# 1989 AHSME Problems/Problem 5

## Problem

Toothpicks of equal length are used to build a rectangular grid as shown. If the grid is 20 toothpicks high and 10 toothpicks wide, then the number of toothpicks used is $[asy] real xscl = 1.2; int[] x = {0,1,2,4,5},y={0,1,3,4,5}; for(int a:x){ for(int b:y) { dot((a*xscl,b)); } } for(int a:x) { pair prev = (a,y); for(int i = 1;i $\textrm{(A)}\ 30\qquad\textrm{(B)}\ 200\qquad\textrm{(C)}\ 410\qquad\textrm{(D)}\ 420\qquad\textrm{(E)}\ 430$

## Solution

There are 21 horizontal lines made of 10 matches, and 11 vertical lines made of 20 matches, and $21\cdot10+11\cdot20=\boxed{430}$.

Alternatively, the frame can be dissected into $20\cdot10$ L shapes, the right-hand border, and the bottom border: $[asy] path p,q; for(int i=0;i<3;++i) { for(int j=3;j<6;++j) { p = (i+0.75,j)--(i,j)--(i,j-0.75); draw(p); dot((i,j)); } } for(int i=0;i<3;++i) { q = (i,0)--(i+0.75,0); draw(q);dot((i,0)); q = (5,5-i)--(5,4.25-i); draw(q);dot((5,5-i)); } draw((4,0)--(4.75,0));dot((4,0)); draw((5,1)--(5,0.25));dot((5,1)); draw((3,2)--(4,1),dotted); [/asy]$ in which case the calculation is $2(20\cdot 10)+20+10=\boxed{430}$.

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