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

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==Solution 2==
 
==Solution 2==
 
The first hexagon has <math>1</math> dot. The second hexagon has <math>1+6</math> dots. The third hexagon <math>1+6+12</math> dots. Following this pattern, we predict that the fourth hexagon will have <math>1+6+12+18=37</math> dots <math>\implies\boxed{\textbf{(B) }37}</math>.<br>
 
The first hexagon has <math>1</math> dot. The second hexagon has <math>1+6</math> dots. The third hexagon <math>1+6+12</math> dots. Following this pattern, we predict that the fourth hexagon will have <math>1+6+12+18=37</math> dots <math>\implies\boxed{\textbf{(B) }37}</math>.<br>
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~ junaidmansuri
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~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]
  
 
==Solution 3==
 
==Solution 3==

Revision as of 18:32, 18 November 2020

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]

Diagram by sircalcsalot

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

Solution

We find the pattern $1, 6, 12, 18, \ldots$. The sum of the first four numbers in this sequence is $\boxed{\textbf{(B) }37}$.

Solution 2

The first hexagon has $1$ dot. The second hexagon has $1+6$ dots. The third hexagon $1+6+12$ dots. Following this pattern, we predict that the fourth hexagon will have $1+6+12+18=37$ dots $\implies\boxed{\textbf{(B) }37}$.
~junaidmansuri

Solution 3

Each band adds $1, 6, 2(6), 3(6), 4(6) \cdots,$ so we have $1+6+2(6)+3(6)=1+6(6)=1+36=\boxed{\textbf{(B) }37}.$

[pog]

Solution 4

Let the hexagon with $1$ dot be $h_0$. Notice that the rest of the terms are generated by the recurrence relation $h_n=h_{n-1}+6n$ for $n> 0$. Using this, we find that $h_1=7,h_2=19,$ and $h_3=\textbf{(B) }37$.

-franzliszt

See also

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

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