Difference between revisions of "2020 AMC 8 Problems/Problem 4"
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/eSxzI8P9_h8 | https://youtu.be/eSxzI8P9_h8 | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/YnwkBZTv5Fw?t=123 | ||
==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}} |
Revision as of 14:27, 18 April 2021
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, and 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)
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 .
Video Solution by WhyMath
~savannahsolver
Video Solution
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=123
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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