2012 AMC 12B Problems/Problem 17

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 P1, P2, P3, and P4, respectively. Let the lines that go through each point be labeled L1, L2, L3, and L4, respectively. Since L1 and L2 go through SP and RQ, respectively, and SP and RQ are opposite sides of the square, we can say that L1 and L2 are parallel with slope m. Similarly, L3 and L4 have slope -1/m. Also, note that since square PQRS lies in the first quadrant, L1 and L2 must have a positive slope.

Using the point-slope form, we can now find the equations of all four lines:

L1: y = m(x-3)

L2: y = m(x-5)

L3: y = (-1/m)(x-7)

L4: y = (-1/m)(x-13)

Note that since PQRS is a square, it follows that the Delta x between points P and Q is equal to the 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.

L1 and L3 intersect at Point P. Setting the equations for L1 and L3 equal to each other and solving for x, we find that they intersect at x = (3m^2 + 7)/(m^2 + 1). Similarly the x-coordinate of Point Q is found to be x = (5m^2 + 7)/(m^2 + 1). Subtracting the two, we get Delta x = (2m^2)/(m^2 + 1).

Substituting the x-coordinate for Point Q found above into the equation for L2, we find that the y-coordinate for Point Q is y = 2m/(m^2+1). Intersecting L2 and L4, we find the x-coordinate of Point R to be x = (5m^2 + 13)/(m^2 + 1). Substituting this into the equation for L2, we find the y-coordinate for Point R is y = 8m/(m^2 + 1). Subtracting the two, we get Delta y = 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 P1 and P2 with slope 3 and the line which goes through the midpoint of P3 and P4 with slope -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 = (-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, the answer is C.

--Jm314 12:51, 26 February 2012 (EST)

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 $\framebox{C}$ .

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2012 AMC 12B Problems/Problem 17

Problem

Solution 1

Solution 2