2020 AMC 8 Problems/Problem 4
Contents
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
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
The first hexagon has dot, the second hexagon has
dots, the third hexagon
dots, and so on. The pattern continues since to go from hexagon
to hexagon
we add a new ring of dots around the outside of the existing ones, with each side of the ring having side length
Thus the number of dots added is
(we subtract
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
dots.
Solution 3 (Variant of Solution 2)
The dots in the next hexagon have four layers. From innermost to outermost:
- The first layer has
dot.
- The second layer has
dots:
dot at each vertex of the hexagon.
- The third layer has
dots:
dot at each vertex of the hexagon and
other dot on each edge of the hexagon.
- The fourth layer has
dots:
dot at each vertex of the hexagon and
other dots on each edge of the hexagon.
Together, the answer is
~MRENTHUSIASM
Solution 4 (Variant of Solution 2)
Let the number of dots in the first hexagon be By the same argument as in Solution 2, we have
for
Using this, we find that
and
Solution 5 (Brute Force)
From the full diagram below, the answer is
size(500); real side1 = 1.5; real side2 = 4.0; real side3 = 6.5; real side4 = 9.0; real pos = 2.5; pair s1 = (-10,-1/2-sqrt(3)); pair s2 = (15,1/2+sqrt(3)); pair s3 = (36,1+2*sqrt(3)); 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)); } fill(circle(origin+s3, 1), grey1); for (int i = 0; i < 6; ++i) { fill(circle(pos*dir(60*i)+s3,1), grey2); fill(circle(2*pos*dir(60*i)+s3,1), grey1); fill(circle(sqrt(3)*pos*dir(60*i+30)+s3,1), grey1); fill(circle(3*pos*dir(60*i)+s3,1), grey2); fill(circle(sqrt(7)*pos*dir(60*i+90-aTan(sqrt(3)/9))+s3,1), grey2); fill(circle(sqrt(7)*pos*dir(60*i-90+aTan(sqrt(3)/9))+s3,1), grey2); draw(side4*dir(60*i)+s3--side4*dir(60*i-60)+s3,linewidth(1.25)); } } (Error making remote request. Unknown error_msg)
~MRENTHUSIASM
Video Solution
https://www.youtube.com/watch?v=_IjQnXnVKeU
Video Solution by WhyMath
~savannahsolver
Video Solution
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=123
~Interstigation
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 |
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