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

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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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==Problem==
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If the pattern in the diagram continues, what fraction of eighth triangle would be shaded?
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<asy>
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unitsize(10);
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draw((0,0)--(12,0)--(6,6sqrt(3))--cycle);
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draw((15,0)--(27,0)--(21,6sqrt(3))--cycle);
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fill((21,0)--(18,3sqrt(3))--(24,3sqrt(3))--cycle,black);
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draw((30,0)--(42,0)--(36,6sqrt(3))--cycle);
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fill((34,0)--(32,2sqrt(3))--(36,2sqrt(3))--cycle,black);
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fill((38,0)--(36,2sqrt(3))--(40,2sqrt(3))--cycle,black);
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fill((36,2sqrt(3))--(34,4sqrt(3))--(38,4sqrt(3))--cycle,black);
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draw((45,0)--(57,0)--(51,6sqrt(3))--cycle);
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fill((48,0)--(46.5,1.5sqrt(3))--(49.5,1.5sqrt(3))--cycle,black);
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fill((51,0)--(49.5,1.5sqrt(3))--(52.5,1.5sqrt(3))--cycle,black);
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fill((54,0)--(52.5,1.5sqrt(3))--(55.5,1.5sqrt(3))--cycle,black);
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fill((49.5,1.5sqrt(3))--(48,3sqrt(3))--(51,3sqrt(3))--cycle,black);
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fill((52.5,1.5sqrt(3))--(51,3sqrt(3))--(54,3sqrt(3))--cycle,black);
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fill((51,3sqrt(3))--(49.5,4.5sqrt(3))--(52.5,4.5sqrt(3))--cycle,black);
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</asy>
  
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[http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2331778&sid=3057ce3af8558814a3f7473fd1629118#p2331778 picture]
 
  
 
<math> \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} </math>
 
<math> \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} </math>
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==Solution==
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In each phase, all small triangles are congruent. The number of shaded triangles follows the pattern:
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<math>0, 1, 3, 6, ...</math>
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which is the pattern of "triangular numbers".  Each time, the number <math>1, 2, 3, 4, 5...</math> is added to the previous term.  Thus, the first eight terms are:
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<math>0, 1, 3, 6, 10, 15, 21, 28</math>
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In the eighth diagram, there will be <math>28</math> shaded triangles.
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The total number of small triangles follows the pattern:
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<math>1, 4, 9, 16, ...</math>
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which is the pattern of "square numbers".  Thus, the eighth triangle will be divided into <math>8^2 = 64</math> small triangles in total.
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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 <math>\frac{28}{64} = \frac{7}{16}</math>, and the correct choice is <math>\boxed{C}</math>
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== See also ==
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{{AJHSME box|year=1998|num-b=22|num-a=24}}
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* [[AJHSME]]
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* [[AJHSME Problems and Solutions]]
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* [[Mathematics competition resources]]
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{{MAA Notice}}

Latest revision as of 14:23, 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

In each phase, all small triangles are congruent. 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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