Difference between revisions of "1991 AJHSME Problems/Problem 25"
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| − | With each change, <math>3 | + | With each change, <math>\frac{3}{4}</math> of the black space from the previous stage remains. Since there are <math>5</math> changes, the fractional part of the triangle that remains black is <math>\left( \frac{3}{4} \right)^5=\frac{243}{1024}\rightarrow \boxed{\text{C}}</math>. |
Latest revision as of 23:57, 14 June 2022
Problem
An equilateral triangle is originally painted black. Each time the triangle is changed, the middle fourth of each black triangle turns white. After five changes, what fractional part of the original area of the black triangle remains black?
![[asy] unitsize(36); fill((0,0)--(2,0)--(1,sqrt(3))--cycle,gray); draw((0,0)--(2,0)--(1,sqrt(3))--cycle,linewidth(1)); fill((4,0)--(6,0)--(5,sqrt(3))--cycle,gray); fill((5,0)--(9/2,sqrt(3)/2)--(11/2,sqrt(3)/2)--cycle,white); draw((5,sqrt(3))--(4,0)--(5,0)--(9/2,sqrt(3)/2)--(11/2,sqrt(3)/2)--(5,0)--(6,0)--cycle,linewidth(1)); fill((8,0)--(10,0)--(9,sqrt(3))--cycle,gray); fill((9,0)--(17/2,sqrt(3)/2)--(19/2,sqrt(3)/2)--cycle,white); fill((17/2,0)--(33/4,sqrt(3)/4)--(35/4,sqrt(3)/4)--cycle,white); fill((9,sqrt(3)/2)--(35/4,3*sqrt(3)/4)--(37/4,3*sqrt(3)/4)--cycle,white); fill((19/2,0)--(37/4,sqrt(3)/4)--(39/4,sqrt(3)/4)--cycle,white); draw((9,sqrt(3))--(35/4,3*sqrt(3)/4)--(37/4,3*sqrt(3)/4)--(9,sqrt(3)/2)--(35/4,3*sqrt(3)/4)--(33/4,sqrt(3)/4)--(35/4,sqrt(3)/4)--(17/2,0)--(33/4,sqrt(3)/4)--(8,0)--(9,0)--(17/2,sqrt(3)/2)--(19/2,sqrt(3)/2)--(9,0)--(19/2,0)--(37/4,sqrt(3)/4)--(39/4,sqrt(3)/4)--(19/2,0)--(10,0)--cycle,linewidth(1)); label("Change 1",(3,3*sqrt(3)/4),N); label("$\Longrightarrow $",(3,5*sqrt(3)/8),S); label("Change 2",(7,3*sqrt(3)/4),N); label("$\Longrightarrow $",(7,5*sqrt(3)/8),S); [/asy]](http://latex.artofproblemsolving.com/d/7/b/d7b30328339e814efdbf433f8b9287f116e6598c.png)
Solution
With each change, 
of the black space from the previous stage remains. Since there are 
changes, the fractional part of the triangle that remains black is 
.
edited by stjwyl
See Also
| 1991 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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. 
