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

m (Video Solution)
(5 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 +
==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?
 
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>
 
<asy>
Line 28: Line 29:
 
</asy>
 
</asy>
  
Diagram by sircalcsalot
+
<math>\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49</math>
  
<math>\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49</math>
+
==Solution 1==
  
==Solution==
+
Looking at the rows of each hexagon, we see that the first hexagon has <math>1</math> dot, the second has <math>2+3+2</math> dots and the third has <math>3+4+5+4+3</math> dots, and given the way the hexagons are constructed, it is clear that this pattern continues. Hence the fourth hexagon has <math>4+5+6+7+6+5+4=\boxed{\textbf{(B) }37}</math> dots.
We find the pattern <math>1, 6, 12, 18, \ldots</math>. The sum of the first four numbers in this sequence is <math>\boxed{\textbf{(B) }37}</math>.
 
  
 
==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, and so on. The pattern continues since to go from hexagon <math>n</math> to hexagon <math>(n+1)</math>, we add a new ring of hexagons around the outside of the existing ones, with each side of the ring having side length <math>(n+1)</math>. Thus the number of hexagons added is <math>6(n+1)-6 = 6n</math> (we subtract <math>6</math> 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 <math>1+6+12+18=\boxed{\textbf{(B) }37}</math> dots.
~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]
 
 
 
==Solution 3==
 
Each band adds <math>1, 6, 2(6), 3(6), 4(6) \cdots,</math> so we have <math>1+6+2(6)+3(6)=1+6(6)=1+36=\boxed{\textbf{(B) }37}.</math>
 
 
 
[pog]
 
 
 
==Solution 4==
 
 
 
Let the hexagon with <math>1</math> dot be <math>h_0</math>. Notice that the rest of the terms are generated by the recurrence relation <math>h_n=h_{n-1}+6n</math> for <math>n> 0</math>. Using this, we find that <math>h_1=7,h_2=19,</math> and <math>h_3=\textbf{(B) }37</math>.
 
 
 
-franzliszt
 
  
==Solution 4==
+
==Solution 3 (variant of Solution 2)==
  
Adding up the dots by rows in each hexagon, we see that the first hexagon has <math>1</math> dot, the second has <math>2+3+2</math> dots and the third has <math>3+4+5+4+3</math> dots. Following the pattern, the fourth hexagon has <math>4+5+6+7+6+5+4=\boxed{\textbf{(B) }37}</math>.
+
Let the number of dots in the first hexagon be <math>h_0 = 1</math>. By the same argument as in Solution 2, we have <math>h_n=h_{n-1}+6n</math> for <math>n > 0</math>. Using this, we find that <math>h_1=7</math>, <math>h_2=19,</math> and <math>h_3=\boxed{\textbf{(B) }37}</math>.
  
-vaporwave
+
==Video Solution==
 +
https://youtu.be/eSxzI8P9_h8
  
 
==See also==  
 
==See also==  
 
{{AMC8 box|year=2020|num-b=3|num-a=5}}
 
{{AMC8 box|year=2020|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:17, 26 November 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/eSxzI8P9_h8

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
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