Difference between revisions of "2013 AIME I Problems/Problem 13"
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{{AIME box|year=2013|n=I|num-b=12|num-a=14}} | {{AIME box|year=2013|n=I|num-b=12|num-a=14}} | ||
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+ | Using Heron's Formula we can get the area of the triangle <math>\Delta AB_0C_0 = 90</math>. Since <math>\Delta AB_0C_0 \sim \Delta B_0C_1C_0 </math> then the scale factor for the dimensions of <math> \Delta B_0C_1C_0 </math> to <math>\Delta AB_0C_0 </math> is <math> \dfrac{17}{25}.</math> | ||
+ | Therefore, the area of <math> \Delta B_0C_1C_0 </math> is <math> (\dfrac{17}{25})^2(90) </math>. Also, the dimensions of the other sides of the <math> \Delta B_0C_1C_0 </math> can be easily computed: <math> \overline{B_0C_1}= \dfrac{17}{25}(12) </math> and <math> \overline{C_1C_0} = \dfrac{17^2}{25} </math>. This allows us to compute one side of the triangle <math>\Delta AB_0C_0 </math>, <math> \overline{AC_1} = 25 - \dfrac{17^2}{25} = \dfrac{25^2 - 17^2}{25} </math>. Therefore, the scale factor <math> \Delta AB_1C_1 </math> to <math>\Delta AB_0C_0 </math> is <math> \dfrac{25^2 - 17^2}{25^2}</math> , which yields the length of <math> \overline{B_1C_1} </math> as <math> \dfrac{25^2 - 17^2}{25^2}(17)</math>. Therefore, the scale factor for <math> \Delta B_1C_2C_1 </math> to <math> \Delta B_0C_1C_0 </math> is <math> \dfrac{25^2 - 17^2}{25^2} </math>. Some more algebraic manipulation will show that <math> \Delta B_nC_{n+1}C_n </math> to <math> \Delta B_{n-1}C_nC_{n-1} </math> is still <math> \dfrac{25^2 - 17^2}{25^2} </math>. Also, since the triangles are disjoint, the area of the union is the sum of the areas. Therefore, the area is the geometric series | ||
+ | <math> \dfrac{90 \cdot 17^2}{25} \sum_{n=0}^{\infty} (\dfrac{25^2-17^2}{25^2})^2 </math> | ||
+ | At this point, it may be wise to "simplify" <math> 25^2 - 17^2 = (25-17)(25+17) = (8)(42) = 336</math>. | ||
+ | So the geometric series converges to | ||
+ | <math>\dfrac{90 \cdot 17^2}{25} \dfrac{1}{1 - \dfrac{336^2}{625^2}} = \dfrac{90 \cdot 17^2}{25} \dfrac{625^2}{625^2 - 336^2}</math>. | ||
+ | Using the diffference of squares, we get <math>\dfrac{90 \cdot 17^2}{25}\dfrac{625^2}{(625 - 336)(625 + 336)</math>, which simplifies to <math> \dfrac{90 \cdot 17^2}{25} \dfrac{625^2}{(289)(961)}</math>. Cancellling all common factors, we get the reduced fraction, <math> = \dfrac{90 \cdot 25^3}{31^2} </math>, yielding the answer <math>961</math>. --[[User:Lmbailey|Lmbailey]] 17:30, 3 April 2013 (EDT) |
Revision as of 16:30, 3 April 2013
Contents
[hide]Problem 13
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
(solution)
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 |
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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 diffference of squares, we get $\dfrac{90 \cdot 17^2}{25}\dfrac{625^2}{(625 - 336)(625 + 336)$ (Error compiling LaTeX. Unknown error_msg), which simplifies to . Cancellling all common factors, we get the reduced fraction, , yielding the answer . --Lmbailey 17:30, 3 April 2013 (EDT)