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

m (→‎Solution 3 (Pattern of the Bands))
m (→‎Solution 3 (Pattern of the Bands))
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<u><b>Remark</b></u>
 
<u><b>Remark</b></u>
  
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For positive integers <math>n,</math> let <math>h_n</math> denotes the number of dots in the <math>n</math>th hexagon. We have <math>h_1=1</math> and <math>h_{n+1}=h_n+6n.</math>
+
For positive integers <math>n,</math> let <math>h_n</math> denote the number of dots in the <math>n</math>th hexagon. We have <math>h_1=1</math> and <math>h_{n+1}=h_n+6n.</math>
  
 
It follows that <math>h_2=7,h_3=19,</math> and <math>h_4=37.</math>
 
It follows that <math>h_2=7,h_3=19,</math> and <math>h_4=37.</math>

Revision as of 01:23, 12 January 2022

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] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(250); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/256, 100/256, 100/256); pen grey2 = rgb(183/256, 183/256, 183/256); for (int i=0; i<7; ++i) { A[i] = rotate(60*i)*(1,0);} path hex = A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--cycle; fill(p,grey1); draw(scale(1.25)*hex,black+linewidth(1.25)); pair S = 6A[0]+2A[1]; fill(shift(S)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(S+2*A[i])*p,grey2);} draw(shift(S)*scale(3.25)*hex,black+linewidth(1.25)); pair T = 16A[0]+4A[1]; fill(shift(T)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(T+2*A[i])*p,grey2); fill(shift(T+4*A[i])*p,grey1); fill(shift(T+2*A[i]+2*A[i+1])*p,grey1); } draw(shift(T)*scale(5.25)*hex,black+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 (Pattern of the Rows)

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. 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 (Pattern of the Bands)

The dots in the next hexagon have four bands. From innermost to outermost:

  1. The first band has $1$ dot.

  2. The second band has $6$ dots: $1$ dot at each vertex of the hexagon.

  3. The third band has $6+6\cdot1=12$ dots: $1$ dot at each vertex of the hexagon and $1$ other dot on each edge of the hexagon.

  4. The fourth band has $6+6\cdot2=18$ dots: $1$ dot at each vertex of the hexagon and $2$ other dots on each edge of the hexagon.

Together, the answer is $1+6+12+18=\boxed{\textbf{(B) }37}.$

~MRENTHUSIASM

Solution 3 (Pattern of the Bands)

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 band of dots around the outside of the existing ones, with each side of the band having side length $(n+1).$ Thus the number of dots added is $6(n+1)-6 = 6n$ (we subtract $6$ as each of the corner hexagons in the band is counted as part of two sides.). We therefore predict that that the fourth hexagon has $1+6+12+18=\boxed{\textbf{(B) }37}$ dots.

Remark

For positive integers $n,$ let $h_n$ denote the number of dots in the $n$th hexagon. We have $h_1=1$ and $h_{n+1}=h_n+6n.$

It follows that $h_2=7,h_3=19,$ and $h_4=37.$

Solution 4 (Brute Force)

From the full diagram below, the answer is $\boxed{\textbf{(B) }37}.$ [asy] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(400); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/256, 100/256, 100/256); pen grey2 = rgb(183/256, 183/256, 183/256); for (int i=0; i<7; ++i) { A[i] = rotate(60*i)*(1,0);} path hex = A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--cycle; fill(p,grey1); draw(scale(1.25)*hex,black+linewidth(1.25)); pair S = 6A[0]+2A[1]; fill(shift(S)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(S+2*A[i])*p,grey2);} draw(shift(S)*scale(3.25)*hex,black+linewidth(1.25)); pair T = 16A[0]+4A[1]; fill(shift(T)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(T+2*A[i])*p,grey2); fill(shift(T+4*A[i])*p,grey1); fill(shift(T+2*A[i]+2*A[i+1])*p,grey1); } draw(shift(T)*scale(5.25)*hex,black+linewidth(1.25)); pair R = 30A[0]+6A[1]; fill(shift(R)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(R+2*A[i])*p,grey2); fill(shift(R+4*A[i])*p,grey1); fill(shift(R+2*A[i]+2*A[i+1])*p,grey1); fill(shift(R+6*A[i+1])*p,grey2); fill(shift(R+2*A[i]+4*A[i+1])*p,grey2); fill(shift(R+4*A[i]+2*A[i+1])*p,grey2); } draw(shift(R)*scale(7.25)*hex,black+linewidth(1.25)); [/asy] ~MRENTHUSIASM

Video Solution by WhyMath

https://youtu.be/szWgrOPNw8c

~savannahsolver

Video Solution by The Learning Royal

https://youtu.be/eSxzI8P9_h8

~The Learning Royal

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=123

~Interstigation

Video Solution by North America Math Contest Go Go Go

https://www.youtube.com/watch?v=_IjQnXnVKeU

~North America Math Contest Go Go Go

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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