Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AIME I Problems/Problem 13"
m
 
(14 intermediate revisions by 6 users not shown)
Line 2: Line 2:
 
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>.
  
−
==Solution 1 (Simple, Sane Solution) ==
+
==Solution 1 ==
−
Well, first draw a good diagram! One is provided below. Convince yourself that every <math>B_nC_n</math> is parallel to each other for any nonnegative <math>n</math>. Next, convince yourself that the area we seek is simply the ratio <math>k=\frac{B_0B_1C_1}{B_0B_1C_1+C_1C_0B_0}</math>, because it repeats in smaller and smaller units. Note that the area of the triangle, by Heron's formula, is 90.
+
Note that every <math>B_nC_n</math> is parallel to each other for any nonnegative <math>n</math>. Also, the area we seek is simply the ratio <math>k=\frac{[B_0B_1C_1]}{[B_0B_1C_1]+[C_1C_0B_0]}</math>, 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 <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>p= \boxed{961}</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 <math>(\frac{336}{625})^2</math>, after seeing that <math>C_1C_0 = \frac{289}{625}</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>, which gives <math>q= \boxed{961}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==
Line 23: Line 23:
 
<math>\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}</math>.
 
<math>\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}</math>.
 
Using the difference of squares, we get <math>\dfrac{90 \cdot 17^2}{25^2}\dfrac{625^2}{(625 - 336)(625 + 336)}</math>, which simplifies to  <math> \dfrac{90 \cdot 17^2}{25^2} \dfrac{625^2}{(289)(961)}</math>. Cancelling all common factors, we get the reduced fraction  <math> = \dfrac{90 \cdot 25^2}{31^2} </math>. So <math>\frac{p}{q}=1-\frac{90 \cdot 25^2}{31^2}=\frac{90 \cdot 336}{961}</math>, yielding the answer <math>\fbox{961}</math>.
 
Using the difference of squares, we get <math>\dfrac{90 \cdot 17^2}{25^2}\dfrac{625^2}{(625 - 336)(625 + 336)}</math>, which simplifies to  <math> \dfrac{90 \cdot 17^2}{25^2} \dfrac{625^2}{(289)(961)}</math>. Cancelling all common factors, we get the reduced fraction  <math> = \dfrac{90 \cdot 25^2}{31^2} </math>. So <math>\frac{p}{q}=1-\frac{90 \cdot 25^2}{31^2}=\frac{90 \cdot 336}{961}</math>, yielding the answer <math>\fbox{961}</math>.
 +
 +
==Solution 3==
 +
For this problem, the key is to find the <math>\frac{[\triangle{B_nB_{n-1}C_n}]}{[\triangle{AB_{n-1}C_{n-1}}]}</math>.
 +
 +
The area of the biggest triangle is <math>90</math> according to the Heron's formula easily
 +
 +
Firstly, we discuss the ratio of  <math>\frac{[\triangle{B_0C_1C_0}]}{[\triangle{AB_0C_0}]}</math>
 +
 +
Since the problem said that two triangles are similar, so <math>\frac{C_1C_0}{B_0C_0}=\frac{17}{25}</math>,
 +
 +
Getting that <math>C_1C_0=\frac{289}{25}</math>, which is not hard to find that <math>AC_1=\frac{336}{25}</math>, Since <math>\frac{AB_1}{AB_0}=\frac{AC_1}{AC_0}=\frac{336}{625}</math>,
 +
 +
we can find the ratio of <math>\frac{[\triangle{B_0B_1C_1}]}{[\triangle{AB_0C_0}]}=\frac{336}{625}*\frac{289}{625}</math>, the common ratio between two similar triangles is <math>(\frac{336}{625})^2</math>, the similar triangles means two consecutive <math>(\triangle{AB_nC_n});(\triangle{AB_{n+1}C_{n+1}})</math>
 +
 +
Now the whole summation of <math>S=1+(\frac{336}{625})^2+(\frac{336}{625})^3+....+(\frac{336}{625})^n=\frac{625^2}{961*289}</math>
 +
 +
The desired answer is <math>90*\frac{336*289*625^2}{625^2*961*289}=\frac{30240}{961}</math> Which our answer is <math>\fbox{961}</math>
 +
 +
~bluesoul
  
 
[[File:AIME13.png]]
 
[[File:AIME13.png]]
 +
 +
==Solution 4==
 +
[[File:2013 AIME I 13.png|530px|right]]
 +
 +
Let <math>k</math> be the coefficient of the similarity of triangles
 +
<cmath>\triangle B_0 C_1 C_0 \sim \triangle AB_0 C_0 \implies k = \frac {B_0 C_0}{AC_0} = \frac {17}{25}.</cmath>
 +
Then area <math>\frac {[B_0 C_1 C_0]}{[AB_0 C_0 ]} = k^2 \implies \frac {[AB_0 C_1]}{[AB_0 C_0]} = 1 – k^2.</math>
 +
 +
The height of triangles <math>\triangle B_0C_1A</math> and <math>\triangle AB_0C_0</math> from <math>B_0</math> is the same <math>\implies \frac {AC_1}{AC_0} = 1 – k^2.</math>
 +
 +
The coefficient of the similarity of triangles <math>\triangle AB_1C_1 \sim \triangle AB_0C_0</math> is  <math> \frac {AC_1}{AC_0} = 1 – k^2 \implies \frac {[B_1C_1C_2 ]}{[AB_0C_0 ]} = k^2 (1 – k^2)^2. </math>
 +
 +
Analogically the coefficient of the similarity of triangles <math>\triangle AB_2C_2 \sim \triangle AB_0C_0</math> is  <math> (1 – k^2)^2 \implies \frac {[B_2C_2C_3]}{[AB_0C_0 ]} = k^2 (1 – k^2)^4 </math> and so on.
 +
 +
The yellow area <math>[Y]</math> is <math>\frac {[Y]}{[AB_0C_0 ]} = k^2 + k^2 (1 – k^2)^2 + k^2 (1 – k^2)^4 +.. = \frac {k^2}{1 – (1 – k^2)^2} = \frac{1}{2 – k^2}.</math>
 +
 +
The required area is <math>[AB_0C_0 ] – [Y] = [AB_0C_0 ] \cdot (1 –  \frac{1}{2 – k^2}) = [AB_0C_0 ] \cdot \frac  {1 –  k^2}{2 – k^2} = [AB_0C_0 ] \cdot \frac {25^2 – 17^2} {2 \cdot 25^2 – 17^2} =  [AB_0C_0 ] \cdot  \frac {336}{961}.</math>
 +
 +
The number <math>961</math> is prime, <math>[AB_0C_0]</math> is integer but not <math>961,</math> therefore the answer is <math>\boxed{961}</math>.
 +
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 +
 +
==Video Solution==
 +
https://youtu.be/IdM24SLrxQw?si=mu5fQ-_rFZM4ud2_
 +
 +
~MathProblemSolvingSkills.com
 +
 +
 +
==Video Solution by mop 2024==
 +
https://youtu.be/byoHYJx40bU
 +
 +
~r00tsOfUnity
  
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:12, 8 June 2024

Problem

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 $q$.

Solution 1

Note that every $B_nC_n$ is parallel to each other for any nonnegative $n$. Also, 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 $(\frac{336}{625})^2$, after seeing that $C_1C_0 = \frac{289}{625}$, . 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 $90\cdot \frac{5^8-336^2-17^2\cdot 5^4}{5^8-336^2}$, which gives $q= \boxed{961}$.

Solution 2

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 difference of squares, we get $\dfrac{90 \cdot 17^2}{25^2}\dfrac{625^2}{(625 - 336)(625 + 336)}$, which simplifies to $\dfrac{90 \cdot 17^2}{25^2} \dfrac{625^2}{(289)(961)}$. Cancelling 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 $\fbox{961}$.

Solution 3

For this problem, the key is to find the $\frac{[\triangle{B_nB_{n-1}C_n}]}{[\triangle{AB_{n-1}C_{n-1}}]}$.

The area of the biggest triangle is $90$ according to the Heron's formula easily

Firstly, we discuss the ratio of $\frac{[\triangle{B_0C_1C_0}]}{[\triangle{AB_0C_0}]}$

Since the problem said that two triangles are similar, so $\frac{C_1C_0}{B_0C_0}=\frac{17}{25}$,

Getting that $C_1C_0=\frac{289}{25}$, which is not hard to find that $AC_1=\frac{336}{25}$, Since $\frac{AB_1}{AB_0}=\frac{AC_1}{AC_0}=\frac{336}{625}$,

we can find the ratio of $\frac{[\triangle{B_0B_1C_1}]}{[\triangle{AB_0C_0}]}=\frac{336}{625}*\frac{289}{625}$, the common ratio between two similar triangles is $(\frac{336}{625})^2$, the similar triangles means two consecutive $(\triangle{AB_nC_n});(\triangle{AB_{n+1}C_{n+1}})$

Now the whole summation of $S=1+(\frac{336}{625})^2+(\frac{336}{625})^3+....+(\frac{336}{625})^n=\frac{625^2}{961*289}$

The desired answer is $90*\frac{336*289*625^2}{625^2*961*289}=\frac{30240}{961}$ Which our answer is $\fbox{961}$

~bluesoul

AIME13.png

Solution 4

2013 AIME I 13.png

Let $k$ be the coefficient of the similarity of triangles \[\triangle B_0 C_1 C_0 \sim \triangle AB_0 C_0 \implies k = \frac {B_0 C_0}{AC_0} = \frac {17}{25}.\] Then area $\frac {[B_0 C_1 C_0]}{[AB_0 C_0 ]} = k^2 \implies \frac {[AB_0 C_1]}{[AB_0 C_0]} = 1 – k^2.$

The height of triangles $\triangle B_0C_1A$ and $\triangle AB_0C_0$ from $B_0$ is the same $\implies \frac {AC_1}{AC_0} = 1 – k^2.$

The coefficient of the similarity of triangles $\triangle AB_1C_1 \sim \triangle AB_0C_0$ is $\frac {AC_1}{AC_0} = 1 – k^2 \implies \frac {[B_1C_1C_2 ]}{[AB_0C_0 ]} = k^2 (1 – k^2)^2.$

Analogically the coefficient of the similarity of triangles $\triangle AB_2C_2 \sim \triangle AB_0C_0$ is $(1 – k^2)^2 \implies \frac {[B_2C_2C_3]}{[AB_0C_0 ]} = k^2 (1 – k^2)^4$ and so on.

The yellow area $[Y]$ is $\frac {[Y]}{[AB_0C_0 ]} = k^2 + k^2 (1 – k^2)^2 + k^2 (1 – k^2)^4 +.. = \frac {k^2}{1 – (1 – k^2)^2} = \frac{1}{2 – k^2}.$

The required area is $[AB_0C_0 ] – [Y] = [AB_0C_0 ] \cdot (1 – \frac{1}{2 – k^2}) = [AB_0C_0 ] \cdot \frac {1 – k^2}{2 – k^2} = [AB_0C_0 ] \cdot \frac {25^2 – 17^2} {2 \cdot 25^2 – 17^2} = [AB_0C_0 ] \cdot \frac {336}{961}.$

The number $961$ is prime, $[AB_0C_0]$ is integer but not $961,$ therefore the answer is $\boxed{961}$.

vladimir.shelomovskii@gmail.com, vvsss


Video Solution

https://youtu.be/IdM24SLrxQw?si=mu5fQ-_rFZM4ud2_

~MathProblemSolvingSkills.com


Video Solution by mop 2024

https://youtu.be/byoHYJx40bU

~r00tsOfUnity

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_13&oldid=220237"