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 == | ||
+ | === 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. | ||
{{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 14:41, 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
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.
2018 AIME II (Problems • Answer Key • Resources) | ||
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