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Difference between revisions of "2011 AIME II Problems/Problem 13"
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==Problem==
 
==Problem==
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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>.
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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 circumcenters of triangles ABP and CDP respectively. Given that AB = 12 and <math>\angle O_{1}PO_{2} = 120^{\circ}</math>, then <math>AP = \sqrt{a} + \sqrt{b}</math>, where a and b are positive integers. Find a + b.
 
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==Solution==
 
==Solution==
 +
<geogebra>7b0d7e3170597705121a87857a112a90dff8cac9</geogebra>
  
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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>.
  
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<geogebra>7b0d7e3170597705121a87857a112a90dff8cac9</geogebra>
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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 [[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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Denote the midpoint of DC be E and the midpoint of AB be F. Because they are the circumcenters, both Os lie on the perpendicular bisectors of AB and CD and these bisectors go through E and F.
 
−
 
 
−
 
 
−
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.
 
−
 
 
  
 
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>.
  
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+
Now, using [[Law of Cosines]] on <math>\triangle ABP</math> and letting <math>x = AP</math>,  
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Now, using Law of Cosines on triangle ABP and letting AP be x,  
 
  
 
<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>
 +
 +
==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 $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$. Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$, then $AP = \sqrt{a} + \sqrt{b}$, where $a$ and $b$ are positive integers. Find $a + b$.

Solution

<geogebra>7b0d7e3170597705121a87857a112a90dff8cac9</geogebra>

Denote the midpoint of $\overline{DC}$ be $E$ and the midpoint of $\overline{AB}$ be $F$. Because they are the circumcenters, both Os lie on the perpendicular bisectors of $AB$ and $CD$ and these bisectors go through $E$ and $F$.

It is given that $\angleO_{1}PO_{2}=120^{\circ}$ (Error compiling LaTeX. Unknown error_msg). Because $O_{1}P$ and $O_{1}B$ are radii of the same circle, the have the same length. This is also true of $O_{2}P$ and $O_{2}D$. Because $m\angle CAB=m\angle ACD=45^{\circ}$, $m\stackrel{\frown}{PD}=m\stackrel{\frown}{PB}=2(45^{\circ})=90^{\circ}$. Thus, $O_{1}PB$ and $O_{2}PD$ are isosceles right triangles. Using the given information above and symmetry, $m\angle DPB = 120^{\circ}$. 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 $O_{1}BF$ and $O_{2}DE$ have measures of 30 degrees. Thus, both triangles $O_{1}BF$ and $O_{2}DE$ 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, $DO_{2}=BO_{1}=4\sqrt{3}$. Because of 45-45-90 right triangles, $PB=PD=4\sqrt{6}$.

Now, using Law of Cosines on $\triangle ABP$ and letting $x = AP$,

$96=144+x^{2}-24x\frac{\sqrt{2}}{2}$

$96=144+x^{2}-12x\sqrt{2}$

$0=x^{2}-12x\sqrt{2}+48$

Using quadratic formula,

$x = \frac{12 \sqrt{2} \pm \sqrt{288-(4)(48)}}{2}$

$x = \frac{12 \sqrt{2} \pm \sqrt{288-192}}{2}$

$x = \frac{12 \sqrt{2} \pm \sqrt{96}}{2}$

$x = \frac{2 \sqrt{72} \pm 2 \sqrt{24}}{2}$

$x = \sqrt{72} \pm \sqrt{24}$


Because it is given that $AP > CP$, $AP>6\sqrt{2}$, so the minus version of the above equation is too small. Thus, $AP=\sqrt{72}+ \sqrt{24}$ and a + b = 24 + 72 = $\framebox[1.5\width]{96.}$

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
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