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 "2012 AMC 12B Problems/Problem 17"

m
(→‎Solution 6)
Line 127: Line 127:
 
label("H",H,NE);</asy>
 
label("H",H,NE);</asy>
  
 +
<math>SP: y = mx - 3m</math>
 +
 +
<math>RQ: y = mx-5m</math>
 +
 +
<math>PR: y = -\frac{1}{m}x + \frac{7}{m}</math>
 +
 +
<math>SR: y = -\frac{1}{m}x + \frac{13}{m}</math>
 +
 +
 +
Let <math>SP = RP = PR = SR = a</math>, <math>\angle GAB = \angle HCD = \theta</math>, the slope of <math>SP</math> be <math>k</math>
 +
 +
<math>\tan \theta = k</math>, <math>\cot \theta = -\frac{1}{k}</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>k = 3</math>
 +
 +
 +
<math>SP: y = 3x - 9</math>
 +
 +
<math>RQ: y = 3x-15</math>
 +
 +
<math>PR: 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>
  
 
== See Also ==
 
== See Also ==

Revision as of 10:57, 26 December 2022

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


~Hithere22702

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]

$SP: y = mx - 3m$

$RQ: y = mx-5m$

$PR: y = -\frac{1}{m}x + \frac{7}{m}$

$SR: y = -\frac{1}{m}x + \frac{13}{m}$


Let $SP = RP = PR = SR = a$, $\angle GAB = \angle HCD = \theta$, the slope of $SP$ be $k$

$\tan \theta = k$, $\cot \theta = -\frac{1}{k}$

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

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


$SP: y = 3x - 9$

$RQ: y = 3x-15$

$PR: y = -\frac{1}{3}x + \frac{7}{3}$

$SR: y = -\frac{1}{3}x + \frac{13}{3}$

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

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

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

See Also

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
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_17&oldid=184904"