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Difference between revisions of "2020 AMC 8 Problems/Problem 4"
(→‎Video Solution by Math-X (First understand the problem!!!))
 
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==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>
 +
// diagram by SirCalcsALot, edited by MRENTHUSIASM
 
size(250);
 
size(250);
−
real side1 = 1.5;
+
path p = scale(0.8)*unitcircle;
−
real side2 = 4.0;
+
pair[] A;
−
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 grey1 = rgb(100/256, 100/256, 100/256);
 
pen grey2 = rgb(183/256, 183/256, 183/256);
 
pen grey2 = rgb(183/256, 183/256, 183/256);
−
fill(circle(origin + s1, 1), grey1);
+
for (int i=0; i<7; ++i) { A[i] = rotate(60*i)*(1,0);}
−
for (int i = 0; i < 6; ++i) {
+
path hex = A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--cycle;
−
draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25));
+
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);
 
}
 
}
−
fill(circle(origin, 1), grey1);
+
draw(shift(T)*scale(5.25)*hex,black+linewidth(1.25));
−
for (int i = 0; i < 6; ++i) {
+
</asy>
−
fill(circle(pos*dir(60*i),1), grey2);
+
 
−
draw(side2*dir(60*i)--side2*dir(60*i-60),linewidth(1.25));
+
<math>\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49</math>
 +
 
 +
==Solution 1 (Pattern of the Rows)==
 +
 
 +
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. 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.
 +
 
 +
== Solution 2 (Pattern of the Bands) ==
 +
The dots in the next hexagon have four bands. From innermost to outermost:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>The first band has <math>1</math> dot.</li><p>
 +
  <li>The second band has <math>6</math> dots: <math>1</math> dot at each vertex of the hexagon.</li><p>
 +
  <li>The third band has <math>6+6\cdot1=12</math> dots: <math>1</math> dot at each vertex of the hexagon and <math>1</math> other dot on each edge of the hexagon.</li><p>
 +
  <li>The fourth band has <math>6+6\cdot2=18</math> dots: <math>1</math> dot at each vertex of the hexagon and <math>2</math> other dots on each edge of the hexagon.</li><p>
 +
</ol>
 +
Together, the answer is <math>1+6+12+18=\boxed{\textbf{(B) }37}.</math>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
==Solution 3 (Pattern of the Bands)==
 +
The first hexagon has <math>1</math> dot, the second hexagon has <math>1+6</math> dots, the third hexagon has <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 band of dots around the outside of the existing ones, with each side of the band having side length <math>(n+1).</math> Thus, the number of dots added is <math>6(n+1)-6 = 6n</math> (we subtract <math>6</math> as each of the corner hexagons in the band is counted as part of two sides.). We therefore predict that the fourth hexagon has <math>1+6+12+18=\boxed{\textbf{(B) }37}</math> dots.
 +
 
 +
<u><b>Remark</b></u>
 +
 
 +
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>
 +
 
 +
== Solution 4 (Brute Force) ==
 +
From the full diagram below, the answer is <math>\boxed{\textbf{(B) }37}.</math>
 +
<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);
 
}
 
}
−
fill(circle(origin+s2, 1), grey1);
+
draw(shift(T)*scale(5.25)*hex,black+linewidth(1.25));
−
for (int i = 0; i < 6; ++i) {
+
 
−
fill(circle(pos*dir(60*i)+s2,1), grey2);
+
pair R = 30A[0]+6A[1];
−
fill(circle(2*pos*dir(60*i)+s2,1), grey1);
+
fill(shift(R)*p,grey1);
−
fill(circle(sqrt(3)*pos*dir(60*i+30)+s2,1), grey1);
+
for (int i=0; i<6; ++i) {  
−
draw(side3*dir(60*i)+s2--side3*dir(60*i-60)+s2,linewidth(1.25));
+
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>
 
</asy>
 +
~MRENTHUSIASM
 +
 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=8hgK6rESdek&t=9s
 +
 +
~NiuniuMaths
  
−
Diagram by sircalcsalot
+
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/UnVo6jZ3Wnk?si=bKY7jHWZFGeYKoM6&t=324
  
−
<math>\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49</math>
+
~Math-X
  
−
==Solution==
+
==Video Solution (🚀Under 2 min🚀)==
−
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>.
+
https://youtu.be/V5EaJihwEMQ
  
−
==Solution 2==
+
~Education, the Study of Everything
−
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>
 
−
~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]
 
  
−
==Solution 3==
+
==Video Solution by WhyMath==
−
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>
+
https://youtu.be/szWgrOPNw8c
  
−
[pog]
+
~savannahsolver
  
−
==Solution 4==
+
==Video Solution by The Learning Royal==
 +
https://youtu.be/eSxzI8P9_h8
  
−
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>.
+
~The Learning Royal
  
−
-franzliszt
+
==Video Solution by Interstigation==
 +
https://youtu.be/YnwkBZTv5Fw?t=123
  
−
==Solution 5==
+
~Interstigation
  
−
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>.
+
== Video Solution by North America Math Contest Go Go Go ==
 +
https://www.youtube.com/watch?v=_IjQnXnVKeU
  
−
-vaporwave
+
~North America Math Contest Go Go Go
  
−
==Video Solution==
+
==Video Solution by STEMbreezy==
−
https://youtu.be/szWgrOPNw8c
+
https://youtu.be/L_vDc-i585o?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=141
  
−
~savannahsolver
+
~STEMbreezy
  
 
==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}}

Latest revision as of 06:14, 24 January 2024

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

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

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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