2011 AIME II Problems/Problem 13

Revision as of 17:39, 1 December 2011 by Hoipalloi9001 (talk | contribs) (Solution 2)


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 1


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. ! Undefined control sequence.). 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 $x = AP$,




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 = $\framebox[1.5\width]{96.}$

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 $\overline{BP} = \overline{DP}$. Construct diagonal $\overline{BD}$ 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 $\overline{AC}$ be $M$, and since the diagonals are perpendicular, both triangle $DMP$ and triangle $BMP$ are 30-60-90 right triangles. Since $\overline{AB} = 12$, $\overline{AC} = \overline{BD} = 12\sqrt{2}$ and $\overline{BM} = \overline{DM} = 6\sqrt{2}$. 30-60-90 triangles' sides are in the ratio $1 : \sqrt{3} : 2$, so $\overline{MP} = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt {6}$. $\overlineAP = \overline{MP} + \overline{BM} = 6\sqrt{2} + 2\sqrt{6} = \sqrt{72} + \sqrt{24}$ (Error compiling LaTeX. ! Undefined control sequence.). Hence, $72 + 24 = \framebox[1.5\width]{96.}$.

See also

2011 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AIME Problems and Solutions
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