Difference between revisions of "1998 AJHSME Problems/Problem 23"
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Revision as of 23:32, 4 July 2013
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]](http://latex.artofproblemsolving.com/5/5/a/55aac6c624146c44f6a45b961bf3a9c3f8fd89c6.png)
Solution
All small triangles are congruent in each iteration of the diagram. The number of shaded triangles follows the pattern:
which is the pattern of "triangular numbers". Each time, the number 
is added to the previous term. Thus, the first eight terms are:
In the eighth diagram, there will be 
shaded triangles.
The total number of small triangles follows the pattern:
which is the pattern of "square numbers". Thus, the eighth triangle will be divided into 
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 
, and the correct choice is 
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. 
