Difference between revisions of "2011 AIME II Problems/Problem 13"
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Directly from that we know that the perpendicular bisectors of <math>AP</math> and <math>CP</math> are also incrementing at the rate of <math>\frac{i}{2}</math>, and since the perpendicular bisectors of <math>AB</math> and <math>CD</math> are unchanging regardless of the location of <math>P</math>, it's easy to see that the circumcenters of triangles <math>ABP</math> and <math>CDP</math> are shifting at the same rate. | Directly from that we know that the perpendicular bisectors of <math>AP</math> and <math>CP</math> are also incrementing at the rate of <math>\frac{i}{2}</math>, and since the perpendicular bisectors of <math>AB</math> and <math>CD</math> are unchanging regardless of the location of <math>P</math>, it's easy to see that the circumcenters of triangles <math>ABP</math> and <math>CDP</math> are shifting at the same rate. | ||
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As <math>P</math> shifts towards <math>C</math>, <math>O_1</math> and <math>O_2</math> shift down along line <math>\overline{M_1M_3}</math>. Both circumcenters are shifting at the same constant rate, so <math>\overline{M_1O_1}=\overline{M_3O_2}</math>. | As <math>P</math> shifts towards <math>C</math>, <math>O_1</math> and <math>O_2</math> shift down along line <math>\overline{M_1M_3}</math>. Both circumcenters are shifting at the same constant rate, so <math>\overline{M_1O_1}=\overline{M_3O_2}</math>. | ||
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Therefore, triangles <math>ABO_1</math> and <math>CDO_2</math> are congruent because they are both isosceles and have congruent heights and bases. If these two triangles are congruent, then both circumscribed circles are congruent, and hence any radii of those circles would also be congruent. From this, triangle <math>O_1O_2P</math> is also isosceles because two of its legs are circumradii. | Therefore, triangles <math>ABO_1</math> and <math>CDO_2</math> are congruent because they are both isosceles and have congruent heights and bases. If these two triangles are congruent, then both circumscribed circles are congruent, and hence any radii of those circles would also be congruent. From this, triangle <math>O_1O_2P</math> is also isosceles because two of its legs are circumradii. |
Revision as of 22:19, 30 November 2011
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
<geogebra>7b0d7e3170597705121a87857a112a90dff8cac9</geogebra>
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
Preliminary Step: Define variables
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
.
Step 1: Prove and thus triangle $\deltaPO_1O_2$ (Error compiling LaTeX. Unknown error_msg) is isosceles
Imagine that is collocated with
, that is that
is the center of square
.
If
,
then
,
and
.
Then, for every increment of along diagonal
toward vertex
,
. If point
is shifting at increment
, then clearly the midpoints of
and
are incrementing at the rate of
.
Directly from that we know that the perpendicular bisectors of and
are also incrementing at the rate of
, and since the perpendicular bisectors of
and
are unchanging regardless of the location of
, it's easy to see that the circumcenters of triangles
and
are shifting at the same rate.
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 have congruent heights and bases. If these two triangles are congruent, then both circumscribed circles are congruent, and hence any radii of those circles would also be congruent. From this, triangle
is also isosceles because two of its legs are circumradii.
Step 2: Set up equations to solve for
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
so
.
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 |