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

m (→‎Problem)
m (→‎Solution 5 (Brute Force))
Line 64: Line 64:
 
real side4 = 9.0;
 
real side4 = 9.0;
 
real pos = 2.5;
 
real pos = 2.5;
−
pair s1 = (-10,-2.19);
+
pair s1 = (-10,-1/2-sqrt(3));
−
pair s2 = (15,2.19);
+
pair s2 = (15,1/2+sqrt(3));
−
pair s3 = (36,4.38);
+
pair s3 = (36,1+2*sqrt(3));
 
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);
+
fill(circle(origin+s1, 1), grey1);
 
for (int i = 0; i < 6; ++i) {
 
for (int i = 0; i < 6; ++i) {
 
draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25));
 
draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25));
Line 91: Line 91:
 
fill(circle(sqrt(3)*pos*dir(60*i+30)+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(3*pos*dir(60*i)+s3,1), grey2);
−
fill(circle(sqrt(7)*pos*dir(60*i+79.1066054)+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-79.1066054)+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));
 
draw(side4*dir(60*i)+s3--side4*dir(60*i-60)+s3,linewidth(1.25));
 +
}
 
}
 
}
 
</asy>
 
</asy>

Revision as of 11:20, 7 October 2021

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] size(250); real side1 = 1.5; real side2 = 4.0; real side3 = 6.5; real pos = 2.5; pair s1 = (-10,-1/2-sqrt(3)); pair s2 = (15,1/2+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)); } [/asy]

$\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49$

Solution 1

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

The first hexagon has $1$ dot, the second hexagon has $1+6$ dots, the third hexagon $1+6+12$ dots, and so on. The pattern continues since to go from hexagon $n$ to hexagon $(n+1),$ we add a new ring of dots around the outside of the existing ones, with each side of the ring 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 ring is counted as part of two sides), confirming the pattern. We therefore predict that that the fourth hexagon has $1+6+12+18=\boxed{\textbf{(B) }37}$ dots.

Solution 3 (Variant of Solution 2)

The dots in the next hexagon have four layers. From innermost to outermost:

  1. The first layer has $1$ dot.

  2. The second layer has $6$ dots: $1$ dot at each vertex of the hexagon.

  3. The third layer 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 layer 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 4 (Variant of Solution 2)

Let the number of dots in the first hexagon be $h_0 = 1.$ By the same argument as in Solution 2, we have $h_n=h_{n-1}+6n$ for $n > 0.$ Using this, we find that $h_1=7,h_2=19,$ and $h_3=\boxed{\textbf{(B) }37}.$

Solution 5 (Brute Force)

From the full diagram below, the answer is $\boxed{\textbf{(B) }37}.$ 

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));
}
}
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~MRENTHUSIASM

Video Solution

https://www.youtube.com/watch?v=_IjQnXnVKeU

Video Solution by WhyMath

https://youtu.be/szWgrOPNw8c

~savannahsolver

Video Solution

https://youtu.be/eSxzI8P9_h8

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. AMC logo.png

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