Difference between revisions of "2013 AIME I Problems/Problem 13"
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==Solution 1 == | ==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. | + | 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 <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>. | + | 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 == | ||
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==Solution 4== | ==Solution 4== | ||
− | [[File:2013 AIME I 13.png| | + | [[File:2013 AIME I 13.png|530px|right]] |
Let <math>k</math> be the coefficient of the similarity of triangles | Let <math>k</math> be the coefficient of the similarity of triangles | ||
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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> | 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 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 ] (1 – \frac{1}{2 – k^2}) = [AB_0C_0 ]\frac {1 – k^2}{2 – k^2} = [AB_0C_0 ]\frac {25^2 – 17^2} {2 \cdot 25^2 – 17^2} = [AB_0C_0 ] \frac {336}{961}.</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>. | + | 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 19:12, 8 June 2024
Contents
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.