The problem statement from the Ole Miss Problems of the Week page is:

Square ABCD is divided and shaded as shown in the diagram to the left. The shaded square is one-fourth the area of square ABCD. If the pattern continued forever and ever, what percentage of the original square would be shaded?

Sounds like with just a small prybar, we could force this to be a computation we could — gasp! who would have thought we could possibly choose this approach — program on our programmable CFX-9850G or FX-7400G!

Let us see. That big shaded square, we are told, is one quarter of the entire square. Then there is another shaded square that is one quarter of one quarter of the entire square. Then there is another shaded square that is one quarter of one quarter of one quarter of the entire square. Hmm. I guess if we wrote a program to add

¼ + ¼^{2} + ¼^{3} + ...

we would have an answer.

We quickly pound out the following program, available here in text or as a download as a text file with .CAT file contents. As with other programs on this site, text starting with a semicolon (“;”) are comments and is not to be entered into the calculator.

; Variables: ; A is the size of the current square ; B is the total shaded so far ; C is the total shaded plus the current coverage ; Symbols: ; -> is assignment arrow ; / is divide operator ; <> is not equal relational 1->A ; Initialize size of square to 1 0->C ; Initialize fraction shaded to 0 Do ; Loop down through smaller squares A/4->A ; One-quarter of current square size is shaded C->B ; Get copy of current shaded coverage A+B->C ; Add in new shaded coverage LpWhile C<>B ; Loop until there is no change 100B ; Report the percentage

Running this program, we quickly get:

33.33333333

So 33.33333333% of the square is shaded, to within round-off error.

Of course, the problem did not say “*approximately* what
percent.”
It said “what percent.”
Now, is that .33333333% actually .33333333%, or is it 1/3 percent?

Let us go back for a minute to one of our preliminary steps. We had the fraction of the square covered as:

¼ + ¼^{2} + ¼^{3} + ...

Thinking hard, we realize this is a *geometric sum* —
each term is a multiple of the term in front of it.
The first term in this example is ¼, and the ratio is
(coincidentally) ¼.
The formula for a geometric sum is:

S = (first term)/(1 - ratio)

where *S* is the sum, *first term* is the first term,
and *ratio* is the ratio.
So the sum — the proportion of the square ABCD that is
shaded — is ¼/(1-¼) = ¼/¾ = 1/3.

Of course, if we happen not to remember the formula for a geometric
sum, we are stuck.
Or are we?
Thinking about it, we realize that since the series of squares
inside squares is infinite, we can add another at the end without
changing anything.
We can add another square at *either* end.
In particular, we can add another square *outside* the
squares and not change anything.
So suppose some proportion *p* is shaded.
The proportion shaded does not change if we make what we
have the inner set of squares — that is, if we add another
set of squares around it.
We add 1 shaded square and 2 white squares to make 4 squares total,
and the ratio *p* is unchanged.
So

p = (1 + p)/4, or 4p = 1 + p, or 3p = 1, or p = 1/3.

So the proportion covered is exactly 1/3, and the percentage covered is exactly 33 1/3%.

Sounds like a plan to me.

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Copyright © 2001 Brian Hetrick

Page last updated 25 November 2001.