Difference between revisions of "2018 AIME II Problems/Problem 7"
(→Solution) |
|||
Line 12: | Line 12: | ||
Solution by ktong | Solution by ktong | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Let <math>T_1</math> stand for <math>AP_1Q_1</math>, and <math>T_k = AP_kQ_k</math>. All triangles <math>T</math> are similar by AAA. Let the area of <math>T_1</math> be <math>x</math>. The next trapezoid will also have an area of <math>x</math>, as given. Therefore, <math>T_k</math> has an area of <math>\sqrt{k}x</math>. The ratio of the areas is equal to the square of the scale factor for any plane figure and its image. Therefore, <math>AP_k=AP_1\cdot \sqrt{k}</math>, and the same if <math>Q</math> is substituted for <math>P</math> throughout. We want the side <math>P_kQ_k</math> to be rational. Setting up proportions: <cmath>5\sqrt{3} : \sqrt{2450}=35\sqrt{2}</cmath> <cmath>\sqrt{6} : 14</cmath> which shows that <math>x=\frac{\sqrt{6}}{14}</math>. In order for <math>\sqrt{k}x</math> to be rational, <math>\sqrt{k}</math> must be some rational multiple of <math>\sqrt{6}</math>. This is achieved at <math>k=\sqrt{6}, 2\sqrt{6}, ... , 20\sqrt{6}</math>. We end there as <math>21\sqrt{6}=\sqrt{2646}</math>. There are 20 numbers from 1 to 20, so there are <math>\boxed{020}</math> solutions. | ||
+ | |||
+ | Solution by [[User:a1b2|a1b2]] | ||
+ | |||
{{AIME box|year=2018|n=II|num-b=6|num-a=8}} | {{AIME box|year=2018|n=II|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:50, 25 March 2018
Contents
[hide]Problem
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.
Solution 1
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
Solution 2
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
Solution 3
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
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.