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

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==Solution 1==
 
==Solution 1==
<geogebra>7b0d7e3170597705121a87857a112a90dff8cac9</geogebra>
 
  
 
Denote the [[midpoint]] of <math>\overline{DC}</math> be <math>E</math> and the midpoint of <math>\overline{AB}</math> be <math>F</math>. Because they are the circumcenters, both Os lie on the [[perpendicular bisector]]s of <math>AB</math> and <math>CD</math> and these bisectors go through <math>E</math> and <math>F</math>.
 
Denote the [[midpoint]] of <math>\overline{DC}</math> be <math>E</math> and the midpoint of <math>\overline{AB}</math> be <math>F</math>. Because they are the circumcenters, both Os lie on the [[perpendicular bisector]]s of <math>AB</math> and <math>CD</math> and these bisectors go through <math>E</math> and <math>F</math>.
  
It is given that <math>\angleO_{1}PO_{2}=120^{\circ}</math>. Because <math>O_{1}P</math> and <math>O_{1}B</math> are [[radius|radii]] of the same circle, the have the same length. This is also true of <math>O_{2}P</math> and <math>O_{2}D</math>. Because <math>m\angle CAB=m\angle ACD=45^{\circ}</math>, <math>m\stackrel{\frown}{PD}=m\stackrel{\frown}{PB}=2(45^{\circ})=90^{\circ}</math>. Thus, <math>O_{1}PB</math> and <math>O_{2}PD</math> are isosceles right triangles. Using the given information above and symmetry, <math>m\angle DPB = 120^{\circ}</math>. 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.
+
It is given that <math>\angle O_{1}PO_{2}=120^{\circ}</math>. Because <math>O_{1}P</math> and <math>O_{1}B</math> are [[radius|radii]] of the same circle, the have the same length. This is also true of <math>O_{2}P</math> and <math>O_{2}D</math>. Because <math>m\angle CAB=m\angle ACD=45^{\circ}</math>, <math>m\stackrel{\frown}{PD}=m\stackrel{\frown}{PB}=2(45^{\circ})=90^{\circ}</math>. Thus, <math>O_{1}PB</math> and <math>O_{2}PD</math> are isosceles right triangles. Using the given information above and symmetry, <math>m\angle DPB = 120^{\circ}</math>. 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 <math>O_{1}BF</math> and <math>O_{2}DE</math> have measures of 30 degrees. Thus, both triangles <math>O_{1}BF</math> and <math>O_{2}DE</math> 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, <math>DO_{2}=BO_{1}=4\sqrt{3}</math>. Because of 45-45-90 right triangles, <math>PB=PD=4\sqrt{6}</math>.
 
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 <math>O_{1}BF</math> and <math>O_{2}DE</math> have measures of 30 degrees. Thus, both triangles <math>O_{1}BF</math> and <math>O_{2}DE</math> 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, <math>DO_{2}=BO_{1}=4\sqrt{3}</math>. Because of 45-45-90 right triangles, <math>PB=PD=4\sqrt{6}</math>.
  
Now, using [[Law of Cosines]] on <math>\triangle ABP</math> and letting <math>x = AP</math>,  
+
Now, letting <math>x = AP</math> and using [[Law of Cosines]] on <math>\triangle ABP</math>, we have
  
<math>96=144+x^{2}-24x\frac{\sqrt{2}}{2}</math>
+
<cmath>96=144+x^{2}-24x\frac{\sqrt{2}}{2}</cmath>
 +
<cmath>0=x^{2}-12x\sqrt{2}+48</cmath>
  
<math>96=144+x^{2}-12x\sqrt{2}</math>
+
Using the quadratic formula, we arrive at
  
<math>0=x^{2}-12x\sqrt{2}+48</math>
+
<cmath>x = \sqrt{72} \pm \sqrt{24}</cmath>
  
Using quadratic formula,
+
Taking the positive root, <math>AP=\sqrt{72}+ \sqrt{24}</math> and the answer is thus <math>\framebox[1.5\width]{096.}</math>
  
<math>x = \frac{12 \sqrt{2} \pm \sqrt{288-(4)(48)}}{2}</math>
+
==Solution 2==
 +
This takes a slightly different route than Solution 1.
  
<math>x = \frac{12 \sqrt{2} \pm \sqrt{288-192}}{2}</math>
+
Solution 1 proves that <math>\angle{DPB}=120^{\circ}</math> and that <math>\overline{BP} = \overline{DP}</math>.
 +
Construct diagonal <math>\overline{BD}</math> and using the two statements above it quickly becomes clear that <math>\angle{BDP} = \angle{DBP} = 30^{\circ}</math> by isosceles triangle base angles.
 +
Let the midpoint of diagonal <math>\overline{AC}</math> be <math>M</math>, and since the diagonals are perpendicular, both triangle <math>DMP</math> and triangle <math>BMP</math> are 30-60-90 right triangles.
 +
Since <math>\overline{AB} = 12</math>, <math>\overline{AC} = \overline{BD} = 12\sqrt{2}</math> and <math>\overline{BM} = \overline{DM} = 6\sqrt{2}</math>.
 +
30-60-90 triangles' sides are in the ratio <math>1 : \sqrt{3} : 2</math>, so <math>\overline{MP} = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt {6}</math>.
 +
<math>\overline{AP} = \overline{MP} + \overline{BM} = 6\sqrt{2} + 2\sqrt{6} = \sqrt{72} + \sqrt{24}</math>.
 +
Hence, <math>72 + 24 = \framebox[1.5\width]{096}</math>.
  
<math>x = \frac{12 \sqrt{2} \pm \sqrt{96}}{2}</math>
+
==Solution 3==
 +
Use vectors. In an <math>xy</math> plane, let <math>(-s,0)</math> be <math>A</math>, <math>(0,s)</math> be <math>B</math>, <math>(s,0)</math> be <math>C</math>, <math>(0,-s)</math> be <math>D</math>, and <math>(p,0)</math> be P, where <math>s=|AB|/\sqrt{2}=6\sqrt{2}</math>. It remains to find <math>p</math>.
  
<math>x = \frac{2 \sqrt{72} \pm 2 \sqrt{24}}{2}</math>
+
The line <math>y=-x</math> is the [[perpendicular bisector]] of <math>AB</math> and <math>CD</math>, so <math>O_1</math> and <math>O_2</math> lies on the line. Now compute the [[perpendicular bisector]] of <math>AP</math>. The center has coordinate <math>(\frac{p-s}{2},0)</math>, and the segment is part of the <math>x</math>-axis, so the perpendicular bisector has equation <math>x=\frac{p-s}{2}</math>. Since <math>O_1</math> is the [[circumcenter]] of triangle <math>ABP</math>, it lies on the perpendicular bisector of both <math>AB</math> and <math>AP</math>, so
 +
<cmath>
 +
O_1=(\frac{p-s}{2},-\frac{p-s}{2})
 +
</cmath>
 +
Similarly,
 +
<cmath>
 +
O_2=(\frac{p+s}{2},-\frac{p+s}{2})
 +
</cmath>
 +
The relation <math>\angle O_1PO_2=120^\circ</math> can now be written using [[Vectors|dot product]] as
 +
<cmath>
 +
\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}|
 +
</cmath>
 +
Computation of both sides yields
 +
<cmath>
 +
\frac{p^2-s^2}{p^2+s^2}=-\frac{1}{2}
 +
</cmath>
 +
Solve for <math>p</math> gives <math>p=s/\sqrt{3}=2\sqrt{6}</math>, so <math>AP=s+p=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}</math>. The answer is 72+24<math>\Rightarrow\boxed{096}</math>
  
<math>x = \sqrt{72} \pm \sqrt{24}</math>
+
==Solution 4==
 +
Translate <math>\triangle{ABP}</math> so that the image of <math>AB</math> coincides <math>DC</math>. Let the image of <math>P</math> be <math>P’</math>.
  
 +
<math>\angle{DPC}=\angle{CPB}</math> by symmetry, and <math>\angle{APB}=\angle{DP’C}</math> because translation preserves angles. Thus <math>\angle{DP’C}+\angle{CPD}=\angle{CPB}+\angle{APB}=180^\circ</math>. Therefore, quadrilateral <math>CPDP’</math> is cyclic. Thus the image of <math>O_1</math> coincides with <math>O_2</math>.
  
Because it is given that <math>AP > CP</math>, <math>AP>6\sqrt{2}</math>, so the minus version of the above equation is too small.
+
<math>O_1P</math> is parallel to <math>O_2P’</math> so <math>\angle{P’O_2P}=\angle{O_1PO_2}=120^\circ</math>, so <math>\angle{PDP’}=60^\circ</math> and <math>\angle{PDC}=15^\circ</math>, thus <math>\angle{ADP}=75^{\circ}</math>.
Thus, <math>AP=\sqrt{72}+ \sqrt{24}</math> and a + b = 24 + 72 = <math>\framebox[1.5\width]{96.}</math>
 
  
==Solution 2==
+
Let <math>M</math> be the foot of the perpendicular from <math>D</math> to <math>AC</math>. Then <math>\triangle{AMD}</math> is a 45-45-90 triangle and <math>\triangle{DMP}</math> is a 30-60-90 triangle. Thus
Let the midpoint of side <math>\overline{AB}</math> be <math>M_1</math>, the midpoint of diagonal <math>\overline{AC}</math> be <math>M_2</math>, and the midpoint of side <math>\overline{CD}</math> be <math>M_3</math>. Consider the general case in which <math>P</math> is collocated with <math>M_2</math>, that is that <math>P</math> is the center of the square. Let <math>d</math> be the half the length of the diagonal of any given square <math>ABCD</math>. Then, for every increment of <math>i</math> along diagonal <math>\overline{AC}</math> toward vertex <math>C</math>, <math>\overline{AP}</math> is equivalent to <math>d+i</math>. From this, we know that both the midpoint of <math>\overline{AP}</math> and the midpoint of <math>\overline{CP}</math> shift <math>\frac{i}{2}</math> for every shift <math>i</math> of point <math>P</math>. Since <math>\overline{AC}</math> is a diagonal of a square, we know that angle <math>BAC</math> is <math>45^{\circ}</math>. So, angle <math>AM_2M_1</math> is also <math>45^{\circ}</math>. From this, the triangle formed by the midpoint of <math>\overline{AP}</math>, <math>M_2</math>, and <math>O_1</math> is and isosceles right triangle. This is also true for the triangle formed by the midpoint of <math>\overline{CP}</math>, <math>O_2</math>, and <math>M_2</math>. Using this idea, Both <math>O_1</math> and <math>O_2</math> are shifting down along line <math>\overline{M_1M_2}</math> at the rate of <math>\frac{d}{\sqrt{2}}-\sqrt{2}(\frac{d}{2}-\frac{i}{2})</math> per <math>i</math> shift of <math>P</math>. So, <math>\overline{M_1O_1}=\overline{M_3O_2}</math>, and since side triangles <math>ABO_1</math> and <math>CDO_2</math> 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 <math>O_1O_2P</math> is also isosceles since the segments from <math>P</math> 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 <math>\theta</math> that represents angle <math>O_1PO_2</math>,  by equal base angles of an isosceles triagnle, angle <math>PO_1O_2=\frac{180-\theta}{2}=90-\frac{\theta}{2}</math>. From earlier, angle <math>O_1M_2A=45^{\circ}</math>, and so angle <math>O_1M_2P=135^{\circ}</math>. It follows that angle <math>O_1PA=180-135-(90-\frac{\theta}{2})=-45+\frac{\theta}{2}</math>. In the triangle formed by the midpoint of <math>\overline{AP}</math>, <math>O_1</math>, and <math>P</math>, <math>tan(-45+\frac{\theta}{2})=\frac{d-\frac{d+i}{2}}{\frac{d+i}{2}}</math>. Simplifying yields <math>tan(-45+\frac{\theta}{2})=\frac{d-i}{d+i}</math>. Substituting <math>120^{\circ}=\theta</math> and <math>6\sqrt{2}=d</math>, we get <math>tan 15=\frac{6\sqrt{2}-i}{6\sqrt{2}+i}</math>. Using the half angle trig identity, <math>tan 15=2-\sqrt{3}</math>, so 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>.
+
 
 +
<math>AM=6\sqrt{2}</math> and <math>MP=\frac{6\sqrt{2}}{\sqrt{3}}</math>.
 +
 
 +
This gives us <math>AP=AM+MP=\sqrt{72}+\sqrt{24}</math>, and the answer is <math>72+24=\boxed{096}.</math>
 +
 
 +
==Solution 5==
 +
 
 +
Reflect <math>O_1</math> across <math>AP</math> to <math>O_1'</math>. By symmetry <math>O_1’</math> is the circumcenter of <math>\triangle{ADP}</math>
 +
 
 +
<math>\angle{DO_1’P}</math> = <math>2*\angle{DAP} = 90^\circ</math>, so <math>\angle{O_1’PD}=45^\circ</math>
 +
 
 +
similarly <math>\angle{DO_2P}</math> = <math>2*\angle{DCP} = 90^\circ</math>, so <math>\angle{O_2PD}=45^\circ</math>
 +
 
 +
Therefore <math>\angle{O_1’PO_2}=90^\circ</math>, so that <math>\angle{O_1’PO_1} =120^\circ - 90^\circ = 30^\circ</math>
 +
 
 +
By symmetry, <math>\angle{O_1'PA} = \angle{APO_1} = 0.5*\angle{O_1’PO_1} = 15^\circ</math>
 +
 
 +
Therefore, since <math>O_1’</math> is the circumcenter of <math>\triangle{ADP}</math>, <math>\angle{ADP}</math> = <math>0.5*(180^\circ - 2*\angle{O_1'PA}) = 75^\circ</math>
 +
 
 +
Therefore <math>\angle{APD} = 180^\circ - 45^\circ - 75^\circ = 60^\circ</math>
 +
 
 +
Using sine rule in <math>\triangle{ADP}</math>, <math>AP = (12 * \sin 75^\circ) / \sin 60^\circ =\sqrt{72}+\sqrt{24}</math>, and the answer is <math>72+24=\boxed{096}.</math>
 +
 
 +
By Kris17
 +
 
 +
==The Simple Way of the Coordinate Plane Solution==
 +
Why not use coordinates? After all, 45 degrees is rather friendly in terms of ordered-pair representation! We can set <math>A=(0, 12)</math>, <math>B=(12,12)</math>, <math>C=(12, 0)</math>, <math>D=(0, 0)</math>. Let this <math>P=(a, 12-a)</math> for some <math>a</math>.  
 +
 
 +
We also know that the circumcenter is the intersection of all perpendicular bisectors of sides, but two will suffice also due to this property. Therefore, we see that <math>O_{1}</math> is the intersection of <math>x=6</math> and, knowing the midpoint of <math>AP</math> to be <math>(\frac{a}{2}, \frac{12-a}{2})</math> and thus the equation to be <math>y=x+(12-a)</math>, we get <math>(6, 18-a)</math>. Likewise for <math>O_{2}</math> it's <math>(6, 6-a)</math>. Now what do we see? <math>O_{1}P=O_{2}P</math> (just look at the coordinates)! So both of those distances are <math>4\sqrt{3}</math>. Solving for <math>a</math> we get it to be <math>6+2\sqrt{3}</math>, since <math>AP>CP</math>. Multiply by <math>\sqrt{2}</math> because we are looking for <math>AP</math> to get the answer of <math>\boxed{096}</math>.
 +
 
 +
==Solution 6 (Pure Angle Chasing)==
 +
Let <math>\angle APD = \theta</math>. Then <math>\angle ADP = 180^{\circ}-45^{\circ}-\theta=135^{\circ}-\theta \implies \angle PDC=\theta-45^{\circ}</math>.
 +
Realize that because <math>O_1</math> is a circumcenter, <math>\angle DO_1P=\angle DCP=45^{\circ} \implies \angle DPO_1=\frac{180^{\circ}-\angle DO_1P}{2}=45^{\circ}</math>. Then <math>\angle O_2PA=75^{\circ}-\theta \implies \angle AOP = 180^{\circ}-2\angle O_2PA = 2\theta+30^{\circ} \implies \angle ABP=\theta + 15^{\circ} \implies \angle PBC=75-\theta</math>.
 +
Now, because <math>P</math> lies on diagonal <math>AC</math>, <math>\triangle PDC \cong \triangle PBC \implies \angle PDC = \angle PBC \implies \theta-45^{\circ}=75^{\circ}-\theta \implies \theta = 60^{\circ}</math>.
 +
To finish, we look at <math>\triangle ADP</math>. Drop a perpendicular from <math>D</math> to <math>AP</math> at <math>E</math>. Then <math>\triangle ADE</math> is a <math>45-45-90</math> and <math>\triangle PDE</math> is a <math>30-60-90</math>. Therefore, <math>DE=EA=6\sqrt{2}, EP=2\sqrt{6}</math>, so <math>AP=AE+EP=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24} \implies \boxed{096}</math>. <math>\blacksquare</math> ~msc
  
 
==See also==
 
==See also==
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[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Revision as of 22:38, 23 November 2020

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, letting $x = AP$ and using Law of Cosines on $\triangle ABP$, we have

\[96=144+x^{2}-24x\frac{\sqrt{2}}{2}\] \[0=x^{2}-12x\sqrt{2}+48\]

Using the quadratic formula, we arrive at

\[x = \sqrt{72} \pm \sqrt{24}\]

Taking the positive root, $AP=\sqrt{72}+ \sqrt{24}$ and the answer is thus $\framebox[1.5\width]{096.}$

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 $72+24=\boxed{096}.$

Solution 5

Reflect $O_1$ across $AP$ to $O_1'$. By symmetry $O_1’$ is the circumcenter of $\triangle{ADP}$

$\angle{DO_1’P}$ = $2*\angle{DAP} = 90^\circ$, so $\angle{O_1’PD}=45^\circ$

similarly $\angle{DO_2P}$ = $2*\angle{DCP} = 90^\circ$, so $\angle{O_2PD}=45^\circ$

Therefore $\angle{O_1’PO_2}=90^\circ$, so that $\angle{O_1’PO_1} =120^\circ - 90^\circ = 30^\circ$

By symmetry, $\angle{O_1'PA} = \angle{APO_1} = 0.5*\angle{O_1’PO_1} = 15^\circ$

Therefore, since $O_1’$ is the circumcenter of $\triangle{ADP}$, $\angle{ADP}$ = $0.5*(180^\circ - 2*\angle{O_1'PA}) = 75^\circ$

Therefore $\angle{APD} = 180^\circ - 45^\circ - 75^\circ = 60^\circ$

Using sine rule in $\triangle{ADP}$, $AP = (12 * \sin 75^\circ) / \sin 60^\circ =\sqrt{72}+\sqrt{24}$, and the answer is $72+24=\boxed{096}.$

By Kris17

The Simple Way of the Coordinate Plane Solution

Why not use coordinates? After all, 45 degrees is rather friendly in terms of ordered-pair representation! We can set $A=(0, 12)$, $B=(12,12)$, $C=(12, 0)$, $D=(0, 0)$. Let this $P=(a, 12-a)$ for some $a$.

We also know that the circumcenter is the intersection of all perpendicular bisectors of sides, but two will suffice also due to this property. Therefore, we see that $O_{1}$ is the intersection of $x=6$ and, knowing the midpoint of $AP$ to be $(\frac{a}{2}, \frac{12-a}{2})$ and thus the equation to be $y=x+(12-a)$, we get $(6, 18-a)$. Likewise for $O_{2}$ it's $(6, 6-a)$. Now what do we see? $O_{1}P=O_{2}P$ (just look at the coordinates)! So both of those distances are $4\sqrt{3}$. Solving for $a$ we get it to be $6+2\sqrt{3}$, since $AP>CP$. Multiply by $\sqrt{2}$ because we are looking for $AP$ to get the answer of $\boxed{096}$.

Solution 6 (Pure Angle Chasing)

Let $\angle APD = \theta$. Then $\angle ADP = 180^{\circ}-45^{\circ}-\theta=135^{\circ}-\theta \implies \angle PDC=\theta-45^{\circ}$. Realize that because $O_1$ is a circumcenter, $\angle DO_1P=\angle DCP=45^{\circ} \implies \angle DPO_1=\frac{180^{\circ}-\angle DO_1P}{2}=45^{\circ}$. Then $\angle O_2PA=75^{\circ}-\theta \implies \angle AOP = 180^{\circ}-2\angle O_2PA = 2\theta+30^{\circ} \implies \angle ABP=\theta + 15^{\circ} \implies \angle PBC=75-\theta$. Now, because $P$ lies on diagonal $AC$, $\triangle PDC \cong \triangle PBC \implies \angle PDC = \angle PBC \implies \theta-45^{\circ}=75^{\circ}-\theta \implies \theta = 60^{\circ}$. To finish, we look at $\triangle ADP$. Drop a perpendicular from $D$ to $AP$ at $E$. Then $\triangle ADE$ is a $45-45-90$ and $\triangle PDE$ is a $30-60-90$. Therefore, $DE=EA=6\sqrt{2}, EP=2\sqrt{6}$, so $AP=AE+EP=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24} \implies \boxed{096}$. $\blacksquare$ ~msc

See also

2011 AIME II (ProblemsAnswer KeyResources)
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