickly becomes clear that
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
.
.
Hence,
.
Solution 3
Use vectors. In an plane, let
be
,
be
,
be
,
be
, and
be P, where
. It remains to find
.
The line is the perpendicular bisector of
and
, so
and
lies on the line. Now compute the perpendicular bisector of
. The center has coordinate
, and the segment is part of the
-axis, so the perpendicular bisector has equation
. Since
is the circumcenter of triangle
, it lies on the perpendicular bisector of both
and
, so
Similarly,
The relation
can now be written using dot product as
Computation of both sides yields
Solve for
gives
, so
. The answer is 72+24
Solution 4
Translate so that the image of
coincides
. Let the image of
be
.
by symmetry, and
because translation preserves angles. Thus
. Therefore, quadrilateral
is cyclic. Thus the image of
coincides with
.
is parallel to
so
, so
and
, thus
.
Let be the foot of the perpendicular from
to
. Then
is a 45-45-90 triangle and
is a 30-60-90 triangle. Thus
and
.
This gives us , and the answer is
Solution 5
Reflect across
to
. By symmetry
is the circumcenter of
=
, so
similarly =
, so
Thus, , so that
By symmetry,
Therefore, since is the circumcenter of
,
=
Therefore
Using sine rule in ,
, and the answer is
By Kris17
The Simple Way of the Coordinate Plane Solution
Why not use coordinates? After all, 45 degrees is rather friendly in terms of ordered-pair representation! We can set ,
,
,
. Let this
for some
.
We also know that the circumcenter is the intersection of all perpendicular bisectors of sides, but two will suffice also due to this property. Therefore, we see that is the intersection of
and, knowing the midpoint of
to be
and thus the equation to be
, we get
. Likewise for
it's
. Now what do we see?
(just look at the coordinates)! So both of those distances are
. Solving for
we get it to be
, since
. Multiply by
because we are looking for
to get the answer of
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.