1998 AHSME Problems/Problem 25
Problem
A piece of graph paper is folded once so that is matched with , and is matched with . Find .
Solution
Solution 1
The line of the fold is the perpendicular bisector of the segment that connects and . The point is the image of the point according to this axis. The situation looks as follows.
Now, we will compute the coordinates of the point , using the following facts:
- The triangles and are congruent.
- is positive
As the triangles and are congruent, their areas are equal. The area of the triangle is of the size of the vector product , and the area of is of the size of .
We get that .
The equality remains valid if we multiply the vector by any constant. In other words, instead of we can use any vector with the same direction.
The axis of symmetry is perpendicular to . Thus its direction is .
We get that .
Substituting the coordinates and we get . This simplifies to .
We just discovered that the coordinates of are . We will now use the second two facts mentioned above to find .
We have and therefore . We know that , and . Simplifying, we get the equation . This has exactly one positive root .
It follows that , and that .
Solution 2
Note that the fold is the perpendicular bisector of and . Thus, the fold goes through the midpoint .
The fold also has a slope of , where the is the slope of the line connecting these two points. We find . Thus, the slope of the fold is and goes through , so the equation of the fold is .
We want the line connecting and to have the same fold as a perpendicular bisector. The slope should be , so we get , which leads to .
We also want to bisect the segment from to . Thus, the midpoint must lie on the line. Plugging into the equation of the line, we find , which simplifies to
Solving the system of two equations in two variables and gives and , for a sum of .
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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