Difference between revisions of "2013 AIME I Problems/Problem 13"
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==Solution 3== | ==Solution 3== | ||
− | For this problem, the key is to find the <math>\frac{[\triangle{ | + | 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 | The area of the biggest triangle is <math>90</math> according to the Heron's formula easily | ||
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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>, | 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{ | + | 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> | 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> |
Revision as of 04:36, 23 December 2021
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
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.