2007 AMC 12B Problems/Problem 22
Contents
Problem 22
Two particles move along the edges of equilateral in the direction starting simultaneously and moving at the same speed. One starts at , and the other starts at the midpoint of . The midpoint of the line segment joining the two particles traces out a path that encloses a region . What is the ratio of the area of to the area of ?
Solution
First, notice that each of the midpoints of ,, and are on the locus. Suppose after some time the particles have each been displaced by a short distance , to new positions and respectively. Consider and drop a perpendicular from to hit at . Then, and . From here, we can use properties of a triangle to determine the lengths and as monomials in . Thus, the locus of the midpoint will be linear between each of the three special points mentioned above. It follows that the locus consists of the only triangle with those three points as vertices. (A cheaper way to find the shape of the region is to look at the answer choices: if it were any sort of conic section then the ratio would not generally be rational.) Comparing inradii between this "midpoint" triangle and the original triangle, the area contained by must be of the total area.
Solution 2
WLOG, let point be located on the origin. Let side be on the -axis, and let point be . When one of the ants is on a vertex, the other will be on the midpoint of the opposite side. Marking these points gives us a rough view of what seem to be the vertices of a smaller equilateral triangle, that is concentric with the larger one. After observing(or imagining) the ants moving, we find that the area enclosed is indeed an equilateral triangle.
Let the midpoint of be . Then we can find the coordinates of using the equations of two lines: the extension of and the extension of . The slope of any line is simply the tangent of the angle between the line and the -axis. The angle between and the -axis is , and .
Therefore the equation of the line containing is
The slope of the line containing is , and it goes through the point . After solving for the -intercept, we get . Setting these equal to each other yields
The midpoint of and the is the bottom-left vertex of the smaller equilateral triangle. Using the distance formula, we get the point
Since the smaller triangle is simply a dilation of the larger one, they both have an axis of symmetry over . This yields the bottom-right vertex of the smaller triangle as
Therefore the side length of the smaller triangle is The problem asks us for the ratio of the areas, and since area is two-dimensional, we simply square the ratio to get
See Also
2007 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.