2008 AMC 10B Problems/Problem 7

Problem

An equilateral triangle of side length $10$ is completely filled in by non-overlapping equilateral triangles of side length $1$ . How many small triangles are required?

$\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000$

Solution

(C) The area of the large triangle is $\frac{10^2\sqrt3}{4}$ , while the area each small triangle is $\frac{1^2\sqrt3}{4}$ . Dividing these two quantities, we get 100, therefore $\boxed{100}$ small triangles can fit in the large one.

Another Solution: $[asy] unitsize(0.5cm); defaultpen(0.8); for (int i=0; i<10; ++i) { draw( (i*dir(60)) -- ( (10,0) + (i*dir(120)) ) ); } for (int i=0; i<10; ++i) { draw( (i*dir(0)) -- ( 10*dir(60) + (i*dir(-60)) ) ); } for (int i=0; i<10; ++i) { draw( ((10-i)*dir(60)) -- ((10-i)*dir(0)) ); } [/asy]$

The number of triangles is $1+3+\dots+19 = \boxed{100}$ .

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: $19+17+15...+3+1 = \frac{19+1}{2}\cdot 10 = \boxed{100}$

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2008 AMC 10B Problems/Problem 7

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