Difference between revisions of "2013 AIME I Problems/Problem 13"
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+ | == Problem == | ||
+ | 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 == | ||
+ | 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. | ||
+ | |||
+ | 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 == |
− | + | 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 | 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 | 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>. | 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 | 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> | + | <math> \dfrac{90 \cdot 17^2}{25^2} \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>. | 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 | 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>. | + | <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 | + | 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}} |
Latest revision as of 18:12, 8 June 2024
Contents
[hide]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
Note that every is parallel to each other for any nonnegative . Also, 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 , which gives .
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 .
Solution 3
For this problem, the key is to find the .
The area of the biggest triangle is according to the Heron's formula easily
Firstly, we discuss the ratio of
Since the problem said that two triangles are similar, so ,
Getting that , which is not hard to find that , Since ,
we can find the ratio of , the common ratio between two similar triangles is , the similar triangles means two consecutive
Now the whole summation of
The desired answer is Which our answer is
~bluesoul
Solution 4
Let be the coefficient of the similarity of triangles Then area
The height of triangles and from is the same
The coefficient of the similarity of triangles is
Analogically the coefficient of the similarity of triangles is and so on.
The yellow area is
The required area is
The number is prime, is integer but not therefore the answer is .
vladimir.shelomovskii@gmail.com, vvsss
Video Solution
https://youtu.be/IdM24SLrxQw?si=mu5fQ-_rFZM4ud2_
~MathProblemSolvingSkills.com
Video Solution by mop 2024
~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.