Difference between revisions of "2013 AIME I Problems/Problem 13"
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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} \dfrac{1}{1 - \dfrac{336^2}{625^2}} = \dfrac{90 \cdot 17^2}{25} \dfrac{625^2}{625^2 - 336^2}</math>. | ||
− | Using the diffference of squares, we get <math>\dfrac{90 \cdot 17^2}{25}\dfrac{625^2}{(625 - 336)(625 + 336)</math>, which simplifies to <math> \dfrac{90 \cdot 17^2}{25} \dfrac{625^2}{(289)(961)}</math>. Cancellling all common factors, we get the reduced fraction, <math> = \dfrac{90 \cdot 25^3}{31^2} </math>, yielding the answer <math>961</math>. | + | Using the diffference of squares, we get <math>\dfrac{90 \cdot 17^2}{25}\dfrac{625^2}{(625 - 336)(625 + 336)}</math>, which simplifies to <math> \dfrac{90 \cdot 17^2}{25} \dfrac{625^2}{(289)(961)}</math>. Cancellling all common factors, we get the reduced fraction, <math> = \dfrac{90 \cdot 25^3}{31^2} </math>, yielding the answer <math>961</math>. |
[[File:AIME13.png]] | [[File:AIME13.png]] |
Revision as of 18:57, 3 April 2013
Problem 13
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
(solution)
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 diffference of squares, we get
, which simplifies to
. Cancellling all common factors, we get the reduced fraction,
, yielding the answer
.
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 |