Difference between revisions of "2011 AIME II Problems/Problem 13"
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In triangle <math>M_4O_1P</math>, <math>tan 75 = \frac{\frac{6\sqrt{2}+i}{2}}{6\sqrt{2}-\frac{6\sqrt{2}+i}{2}}</math>. | In triangle <math>M_4O_1P</math>, <math>tan 75 = \frac{\frac{6\sqrt{2}+i}{2}}{6\sqrt{2}-\frac{6\sqrt{2}+i}{2}}</math>. | ||
− | + | And so <math>2+\sqrt{3} = \frac{6\sqrt{2}+i}{6\sqrt{2}-i}</math>. | |
− | Solving for <math>i</math> | + | Solving for <math>i</math> gives <math>i = 2\sqrt{6}</math>. To find the total length <math>\overline{AP}</math>, we add <math>6\sqrt{2}+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 16:05, 1 December 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. ! Undefined control sequence.). 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. ! Undefined control sequence.) 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 , . And 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 |