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 $\angleO_{1}PO_{2}=120^{\circ}$ (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 = $\framebox[1.5\width]{96.}$

Solution 2

Let the midpoint of side $\overline{AB}$ be $M_1$ , the midpoint of diagonal $\overline{AC}$ be $M_2$ , and the midpoint of side $\overline{CD}$ be $M_3$ . Consider the special case in which $P$ is collocated with $M_2$ , that is that $P$ is the center of the square. Let $d$ be the half the length of the diagonal of any given square $ABCD$ . Then, for every increment of $i$ along diagonal $\overline{AC}$ toward vertex $C$ , $\overline{AP}$ is equivalent to $d+i$ . From this, we know that both the midpoint of $\overline{AP}$ and the midpoint of $\overline{CP}$ shift $\frac{i}{2}$ for every shift $i$ of point $P$ . Since $\overline{AC}$ is a diagonal of a square, we know that angle $BAC$ is $45^{\circ}$ . So, angle $AM_2M_1$ is also $45^{\circ}$ . From this, the triangle formed by the midpoint of $\overline{AP}$ , $M_2$ , and $O_1$ is and isosceles right triangle. This is also true for the triangle formed by the midpoint of $\overline{CP}$ , $O_2$ , and $M_2$ . Using this idea, Both $O_1$ and $O_2$ are shifting down along line $\overline{M_1M_2}$ at the rate of $\frac{d}{\sqrt{2}}-\sqrt{2}(\frac{d}{2}-\frac{i}{2})$ per $i$ shift of $P$ . So, $\overline{M_1O_1}=\overline{M_3O_2}$ , and since side triangles $ABO_1$ and $CDO_2$ share that height, congruent bases, and the fact that they are both isosceles (the sets of two congruent legs are circumradii of the same circle), they are congruent. This means that triangle $O_1O_2P$ is also isosceles since the segments from $P$ to each of the circumcenters are both radii of the congruent circles (the two triangles are congruent, so the radii of each circle is congruent, so the circles are congruent). Given any angle $\theta$ that represents angle $O_1PO_2$ , by equal base angles of an isosceles triagnle, angle $PO_1O_2=\frac{180-\theta}{2}=90-\frac{\theta}{2}$ . From earlier, angle $O_1M_2A=45^{\circ}$ , and so angle $O_1M_2P=135^{\circ}$ . It follows that angle $O_1PA=180-135-(90-\frac{\theta}{2})=-45+\frac{\theta}{2}$ . In the triangle formed by the midpoint of $\overline{AP}$ , $O_1$ , and $P$ , $tan(-45+\frac{\theta}{2})=\frac{d-\frac{d+i}{2}}{\frac{d+i}{2}}$ . Simplifying yields $tan(-45+\frac{\theta}{2})=\frac{d-i}{d+i}$ . Substituting $120^{\circ}=\theta$ and $6\sqrt{2}=d$ , we get $tan 15=\frac{6\sqrt{2}-i}{6\sqrt{2}+i}$ . Using the half angle trig identity, $tan 15=2-\sqrt{3}$ , so solving for $i$ gives $i=2\sqrt{6}$ . To find the total length $\overline{AP}$ , we add $d+i=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}$ . Hence, $72+24 = \framebox[1.5\width]{96.}$ .

2011 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2011 AIME II Problems/Problem 13

Contents

Problem

Solution 1

Solution 2

See also