Difference between revisions of "2012 AMC 12B Problems/Problem 17"

Latest revision as of 10:49, 10 October 2024

Problem

Square $PQRS$ lies in the first quadrant. Points $(3,0), (5,0), (7,0),$ and $(13,0)$ lie on lines $SP, RQ, PQ$ , and $SR$ , respectively. What is the sum of the coordinates of the center of the square $PQRS$ ?

$\textbf{(A)}\ 6\qquad\textbf{(B) }\frac{31}5\qquad\textbf{(C) }\frac{32}5\qquad\textbf{(D) }\frac{33}5\qquad\textbf{(E) }\frac{34}5$

Diagram

$[asy] size(7cm); pair A=(0,0),B=(1,1.5),D=B*dir(-90),C=B+D-A; draw((-4,-2)--(8,-2), Arrows); draw(A--B--C--D--cycle); pair AB = extension(A,B,(0,-2),(1,-2)); pair BC = extension(B,C,(0,-2),(1,-2)); pair CD = extension(C,D,(0,-2),(1,-2)); pair DA = extension(D,A,(0,-2),(1,-2)); draw(A--AB--B--BC--C--CD--D--DA--A, dotted); dot(AB^^BC^^CD^^DA);[/asy]$

(diagram by MSTang)

Solution 1

$[asy] size(14cm); pair A=(3,0),B=(5,0),C=(7,0),D=(13,0),EE=(4,0),F=(10,0),P=(3.4,1.2),Q=(5.2,0.6),R=(5.8,2.4),SS=(4,3),M=(4.6,1.8),G=(3.2,0.6),H=(7.6,1.8); dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H); draw(A--SS--D--cycle); draw(P--Q--R^^B--Q--C); draw(EE--M--F^^G--B^^C--H,dotted); label("A",A,SW); label("B",B,S); label("C",C,S); label("D",D,SE); label("E",EE,S); label("F",F,S); label("P",P,W); label("Q",Q,NW); label("R",R,NE); label("S",SS,N); label("M",M,S); label("G",G,W); label("H",H,NE);[/asy]$

Construct the midpoints $E=(4,0)$ and $F=(10,0)$ and triangle $\triangle EMF$ as in the diagram, where $M$ is the center of square $PQRS$ . Also construct points $G$ and $H$ as in the diagram so that $BG\parallel PQ$ and $CH\parallel QR$ .

Observe that $\triangle AGB\sim\triangle CHD$ while $PQRS$ being a square implies that $GB=CH$ . Furthermore, $CD=6=3\cdot AB$ , so $\triangle CHD$ is 3 times bigger than $\triangle AGB$ . Therefore, $HD=3\cdot GB=3\cdot HC$ . In other words, the longer leg is 3 times the shorter leg in any triangle similar to $\triangle AGB$ .

Let $K$ be the foot of the perpendicular from $M$ to $EF$ , and let $x=EK$ . Triangles $\triangle EKM$ and $\triangle MKF$ , being similar to $\triangle AGB$ , also have legs in a 1:3 ratio, therefore, $MK=3x$ and $KF=9x$ , so $10x=EF=6$ . It follows that $EK=0.6$ and $MK=1.8$ , so the coordinates of $M$ are $(4+0.6,1.8)=(4.6,1.8)$ and so our answer is $4.6+1.8 = 6.4 =$ $\boxed{\mathbf{(C)}\ 32/5}$ .

Solution 2

$[asy] size(7cm); pair A=(0,0),B=(1,1.5),D=B*dir(-90),C=B+D-A; draw((-4,-2)--(8,-2), Arrows); draw(A--B--C--D--cycle); pair AB = extension(A,B,(0,-2),(1,-2)); pair BC = extension(B,C,(0,-2),(1,-2)); pair CD = extension(C,D,(0,-2),(1,-2)); pair DA = extension(D,A,(0,-2),(1,-2)); draw(A--AB--B--BC--C--CD--D--DA--A, dotted); dot(AB^^BC^^CD^^DA);[/asy]$

Let the four points be labeled $P_1$ , $P_2$ , $P_3$ , and $P_4$ , respectively. Let the lines that go through each point be labeled $L_1$ , $L_2$ , $L_3$ , and $L_4$ , respectively. Since $L_1$ and $L_2$ go through $SP$ and $RQ$ , respectively, and $SP$ and $RQ$ are opposite sides of the square, we can say that $L_1$ and $L_2$ are parallel with slope $m$ . Similarly, $L_3$ and $L_4$ have slope $-\frac{1}{m}$ . Also, note that since square $PQRS$ lies in the first quadrant, $L_1$ and $L_2$ must have a positive slope. Using the point-slope form, we can now find the equations of all four lines: $L_1: y = m(x-3)$ , $L_2: y = m(x-5)$ , $L_3: y = -\frac{1}{m}(x-7)$ , $L_4: y = -\frac{1}{m}(x-13)$ .

Since $PQRS$ is a square, it follows that $\Delta x$ between points $P$ and $Q$ is equal to $\Delta y$ between points $Q$ and $R$ . Our approach will be to find $\Delta x$ and $\Delta y$ in terms of $m$ and equate the two to solve for $m$ . $L_1$ and $L_3$ intersect at point $P$ . Setting the equations for $L_1$ and $L_3$ equal to each other and solving for $x$ , we find that they intersect at $x = \frac{3m^2 + 7}{m^2 + 1}$ . $L_2$ and $L_3$ intersect at point $Q$ . Intersecting the two equations, the $x$ -coordinate of point $Q$ is found to be $x = \frac{5m^2 + 7}{m^2 + 1}$ . Subtracting the two, we get $\Delta x = \frac{2m^2}{m^2 + 1}$ . Substituting the $x$ -coordinate for point $Q$ found above into the equation for $L_2$ , we find that the $y$ -coordinate of point $Q$ is $y = \frac{2m}{m^2+1}$ . $L_2$ and $L_4$ intersect at point $R$ . Intersecting the two equations, the $y$ -coordinate of point $R$ is found to be $y = \frac{8m}{m^2 + 1}$ . Subtracting the two, we get $\Delta y = \frac{6m}{m^2 + 1}$ . Equating $\Delta x$ and $\Delta y$ , we get $2m^2 = 6m$ which gives us $m = 3$ . Finally, note that the line which goes though the midpoint of $P_1$ and $P_2$ with slope $3$ and the line which goes through the midpoint of $P_3$ and $P_4$ with slope $-\frac{1}{3}$ must intersect at at the center of the square. The equation of the line going through $(4,0)$ is given by $y = 3(x-4)$ and the equation of the line going through $(10,0)$ is $y = -\frac{1}{3}(x-10)$ . Equating the two, we find that they intersect at $(4.6, 1.8)$ . Adding the $x$ and $y$ -coordinates, we get $6.4 = 32/5$ . Thus, answer choice $\boxed{\textbf{(C)}}$ is correct.

Solution 3

Note that the center of the square lies along a line that has an $x-$ intercept of $\frac{3+5}{2}=4$ , and also along another line with $x-$ intercept $\frac{7+13}{2}=10$ . Since these 2 lines are parallel to the sides of the square, they are perpendicular (since the sides of a square are). Let $m$ be the slope of the first line. Then $-\frac{1}{m}$ is the slope of the second line. We may use the point-slope form for the equation of a line to write $l_1:y=m(x-4)$ and $l_2:y=-\frac{1}{m}(x-10)$ . We easily calculate the intersection of these lines using substitution or elimination to obtain $\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)$ as the center or the square. Let $\theta$ denote the (acute) angle formed by $l_1$ and the $x-$ axis. Note that $\tan\theta=m$ . Let $s$ denote the side length of the square. Then $\sin\theta=s/2$ . On the other hand the acute angle formed by $l_2$ and the $x-$ axis is $90-\theta$ so that $\cos\theta=\sin(90-\theta)=s/6$ . Then $m=\tan\theta=3$ . Substituting into $\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)$ we obtain $\left(\frac{23}{5},\frac{9}{5}\right)$ so that the sum of the coordinates is $\frac{32}{5}=6.4$ . Hence the answer is $\framebox{C}$ .

Solution 4 (Fast)

Suppose

$SP: y=m(x-3)$ $RQ: y=m(x-5)$ $PQ: -my=x-7$ $SR: -my=x-13$

where $m >0$ .

Recall that the distance between two parallel lines $Ax+By+C=0$ and $Ax+By+C_1=0$ is $|C-C_1|/\sqrt{A^2+B^2}$ , we have distance between $SP$ and $RQ$ equals to $2m/\sqrt{1+m^2}$ , and the distance between $PQ$ and $SR$ equals to $6/\sqrt{1+m^2}$ . Equating them, we get $m=3$ .

Then, the center of the square is just the intersection between the following two "mid" lines:

$L_1: y=3(x-4)$ $L_2: -3y = x-10$

The solution is $(4.6,1.8)$ , so we get the answer $4.6+1.8=6.4$ . $\framebox{C}$ .

Solution 5 (Trigonometry)

Using the diagram shown in Solution 1, we can set angle $BCQ$ as $\theta$ . We know that $AB=2$ and $BC=2$ . Now using $AA$

similarity, we know that $\triangle BCQ\sim\triangle ACP$ in a $1:2$ ratio. Now we can see that $CQ=-2$ $\cos\theta$ , therefore,

meaning that $PQ=-2$ $\cos\theta$ . $PQRS$ is a square, so $QR=-2$ $\cos\theta$ . We also know that $QCHR$ is also a square since its

angles are $90^\circ$ and all of its sides are equal. Because squares $PQRS$ and $QCHR$ have equal side lengths, they are

congruent leading to the conclusion that side $CH=-2$ $\cos\theta$ . Since $PQRS$ is a square, lines $PQ$ and $SR$ are parallel

meaning that angle $CDH$ and angle $BCQ$ are congruent. We can easily calculate that the length of $CD=6$ and furthermore that

$CH=6$ $\sin\theta$ . Setting $6\sin\theta=-2\cos\theta$ , we get that $\tan\theta=-1/3$ . This means $-1/3$ is the slope of line $PQ$

and the lines parallel to it. This is good news because we are dealing with easy numbers. We can solve for the coordinates of

points $E$ and $F$ because they are the midpoints. This will make solving for the center of square $PQRS$ easier. $E=(4,0)$ and

$F=(10,0)$ . We know the slopes of lines $MF$ and $ME$ , which are $-1/3$ and $3$ respectively. Now we can get the two equations.

$\left\{\begin{array}{l}y=-1/3x+10/3\\y=3x-12\end{array}\right.$

By solving:

$-1/3x+10/3=3x-12,$

we find that $x=4.6$ . Then plugging $x$ back into one of the first equations, we can find $y$ and the final coordinate turns out to be $(4.6,1.8)$ . Summing up the values of $x$ and $y$ , you get $4.6+1.8=6.4=32/5$ . $\boxed{\mathbf{(C)}\ 32/5}$ .

~kempwood

Solution 6

$[asy] size(14cm); pair A=(3,0),B=(5,0),C=(7,0),D=(13,0),EE=(4,0),F=(10,0),P=(3.4,1.2),Q=(5.2,0.6),R=(5.8,2.4),SS=(4,3),M=(4.6,1.8),G=(3.2,0.6),H=(7.6,1.8); dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H); draw(A--SS--D--cycle); draw(P--Q--R^^B--Q--C); draw(EE--M--F^^G--B^^C--H,dotted); label("A",A,SW); label("B",B,S); label("C",C,S); label("D",D,SE); label("E",EE,S); label("F",F,S); label("P",P,W); label("Q",Q,NW); label("R",R,NE); label("S",SS,N); label("M",M,S); label("G",G,W); label("H",H,NE);[/asy]$

$SP: y = mx - 3m$ , $RQ: y = mx-5m$ , $PQ: y = -\frac{1}{m}x + \frac{7}{m}$ , $SR: y = -\frac{1}{m}x + \frac{13}{m}$

Let $SP = RP = PQ = SR = a$ , $\angle GAB = \angle HCD = \theta$ , and the slope of $SP$ be $m$ .

When the slope of $SP$ is $m$ , the slope of $SR$ is $-\frac{1}{m}$ , $\tan \theta = m$ , $\cot \theta = -\frac{1}{m}$

$\sin \theta = \frac{GB}{AB} = \frac{a}{2}$ , $\cos \theta = \frac{HC}{CD} = \frac{a}{6}$

As $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{a}{2}}{\frac{a}{6}} = 3$ , $m = 3$

$SP: y = 3x - 9$ , $RQ: y = 3x-15$ , $PQ: y = -\frac{1}{3}x + \frac{7}{3}$ , $SR: y = -\frac{1}{3}x + \frac{13}{3}$

$3x - 9 = -\frac{1}{3}x + \frac{13}{3}$ , $x = 4$ , $y = 3$ , $S = (4, 3)$

$3x-15 = -\frac{1}{3}x + \frac{7}{3}$ , $x = 5.2$ , $y = 0.6$ , $Q = (5.2, 0.6)$

$M = (4.6, 1.8)$ , $4.6 + 1.8 = \boxed{\mathbf{(C)}\ 32/5}$

~isabelchen

Solution 7 (Pure Trig and Similarity)

$[asy] unitsize(1 cm); pair P,Q,R,S,A,B,C,D,E,F,F1,F2,F3,M; P = (3.4,1.2); Q = (5.2,0.6); R = (5.8,2.4); S = (4,3); A = (3,0); B = (5,0); C = (7,0); D = (13,0); E = (3.2,0.6); F = (7.6,1.8); F1 = (3.2,0); F2 = (3.4,0); F3 = (5.8,0); M = (4.6,1.8); dot(P); dot(Q); dot(R); dot(S); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(F1); dot(F2); dot(F3); dot(M); draw(P--Q--R--S--cycle); draw(P--A); draw(Q--B); draw(R--D); draw(Q--C); draw(A--D); draw(B--E, dashed); draw(C--F, dashed); draw(E--F1, dashed); draw(P--F2, dashed); draw(R--F3, dashed); draw(P--R, dashed); label("$A$",A,W); label("$B$",B,SW); label("$C$",C,SW); label("$D$",D,SW); label("$P$",P,W); label("$Q$",Q,ENE); label("$R$",R,N); label("$S$",S,N); label("$E$",E,NW); label("$F$",F,NE); label("$F_1$",F1,SSW); label("$F_2$",F2,SE); label("$F_3$",F3,SW); label("$M$",M,NW); [/asy]$ Let $PQ=x$ and $\angle PCA=\theta.$ Draw the line $BE$ such that $E$ is on $AP$ and $BE\parallel PQ.$ Also, Draw the line $CF$ such that $F$ is on $DR$ and $CF\parallel RQ.$ Then $EB=FC=x$ and $\angle EBA=\angle FDC=\theta.$ Also, note that $AB=2$ and $CD=6.$ Hence: $\cos(\theta)=\frac{x}{2},\sin(\theta)=\frac{x}{6}.$ Thus $\cos(\theta)=3\sin(\theta).$ Since $\theta<\frac{\pi}{2},$ $\sin(\theta)=\sqrt{1-\cos^2(\theta)},$ so $\cos(\theta)=3\sqrt{1-\cos^2(\theta)}.$ Hence, $\cos(\theta)=\frac{3}{\sqrt{10}}$ and $x=\frac{6}{\sqrt{10}}.$ Draw the perpendicular lines $EF_1\perp AD,PF_2\perp AD,RF_3\perp AD.$ Note that: $F_1B=BE\cdot\cos(\theta)=\frac{6}{\sqrt{10}}\cdot\frac{3}{\sqrt{10}}=\frac{9}{5}.$ Hence: $EF_1=\sqrt{EB^2-EF_1^2}=\sqrt{\frac{18}{5}-\frac{81}{25}}=\frac{3}{5}.$ Note that $\triangle EF_1B\sim\triangle PF_2C,$ so: $\frac{PF_2}{EF_1}=\frac{F_2C}{F_1B}=\frac{AC}{AB}=2.$ Hence: $PF_2=\frac{6}{5}, F_2C=\frac{18}{5}.$ So $P$ has coordinates: $\left(7-\frac{18}{5},\frac{6}{5}\right)=\left(\frac{17}{5},\frac{6}{5}\right).$ Also note that $\triangle EF_1B\sim\triangle RF_3D,$ so: $\frac{RF_3}{EF_1}=\frac{F_3D}{F_1B}=\frac{BD}{AB}=4.$ Hence: $RF_3=\frac{12}{5}, F_3D=\frac{36}{5}.$ So $R$ has coordinates: $\left(13-\frac{36}{5},\frac{12}{5}\right)=\left(\frac{29}{5},\frac{12}{5}\right).$ Hence, the center of square $PQRS,$ which is also the midpoint of $PR,$ has coordinates: $\left(\frac{\frac{17}{5}+\frac{29}{5}}{2},\frac{\frac{6}{5}+\frac{12}{5}}{2}\right)=\left(\frac{23}{5},\frac{9}{5}\right).$ We thus see that the answer is: $\frac{23}{5}+\frac{9}{5}=\boxed{\text{(C)}\frac{32}{5}}.$

@@ Line 4: / Line 4: @@
 <math> \textbf{(A)}\ 6\qquad\textbf{(B) }\frac{31}5\qquad\textbf{(C) }\frac{32}5\qquad\textbf{(D) }\frac{33}5\qquad\textbf{(E) }\frac{34}5 </math>
-==Solutions==
+==Diagram==
 <asy> size(7cm); pair A=(0,0),B=(1,1.5),D=B*dir(-90),C=B+D-A; draw((-4,-2)--(8,-2), Arrows); draw(A--B--C--D--cycle); pair AB = extension(A,B,(0,-2),(1,-2)); pair BC = extension(B,C,(0,-2),(1,-2)); pair CD = extension(C,D,(0,-2),(1,-2)); pair DA = extension(D,A,(0,-2),(1,-2)); draw(A--AB--B--BC--C--CD--D--DA--A, dotted); dot(AB^^BC^^CD^^DA);</asy>
@@ Line 10: / Line 11: @@
 (diagram by MSTang)
-===Solution 1===
+==Solution 1==
 <asy> size(14cm);
@@ Line 41: / Line 42: @@
-===Solution 2===
+==Solution 2==
 <asy> size(7cm); pair A=(0,0),B=(1,1.5),D=B*dir(-90),C=B+D-A; draw((-4,-2)--(8,-2), Arrows); draw(A--B--C--D--cycle); pair AB = extension(A,B,(0,-2),(1,-2)); pair BC = extension(B,C,(0,-2),(1,-2)); pair CD = extension(C,D,(0,-2),(1,-2)); pair DA = extension(D,A,(0,-2),(1,-2)); draw(A--AB--B--BC--C--CD--D--DA--A, dotted); dot(AB^^BC^^CD^^DA);</asy>
@@ Line 50: / Line 51: @@
 Since <math>PQRS</math> is a square, it follows that <math>\Delta x</math> between points <math>P</math> and <math>Q</math> is equal to <math>\Delta y</math> between points <math>Q</math> and <math>R</math>. Our approach will be to find <math>\Delta x</math> and <math>\Delta y</math> in terms of <math>m</math> and equate the two to solve for <math>m</math>. <math>L_1</math> and <math>L_3</math> intersect at point <math>P</math>. Setting the equations for <math>L_1</math> and <math>L_3</math> equal to each other and solving for <math>x</math>, we find that they intersect at <math>x = \frac{3m^2 + 7}{m^2 + 1}</math>. <math>L_2</math> and <math>L_3</math> intersect at point <math>Q</math>. Intersecting the two equations, the <math>x</math>-coordinate of point <math>Q</math> is found to be <math>x = \frac{5m^2 + 7}{m^2 + 1}</math>. Subtracting the two, we get <math>\Delta x = \frac{2m^2}{m^2 + 1}</math>. Substituting the <math>x</math>-coordinate for point <math>Q</math> found above into the equation for <math>L_2</math>, we find that the <math>y</math>-coordinate of point <math>Q</math> is <math>y = \frac{2m}{m^2+1}</math>. <math>L_2</math> and <math>L_4</math> intersect at point <math>R</math>. Intersecting the two equations, the <math>y</math>-coordinate of point <math>R</math> is found to be <math>y = \frac{8m}{m^2 + 1}</math>. Subtracting the two, we get <math>\Delta y = \frac{6m}{m^2 + 1}</math>. Equating <math>\Delta x</math> and <math>\Delta y</math>, we get <math>2m^2 = 6m</math> which gives us <math>m = 3</math>. Finally, note that the line which goes though the midpoint of <math>P_1</math> and <math>P_2</math> with slope <math>3</math> and the line which goes through the midpoint of <math>P_3</math> and <math>P_4</math> with slope <math>-\frac{1}{3}</math> must intersect at at the center of the square. The equation of the line going through <math>(4,0)</math> is given by <math>y = 3(x-4)</math> and the equation of the line going through <math>(10,0)</math> is <math>y = -\frac{1}{3}(x-10)</math>. Equating the two, we find that they intersect at <math>(4.6, 1.8)</math>. Adding the <math>x</math> and <math>y</math>-coordinates, we get <math>6.4 = 32/5</math>. Thus, answer choice <math>\boxed{\textbf{(C)}}</math> is correct.
-===Solution 3===
+==Solution 3==
 Note that the center of the square lies along a line that has an <math>x-</math>intercept of <math>\frac{3+5}{2}=4</math>, and also along another line with <math>x-</math>intercept <math>\frac{7+13}{2}=10</math>. Since these 2 lines are parallel to the sides of the square, they are perpendicular (since the sides of a square are). Let <math>m</math> be the slope of the first line. Then <math>-\frac{1}{m}</math> is the slope of the second line. We may use the point-slope form for the equation of a line to write <math>l_1:y=m(x-4)</math> and <math>l_2:y=-\frac{1}{m}(x-10)</math>. We easily calculate the intersection of these lines using substitution or elimination to obtain <math>\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)</math> as the center or the square. Let <math>\theta</math> denote the (acute) angle formed by <math>l_1</math> and the <math>x-</math>axis. Note that <math>\tan\theta=m</math>. Let <math>s</math> denote the side length of the square. Then <math>\sin\theta=s/2</math>. On the other hand the acute angle formed by <math>l_2</math> and the <math>x-</math>axis is <math>90-\theta</math> so that <math>\cos\theta=\sin(90-\theta)=s/6</math>. Then <math>m=\tan\theta=3</math>. Substituting into <math>\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)</math> we obtain <math>\left(\frac{23}{5},\frac{9}{5}\right)</math> so that the sum of the coordinates is <math>\frac{32}{5}=6.4</math>. Hence the answer is <math>\framebox{C}</math>.
-===Solution 4 (Fast)===
+==Solution 4 (Fast)==
 Suppose
@@ Line 73: / Line 74: @@
 The solution is <math>(4.6,1.8)</math>, so we get the answer <math>4.6+1.8=6.4</math>.  <math>\framebox{C}</math>.
-===Solution 5 (Trigonometry)===
+==Solution 5 (Trigonometry)==
 Using the diagram shown in Solution 1, we can set angle <math>BCQ</math> as <math>\theta</math>. We know that <math>AB=2</math> and <math>BC=2</math>. Now using <math>AA</math>
@@ Line 100: / Line 101: @@
-~Hithere22702
+~kempwood
 == Solution 6 ==
@@ Line 126: / Line 127: @@
 label("H",H,NE);</asy>
+<math>SP: y = mx - 3m</math>, <math>RQ: y = mx-5m</math>, <math>PQ: y = -\frac{1}{m}x + \frac{7}{m}</math>, <math>SR: y = -\frac{1}{m}x + \frac{13}{m}</math>
+Let <math>SP = RP = PQ = SR = a</math>, <math>\angle GAB = \angle HCD = \theta</math>, and the slope of <math>SP</math> be <math>m</math>.
+When the slope of <math>SP</math> is <math>m</math>, the slope of <math>SR</math> is <math>-\frac{1}{m}</math>, <math>\tan \theta = m</math>, <math>\cot \theta = -\frac{1}{m}</math>
+<math>\sin \theta = \frac{GB}{AB} = \frac{a}{2}</math>, <math>\cos \theta = \frac{HC}{CD} = \frac{a}{6}</math>
+As <math>\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{a}{2}}{\frac{a}{6}} = 3</math>, <math>m = 3</math>
+<math>SP: y = 3x - 9</math>, <math>RQ: y = 3x-15</math>, <math>PQ: y = -\frac{1}{3}x + \frac{7}{3}</math>, <math>SR: y = -\frac{1}{3}x + \frac{13}{3}</math>
+<math>3x - 9 = -\frac{1}{3}x + \frac{13}{3}</math>, <math>x = 4</math>, <math>y = 3</math>, <math>S = (4, 3)</math>
+<math>3x-15 = -\frac{1}{3}x + \frac{7}{3}</math>, <math>x = 5.2</math>, <math>y = 0.6</math>, <math>Q = (5.2, 0.6)</math>
+<math>M = (4.6, 1.8)</math>, <math>4.6 + 1.8 = \boxed{\mathbf{(C)}\ 32/5}</math>
+~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
+== Solution 7 (Pure Trig and Similarity) ==
+<asy>
+unitsize(1 cm);
+pair P,Q,R,S,A,B,C,D,E,F,F1,F2,F3,M;
+P = (3.4,1.2);
+Q = (5.2,0.6);
+R = (5.8,2.4);
+S = (4,3);
+A = (3,0);
+B = (5,0);
+C = (7,0);
+D = (13,0);
+E = (3.2,0.6);
+F = (7.6,1.8);
+F1 = (3.2,0);
+F2 = (3.4,0);
+F3 = (5.8,0);
+M = (4.6,1.8);
+dot(P);
+dot(Q);
+dot(R);
+dot(S);
+dot(A);
+dot(B);
+dot(C);
+dot(D);
+dot(E);
+dot(F);
+dot(F1);
+dot(F2);
+dot(F3);
+dot(M);
+draw(P--Q--R--S--cycle);
+draw(P--A);
+draw(Q--B);
+draw(R--D);
+draw(Q--C);
+draw(A--D);
+draw(B--E, dashed);
+draw(C--F, dashed);
+draw(E--F1, dashed);
+draw(P--F2, dashed);
+draw(R--F3, dashed);
+draw(P--R, dashed);
+label("$A$",A,W);
+label("$B$",B,SW);
+label("$C$",C,SW);
+label("$D$",D,SW);
+label("$P$",P,W);
+label("$Q$",Q,ENE);
+label("$R$",R,N);
+label("$S$",S,N);
+label("$E$",E,NW);
+label("$F$",F,NE);
+label("$F_1$",F1,SSW);
+label("$F_2$",F2,SE);
+label("$F_3$",F3,SW);
+label("$M$",M,NW);
+</asy>
+Let <math>PQ=x</math> and <math>\angle PCA=\theta.</math> Draw the line <math>BE</math> such that <math>E</math> is on <math>AP</math> and <math>BE\parallel PQ.</math> Also, Draw the line <math>CF</math> such that <math>F</math> is on <math>DR</math> and <math>CF\parallel RQ.</math> Then <math>EB=FC=x</math> and <math>\angle EBA=\angle FDC=\theta.</math> Also, note that <math>AB=2</math> and <math>CD=6.</math> Hence:
+<cmath>\cos(\theta)=\frac{x}{2},\sin(\theta)=\frac{x}{6}.</cmath>
+Thus <math>\cos(\theta)=3\sin(\theta).</math> Since <math>\theta<\frac{\pi}{2},</math> <math>\sin(\theta)=\sqrt{1-\cos^2(\theta)},</math> so <math>\cos(\theta)=3\sqrt{1-\cos^2(\theta)}.</math> Hence, <math>\cos(\theta)=\frac{3}{\sqrt{10}}</math> and <math>x=\frac{6}{\sqrt{10}}.</math> Draw the perpendicular lines <math>EF_1\perp AD,PF_2\perp AD,RF_3\perp AD.</math> Note that:
+<cmath>F_1B=BE\cdot\cos(\theta)=\frac{6}{\sqrt{10}}\cdot\frac{3}{\sqrt{10}}=\frac{9}{5}.</cmath>
+Hence:
+<cmath>EF_1=\sqrt{EB^2-EF_1^2}=\sqrt{\frac{18}{5}-\frac{81}{25}}=\frac{3}{5}.</cmath>
+Note that <math>\triangle EF_1B\sim\triangle PF_2C,</math> so:
+<cmath>\frac{PF_2}{EF_1}=\frac{F_2C}{F_1B}=\frac{AC}{AB}=2.</cmath>
+Hence:
+<cmath>PF_2=\frac{6}{5}, F_2C=\frac{18}{5}.</cmath>
+So <math>P</math> has coordinates:
+<cmath>\left(7-\frac{18}{5},\frac{6}{5}\right)=\left(\frac{17}{5},\frac{6}{5}\right).</cmath>
+Also note that <math>\triangle EF_1B\sim\triangle RF_3D,</math> so:
+<cmath>\frac{RF_3}{EF_1}=\frac{F_3D}{F_1B}=\frac{BD}{AB}=4.</cmath>
+Hence:
+<cmath>RF_3=\frac{12}{5}, F_3D=\frac{36}{5}.</cmath>
+So <math>R</math> has coordinates:
+<cmath>\left(13-\frac{36}{5},\frac{12}{5}\right)=\left(\frac{29}{5},\frac{12}{5}\right).</cmath>
+Hence, the center of square <math>PQRS,</math> which is also the midpoint of <math>PR,</math> has coordinates:
+<cmath>\left(\frac{\frac{17}{5}+\frac{29}{5}}{2},\frac{\frac{6}{5}+\frac{12}{5}}{2}\right)=\left(\frac{23}{5},\frac{9}{5}\right).</cmath>
+We thus see that the answer is:
+<cmath>\frac{23}{5}+\frac{9}{5}=\boxed{\text{(C)}\frac{32}{5}}.</cmath>
 == See Also ==

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 12 Problems and Solutions

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