Difference between revisions of "2013 AIME I Problems/Problem 13"
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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> | 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 | ||
Revision as of 02:08, 1 February 2014
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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
