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| | == See also == | | == See also == |
| | {{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>
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| − | 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
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| − | <math> \dfrac{90 \cdot 17^2}{25} \sum_{n=0}^{\infty} (\dfrac{25^2-17^2}{25^2})^2 </math>
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| − | At this point, it may be wise to "simplify" <math> 25^2 - 17^2 = (25-17)(25+17) = (8)(42) = 336</math>.
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| − | So the geometric series converges to
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| − | <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>.
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| − | 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)
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Revision as of 16:31, 3 April 2013
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 
