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

 
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Square <math>PQRS</math> lies in the first quadrant. Points <math>(3,0), (5,0), (7,0),</math> and <math>(13,0)</math> lie on lines <math>SP, RQ, PQ</math>, and <math>SR</math>, respectively. What is the sum of the coordinates of the center of the square <math>PQRS</math>?
 
Square <math>PQRS</math> lies in the first quadrant. Points <math>(3,0), (5,0), (7,0),</math> and <math>(13,0)</math> lie on lines <math>SP, RQ, PQ</math>, and <math>SR</math>, respectively. What is the sum of the coordinates of the center of the square <math>PQRS</math>?
  
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 6.2\qquad\textbf{(C)}\ 6.4\qquad\textbf{(D)}\ 6.6\qquad\textbf{(E)}\ 6.8 </math>
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<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>
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==Solutions==
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<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>
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(diagram by MSTang)
  
==Solutions==
 
 
===Solution 1===
 
===Solution 1===
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<asy> size(14cm);
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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);
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dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H);
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draw(A--SS--D--cycle);
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draw(P--Q--R^^B--Q--C);
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draw(EE--M--F^^G--B^^C--H,dotted);
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label("A",A,SW);
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label("B",B,S);
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label("C",C,S);
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label("D",D,SE);
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label("E",EE,S);
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label("F",F,S);
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label("P",P,W);
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label("Q",Q,NW);
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label("R",R,NE);
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label("S",SS,N);
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label("M",M,S);
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label("G",G,W);
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label("H",H,NE);</asy>
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Construct the midpoints <math>E=(4,0)</math> and <math>F=(10,0)</math> and triangle <math>\triangle EMF</math> as in the diagram, where <math>M</math> is the center of square <math>PQRS</math>. Also construct points <math>G</math> and <math>H</math> as in the diagram so that <math>BG\parallel PQ</math> and <math>CH\parallel QR</math>.
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Observe that <math>\triangle AGB\sim\triangle CHD</math> while <math>PQRS</math> being a square implies that <math>GB=CH</math>. Furthermore, <math>CD=6=3\cdot AB</math>, so <math>\triangle CHD</math> is 3 times bigger than <math>\triangle AGB</math>. Therefore, <math>HD=3\cdot GB=3\cdot HC</math>. In other words, the longer leg is 3 times the shorter leg in any triangle similar to <math>\triangle AGB</math>.
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Let <math>K</math> be the foot of the perpendicular from <math>M</math> to <math>EF</math>, and let <math>x=EK</math>. Triangles <math>\triangle EKM</math> and <math>\triangle MKF</math>, being similar to <math>\triangle AGB</math>, also have legs in a 1:3 ratio, therefore, <math>MK=3x</math> and <math>KF=9x</math>, so <math>10x=EF=6</math>. It follows that <math>EK=0.6</math> and <math>MK=1.8</math>, so the coordinates of <math>M</math> are <math>(4+0.6,1.8)=(4.6,1.8)</math> and so our answer is <math>4.6+1.8 = 6.4 =</math> <math>\boxed{\mathbf{(C)}\ 32/5}</math>.
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===Solution 2===
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<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 <math>P_1</math>, <math>P_2</math>, <math>P_3</math>, and <math>P_4</math>, respectively. Let the lines that go through each point be labeled <math>L_1</math>, <math>L_2</math>, <math>L_3</math>, and <math>L_4</math>, respectively. Since <math>L_1</math> and <math>L_2</math> go through <math>SP</math> and <math>RQ</math>, respectively, and <math>SP</math> and <math>RQ</math> are opposite sides of the square, we can say that <math>L_1</math> and <math>L_2</math> are parallel with slope <math>m</math>. Similarly, <math>L_3</math> and <math>L_4</math> have slope <math>-\frac{1}{m}</math>. Also, note that since square <math>PQRS</math> lies in the first quadrant, <math>L_1</math> and <math>L_2</math> must have a positive slope. Using the point-slope form, we can now find the equations of all four lines: <math>L_1: y = m(x-3)</math>, <math>L_2: y = m(x-5)</math>, <math>L_3: y = -\frac{1}{m}(x-7)</math>, <math>L_4: y = -\frac{1}{m}(x-13)</math>.
 
Let the four points be labeled <math>P_1</math>, <math>P_2</math>, <math>P_3</math>, and <math>P_4</math>, respectively. Let the lines that go through each point be labeled <math>L_1</math>, <math>L_2</math>, <math>L_3</math>, and <math>L_4</math>, respectively. Since <math>L_1</math> and <math>L_2</math> go through <math>SP</math> and <math>RQ</math>, respectively, and <math>SP</math> and <math>RQ</math> are opposite sides of the square, we can say that <math>L_1</math> and <math>L_2</math> are parallel with slope <math>m</math>. Similarly, <math>L_3</math> and <math>L_4</math> have slope <math>-\frac{1}{m}</math>. Also, note that since square <math>PQRS</math> lies in the first quadrant, <math>L_1</math> and <math>L_2</math> must have a positive slope. Using the point-slope form, we can now find the equations of all four lines: <math>L_1: y = m(x-3)</math>, <math>L_2: y = m(x-5)</math>, <math>L_3: y = -\frac{1}{m}(x-7)</math>, <math>L_4: y = -\frac{1}{m}(x-13)</math>.
  
  
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</math>. Thus, answer choice <math>\boxed{\textbf{(C)}}</math> is correct.
+
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 2===
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===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 perpindicular (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>. Using <math>\cos\theta=\sqrt{1-\sin^2\theta}</math> (for acute <math>\theta</math>) we have <math>\frac{s}{6}=\sqrt{1-\left(\frac{s}{2}\right)^2}</math> where upon <math>s=\frac{3\sqrt{10}}{5}</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>.
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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 3 (Fast)===
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===Solution 4 (Fast)===
 
Suppose
 
Suppose
  
Line 35: Line 72:
  
 
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>.
 
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>.
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===Solution 5 (Trigonometry)===
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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>
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similarity, we know that <math>\triangle BCQ\sim\triangle ACP</math> in a <math>1:2</math> ratio. Now we can see that <math>CQ=-2</math><math>\cos\theta</math>, therefore,
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 +
meaning that <math>PQ=-2</math><math>\cos\theta</math>. <math>PQRS</math> is a square, so <math>QR=-2</math><math>\cos\theta</math>. We also know that <math>QCHR</math> is also a square since its
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angles are <math>90^\circ</math> and all of its sides are equal. Because squares <math>PQRS</math> and <math>QCHR</math> have equal side lengths, they are
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congruent leading to the conclusion that side <math>CH=-2</math><math>\cos\theta</math>. Since <math>PQRS</math> is a square, lines <math>PQ</math> and <math>SR</math> are parallel
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meaning that angle <math>CDH</math> and angle <math>BCQ</math> are congruent. We can easily calculate that the length of <math>CD=6</math> and furthermore that
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<math>CH=6</math><math>\sin\theta</math>. Setting <math>6\sin\theta=-2\cos\theta</math>, we get that <math>\tan\theta=-1/3</math>. This means <math>-1/3</math> is the slope of line <math>PQ</math>
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and the lines parallel to it. This is good news because we are dealing with easy numbers. We can solve for the coordinates of
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points <math>E</math> and <math>F</math> because they are the midpoints. This will make solving for the center of square <math>PQRS</math> easier. <math>E=(4,0)</math> and
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<math>F=(10,0)</math>. We know the slopes of lines <math>MF</math> and <math>ME</math>, which are <math>-1/3</math> and <math>3</math> respectively. Now we can get the two equations.
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<cmath>\left\{\begin{array}{l}y=-1/3x+10/3\\y=3x-12\end{array}\right.</cmath>
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By solving: <center><math> -1/3x+10/3=3x-12, </math></center>we find that <math>x=4.6</math>. Then plugging <math>x</math> back into one of the first equations, we can find <math>y</math> and the final coordinate turns out to be <math>(4.6,1.8)</math>. Summing up the values of <math>x</math> and <math>y</math>, you get <math>4.6+1.8=6.4=32/5</math>.  <math>\boxed{\mathbf{(C)}\ 32/5}</math>.
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~Hithere22702
  
 
== See Also ==
 
== See Also ==

Latest revision as of 18:54, 10 August 2020

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 $PQRS$?

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

Solutions

[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 $4.6+1.8=6.4=32/5$. $\boxed{\mathbf{(C)}\ 32/5}$.


~Hithere22702

See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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