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 "1998 AHSME Problems/Problem 25"
(Solution 2)
Line 5: Line 5:
  
 
== Solution ==
 
== Solution ==
 +
 +
===Solution 1 ===
  
 
The line of the fold is the perpendicular bisector of the segment that connects <math>(0,2)</math> and <math>(4,0)</math>.  
 
The line of the fold is the perpendicular bisector of the segment that connects <math>(0,2)</math> and <math>(4,0)</math>.  
Line 58: Line 60:
  
 
It follows that <math>D=(0.6,6.2)</math>, and that <math>m+n = 6.2 + 0.6 = \boxed{6.8}</math>.
 
It follows that <math>D=(0.6,6.2)</math>, and that <math>m+n = 6.2 + 0.6 = \boxed{6.8}</math>.
 +
 +
 +
===Solution 2===
 +
 +
Note that the fold is the perpendicular bisector of <math>(0,2)</math> and <math>(4,0)</math>.  Thus, the fold goes through the midpoint <math>(2,1)</math>.
 +
 +
The fold also has a slope of <math>-\frac{1}{m}</math>, where the <math>m</math> is the slope of the line connecting these two points.  We find <math>m = \frac{0 - 2}{4 - 0} = -\frac{1}{2}</math>. Thus, the slope of the fold is <math>2</math> and goes through <math>(2,1)</math>, so the equation of the fold is <math>y = 2x - 3</math>.
 +
 +
We want the line connecting <math>(7,3)</math> and <math>(m,n)</math> to have the same fold as a perpendicular bisector. The slope should be <math>-\frac{1}{2}</math>, so we get <math>\frac{n - 3}{m - 7} = -\frac{1}{2}</math>, which leads to <math>m = 13 -2n</math>.
 +
 +
We also want <math>y = 2x - 3</math> to bisect the segment from <math>(7,3)</math> to <math>(m,n)</math>. Thus, the midpoint <math>(x,y) = (\frac{7 + m}{2}, \frac{3 + n}{2})</math> must lie on the line. Plugging into the equation of the line, we find <math>\frac{3 + n}{2} = (7 + m) - 3</math>, which simplifies to <math>n = 5 + 2m</math>
 +
 +
Solving the system of two equations in two variables <math>m = 13 - 2n</math> and <math>n = 5 + 2m</math> gives <math>m = \frac{3}{5}</math> and <math>n = \frac{31}{5}</math> , for a sum of <math>\boxed{\mathrm{(B) \ }6.8}</math>.
 +
  
 
== See also ==
 
== See also ==

Revision as of 02:26, 17 November 2013

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 $\boxed{\mathrm{(B) \ }6.8}$.


See also

1998 AHSME (Problems • Answer Key • Resources)
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
All AHSME 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=1998_AHSME_Problems/Problem_25&oldid=57727"