Difference between revisions of "2020 AMC 8 Problems/Problem 4"
MRENTHUSIASM (talk | contribs) (Added in Sol 4.) |
MRENTHUSIASM (talk | contribs) (→Solution 4 (Brute Force)) |
||
Line 46: | Line 46: | ||
From the full diagram below, the answer is <math>\boxed{\textbf{(B) }37}.</math> | From the full diagram below, the answer is <math>\boxed{\textbf{(B) }37}.</math> | ||
<asy> | <asy> | ||
− | size( | + | size(350); |
real side1 = 1.5; | real side1 = 1.5; | ||
real side2 = 4.0; | real side2 = 4.0; |
Revision as of 05:24, 7 September 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
.
Solution 4 (Brute Force)
From the full diagram below, the answer is
Video Solution 1
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.