Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
(→‎Solution 2)
Line 36: Line 36:
  
 
==Solution 2==
 
==Solution 2==
−
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 special 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>.
+
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>, the midpoint of side <math>\overline{CD}</math> be <math>M_3</math>, the midpoint of segment <math>\overline{AP}</math> be <math>M_4</math>, and the midpoint of <math>\overline{CP}</math> be <math>M_5</math>. Consider the special 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}=d+i</math>. If point <math>P</math> is shifting at increment <math>i</math>, then <math>M_4</math> and <math>M_5</math> 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 <math>\overline{AP}</math> and <math>\overline{CP}</math> are shifting at the same rate. Circumcenters are formed by the perpendicular bisectors of the legs, so <math>O_1</math> and <math>O_2</math> must also be shifting at the same constant rate. For any location of <math>P</math>, both <math>O_1</math> and <math>O_2</math> will lie on line <math>\overline{M_1M_3}</math> because the perpendicular bisector of the side of the square is not changing. When <math>P</math> is located at <math>M_2</math>, <math>O_1</math> and <math>O_2</math> are located at <math>M_1</math> and <math>M_3</math>, respectively. As <math>P</math> shifts towards <math>C</math>, <math>O_1</math> and <math>O_2</math> shift down along line <math>\overline{M_1M_3}</math>. Both circumcenters are shifting at the same constant rate, so <math>\overline{M_1O_1}=\overline{M_3O_2}</math>. Therefore, triangles <math>ABO_1</math> and <math>CDO_2</math> 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 <math>O_1O_2P</math> is also isosceles because two of its legs are circumradii. Now, given angle <math>O_1PO_2=120^{\circ}</math>, angle <math>PO_1O_2=30^{\circ}</math>. We know that angle <math>M_1AM_2=AM_2M_1=M_2O_1M_4=45^{\circ}</math>. Therefore, angle <math>M_4O_1P=30+45=75^{\circ}</math>. In triangle <math>\deltaM_4O_1P</math>, <math>tan 75 = \frac{\frac{d+i}{2}}{d-\frac{d+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>. Solving for <math>i</math> by subsituting <math>6\sqrt{2}=d</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>.
  
 
==See also==
 
==See also==

Revision as of 17:04, 28 November 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. 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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_13&oldid=43531"