Difference between revisions of "2020 AMC 8 Problems/Problem 4"

Revision as of 15:48, 14 December 2020

Problem

Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon? $[asy] size(250); real side1 = 1.5; real side2 = 4.0; real side3 = 6.5; real pos = 2.5; pair s1 = (-10,-2.19); pair s2 = (15,2.19); pen grey1 = rgb(100/256, 100/256, 100/256); pen grey2 = rgb(183/256, 183/256, 183/256); fill(circle(origin + s1, 1), grey1); for (int i = 0; i < 6; ++i) { draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25)); } fill(circle(origin, 1), grey1); for (int i = 0; i < 6; ++i) { fill(circle(pos*dir(60*i),1), grey2); draw(side2*dir(60*i)--side2*dir(60*i-60),linewidth(1.25)); } fill(circle(origin+s2, 1), grey1); for (int i = 0; i < 6; ++i) { fill(circle(pos*dir(60*i)+s2,1), grey2); fill(circle(2*pos*dir(60*i)+s2,1), grey1); fill(circle(sqrt(3)*pos*dir(60*i+30)+s2,1), grey1); draw(side3*dir(60*i)+s2--side3*dir(60*i-60)+s2,linewidth(1.25)); } [/asy]$

$\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49$

Solution 1

Looking at the rows of each hexagon, we see that the first hexagon has $1$ dot, the second has $2+3+2$ dots and the third has $3+4+5+4+3$ dots, and given the way the hexagons are constructed, it is clear that this pattern continues. Hence the fourth hexagon has $4+5+6+7+6+5+4=\boxed{\textbf{(B) }37}$ dots.

Solution 2

The first hexagon has $1$ dot, the second hexagon has $1+6$ dots, the third hexagon $1+6+12$ dots, and so on. The pattern continues since to go from hexagon $n$ to hexagon $(n+1)$ , we add a new ring of hexagons around the outside of the existing ones, with each side of the ring having side length $(n+1)$ . Thus the number of hexagons added is $6(n+1)-6 = 6n$ (we subtract $6$ as each of the corner hexagons in the ring is counted as part of two sides), confirming the pattern. We therefore predict that that the fourth hexagon has $1+6+12+18=\boxed{\textbf{(B) }37}$ dots.

Solution 3 (variant of Solution 2)

Let the number of dots in the first hexagon be $h_0 = 1$ . By the same argument as in Solution 2, we have $h_n=h_{n-1}+6n$ for $n > 0$ . Using this, we find that $h_1=7$ , $h_2=19,$ and $h_3=\boxed{\textbf{(B) }37}$ .

Video Solution

https://youtu.be/SPNobOd4t1c (Includes all the problems and has a free class update)

https://youtu.be/eSxzI8P9_h8

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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 AJHSME/AMC 8 Problems and Solutions

Page

Toolbox

Search