Difference between revisions of "2011 AIME II Problems/Problem 13"
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==Solution 2== | ==Solution 2== | ||
− | + | This takes a slightly different route than Solution 1. | |
− | + | Solution 1 proves that <math>\angleDPB=120^{\circ}</math> and that <math>\overline{BP} = \overline{DP}</math>. | |
− | + | Construct diagonal <math>\overline{BD}</math> and using the two statements above it quickly becomes clear that <math>\angleBDP = \angleDBP = 30^{\circ}</math> by isosceles triangle base angles. | |
− | + | Let the midpoint of diagonal <math>\overline{AC}</math> be <math>M</math>, and since the diagonals are perpendicular, both triangle <math>DMP</math> and triangle <math>BMP</math> are 30-60-90 right triangles. | |
− | + | Since <math>\overline{AB} = 12</math>, <math>\overline{AC} = \overline{BD} = 12\sqrt{2}</math> and <math>\overline{BM} = \overline{DM} = 6\sqrt{2}</math>. | |
− | + | 30-60-90 triangles' sides are in the ratio <math>1 : \sqrt{3} : 2</math>, so <math>\overline{MP} = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt {6}</math>. | |
− | + | <math>\overlineAP = \overline{MP} + \overline{BM} = 6\sqrt{2} + 2\sqrt{6} = \sqrt{72} + \sqrt{24}</math>. | |
− | + | Hence, <math>72 + 24 = 96</math>. | |
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==See also== | ==See also== |
Revision as of 16:39, 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
This takes a slightly different route than Solution 1.
Solution 1 proves that $\angleDPB=120^{\circ}$ (Error compiling LaTeX. ! Undefined control sequence.) and that . Construct diagonal and using the two statements above it quickly becomes clear that $\angleBDP = \angleDBP = 30^{\circ}$ (Error compiling LaTeX. ! Undefined control sequence.) 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 . $\overlineAP = \overline{MP} + \overline{BM} = 6\sqrt{2} + 2\sqrt{6} = \sqrt{72} + \sqrt{24}$ (Error compiling LaTeX. ! Undefined control sequence.). 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 |