Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Video Solution)
(→‎Video Solution by Math-X (First understand the problem!!!))
 
(53 intermediate revisions by 14 users not shown)
Line 1: Line 1:
 
==Problem==
 
==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
  
−
<math>\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49</math>
+
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=8hgK6rESdek&t=9s
 +
 
 +
~NiuniuMaths
  
−
==Solution 1==
+
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/UnVo6jZ3Wnk?si=bKY7jHWZFGeYKoM6&t=324
  
−
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.
+
~Math-X
  
−
==Solution 2==
+
==Video Solution (🚀Under 2 min🚀)==
−
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.
+
https://youtu.be/V5EaJihwEMQ
  
−
==Solution 3 (variant of Solution 2)==
+
~Education, the Study of Everything
  
−
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>.
+
==Video Solution by WhyMath==
 +
https://youtu.be/szWgrOPNw8c
  
−
==Video Solution==
+
~savannahsolver
−
https://www.youtube.com/channel/UCUf37EvvIHugF9gPiJ1yRzQ
 
  
 +
==Video Solution by The Learning Royal==
 
https://youtu.be/eSxzI8P9_h8
 
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==  
 
==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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_4&oldid=212543"