2018 AIME II Problems/Problem 7
Triangle has side lengths , , and . Points are on segment with between and for , and points are on segment with between and for . Furthermore, each segment , , is parallel to . The segments cut the triangle into regions, consisting of trapezoids and triangle. Each of the regions has the same area. Find the number of segments , , that have rational length.
For each between and , the area of the trapezoid with as its bottom base is the difference between the areas of two triangles, both similar to . Let be the length of segment . The area of the trapezoid with bases and is times the area of . (This logic also applies to the topmost triangle if we notice that .) However, we also know that the area of each shape is times the area of . We then have . Simplifying, . However, we know that , so , and in general, and . The smallest that gives a rational is , so is rational if and only if for some integer .The largest such that is less than is , so has possible values.
Solution by zeroman
We have that there are trapezoids and triangle of equal area, with that one triangle being . Notice, if we "stack" the trapezoids on top of the way they already are, we'd create a similar triangle, all of which are similar to , and since the trapezoids and have equal area, each of these similar triangles, have area , and so . We want the ratio of the side lengths . Since area is a 2-dimensional unit of measurement, and side lengths are 1-dimensional, the ratio is simply the square root of the areas, or , so there are solutions.
Solution by ktong
Let stand for , and . All triangles are similar by AAA. Let the area of be . The next trapezoid will also have an area of , as given. Therefore, has an area of . The ratio of the areas is equal to the square of the scale factor for any plane figure and its image. Therefore, , and the same if is substituted for throughout. We want the side to be rational. Setting up proportions: which shows that . In order for to be rational, must be some rational multiple of . This is achieved at . We end there as . There are 20 numbers from 1 to 20, so there are solutions.
Solution by a1b2
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