Difference between revisions of "2011 AIME II Problems/Problem 13"
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==Problem== | ==Problem== | ||
− | + | Point <math>P</math> lies on the diagonal <math>AC</math> of [[square]] <math>ABCD</math> with <math>AP > CP</math>. Let <math>O_{1}</math> and <math>O_{2}</math> be the [[circumcenter]]s of triangles <math>ABP</math> and <math>CDP</math> respectively. Given that <math>AB = 12</math> and <math>\angle O_{1}PO_{2} = 120^{\circ}</math>, then <math>AP = \sqrt{a} + \sqrt{b}</math>, where <math>a</math> and <math>b</math> are positive integers. Find <math>a + b</math>. | |
− | Point P lies on the diagonal AC of square ABCD with AP > CP. Let <math>O_{1}</math> and <math>O_{2}</math> be the | ||
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==Solution== | ==Solution== | ||
+ | <geogebra>7b0d7e3170597705121a87857a112a90dff8cac9</geogebra> | ||
+ | Denote the [[midpoint]] of <math>\overline{DC}</math> be <math>E</math> and the midpoint of <math>\overline{AB}</math> be <math>F</math>. Because they are the circumcenters, both Os lie on the [[perpendicular bisector]]s of <math>AB</math> and <math>CD</math> and these bisectors go through <math>E</math> and <math>F</math>. | ||
− | + | It is given that <math>\angleO_{1}PO_{2}=120^{\circ}</math>. Because <math>O_{1}P</math> and <math>O_{1}B</math> are [[radius|radii]] of the same circle, the have the same length. This is also true of <math>O_{2}P</math> and <math>O_{2}D</math>. Because <math>m\angle CAB=m\angle ACD=45^{\circ}</math>, <math>m\stackrel{\frown}{PD}=m\stackrel{\frown}{PB}=2(45^{\circ})=90^{\circ}</math>. Thus, <math>O_{1}PB</math> and <math>O_{2}PD</math> are isosceles right triangles. Using the given information above and symmetry, <math>m\angle DPB = 120^{\circ}</math>. 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. | |
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− | It is given that <math>\angleO_{1}PO_{2}=120^{\circ}</math>. Because <math>O_{1}P</math> and <math>O_{1}B</math> are radii of the same circle, the have the same length. This is also true of <math>O_{2}P</math> and <math>O_{2}D</math>. Because <math>m\angle CAB=m\angle ACD=45^{\circ}</math>, <math>m\stackrel{\frown}{PD}=m\stackrel{\frown}{PB}=2(45^{\circ})=90^{\circ}</math>. Thus, <math>O_{1}PB</math> and <math>O_{2}PD</math> are isosceles right triangles. Using the given information above and symmetry, <math>m\angle DPB = 120^{\circ}</math>. 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. | ||
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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 <math>O_{1}BF</math> and <math>O_{2}DE</math> have measures of 30 degrees. Thus, both triangles <math>O_{1}BF</math> and <math>O_{2}DE</math> 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, <math>DO_{2}=BO_{1}=4\sqrt{3}</math>. Because of 45-45-90 right triangles, <math>PB=PD=4\sqrt{6}</math>. | 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 <math>O_{1}BF</math> and <math>O_{2}DE</math> have measures of 30 degrees. Thus, both triangles <math>O_{1}BF</math> and <math>O_{2}DE</math> 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, <math>DO_{2}=BO_{1}=4\sqrt{3}</math>. Because of 45-45-90 right triangles, <math>PB=PD=4\sqrt{6}</math>. | ||
− | + | Now, using [[Law of Cosines]] on <math>\triangle ABP</math> and letting <math>x = AP</math>, | |
− | Now, using Law of Cosines on triangle ABP and letting AP | ||
<math>96=144+x^{2}-24x\frac{\sqrt{2}}{2}</math> | <math>96=144+x^{2}-24x\frac{\sqrt{2}}{2}</math> | ||
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<math>0=x^{2}-12x\sqrt{2}+48</math> | <math>0=x^{2}-12x\sqrt{2}+48</math> | ||
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Using quadratic formula, | Using quadratic formula, | ||
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<math>x = \frac{12 \sqrt{2} \pm \sqrt{288-(4)(48)}}{2}</math> | <math>x = \frac{12 \sqrt{2} \pm \sqrt{288-(4)(48)}}{2}</math> | ||
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Because it is given that <math>AP > CP</math>, <math>AP>6\sqrt{2}</math>, so the minus version of the above equation is too small. | Because it is given that <math>AP > CP</math>, <math>AP>6\sqrt{2}</math>, so the minus version of the above equation is too small. | ||
Thus, <math>AP=\sqrt{72}+ \sqrt{24}</math> and a + b = 24 + 72 = <math>\framebox[1.5\width]{96.}</math> | Thus, <math>AP=\sqrt{72}+ \sqrt{24}</math> and a + b = 24 + 72 = <math>\framebox[1.5\width]{96.}</math> | ||
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+ | ==See also== | ||
+ | {{AIME box|year=2011|n=II|num-b=12|num-a=14}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 21:21, 22 August 2011
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
<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 =
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 |