Difference between revisions of "2008 AMC 10B Problems/Problem 7"
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The number of triangles is <math>1+3+\dots+19 = \boxed{100}</math>. | The number of triangles is <math>1+3+\dots+19 = \boxed{100}</math>. | ||
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| + | Also, another way to do it is to notice that as you go row by row (from the bottom), the number of triangles decrease by 2 from 19, so we have: | ||
| + | <math>19+17+15...+3+1 = \frac{19+1}{2}\cdot 10 = \boxed{100}</math> | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=6|num-a=8}} | {{AMC10 box|year=2008|ab=B|num-b=6|num-a=8}} | ||
Revision as of 22:46, 1 February 2011
Problem
An equilateral triangle of side length 
is completely filled in by non-overlapping equilateral triangles of side length 
. How many small triangles are required?
Solution
The area of the large triangle is 
, while the area each small triangle is 
. Dividing these two quantities, we get 100, therefore 
small triangles can fit in the large one.
Another Solution:
![[asy] unitsize(0.5cm); defaultpen(0.8); for (int i=0; i<10; ++i) { draw( (i*dir(60)) -- ( (10,0) + (i*dir(120)) ) ); } for (int i=0; i<10; ++i) { draw( (i*dir(0)) -- ( 10*dir(60) + (i*dir(-60)) ) ); } for (int i=0; i<10; ++i) { draw( ((10-i)*dir(60)) -- ((10-i)*dir(0)) ); } [/asy]](http://latex.artofproblemsolving.com/2/b/6/2b6049d158759587caa572b1d8911fc0deb0b10b.png)
The number of triangles is 
.
Also, another way to do it is to notice that as you go row by row (from the bottom), the number of triangles decrease by 2 from 19, so we have:
See also
| 2008 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
