2011 AIME II Problems/Problem 13

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 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 $\angle O_{1}PO_{2}=120^{\circ}$ . 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 = $\framebox[1.5\width]{096.}$

Solution 2

This takes a slightly different route than Solution 1.

Solution 1 proves that $\angle{DPB}=120^{\circ}$ and that $\overline{BP} = \overline{DP}$ . Construct diagonal $\overline{BD}$ and using the two statements above it quickly becomes clear that $\angle{BDP} = \angle{DBP} = 30^{\circ}$ 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 $\overline{MP} = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt {6}$ . $\overline{AP} = \overline{MP} + \overline{BM} = 6\sqrt{2} + 2\sqrt{6} = \sqrt{72} + \sqrt{24}$ . Hence, $72 + 24 = \framebox[1.5\width]{096}$ .

Solution 3

Use vectors. In an $xy$ plane, let $(-s,0)$ be $A$ , $(0,s)$ be $B$ , $(s,0)$ be $C$ , $(0,-s)$ be $D$ , and $(p,0)$ be P, where $s=|AB|/\sqrt{2}=6\sqrt{2}$ . It remains to find $p$ .

The line $y=-x$ is the perpendicular bisector of $AB$ and $CD$ , so $O_1$ and $O_2$ lies on the line. Now compute the perpendicular bisector of $AP$ . The center has coordinate $(\frac{p-s}{2},0)$ , and the segment is part of the $x$ -axis, so the perpendicular bisector has equation $x=\frac{p-s}{2}$ . Since $O_1$ is the circumcenter of triangle $ABP$ , it lies on the perpendicular bisector of both $AB$ and $AP$ , so $O_1=(\frac{p-s}{2},-\frac{p-s}{2})$ Similarly, $O_2=(\frac{p+s}{2},-\frac{p+s}{2})$ The relation $\angle O_1PO_2=120^\circ$ can now be written using dot product as $\vec{PO_1}\cdot\vec{PO_2}=|\vec{PO_1}|\cdot|\vec{PO_2}|\cos 120^\circ=-\frac{1}{2}|\vec{PO_1}|\cdot|\vec{PO_2}|$ Computation of both sides yields $\frac{p^2-s^2}{p^2+s^2}=-\frac{1}{2}$ Solve for $p$ gives $p=s/\sqrt{3}=2\sqrt{6}$ , so $AP=s+p=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}$ . The answer is 72+24 $\Rightarrow\boxed{096}$

Solution 4

Translate $\triangle{ABP}$ so that the image of $AB$ coincides $DC$ . Let the image of $P$ be $P’$ .

$\angle{DPC}=\angle{CPB}$ by symmetry, and $\angle{APB}=\angle{DP’C}$ because translation preserves angles. Thus $\angle{DP’C}+\angle{CPD}=\angle{CPB}+\angle{APB}=180^\circ$ . Therefore, quadrilateral $CPDP’$ is cyclic. Thus the image of $O_1$ coincides with $O_2$ .

$O_1P$ is parallel to $O_2P’$ so $\angle{P’O_2P}=\angle{O_1PO_2}=120^\circ$ , so $\angle{PDP’}=60^\circ$ and $\angle{PDC}=15^\circ$ , thus $\angle{ADP}=75^circ$ .

Let $M$ be the foot of the perpendicular from $D$ to $AC$ . Then $\triangle{AMD}$ is a 45-45-90 triangle and $\triangle{DMP}$ is a 30-60-90 triangle. Thus

$AM=6\sqrt{2}$ and $MP=\frac{6\sqrt{2}}{\sqrt{3}}$ .

This gives us $AP=AM+MP=\sqrt{72}+\sqrt{24}$ , and the answer is $72+24=\boxed{96}.

==Solution 5==

Reflect$ (Error compiling LaTeX. Unknown error_msg)O_1 $across$ AP $to$ O_1’ $. By symmetry$ O_1’ $is the orthocenter of$ \triangle{ADP}$$ (Error compiling LaTeX. Unknown error_msg)\angle{DO_1’P}=2*\angle{DAP} $= 90^circ$ , so $\angle{O_1’PD}=45^circ$ similarly $\angle{DO_2P}=2*\angle{DCP}$ = 90^circ $, so$ \angle{O_2PD}=45^circ $Thus,$ \angle{O_1’PO_2}=90^circ $, so that$ \angle{O_1’PO_1}=120^circ - 90^circ = 30^circ$

2011 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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