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

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==Yi ga BABA GUO FEN LA!==
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==Problem==
 
If the pattern in the diagram continues, what fraction of the interior would be shaded in the eighth triangle?
 
If the pattern in the diagram continues, what fraction of the interior would be shaded in the eighth triangle?
  

Revision as of 18:54, 31 October 2016

Problem

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

[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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