1998 AHSME Problems/Problem 25

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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

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.

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...

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 $m+n = 6.2 + 0.6 = \boxed{6.8}$ .

1998 AHSME (Problems)
Preceded by Problem 24	Followed by Problem 26
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 • 26 • 27 • 28 • 29 • 30

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1998 AHSME Problems/Problem 25

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