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Difference between revisions of "1998 AJHSME Problems/Problem 23"

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==Problem==
 
==Problem==
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If the pattern in the diagram continues, what fraction of the interior would be shaded in the eighth triangle?
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If the pattern in the diagram continues, what fraction of eighth triangle would be shaded?
  
 
<asy>
 
<asy>

Revision as of 14:19, 29 May 2021

Problem

If the pattern in the diagram continues, what fraction of eighth triangle would be shaded?

[asy] unitsize(10); draw((0,0)--(12,0)--(6,6sqrt(3))--cycle); draw((15,0)--(27,0)--(21,6sqrt(3))--cycle); fill((21,0)--(18,3sqrt(3))--(24,3sqrt(3))--cycle,black); draw((30,0)--(42,0)--(36,6sqrt(3))--cycle); fill((34,0)--(32,2sqrt(3))--(36,2sqrt(3))--cycle,black); fill((38,0)--(36,2sqrt(3))--(40,2sqrt(3))--cycle,black); fill((36,2sqrt(3))--(34,4sqrt(3))--(38,4sqrt(3))--cycle,black); draw((45,0)--(57,0)--(51,6sqrt(3))--cycle); fill((48,0)--(46.5,1.5sqrt(3))--(49.5,1.5sqrt(3))--cycle,black); fill((51,0)--(49.5,1.5sqrt(3))--(52.5,1.5sqrt(3))--cycle,black); fill((54,0)--(52.5,1.5sqrt(3))--(55.5,1.5sqrt(3))--cycle,black); fill((49.5,1.5sqrt(3))--(48,3sqrt(3))--(51,3sqrt(3))--cycle,black); fill((52.5,1.5sqrt(3))--(51,3sqrt(3))--(54,3sqrt(3))--cycle,black); fill((51,3sqrt(3))--(49.5,4.5sqrt(3))--(52.5,4.5sqrt(3))--cycle,black); [/asy]


$\text{(A)}\ \frac{3}{8}\qquad\text{(B)}\ \frac{5}{27}\qquad\text{(C)}\ \frac{7}{16}\qquad\text{(D)}\ \frac{9}{16}\qquad\text{(E)}\ \frac{11}{45}$

Solution

All small triangles are congruent in each iteration of the diagram. The number of shaded triangles follows the pattern:

$0, 1, 3, 6, ...$

which is the pattern of "triangular numbers". Each time, the number $1, 2, 3, 4, 5...$ is added to the previous term. Thus, the first eight terms are:

$0, 1, 3, 6, 10, 15, 21, 28$

In the eighth diagram, there will be $28$ shaded triangles.

The total number of small triangles follows the pattern:

$1, 4, 9, 16, ...$

which is the pattern of "square numbers". Thus, the eighth triangle will be divided into $8^2 = 64$ small triangles in total.

The ratio of shaded to total triangles will be the fraction of the whole figure that's shaded, since all triangles are congruent. Thus, the answer is $\frac{28}{64} = \frac{7}{16}$, and the correct choice is $\boxed{C}$

See also

1998 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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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