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Difference between revisions of "2012 AMC 12B Problems/Problem 17"
(→‎Solution 3 (Fast))
Line 31: Line 31:
  
 
<cmath>L_1: y=3(x-4)</cmath>
 
<cmath>L_1: y=3(x-4)</cmath>
−
<cmath>L_2: -3y = x-6</cmath>
+
<cmath>L_2: -3y = x-10</cmath>
  
 
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>.

Revision as of 13:48, 17 January 2013

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 $PQRS$?

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

Solution 1

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 2

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 perpindicular (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 $\cos\theta=\sin(90-\theta)=s/6$. Using $\cos\theta=\sqrt{1-\sin^2\theta}$ (for acute $\theta$) we have $\frac{s}{6}=\sqrt{1-\left(\frac{s}{2}\right)^2}$ where upon $s=\frac{3\sqrt{10}}{5}$. 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 3 (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 $4.6+1.8=6.4$. $\framebox{C}$.

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
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