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Difference between revisions of "2013 AIME I Problems/Problem 13"

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==Problem==
 
== Problem 13 ==
 
== Problem 13 ==
 
Triangle <math>AB_0C_0</math> has side lengths <math>AB_0 = 12</math>, <math>B_0C_0 = 17</math>, and <math>C_0A = 25</math>. For each positive integer <math>n</math>, points <math>B_n</math> and <math>C_n</math> are located on <math>\overline{AB_{n-1}}</math> and <math>\overline{AC_{n-1}}</math>, respectively, creating three similar triangles <math>\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}</math>. The area of the union of all triangles <math>B_{n-1}C_nB_n</math> for <math>n\geq1</math> can be expressed as <math>\tfrac pq</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>q</math>.
 
Triangle <math>AB_0C_0</math> has side lengths <math>AB_0 = 12</math>, <math>B_0C_0 = 17</math>, and <math>C_0A = 25</math>. For each positive integer <math>n</math>, points <math>B_n</math> and <math>C_n</math> are located on <math>\overline{AB_{n-1}}</math> and <math>\overline{AC_{n-1}}</math>, respectively, creating three similar triangles <math>\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}</math>. The area of the union of all triangles <math>B_{n-1}C_nB_n</math> for <math>n\geq1</math> can be expressed as <math>\tfrac pq</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>q</math>.

Revision as of 14:42, 9 August 2018

Problem

Problem 13

Triangle $AB_0C_0$ has side lengths $AB_0 = 12$, $B_0C_0 = 17$, and $C_0A = 25$. For each positive integer $n$, points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$, respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}$. The area of the union of all triangles $B_{n-1}C_nB_n$ for $n\geq1$ can be expressed as $\tfrac pq$, where $p$ and $q$ are relatively prime positive integers. Find $q$.

Simple, Sane Solution

Well, first draw a good diagram! One is provided below. Convince yourself that every $B_nC_n$ is parallel to each other for any nonnegative $n$. Next, convince yourself that the area we seek is simply the ratio $k=\frac{B_0B_1C_1}{B_0B_1C_1+C_1C_0B_0}$, 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 $AB_0C_0$. For example, if I say some area has ratio $\frac{1}{2}$, that means its area is 45.

Now note that $k=$ 1 minus ratio of $B_1C_1A$ minus ratio $B_0C_0C_1$. We see by similar triangles given that ratio $B_0C_0C_1$ is $\frac{17^2}{25^2}$. Ratio $B_1C_1A$ is, after seeing that $C_1C_0 = \frac{289}{625}$, $(\frac{336}{625})^2$. Now it suffices to find 90 times ratio $B_0B_1C_1$, which is given by 1 minus the two aforementioned ratios. Substituting these ratios to find $k$ and clearing out the $5^8$, we see that the answer is $90 * \frac{5^8-336^2-17^2*5^4}{5^8-336^2}$. Calculation might take two minutes, but you've solved the problem! Denominator: $\boxed{961}$.

Solution

Using Heron's Formula we can get the area of the triangle $\Delta AB_0C_0 = 90$.

Since $\Delta AB_0C_0 \sim \Delta B_0C_1C_0$ then the scale factor for the dimensions of $\Delta B_0C_1C_0$ to $\Delta AB_0C_0$ is $\dfrac{17}{25}.$

Therefore, the area of $\Delta B_0C_1C_0$ is $(\dfrac{17}{25})^2(90)$. Also, the dimensions of the other sides of the $\Delta B_0C_1C_0$ can be easily computed: $\overline{B_0C_1}= \dfrac{17}{25}(12)$ and $\overline{C_1C_0} = \dfrac{17^2}{25}$. This allows us to compute one side of the triangle $\Delta AB_0C_0$, $\overline{AC_1} = 25 - \dfrac{17^2}{25} = \dfrac{25^2 - 17^2}{25}$. Therefore, the scale factor $\Delta AB_1C_1$ to $\Delta AB_0C_0$ is $\dfrac{25^2 - 17^2}{25^2}$ , which yields the length of $\overline{B_1C_1}$ as $\dfrac{25^2 - 17^2}{25^2}(17)$. Therefore, the scale factor for $\Delta B_1C_2C_1$ to $\Delta B_0C_1C_0$ is $\dfrac{25^2 - 17^2}{25^2}$. Some more algebraic manipulation will show that $\Delta B_nC_{n+1}C_n$ to $\Delta B_{n-1}C_nC_{n-1}$ is still $\dfrac{25^2 - 17^2}{25^2}$. Also, since the triangles are disjoint, the area of the union is the sum of the areas. Therefore, the area is the geometric series $\dfrac{90 \cdot 17^2}{25^2} \sum_{n=0}^{\infty} (\dfrac{25^2-17^2}{25^2})^2$ At this point, it may be wise to "simplify" $25^2 - 17^2 = (25-17)(25+17) = (8)(42) = 336$. So the geometric series converges to $\dfrac{90 \cdot 17^2}{25^2} \dfrac{1}{1 - \dfrac{336^2}{625^2}} = \dfrac{90 \cdot 17^2}{25^2} \dfrac{625^2}{625^2 - 336^2}$. Using the diffference of squares, we get $\dfrac{90 \cdot 17^2}{25^2}\dfrac{625^2}{(625 - 336)(625 + 336)}$, which simplifies to $\dfrac{90 \cdot 17^2}{25^2} \dfrac{625^2}{(289)(961)}$. Cancellling all common factors, we get the reduced fraction $= \dfrac{90 \cdot 25^2}{31^2}$. So $\frac{p}{q}=1-\frac{90 \cdot 25^2}{31^2}=\frac{90 \cdot 336}{961}$, yielding the answer $\fbox{961}$.

AIME13.png

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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