2013 AMC 10A Problems/Problem 16
Contents
Problem
A triangle with vertices , , and is reflected about the line to create a second triangle. What is the area of the union of the two triangles?
Solution 1
Let be at , B be at , and be at . Reflecting over the line , we see that , (as the x-coordinate of B is 8), and . Line can be represented as , so we see that is on line .
We see that if we connect to , we get a line of length (between and ). The area of is equal to .
Now, let the point of intersection between and be . If we can just find the area of and subtract it from , we are done.
We realize that because the diagram is symmetric over , the intersection of lines and should intersect at an x-coordinate of . We know that the slope of is . Thus, we can represent the line going through and as . Plugging in , we find that the y-coordinate of F is . Thus, the height of is . Using the formula for the area of a triangle, the area of is .
To get our final answer, we must subtract this from .
Solution 2
First, realize that is the midpoint of and is the midpoint of . Connect to to form . Let the midpoint of be . Connect to . is a median of .
Because is isosceles, is also an altitude of . We know the length of and from the given coordinates. The area of is .
Let the intesection of , and be . is the centroid of . Therefore, it splits into and . The area of quadrilateral
~Zeric Hang
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.