1991 AJHSME Problems/Problem 25

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]

$\text{(A)}\ \frac{1}{1024} \qquad \text{(B)}\ \frac{15}{64} \qquad \text{(C)}\ \frac{243}{1024} \qquad \text{(D)}\ \frac{1}{4} \qquad \text{(E)}\ \frac{81}{256}$

Solution

With each change, $3/4$ of the black space from the previous stage remains. Since there are $5$ changes, the fractional part of the triangle that remains black is $(3/4)^5=243/1024\rightarrow \boxed{\text{C}}$.

See Also

1991 AJHSME (ProblemsAnswer KeyResources)
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