What Is The Area Of This Infinitely Repeating Shape?

If the first square has a side length equal to 1, what is the sum of all blue areas? Can you solve this fascinating problem?

0:00 Problem
1:11 Hint
3:58 Solution

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[xnick_uy]

Cool fact that follows from the calculation: the blue area in just one of the squares is exactly the same as the blue area inside the circumscribed circle.

[NestorAbad]

Hi, thanks for sharing!
You don't really need any series to solve this problem.
Let's think on proportions: consider the first square with the shaded area b₁ inside. Then, remove the second square, which has area 1/2. This figure (the region between these two squares) has area 1-1/2=1/2, and the blue area b₁ inside it has area 1-π/4, as you showed. Then, in this particular picture, the proportion of blue area (that's it, blue/total) is (1-π/4)/(1/2)=2-π/2.
Now, if you go on with the picture you will see that the next area b₂ is enclosed inside the same kind of shape as before: one square of area 1/2 from which we remove a smaller square of area 1/4. So, by similarity (note that we have scaled the previous picture) the fraction of blue area is exactly the same as before: 2-π/2. This means that, no matter how many pieces we consider, the fraction of blue area remains constant to 2-π/2, with respect to the total area considered. In the limit, the total area of the complete picture is 1 (the big square), so the total blue area must be 2-π/2.
I hope the reasoning is clear enough without any pictures!

[josecossich]

I have two things to say:
- 1) I just graduated freshman year and I've been studying math intensely for the last month, having finished Geometry and Algebra 1, now working on Algebra 2.

2) I found this question extremely time consuming and I think that's what made it seem hard, even if the content required itself wasn't too hard. I worked on this for like 3-4 hours and finally got the answer. I feel both proud and humbled by the fact that I was able to answer this question and by how long it took me to do so/the comments, respectively

[azurestar]

This is similar to that joke math teachers will often say to their students

"An infinite number of people show up to a restaurant. The first person orders one pie, the second person then says they want half of what the first person ordered, the third person then says they want have of what the second person ordered, and so on. The cook, a mathematician, stops the infinite group of people from ordering and give them two pies."

[Shrubbist]

I found it easier as follows: subtract out the inner square, so you have a blue region of area 1-pi/4 in a region of area 1/2. So a fraction 2-pi/2 of that region is shaded blue. But the original shape can be constructed by repeating that shape inside itself (reducing in scale each time) to fill space. So the total fraction shaded is still 2-pi/2. the total area is 1 so that is also the total area of blue.
The intermediate step is basically the same: (1-pi/4)*[a sum that ends up equal to 2* the area of the square] so maybe the way I did it was just a sloppy version of the same calculation, without having to remember that geometric series exist.

[zomorion]

Ngl, this was pretty easy. A good refresher from the harder maths.

[Aiden-xnwo]

This one was rather easy. It has a nice solution, though.

[babyli]

It is SUCH a beautiful answer! Well done!! Your videos really make everyone develop a genuine love for maths, thank you so much for that!

[KnowledgeEducationIndia]

I really can't believe that I have solved this question correctly without seeing solution. 😄😄😄😃😃

[Hertog_von_Berkshire]

If like me, you did new-fangled modern maths (SMP) then the ratio of areas between each motif and the next smaller is very simple to work out once you realise that each motif's side length is the same as inner diagonal of the next smaller motif (as explained at 4:53). The Scale Factor is 1:1/sqrt(2) therefore the Area Factor is 1:½. This ratio holds for blue parts, the white parts, any other corresponding parts. You still need to derive and apply the telescope formula but the exercise becomes that little bit simpler. Long live Bryan Thwaites.

[cpsof]

Area of the whole square: 1^2=1
Area of the circle: π*(1/2)^2 = π/4
Area of the first inscribed square: (1/sqrt(2))^2 = 1/2

Ignore the area inside the first inscribed square.
Blue area: 1 - π/4
Whole area: 1 - 1/2 = 1/2

Proportion of the blue area: (1 - π/4) / (1/2) = 2 - π/2

The same pattern repeats inside the smaller squares, so the blue area is 2 - π/2.

[rafciopranks]

1:20 - 4:00 you explained it better in 3 minutes than my teatcher did in 45 minutes

[mohdfaik]

Nice refresher. Thanks MYD. For simplicity, I wrote 2 series, one for the sum of squares’ areas, total area is 2, another series for the sum of circles’ areas, total is Pi/2, net blue areas=2-pi/2=(4-pi)/2

[ravimanju]

The closest and similar questions asked in many Indian entrance tests like CAT, XAT even in SSCs... This one was better flavoured..

[rutgerdekok]

If you rovtate every other square by 45 degrees an easy patern emerges. You realy only have to examine one triangle on the outside. The area of the triangle is one eighth of the the square. And the ratio between blue and white areas inside the triangle is the same as for the whole square. The area of the blue part is in one triangle is:
1/4(1 - pi(1/4)) then multiply that by 8 to have your answer. 1 - pi(1/2)

[scarletevans]

Thank you for a video!
One can even simplify the problem by skipping the area of b2 the way you did, if they know that the scale factor for area is found by squaring the scale factor for length! 😀
So, as we have the second square having a side length of 1/sqrt(2), the area is exactly (1/sqrt(2))^2=1/2 times smaller 🙂And through the similarity it follows for every next area (b3, b4, b5, ...) being also half of the previous one. And what's great about this method is that it works for every shape you want to inscribe within some other one! (i.e. find 1-dimensional scale factor and just square it). Sometimes it's easier or faster to just find the scaling factor by dividing two distances, than being able to figure out the next area we are looking for 🙂
But yeah, I can also imagine the opposite, with areas being easier to find, so actually good to think in both of these ways!

[ebubechukwudi]

I'm really proud of myself after solving this without the solution

[rysea]

Finally managed to get the correct answer on one of your videos! Was really surprised to find out that every subsequent layer had exactly half the area of the previous one

[ronbannon]

I found a similar problem on a 1934 Annual Examination, Upper School Algebra, Ontario, Canada. The exams back then were way more difficult than present-day exams.

[gonzalotapia]

1+1/2+1/4+1/8... = 2.
Now you can multiply each term by the area factor, that is 1-pi/4.
Total area is 2(1-pi/4) or 2-pi/2