2003 AMC 10A Problems/Problem 23
Problem
A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have 
rows of small congruent equilateral triangles, with 
small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of 
small equilateral triangles?
![[asy] unitsize(15mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair Ap=(0,0), Bp=(1,0), Cp=(2,0), Dp=(3,0), Gp=dir(60); pair Fp=shift(Gp)*Bp, Ep=shift(Gp)*Cp; pair Hp=shift(Gp)*Gp, Ip=shift(Gp)*Fp; pair Jp=shift(Gp)*Hp; pair[] points={Ap,Bp,Cp,Dp,Ep,Fp,Gp,Hp,Ip,Jp}; draw(Ap--Dp--Jp--cycle); draw(Gp--Bp--Ip--Hp--Cp--Ep--cycle); for(pair p : points) { fill(circle(p, 0.07),white); } pair[] Cn=new pair[5]; Cn[0]=centroid(Ap,Bp,Gp); Cn[1]=centroid(Gp,Bp,Fp); Cn[2]=centroid(Bp,Fp,Cp); Cn[3]=centroid(Cp,Fp,Ep); Cn[4]=centroid(Cp,Ep,Dp); label("$1$",Cn[0]); label("$2$",Cn[1]); label("$3$",Cn[2]); label("$4$",Cn[3]); label("$5$",Cn[4]); for (pair p : Cn) { draw(circle(p,0.1)); }[/asy]](http://latex.artofproblemsolving.com/1/6/c/16ca1f0305d713b2b9e5a0c53658d11829856ec7.png)
Solution
Solution 1
There are 
small equilateral triangles.
Each small equilateral triangle needs toothpicks to make it.
But, each toothpick that isn't one of the 
toothpicks on the outside of the large equilateral triangle is a side for 
small equilateral triangles.
So, the number of toothpicks on the inside of the large equilateral triangle is 
Therefore the total number of toothpicks is 
~dolphin7
Solution 2
The first row of triangles has 
upward-facing triangle, the second row has upward-facing triangles, the third row has upward-facing triangles, and so on having 
upward-facing triangles in the 
row. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are 
~mathpro12345
See Also
| 2003 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
