Difference between revisions of "2018 AIME II Problems/Problem 7"
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Triangle <math>ABC</math> has side lengths <math>AB = 9</math>, <math>BC =</math> <math>5\sqrt{3}</math>, and <math>AC = 12</math>. Points <math>A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B</math> are on segment <math>\overline{AB}</math> with <math>P_{k}</math> between <math>P_{k-1}</math> and <math>P_{k+1}</math> for <math>k = 1, 2, ..., 2449</math>, and points <math>A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C</math> are on segment <math>\overline{AC}</math> with <math>Q_{k}</math> between <math>Q_{k-1}</math> and <math>Q_{k+1}</math> for <math>k = 1, 2, ..., 2449</math>. Furthermore, each segment <math>\overline{P_{k}Q_{k}}</math>, <math>k = 1, 2, ..., 2449</math>, is parallel to <math>\overline{BC}</math>. The segments cut the triangle into <math>2450</math> regions, consisting of <math>2449</math> trapezoids and <math>1</math> triangle. Each of the <math>2450</math> regions has the same area. Find the number of segments <math>\overline{P_{k}Q_{k}}</math>, <math>k = 1, 2, ..., 2450</math>, that have rational length. | Triangle <math>ABC</math> has side lengths <math>AB = 9</math>, <math>BC =</math> <math>5\sqrt{3}</math>, and <math>AC = 12</math>. Points <math>A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B</math> are on segment <math>\overline{AB}</math> with <math>P_{k}</math> between <math>P_{k-1}</math> and <math>P_{k+1}</math> for <math>k = 1, 2, ..., 2449</math>, and points <math>A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C</math> are on segment <math>\overline{AC}</math> with <math>Q_{k}</math> between <math>Q_{k-1}</math> and <math>Q_{k+1}</math> for <math>k = 1, 2, ..., 2449</math>. Furthermore, each segment <math>\overline{P_{k}Q_{k}}</math>, <math>k = 1, 2, ..., 2449</math>, is parallel to <math>\overline{BC}</math>. The segments cut the triangle into <math>2450</math> regions, consisting of <math>2449</math> trapezoids and <math>1</math> triangle. Each of the <math>2450</math> regions has the same area. Find the number of segments <math>\overline{P_{k}Q_{k}}</math>, <math>k = 1, 2, ..., 2450</math>, that have rational length. | ||
− | + | == Solution 1 == | |
− | |||
For each <math>k</math> between <math>2</math> and <math>2450</math>, the area of the trapezoid with <math>\overline{P_kQ_k}</math> as its bottom base is the difference between the areas of two triangles, both similar to <math>\triangle{ABC}</math>. Let <math>d_k</math> be the length of segment <math>\overline{P_kQ_k}</math>. The area of the trapezoid with bases <math>\overline{P_{k-1}Q_{k-1}}</math> and <math>P_kQ_k</math> is <math>(\frac{d_k}{5\sqrt{3}})^2 - (\frac{d_{k-1}}{5\sqrt{3}})^2 = \frac{d_k^2-d_{k-1}^2}{75}</math> times the area of <math>\triangle{ABC}</math>. (This logic also applies to the topmost triangle if we notice that <math>d_0 = 0</math>.) However, we also know that the area of each shape is <math>\frac{1}{2450}</math> times the area of <math>\triangle{ABC}</math>. We then have <math>\frac{d_k^2-d_{k-1}^2}{75} = \frac{1}{2450}</math>. Simplifying, <math>d_k^2-d_{k-1}^2 = \frac{3}{98}</math>. However, we know that <math>d_0^2 = 0</math>, so <math>d_1^2 = \frac{3}{98}</math>, and in general, <math>d_k^2 = \frac{3k}{98}</math> and <math>d_k = \frac{\sqrt{\frac{3k}{2}}}{7}</math>. The smallest <math>k</math> that gives a rational <math>d_k</math> is <math>6</math>, so <math>d_k</math> is rational if and only if <math>k = 6n^2</math> for some integer <math>n</math>.The largest <math>n</math> such that <math>6n^2</math> is less than <math>2450</math> is <math>20</math>, so <math>k</math> has <math>\boxed{020}</math> possible values. | For each <math>k</math> between <math>2</math> and <math>2450</math>, the area of the trapezoid with <math>\overline{P_kQ_k}</math> as its bottom base is the difference between the areas of two triangles, both similar to <math>\triangle{ABC}</math>. Let <math>d_k</math> be the length of segment <math>\overline{P_kQ_k}</math>. The area of the trapezoid with bases <math>\overline{P_{k-1}Q_{k-1}}</math> and <math>P_kQ_k</math> is <math>(\frac{d_k}{5\sqrt{3}})^2 - (\frac{d_{k-1}}{5\sqrt{3}})^2 = \frac{d_k^2-d_{k-1}^2}{75}</math> times the area of <math>\triangle{ABC}</math>. (This logic also applies to the topmost triangle if we notice that <math>d_0 = 0</math>.) However, we also know that the area of each shape is <math>\frac{1}{2450}</math> times the area of <math>\triangle{ABC}</math>. We then have <math>\frac{d_k^2-d_{k-1}^2}{75} = \frac{1}{2450}</math>. Simplifying, <math>d_k^2-d_{k-1}^2 = \frac{3}{98}</math>. However, we know that <math>d_0^2 = 0</math>, so <math>d_1^2 = \frac{3}{98}</math>, and in general, <math>d_k^2 = \frac{3k}{98}</math> and <math>d_k = \frac{\sqrt{\frac{3k}{2}}}{7}</math>. The smallest <math>k</math> that gives a rational <math>d_k</math> is <math>6</math>, so <math>d_k</math> is rational if and only if <math>k = 6n^2</math> for some integer <math>n</math>.The largest <math>n</math> such that <math>6n^2</math> is less than <math>2450</math> is <math>20</math>, so <math>k</math> has <math>\boxed{020}</math> possible values. | ||
Solution by zeroman | Solution by zeroman | ||
+ | |||
+ | ==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 <math>\frac{P_kQ_k}{BC}=\sqrt{\frac{k}{2450}} \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}} \implies k=6n^2<2450 \implies 0<n\leq 20</math>, so there are <math>\boxed{020}</math> solutions. | ||
+ | |||
+ | Solution by ktong | ||
{{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 16:17, 24 March 2018
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
2018 AIME II (Problems • Answer Key • Resources) | ||
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