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

(→‎Solution 5 (Using the Sigma Function))
(→‎Solution 5 (Using the Sigma Function))
Line 102: Line 102:
 
Calculating this sum, we find:
 
Calculating this sum, we find:
  
−
For n=1: 1
+
For <math>n=1</math>: 1<math>
−
For n=2: 2
+
</math>For n=2: 2<math>
−
For n=3: 3
+
</math>For n=3: 3<math>
−
For n=4: 4
+
</math>For n=4: 4<math>
 
Now, we can add these values together:
 
Now, we can add these values together:
  
−
1+2+3+4=10
+
</math>1+2+3+4=10<math>
  
 
However, we need to account for the additional increments of 6 for each term. The total contribution from the increments is:
 
However, we need to account for the additional increments of 6 for each term. The total contribution from the increments is:
  
−
First term: 1
+
</math>First term: 1<math>
−
Second term: 6
+
</math>Second term: 6<math>
−
Third term: 12
+
</math>Third term: 12<math>
−
Fourth term: 18
+
</math>Fourth term: 18<math>
 
Adding these adjusted terms gives us:
 
Adding these adjusted terms gives us:
  
−
1+16+12+18
+
</math>1+16+12+18<math>
  
−
Thus, the final answer is:
+
Thus, our final answer is </math>\boxed{\textbf{(B) }37}$ dots.
−
 
 
−
boxed(B) 37
 
 
​
 
​
  

Revision as of 20:31, 20 September 2025

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 has $1+6+12$ 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 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

Solution 5 (Using the Sigma Function)

To solve this problem, we can utilize the sigma function from calculus. From our analysis in Solution 2, we identified a pattern where each term in the sequence is derived from an increasing difference arithmetic sequence starting at 1. Specifically, each term is incremented by 6, and the difference itself increases by 6 with each subsequent term.

We can express this mathematically using the sigma notation: For n=1 \[\sum\limits^{1}_{n=1}n\] For n=2,3 and 4 \[\sum\limits^{4}_{n=1}6n\] Calculating this sum, we find:

For $n=1$: 1$$ (Error compiling LaTeX. Unknown error_msg)For n=2: 2$$ (Error compiling LaTeX. Unknown error_msg)For n=3: 3$$ (Error compiling LaTeX. Unknown error_msg)For n=4: 4$Now, we can add these values together:$1+2+3+4=10$However, we need to account for the additional increments of 6 for each term. The total contribution from the increments is:$First term: 1$$ (Error compiling LaTeX. Unknown error_msg)Second term: 6$$ (Error compiling LaTeX. Unknown error_msg)Third term: 12$$ (Error compiling LaTeX. Unknown error_msg)Fourth term: 18$Adding these adjusted terms gives us:$1+16+12+18$Thus, our final answer is$\boxed{\textbf{(B) }37}$ dots. ​

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=8hgK6rESdek&t=9s

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=bKY7jHWZFGeYKoM6&t=324

~Math-X

Video Solution (🚀Under 2 min🚀)

https://youtu.be/V5EaJihwEMQ

~Education, the Study of Everything

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

Video Solution by STEMbreezy

https://youtu.be/L_vDc-i585o?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=141

~STEMbreezy

Video Solution by TheNeuralMathAcademy

https://youtu.be/mGQw1yALXYM&t=320s

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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