Difference between revisions of "2011 AIME II Problems/Problem 13"
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==Solution 2== | ==Solution 2== | ||
− | Let the midpoint of side <math>\overline{AB}</math> be <math>M_1</math>, the midpoint of diagonal <math>\overline{AC}</math> be <math>M_2</math>, | + | Let the midpoint of side <math>\overline{AB}</math> be <math>M_1</math>, the midpoint of diagonal <math>\overline{AC}</math> be <math>M_2</math>, the midpoint of side <math>\overline{CD}</math> be <math>M_3</math>, the midpoint of segment <math>\overline{AP}</math> be <math>M_4</math>, and the midpoint of <math>\overline{CP}</math> be <math>M_5</math>. Consider the special case in which <math>P</math> is collocated with <math>M_2</math>, that is that <math>P</math> is the center of the square. Let <math>d</math> be the half the length of the diagonal of any given square <math>ABCD</math>. Then, for every increment of <math>i</math> along diagonal <math>\overline{AC}</math> toward vertex <math>C</math>, <math>\overline{AP}=d+i</math>. If point <math>P</math> is shifting at increment <math>i</math>, then <math>M_4</math> and <math>M_5</math> 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 <math>\overline{AP}</math> and <math>\overline{CP}</math> are shifting at the same rate. Circumcenters are formed by the perpendicular bisectors of the legs, so <math>O_1</math> and <math>O_2</math> must also be shifting at the same constant rate. For any location of <math>P</math>, both <math>O_1</math> and <math>O_2</math> will lie on line <math>\overline{M_1M_3}</math> because the perpendicular bisector of the side of the square is not changing. When <math>P</math> is located at <math>M_2</math>, <math>O_1</math> and <math>O_2</math> are located at <math>M_1</math> and <math>M_3</math>, respectively. 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>. Therefore, triangles <math>ABO_1</math> and <math>CDO_2</math> 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 <math>O_1O_2P</math> is also isosceles because two of its legs are circumradii. Now, given angle <math>O_1PO_2=120^{\circ}</math>, angle <math>PO_1O_2=30^{\circ}</math>. We know that angle <math>M_1AM_2=AM_2M_1=M_2O_1M_4=45^{\circ}</math>. Therefore, angle <math>M_4O_1P=30+45=75^{\circ}</math>. In triangle <math>\deltaM_4O_1P</math>, <math>tan 75 = \frac{\frac{d+i}{2}}{d-\frac{d+i}{2}}</math>. Simplifying yields <math>tan 15 = \frac{d+i}{d-i}</math>. The half-angle identity gives <math>tan 15 = \frac{sin 30}{1+cos 30} = 2-\sqrt{3}</math>. Solving for <math>i</math> by subsituting <math>6\sqrt{2}=d</math> gives <math>i=2\sqrt{6}</math>. To find the total length <math>\overline{AP}</math>, we add <math>d+i=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}</math>. Hence, <math>72+24 = \framebox[1.5\width]{96.}</math>. |
==See also== | ==See also== |
Revision as of 17:04, 28 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
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 |