Vanishing point
How can visual patterns be used to prove sums of series?
Age
14 to 18
Challenge level
You may wish to take a look at Diminishing Returns before working on this problem.
In the interactive environment below, you can move the slider to see how the image is built up.
At each stage of the process, what proportion of the image is coloured blue?
What would happen if we continued the process forever?
Once you've had a chance to think about these questions, click below to see some different approaches to understanding the process.
Here are the first three stages of the process:
In the first picture, the blue area is $\frac12$.
In the second picture, the blue has been partly covered by a purple square with area $\frac14$ so the blue area is $\frac12-\frac14$.
In the third picture, a blue square with area $\frac18$ has been added. As the pattern continues, the blue area will be $\frac12-\frac14+\frac18-\frac{1}{16}+\frac{1}{32}-...$
Instead of looking at how the pattern builds up, we could divide the completed pattern into the nested sections:
This shows the outer ring of the pattern, in which there are four larger purple triangles and four smaller blue triangles with half the area. So the ring is made up of twice as much purple as blue.
The same will be true for each ring added inside, therefore the overall pattern has twice as much purple as blue.
Here's another way to look at the completed pattern:
$\frac23$ of the highlighted shape's area is purple, and $\frac13$ is blue.
The whole pattern can be subdivided into pieces like this, so that means $\frac23$ of the total area of the pattern must be purple, and $\frac13$ of the total area must be blue.
Image
Image
Image
In the first picture, the blue area is $\frac12$.
In the second picture, the blue has been partly covered by a purple square with area $\frac14$ so the blue area is $\frac12-\frac14$.
In the third picture, a blue square with area $\frac18$ has been added. As the pattern continues, the blue area will be $\frac12-\frac14+\frac18-\frac{1}{16}+\frac{1}{32}-...$
Instead of looking at how the pattern builds up, we could divide the completed pattern into the nested sections:
Image
This shows the outer ring of the pattern, in which there are four larger purple triangles and four smaller blue triangles with half the area. So the ring is made up of twice as much purple as blue.
The same will be true for each ring added inside, therefore the overall pattern has twice as much purple as blue.
Here's another way to look at the completed pattern:
Image
$\frac23$ of the highlighted shape's area is purple, and $\frac13$ is blue.
The whole pattern can be subdivided into pieces like this, so that means $\frac23$ of the total area of the pattern must be purple, and $\frac13$ of the total area must be blue.
Below is a collection of images created using repeating processes.
For each pattern, can you find a sequence of fractions whose sum represents the blue area?
For each pattern, can you find different ways to work out the total proportion shaded blue?
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Image
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Can you come up with your own pictures to represent the sums of other series?
You may also be interested in the other problems in our Journeys of Discovery Feature.
Try making some of the patterns by cutting up two squares of paper in contrasting colours.
How does this help you work out what proportion of the total area of the square is taken up by each colour?
Navjot from Sherborne Qatar answered the questions about the first image in detail. This is Navjot's work:
Assume that its side length is $1$ unit.
Stage 1: 0, no part of the square is blue.
Stage 2: The side of the rotated blue square creates an isosceles right angled triangle with two sides being $0.5$ units (since the blue square's side touches the outer square at its side's midpoint).
So, the length of the blue square's side is: $$\sqrt{0.5^2+0.5^2}=\tfrac{\sqrt2}2$$
Blue area $=\left(\frac{\sqrt2}2\right)^2=\frac24$ of square.
Stage 3: To find the side length of the [purple] square inside the blue square, we can divide $\frac{\sqrt2}2$ by $2,$ [which gives] $\frac{\sqrt2}4$ and use this length to find the side length
by using Pythagoras' Theorem again.
Side length of innermost purple square $=\frac12,$ purple area [of the innermost purple square] $= \left(\frac12\right)^2= \frac14,$ total blue area $= \frac24 -\frac14 = \frac14$ of square.
Stage 4: Side length of innermost blue square $= \frac{\sqrt2}4,$ blue area $=\left(\frac{\sqrt2}4\right)^2=\frac2{16}$ of square.
Stage 5: Side length of innermost purple square $= \frac14,$ purple area $=
\left(\frac14\right)^2 = \frac1{16},$ total blue area $=\frac14 + \left(\frac2{16} - \frac1{16}\right) = \frac14 + \frac1{16}$
If we keep going on we will see that the area of the blue squares keep adding up in the form $\sum{\left(\frac14\right)^n}.$
Being a geometric series with a common ratio $<1,$ the summation will tend to a certain value.
Navjot then used a formula to calculate the sum of the series, and got an answer of $\frac13$. This formula was used by most people who sent in solutions to this problem and can be seen below.
For the next pattern, Alicia from Roedean School used the 'ring' method suggested in the problem. This is Alicia's work:
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This shows the outer ring of the pattern. The blue rectangle is twice the area of the pink square, so there is twice as much blue as pink - $\frac13$ of the area is pink and $\frac23$ is blue. The same will be true for each ring added, therefore the overall pattern has twice as much blue as pink.
Sichao from Bangkok Patana School in Thailand found a sequence of fractions and added them up:
Image
Therefore we know that the areas are:
$\frac 12, \frac18, \frac1{32}, \frac1{128}, ..., \frac1{2\times4^n}$
This [infinite series] we [could write as] $\sum_{n=0}^\infty{\frac1{2\times4^n}}$
To calculate the sum of terms we could rearrange the sum.
Let $S_n$ be the sum of terms.
$S_n=\tfrac12+\tfrac18+\tfrac1{32}+\tfrac1{128}+...$
$\Rightarrow 4S_n=\tfrac 42 +\tfrac48+\tfrac4{32}+\tfrac4{128}+...+\tfrac4{2\times4^n}$
$=2+S_n-\frac2{2\times4^n}$ (by substituting $S_n$)
$\Rightarrow 3S_n=2-\frac2{2\times4^n}$
As $n\rightarrow\infty,$ $\frac2{2\times4^n}\rightarrow0$ because $4^n\rightarrow\infty$
Hence $3S_n=2\Rightarrow S_n=\frac23$
Vignesh from Hymers College in the UK did this in a general case to derive a formula:
We know that the sum of an infinite geometric series is given by the equation:
$S=\frac{a}{1-r}$ (where $a$ is the first term and $r$ is the ratio)
This can be proven:
let, $$S=1+r+r^2....$$
Thus, $$ S\times r=r+r^2+r^3...\\ \Rightarrow S-S\times r=1\\
\Rightarrow S(1-r)=1\\
\Rightarrow S=\tfrac1{1-r}$$
(where $1$ is the first term and $r$ is the ratio)
This can then be applied into each of the patterns.
If the first term is $a$ instead of $1$, then the sequence is $a, ar, ar^2, ...$ and so all of the terms are multiplied by $a$. Multiplying all of the terms by $a$ has the same effect as multiplying the sum by $a$, so the formula is $S=\frac a{1-r}.$
Navjot and Nathan from Sherborne Qatar used this formula more directly. This is Nathan's work for one of the other patterns, including a description of the general strategy:
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Nathan and Navjot continued to find the total area shaded blue using sequences of fractions. Alicia continued to use the 'ring' method, but also the summing formula to check the answers. This is Alicia's work on the other patterns:
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In this pattern, each smaller blue square is $\frac14$ of the area of the larger square before it. Now we just need to find the area of the first blue square. Side length is $\frac12$ of the larger square so $\frac12 \times \frac12 = \frac14.$ So the sequence of fractions whose sum represents the blue area is: $\frac14+\frac1{16}+\frac1{64}+\frac1{256} + ...$
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In this pattern, each T shape is $\frac6{25}$ of the area of the square around it. Each smaller ring of the pattern is $\frac1{25}$ of the area of the larger ring. So the sequence of fractions whose sum represents the blue area is: $\frac6{25} + \frac6{625} + \frac6{15625} + ...$
Image
The sequence of fractions whose sum represents the blue area is a geometric sequence with $a = \frac6{25}$ and $r = \frac1{25}.$ So the sum to infinity is equal to $\dfrac{\frac6{25}}{1-\frac1{25}} = \frac14.$
Alicia has used the image and the formula as two separate ways of looking at the problem. How are they linked?
- Could Alicia's work be used to derive the formula?
- Could Sichao's or Vignesh's derivation be interpreted using the 'ring' method?
Why do this problem
When meeting geometric series for the first time, there's a temptation to look at the algebraic definitions first and derive the summation formula without giving students the opportunity to get to grips with why some sums converge and some diverge. In this problem, by offering a geometrical representation, we hope it will be clear to students why these series converge, while providing a hook to inspire curiosity into exploring other convergent series.
Possible approach
You may wish to spend some time with students working on Diminishing Returns before introducing this problem.
Show the interactivity from the problem, and gradually move the slider back and forth a few times to show the pattern growing.
"What proportion of the image is coloured in each colour at each stage?"
"How might the pattern continue?"
Give students some time to discuss and work out their answers, and then bring the class together to agree on a sum that represents the shades areas.
"What would happen if the pattern carried on forever?"
Again, give students some thinking time, and then share different ways of perceiving that $\frac13$ of the final pattern would be shaded blue and $\frac23$ purple. There are some suggestions in the problem of how students might convince themselves, but there are also many other ways of visualising the sum.
Next, show the other patterns and invite students to carry out the same process of working out the sum of fractions represented by the blue shaded areas at each stage, and the total area shaded blue in the completed pattern. Again, encourage a variety of different ways of visualising the sum.
Finally, after discussing the sums and limits for the patterns given in the problem, you might like to invite students to create their own images to represent different sums of fractions.
Show the interactivity from the problem, and gradually move the slider back and forth a few times to show the pattern growing.
"What proportion of the image is coloured in each colour at each stage?"
"How might the pattern continue?"
Give students some time to discuss and work out their answers, and then bring the class together to agree on a sum that represents the shades areas.
"What would happen if the pattern carried on forever?"
Again, give students some thinking time, and then share different ways of perceiving that $\frac13$ of the final pattern would be shaded blue and $\frac23$ purple. There are some suggestions in the problem of how students might convince themselves, but there are also many other ways of visualising the sum.
Next, show the other patterns and invite students to carry out the same process of working out the sum of fractions represented by the blue shaded areas at each stage, and the total area shaded blue in the completed pattern. Again, encourage a variety of different ways of visualising the sum.
Finally, after discussing the sums and limits for the patterns given in the problem, you might like to invite students to create their own images to represent different sums of fractions.
Key questions
What fraction of the square is blue at each stage?
How can you visualise the pattern to help you to find the limit?
Possible extension
The patterns could be used as a brief introduction to the idea of self-similarity in fractal patterns.
Possible support
There are suggestions in the Teachers' Resources to Diminishing Returns which might be useful to support students who are struggling with this task.