1998 AHSME Problems/Problem 25

Problem

A piece of graph paper is folded once so that $(0,2)$ is matched with $(4,0)$ , and $(7,3)$ is matched with $(m,n)$ . Find $m+n$ .

$\mathrm{(A) \ }6.7 \qquad \mathrm{(B) \ }6.8 \qquad \mathrm{(C) \ }6.9 \qquad \mathrm{(D) \ }7.0 \qquad \mathrm{(E) \ }8.0$

Solution

Solution 1

The line of the fold is the perpendicular bisector of the segment that connects $(0,2)$ and $(4,0)$ . The point $(m,n)$ is the image of the point $(7,3)$ according to this axis. The situation looks as follows.

$[asy] size(200); defaultpen(0.8); pair A=(0,2), B=(4,0), C=(7,3); pair u=(1,2), S=(A+B)*0.5; pair T = intersectionpoint(S -- (S+4*u), C -- (C+4*(A-B)) ); pair D = 2*T-C; draw (A--B); draw ( (S-u) -- (S+4*u), dashed ); draw ( S -- C, black, Arrow ); draw ( S -- D, black, Arrow ); draw ( C -- D, Dotted ); dot(A); dot(B); dot(C); dot(D); label("$A(0,2)$",A,SW); label("$B(4,0)$",B,S); label("$S$",S,WSW*1.3); label("$T$",T,ENE*1.3); label("$C(7,3)$",C,SE); label("$D(m,n)$",D,NE); [/asy]$

Now, we will compute the coordinates of the point $D$ , using the following facts:

The triangles $SCT$ and $SDT$ are congruent.
$|SD| = |SC|$
$m$ is positive

As the triangles $SCT$ and $SDT$ are congruent, their areas are equal. The area of the triangle $SCT$ is $1/2$ of the size of the vector product $\overrightarrow{SC}\times\overrightarrow{ST}$ , and the area of $SDT$ is $1/2$ of the size of $\overrightarrow{ST}\times\overrightarrow{SD}$ .

We get that $\overrightarrow{SC}\times\overrightarrow{ST} = \overrightarrow{ST}\times\overrightarrow{SD}$ .

The equality remains valid if we multiply the vector $\overrightarrow{ST}$ by any constant. In other words, instead of $\overrightarrow{ST}$ we can use any vector with the same direction.

The axis of symmetry is perpendicular to $\overrightarrow{AB}$ . Thus its direction is $(2,4)=(1,2)$ .

We get that $\overrightarrow{SC}\times (1,2) = (1,2) \times\overrightarrow{SD}$ .

Substituting the coordinates $\overrightarrow{SC}=(5,2)$ and $\overrightarrow{SD}=(m-2,n-1)$ we get $5\cdot 2 - 2\cdot 1 = 1\cdot (n-1) - 2\cdot(m-2)$ . This simplifies to $n=2m+5$ .

We just discovered that the coordinates of $D$ are $(m,2m+5)$ . We will now use the second two facts mentioned above to find $m$ .

We have $|SD| = |SC|$ and therefore $|SD|^2 = |SC|^2$ . We know that $|SC|^2 = 5^2 + 2^2 = 29$ , and $|SD|^2 = (m-2)^2 + (2m+4)^2$ . Simplifying, we get the equation $5m^2 + 12m - 9 = 0$ . This has exactly one positive root $m=0.6$ .

It follows that $D=(0.6,6.2)$ , and that $m+n = 6.2 + 0.6 = \boxed{6.8}$ .

Solution 2

Note that the fold is the perpendicular bisector of $(0,2)$ and $(4,0)$ . Thus, the fold goes through the midpoint $(2,1)$ .

The fold also has a slope of $-\frac{1}{m}$ , where the $m$ is the slope of the line connecting these two points. We find $m = \frac{0 - 2}{4 - 0} = -\frac{1}{2}$ . Thus, the slope of the fold is $2$ and goes through $(2,1)$ , so the equation of the fold is $y = 2x - 3$ .

We want the line connecting $(7,3)$ and $(m,n)$ to have the same fold as a perpendicular bisector. The slope should be $-\frac{1}{2}$ , so we get $\frac{n - 3}{m - 7} = -\frac{1}{2}$ , which leads to $m = 13 -2n$ .

We also want $y = 2x - 3$ to bisect the segment from $(7,3)$ to $(m,n)$ . Thus, the midpoint $(x,y) = (\frac{7 + m}{2}, \frac{3 + n}{2})$ must lie on the line. Plugging into the equation of the line, we find $\frac{3 + n}{2} = (7 + m) - 3$ , which simplifies to $n = 5 + 2m$

Solving the system of two equations in two variables $m = 13 - 2n$ and $n = 5 + 2m$ gives $m = \frac{3}{5}$ and $n = \frac{31}{5}$ , for a sum of $\boxed{\mathrm{(B) \ }6.8}$ .

1998 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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1998 AHSME Problems/Problem 25

Contents

Problem

Solution

Solution 1

Solution 2

See also