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

Revision as of 08:35, 20 September 2019

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 $PQRS$ ?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 6.2\qquad\textbf{(C)}\ 6.4\qquad\textbf{(D)}\ 6.6\qquad\textbf{(E)}\ 6.8$

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 $\boxed{\mathbf{(C)}\ 6.4}$ .

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 $6.4$ . 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 $4.6+1.8=6.4$ . $\framebox{C}$ .

@@ Line 9: / Line 9: @@
 <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 [i] MSTang [/i])
+(diagram by MSTang)
 ===Solution 1===

2012 AMC 12B (Problems • Answer Key • Resources)
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
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