4 Overlapping Circles Puzzle

What is the area between a large circle and 4 smaller circles as shown in the diagram? The large circle's diameter is 84. This problem was given to students in Singapore. Can you figure it out? Watch the video for a solution.

Correction: at 4:48, the middle circle should have a diameter of 42 (the value 41 was a typo, sorry).

My blog post for this video

Credit for problem solutions 1, 3 and many similar maths problems

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[MindYourDecisions]

Correction: at 4:48, the middle circle should have a diameter of 42 (the value 41 was a typo, sorry).

[Kihidokid]

when he drew the quare i nearly shit myself thats so obvious

[jensraab]

There's actually a quicker way that I figured out after a while:
The small circles have half the diameter of the big circle. Therefore the big circle has four times the area of one small circle. Since we have four small circles, they have exactly the same area as the big circle!
Of course, they overlap. The area of the overlapping regions must therefore be equal to the shaded region.
That means we only have to figure out how to calculate one of the overlapping regions, then multiply by four (because there's four of them). Done! :-)

[jeffreycanfield]

Turn on auto captions.

"Hey, this is brush tool locker."

[jihongzhi]

How do you know that the small circles intersect at antipodes? (This is required since we are assuming the remaining pieces are perfect semi-circles.) It appears plausible, but it isn't obvious to me.

[henk-ottolimburg]

I kind of miss the proof that the intersection of the smaler circles is at exactly the middle.
It is true, I can proof it, but...."are you able to proof it?"

[olafdegraaff]

I feel so stupid that I couldn't figure it

[shikikan]

I saw the square pretty easily, it does take some attention though.

[tearlach]

My answer is
(42^2*pi) - 3(21^2*pi) - (42^2-21^2*pi)
Area of large circle
minus
area of 2 small circles alone with the top half of the north circle combined with the bottom half of the south circle, effectively combining to make a third whole circle.
minus
area of a square minus small circle of the same diameter. I do this last part because the area that we have left to subtract (The top part of the south circle and the bottom part of the north circle, which have already had those "eye" shapes subtracted, make up the same space as would be in a square minus circle area).

[cartman]

d of bigger circle=84
d of smaller circles=42
sides of square=42
area of shaded region
AREA OF BIGGER CIRCLE-AREA OF SQUARE-2 TIMES AREA OF SMALLER CIRCLE
Then,
Area of shaded region:
PI 42^2 - 42^2 -2(PI 21^2)
=22/7*42*42 - 42*42 - 2(22/7*21*21)
=5544-1764-2772
=1008

[arimago]

attempt (paused 0:44)
i teach this stuff so if i get this wrong it will be _fucking embarrassing_

A(shaded) = A(large circle) – 4A(small circle) + 4A(leaf)

D(large circle) = 84 ∴ r(large circle) = 42 = D(small circle)
A(large circle) = 1764π
4A(small circle) = 4(441π) = 1764π
∴ A(shaded) = 4A(leaf) //holy shit!!

a "leaf" is twice the area between the small circumference and a quarter-circle chord. i don't know how to get that, but i can get the area of a circumscribed square, and the diagonal = D(small circle) = 42.
side(square) = (42√2)/2 ∴ area(square) = ((42√2)/2)^2 = ((1764*2)/4) = 882

A(small circle) = 441π
A(square) + 2A(leaf) = A(small circle)
2A(leaf) = 441π – 882
4A(leaf) = 882π – 1764 ∴ A(shaded) = *882π – 1764*

[sergeyg]

#1 is by far the coolest method, as it just ignores the overlap. Another method: compute the area of half of each overlap as the difference between a quarter of a small circle and a right triangle, that has the right angle at the center of the small circle (i.e. 90 deg sector minus the triangle the sector forms when you leave off the segment): R^2(pi/2-1). Also cool to note that the area of each connected shaded orange area is equal to the area of each overlap.

[nathan]

Off the top of my head, (42² * π) - 42² - 4((21² * π) / 2).

[christianfunintuscany]

While solving I noticed that the area of the central blue flower actually equals the area we are looking for ... this is a pretty nice thing to remember for this type of problems ...

[benreinicke]

Dang this Singaporean education sounds fun but im stuck here in America ;/

[parepidemosproductions]

For method 2, I was really confused at the final step because I didn't understand how removing the overlapping circles balanced out the double counting of that region but listening and watching for the second time made the difference. Thanks for showing the geometric method (which was the one I came to on my own because of it's simple characteristics) and explicit version which showed using circle qualities like sectors

[WayneLinnlikestouseGeoGebra]

I went through Singapore Primary/elementary school education and this approach is widely used and there are many variations of this problem with the same approach. It's not hard really.

[rahvithecolorful]

People saying that it should be harder for elementary schools kids forget that it's not about age, but about what and how you're taught. If you have good teachers and a good material, if anything the younger you are the easier it is to learn and to think outside the box.
This kind of problem is way harder for people who have been out of school for a long time and haven't used geometry for pretty much nothing in daily life for years, takes a little while to go dig for that information in the depths of your brain.

[jonathanryals]

Hey I actually got this one! My method was similar to #3, but I considered a quarter circle and computated 1/8 of the total overlapping region... I boiled it down to a problem of computating the area of a vessica then simple addition and subtraction of known terms... I toyed with the idea of using radians and calculus to find the area before I realized i could use the simple circle minus triangle method.

[Anonymous-kwls]

yes it was a easy problem..
the area of 4 inner circles is same as the area of one big circle just becoz some area is being overlapped.. what I did is just draw one small circle in a corner.. and as you already have two overlapped parts in the small circle. add two more same design in that small circle.. now you have totally 4designs in that small circle.. now draw a square outside a circle.. such that it's sides are tangent to the circle.. now calculate the area of that 4designs.. for that u have to find the area of 4quadrants and subtract this area from square area.. then subtract this area from the area of this small circle of diameter 42.. u will get the area of 4designs.. and this area is ur required ans.