Difference between revisions of "2018 AIME II Problems/Problem 7"
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==Solution 2== | ==Solution 2== | ||
− | We have that there are <math>2449</math> trapezoids and <math>1</math> triangle of equal area, with that one triangle being <math>AP_1Q_1</math>. Notice, if we "stack" the trapezoids on top of <math>\bigtriangleup AP_1Q_1</math> the way they already are, we'd create a similar triangle, all of which are similar to <math>\bigtriangleup ABC</math>, and since the trapezoids and <math>\bigtriangleup AP_1Q_1</math> have equal area, each of these similar triangles, <math>AP_kQ_k</math> have area <math>\frac{k}{2450}\left[ ABC\right]</math>, and so <math>\frac{\left[ AP_kQ_k\right]}{\left[ABC\right]}=\frac{k}{2450}</math>. We want the ratio of the side lengths <math>P_kQ_k:BC</math>. 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 < | + | We have that there are <math>2449</math> trapezoids and <math>1</math> triangle of equal area, with that one triangle being <math>AP_1Q_1</math>. Notice, if we "stack" the trapezoids on top of <math>\bigtriangleup AP_1Q_1</math> the way they already are, we'd create a similar triangle, all of which are similar to <math>\bigtriangleup ABC</math>, and since the trapezoids and <math>\bigtriangleup AP_1Q_1</math> have equal area, each of these similar triangles, <math>AP_kQ_k</math> have area <math>\frac{k}{2450}\left[ ABC\right]</math>, and so <math>\frac{\left[ AP_kQ_k\right]}{\left[ABC\right]}=\frac{k}{2450}</math>. We want the ratio of the side lengths <math>P_kQ_k:BC</math>. 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 <cmath>\frac{P_kQ_k}{BC}=\sqrt{\frac{k}{2450}}</cmath> <cmath>\implies P_kQ_k=BC\cdot \sqrt{\frac{k}{2450}}=5\sqrt{3}\cdot\sqrt{\frac{k}{2450}}=\frac{1}{7}\cdot \sqrt{\frac{3k}{2}}=\frac{3}{7}\sqrt{\frac{k}{6}}</cmath> <cmath>\implies k=6n^2<2450 </cmath> <cmath>\implies 0<n\leq 20</cmath> so there are <math>\boxed{020}</math> solutions. |
− | Solution by ktong | + | ~Solution by ktong |
+ | |||
+ | ~Beautified by jdong2006 | ||
==Solution 3== | ==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 AA. 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> | + | 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 AA. 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>kx</math>. The ratio of the areas is equal to the square of the scale factor for any plane figure and its image. Therefore, <math>P_k Q_k=P_1 Q_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_k Q_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>P_1 Q_1=\frac{\sqrt{6}}{14}</math>. In order for <math>\sqrt{k} P_1 Q_1</math> to be rational, <math>\sqrt{k}</math> must be some rational multiple of <math>\sqrt{6}</math>. This is achieved at <math>\sqrt{k}=\sqrt{6}, 2\sqrt{6}, \ldots, 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 a1b2 | Solution by a1b2 |
Latest revision as of 00:12, 3 September 2021
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.
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
~Beautified by jdong2006
Solution 3
Let stand for , and . All triangles are similar by AA. 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
See Also
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 |
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