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2011 AIME II Problems/Problem 13

Revision as of 16:04, 28 November 2011 by Hoipalloi9001 (talk | contribs) (→‎Solution 2)

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

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

Solution 2

Let the midpoint of side $\overline{AB}$ be $M_1$, the midpoint of diagonal $\overline{AC}$ be $M_2$, the midpoint of side $\overline{CD}$ be $M_3$, the midpoint of segment $\overline{AP}$ be $M_4$, and the midpoint of $\overline{CP}$ be $M_5$. 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 $C$, $\overline{AP}=d+i$. If point $P$ is shifting at increment $i$, then $M_4$ and $M_5$ are incrementing at the same rate (one is growing at a certain rate, the other is shortening at that same rate). This also means that the perpendicular bisectors of both $\overline{AP}$ and $\overline{CP}$ are shifting at the same rate. Circumcenters are formed by the perpendicular bisectors of the legs, so $O_1$ and $O_2$ must also be shifting at the same constant rate. For any location of $P$, both $O_1$ and $O_2$ will lie on line $\overline{M_1M_3}$ because the perpendicular bisector of the side of the square is not changing. When $P$ is located at $M_2$, $O_1$ and $O_2$ are located at $M_1$ and $M_3$, respectively. As $P$ shifts towards $C$, $O_1$ and $O_2$ shift down along line $\overline{M_1M_3}$. Both circumcenters are shifting at the same constant rate, so $\overline{M_1O_1}=\overline{M_3O_2}$. Therefore, triangles $ABO_1$ and $CDO_2$ are congruent because they are both isosceles and both have congruent heights and bases. If these two triangles are congruent, then both circumscribed circles are congruent. Triangle $O_1O_2P$ is also isosceles because two of its legs are circumradii. Now, given angle $O_1PO_2=120^{\circ}$, angle $PO_1O_2=30^{\circ}$. We know that angle $M_1AM_2=AM_2M_1=M_2O_1M_4=45^{\circ}$. Therefore, angle $M_4O_1P=30+45=75^{\circ}$. In triangle $\deltaM_4O_1P$ (Error compiling LaTeX. Unknown error_msg), $tan 75 = \frac{\frac{d+i}{2}}{d-\frac{d+i}{2}}$. Simplifying yields $tan 15 = \frac{d+i}{d-i}$. The half-angle identity gives $tan 15 = \frac{sin 30}{1+cos 30} = 2-\sqrt{3}$. Solving for $i$ by subsituting $6\sqrt{2}=d$ 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, $72+24 = \framebox[1.5\width]{96.}$.

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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