Difference between revisions of "2011 AIME II Problems/Problem 13"

(Solution 2)
(Solution 2)
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In triangle <math>M_4O_1P</math>, <math>tan 75 = \frac{\frac{6\sqrt{2}+i}{2}}{6\sqrt{2}-\frac{6\sqrt{2}+i}{2}}</math>.  
 
In triangle <math>M_4O_1P</math>, <math>tan 75 = \frac{\frac{6\sqrt{2}+i}{2}}{6\sqrt{2}-\frac{6\sqrt{2}+i}{2}}</math>.  
Simplifying yields <math>tan 15 = \frac{d+i}{d-i}</math>. The half-angle identity gives <math>tan 15 = \frac{sin 30}{1+cos 30} = 2-\sqrt{3}</math> so <math>2-\sqrt{3} = \frac{d+i}{d-i}</math>.  
+
And so <math>2+\sqrt{3} = \frac{6\sqrt{2}+i}{6\sqrt{2}-i}</math>.
  
Solving for <math>i</math> gives <math>i=2\sqrt{6}</math>. To find the total length <math>\overline{AP}</math>, we add <math>d+i=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}</math>. Hence, <math>72+24 = \framebox[1.5\width]{96.}</math>.
+
Solving for <math>i</math> gives <math>i = 2\sqrt{6}</math>. To find the total length <math>\overline{AP}</math>, we add <math>6\sqrt{2}+i=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}</math>. Hence, <math>72+24 = \framebox[1.5\width]{96.}</math>.
  
 
==See also==
 
==See also==

Revision as of 16:05, 1 December 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 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. ! 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 $x = AP$,

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

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

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

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

Preliminary Step: Define variables

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 $M_5$.

Step 1: Prove $PO_1=PO_2$ and thus triangle $\deltaPO_1O_2$ (Error compiling LaTeX. ! Undefined control sequence.) is isosceles

Imagine that $P$ is collocated with $M_2$, that is that $P$ is the center of square $ABCD$. If $AB=12$, then $AC=12\sqrt{2}$, and $AM_2=AP=6\sqrt{2}$.

Then, for every increment of $i$ along diagonal $\overline{AC}$ toward vertex $C$, $\overline{AP}=6\sqrt{2}+i$. If point $P$ is shifting at increment $i$, then clearly the midpoints of $AP$ and $CP$ are incrementing at the rate of $\frac{i}{2}$.

Directly from that we know that the perpendicular bisectors of $AP$ and $CP$ are also incrementing at the rate of $\frac{i}{2}$, and since the perpendicular bisectors of $AB$ and $CD$ are unchanging regardless of the location of $P$, it's easy to see that the circumcenters of triangles $ABP$ and $CDP$ are shifting at the same rate.


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 have congruent heights and bases. If these two triangles are congruent, then both circumscribed circles are congruent, and hence any radii of those circles would also be congruent. From this, triangle $O_1O_2P$ is also isosceles because two of its legs are circumradii.

Step 2: Set up equations to solve for $i$

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 $M_4O_1P$, $tan 75 = \frac{\frac{6\sqrt{2}+i}{2}}{6\sqrt{2}-\frac{6\sqrt{2}+i}{2}}$. And so $2+\sqrt{3} = \frac{6\sqrt{2}+i}{6\sqrt{2}-i}$.

Solving for $i$ gives $i = 2\sqrt{6}$. To find the total length $\overline{AP}$, we add $6\sqrt{2}+i=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}$. Hence, $72+24 = \framebox[1.5\width]{96.}$.

See also

2011 AIME II (ProblemsAnswer KeyResources)
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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