Difference between revisions of "2020 AMC 8 Problems/Problem 4"
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==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); | ||
− | + | path p = scale(0.8)*unitcircle; | |
− | + | pair[] A; | |
− | |||
− | |||
− | |||
− | pair | ||
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( | + | 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( | + | 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)); | |
− | for (int i = 0; i < | + | </asy> |
− | fill( | + | |
− | draw( | + | <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); | ||
} | } | ||
− | + | draw(shift(T)*scale(5.25)*hex,black+linewidth(1.25)); | |
− | for (int i = 0; i < 6; ++i) { | + | |
− | fill( | + | pair R = 30A[0]+6A[1]; |
− | fill( | + | fill(shift(R)*p,grey1); |
− | fill( | + | 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> | </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 | |
− | |||
− | ==Solution | + | ==Video Solution by WhyMath== |
+ | https://youtu.be/szWgrOPNw8c | ||
− | + | ~savannahsolver | |
− | ==Video Solution== | + | ==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 07:14, 24 January 2024
Contents
- 1 Problem
- 2 Solution 1 (Pattern of the Rows)
- 3 Solution 2 (Pattern of the Bands)
- 4 Solution 3 (Pattern of the Bands)
- 5 Solution 4 (Brute Force)
- 6 Video Solution by NiuniuMaths (Easy to understand!)
- 7 Video Solution by Math-X (First understand the problem!!!)
- 8 Video Solution (🚀Under 2 min🚀)
- 9 Video Solution by WhyMath
- 10 Video Solution by The Learning Royal
- 11 Video Solution by Interstigation
- 12 Video Solution by North America Math Contest Go Go Go
- 13 Video Solution by STEMbreezy
- 14 See also
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?
Solution 1 (Pattern of the Rows)
Looking at the rows of each hexagon, we see that the first hexagon has dot, the second has
dots, and the third has
dots. Given the way the hexagons are constructed, it is clear that this pattern continues. Hence, the fourth hexagon has
dots.
Solution 2 (Pattern of the Bands)
The dots in the next hexagon have four bands. From innermost to outermost:
- The first band has
dot.
- The second band has
dots:
dot at each vertex of the hexagon.
- The third band has
dots:
dot at each vertex of the hexagon and
other dot on each edge of the hexagon.
- The fourth band has
dots:
dot at each vertex of the hexagon and
other dots on each edge of the hexagon.
Together, the answer is
~MRENTHUSIASM
Solution 3 (Pattern of the Bands)
The first hexagon has dot, the second hexagon has
dots, the third hexagon has
dots, and so on. The pattern continues since to go from hexagon
to hexagon
we add a new band of dots around the outside of the existing ones, with each side of the band having side length
Thus, the number of dots added is
(we subtract
as each of the corner hexagons in the band is counted as part of two sides.). We therefore predict that the fourth hexagon has
dots.
Remark
For positive integers let
denote the number of dots in the
th hexagon. We have
and
It follows that and
Solution 4 (Brute Force)
From the full diagram below, the answer is
~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🚀)
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by The Learning Royal
~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.