Difference between revisions of "2013 AIME I Problems/Problem 13"
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For ease, all ratios I will use to solve this problem are with respect to the area of <math>AB_0C_0</math>. For example, if I say some area has ratio <math>\frac{1}{2}</math>, that means its area is 45. | For ease, all ratios I will use to solve this problem are with respect to the area of <math>AB_0C_0</math>. For example, if I say some area has ratio <math>\frac{1}{2}</math>, that means its area is 45. | ||
− | Now note that <math>k=</math> 1 minus ratio of <math>B_1C_1A</math> minus ratio <math>B_0C_0C_1</math>. We see by similar triangles given that ratio <math>B_0C_0C_1</math> is <math>\frac{17^2}{25^2}</math>. Ratio <math>B_1C_1A</math> is, after seeing that <math>C_1C_0 = \frac{289}{625}</math>, <math>(\frac{336}{625})^2</math>. Now it suffices to find 90 times ratio <math>B_0B_1C_1</math>, which is given by 1 minus the two aforementioned ratios. Substituting these ratios to find <math>k</math> and clearing out the <math>5^8</math>, we see that the answer is <math>90\cdot \frac{5^8-336^2-17^2\cdot 5^4}{5^8-336^2}</math>. Calculation might take some time, but you've solved the problem! <math> | + | Now note that <math>k=</math> 1 minus ratio of <math>B_1C_1A</math> minus ratio <math>B_0C_0C_1</math>. We see by similar triangles given that ratio <math>B_0C_0C_1</math> is <math>\frac{17^2}{25^2}</math>. Ratio <math>B_1C_1A</math> is, after seeing that <math>C_1C_0 = \frac{289}{625}</math>, <math>(\frac{336}{625})^2</math>. Now it suffices to find 90 times ratio <math>B_0B_1C_1</math>, which is given by 1 minus the two aforementioned ratios. Substituting these ratios to find <math>k</math> and clearing out the <math>5^8</math>, we see that the answer is <math>90\cdot \frac{5^8-336^2-17^2\cdot 5^4}{5^8-336^2}</math>. Calculation might take some time, but you've solved the problem! <math>q= \boxed{961}</math>. |
== Solution 2 == | == Solution 2 == |
Latest revision as of 00:37, 2 March 2021
Problem
Triangle has side lengths , , and . For each positive integer , points and are located on and , respectively, creating three similar triangles . The area of the union of all triangles for can be expressed as , where and are relatively prime positive integers. Find .
Solution 1 (Simple, Sane Solution)
Well, first draw a good diagram! One is provided below. Convince yourself that every is parallel to each other for any nonnegative . Next, convince yourself that the area we seek is simply the ratio , because it repeats in smaller and smaller units. Note that the area of the triangle, by Heron's formula, is 90.
For ease, all ratios I will use to solve this problem are with respect to the area of . For example, if I say some area has ratio , that means its area is 45.
Now note that 1 minus ratio of minus ratio . We see by similar triangles given that ratio is . Ratio is, after seeing that , . Now it suffices to find 90 times ratio , which is given by 1 minus the two aforementioned ratios. Substituting these ratios to find and clearing out the , we see that the answer is . Calculation might take some time, but you've solved the problem! .
Solution 2
Using Heron's Formula we can get the area of the triangle .
Since then the scale factor for the dimensions of to is
Therefore, the area of is . Also, the dimensions of the other sides of the can be easily computed: and . This allows us to compute one side of the triangle , . Therefore, the scale factor to is , which yields the length of as . Therefore, the scale factor for to is . Some more algebraic manipulation will show that to is still . Also, since the triangles are disjoint, the area of the union is the sum of the areas. Therefore, the area is the geometric series At this point, it may be wise to "simplify" . So the geometric series converges to . Using the difference of squares, we get , which simplifies to . Cancelling all common factors, we get the reduced fraction . So , yielding the answer .
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.