A very challenging question

Thanks to Aditya for the suggestion! A version of this problem given to students in class 10 in India (ages 14 - 17). Thanks to Tony for alerting me of an error in my original video! This one was so tricky that I missed it on my first try :)

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[MichaelBuck]

FYI: For those wondering about how he found the Radius of the large triangle.
We use the 30 60 90 triangle sides formulas: Based on the Pythagorean formula of A2 + B2 = C2
In particular a 30 60 90 triangle has a consistant ratio as follows:

1 : 2 : 3 for angles (30° : 60° : 90°)
1 : √3 : 2 for sides (a : a√3 : 2a)

In this case we know side "b", the side between 90 & 30 degrees

If we know the length b, we can find out that: a= b/√3 & c = 2a. Note: a is derived from b = a√3

a=b/√3 = 2/√3
b = 2
c = 2(2/√3) = 4/√3 = ~2.3
Radius = 2 + ~2.3 = ~4.3
Area = Pr2 = ~3.14 * (~4.3)2 = ~3.14 + ~18.49 = ~21.63

[wwoods]

Save a couple of steps by noting that the three blue sectors add up to half a grey circle. So you don't need to subtract three grey circles and then add back the widget in the middle. Just subtract 2.5 grey circles: 5/2 π 2^2 = 10π.

[lorenzcangiuz]

For those interested, you find the area of the triangle, you find the height of the triangle to then find the center of the biggest circle, the length from the center of the circle to the center of the big one in 2/3 of the height of the triangle, then you add 2 and you have the radius of the big circle, then you calculate the area minus 2.5 areas of the circles, because the three sectors make a semi circle

[geoninja]

I started out going for the exact answer, but my brain started aching, so went for and was happy to get it!

[Ndiedddd]

Took me some time, but I got the answer. Really rewarding once you solve a question from this channel lol

[jomertomale]

Easier method:
Area = Area of Big Circle - 3*(Area of 3 pacmans, 5/6 circle) - Area of Triangle + 3*(Area of 3 pies, 1/6 circle)

[pineapplegodguy]

i'm just gonna address the bothersome part of finding the area of the outside circle, since the rest is generally well understood.

by trig, and since it is an equilateral triangle, the distance between the base of the triangle and its midpoint is
sin(30 degrees)*2/cos(30 degrees) = 2/sqrt(3)

consider the upper circle (indifferent, but easier to visualize). since it can be found from pythagoras that the height of the triangle is 2sqrt(3), the distance between the center of the circle and the midpoint of the triangle is 2sqrt(3) - 2/sqrt(3)

the distance between the top vertex of the triangle and the outside circumference is 2.

in order to find the radius of the outside circle, you just have to add those components: 2 + 2sqrt(3) - 2/sqrt(3)

now you plug this number into the equation for the area of the circle pi*(2 + 2sqrt(3) - 2/sqrt(3))^2

[mosab]

The center of the triangle and the large circle is in the two thirds of the triangle hight from the top, find the hight of the triangle the take two thirds from it then add 2( the radius of the small circle) and you have the radius of the large circle.

[atingupta]

When you write everything together in the same equation, you get:

Largecircle - 3*smallcircle - triangle + 2*3sectors

Here, 2*3sectors has the same area as 1 small circle

So this expression becomes:
Largecircle - 2*smallcircle - triangle

Here you don't need to calculate the area of the sectors and the equation gets even more simplified. thought this would be helpful to someone

[markstahl]

Another way to think of the area addition, since you add the semicircle twice is “Big Circle - 2 Small Circles - Triangle”.

[zona]

This is the first geometry problem from your videos that I could solve without any help. It's such a great feeling! 15/yo

[-zero-]

the area of inner triangle shade can be calculated by 3*( pi*2²/2) since its just a semicircle, since the sum of all angles is 180°

[luiscarlosvieira]

this is an unique channel, you should do a video of your statistics, like where are your viewers from... I am curious about that...

[GeeeDann]

You can also fine the radius of the large circle by using sine rule

[kacubemember]

This video was awesome too, as always :)
I can't describe how much your videos have helped me prepare for competitive exams. I was always scared of Geometry, but now I love solving them :D
Love from India

[justablank]

The middle triangle is an equilibrial one (idk if it's the right name but basically a triangle has 3 equal sides) so each angle would be 60 degrees. Then we are going to have a ratio of 60:360 =1:6 therefore we gonna get 1/6 times 3 the area of each circle.

[anandk]

Solved it bang on right in first attempt, in 35-40 minutes (just for my remembrance - on 12th May 2022, from 6:46 pm to 7:25 pm or so) !!!

I'm really shocked to see that no one (including uploader) has cared to think of proving center of largest circle to be exactly the same as circumcenter of equilateral triangle. But proving that is absolutely necessary, otherwise further calculations based on equilateral triangle cannot be considered valid. So here's the detailed solution.


EDIT :

Some people feel that finding radius of largest circle was the most challenging step. I thoroughly disagree this.

The most challenging thing ever, I felt, was to PROVE THAT THE CENTER OF THE LARGEST CIRCLE IS EXACTLY THE SAME AS THE CIRCUMCENTER OF THE EQUILATERAL TRIANGLE.

Many people may assume this to be true due to symmetry. But since Geometry never allows any assumptions, I spent almost 20-25 minutes just thinking about this. And was able to successfully prove it. Here's what I did.

Let O be center of the largest circle and A, B, C respectively be the centers of 3 inner circles from top to bottom.

Now after forming a vertical line segment through A till top and all the way through triangle till bottom of largest circle, we will get the diameter of the largest circle ( this seems the most challenging step here, WHY SHOULD THE LINE SEGMENT PASS THROUGH CENTER OF THE LARGEST CIRCLE ???

Again spare a conscious thought here - The vertical line segment is the diameter only because THE STRAIGHT LINE FROM TOP THROUGH THE TRIANGLE ALL THE WAY TILL LOWERMOST POINT, WILL BE PERPENDICULAR BECAUSE THE TOP INNER CIRCLE AND LARGEST CIRCLE SHARE SAME POINT OF CONTACT, SO THE CHORD PERPENDICULAR TO THE TANGENT THROUGH POINT OF CONTACT MUST PASS THROUGH THE CENTER !!!

THIS WAS HONESTLY CRAZY )

Since the diameter passes through A till midpoint of BC, it's clearly the perpendicular bisector of BC.
[ Again you must question here why exactly should the diameter pass through midpoint of triangle base. That's because it just takes us to know that vertical join through common point of contact will pass through center, and since altitude of equilateral triangle is also perpendicular to the base of triangle, hence both of them (i.e. triangle base and tangent at top) will be parallel, which means that the vertical join will be a straight line segment through center, essentially the diameter. ]


But O also lies on perpendicular bisector. Hence vertices B and C will be equidistant from O.

Similarly we can prove that vertices (A, B) and (A, C) each will be equidistant from point O.

When does this happen in a triangle ???

It happens ONLY WHEN SUCH POINT IS THE POINT OF CONCURRENCE OF PERPENDICULAR BISECTORS OF SIDES (KNOWN AS THE CIRCUMCENTER) OF THE TRIANGLE.

That means O is the circumcenter of equilateral triangle ABC.

But O is the center of largest circle.

THIS CLEARLY MEANS AND PROVES THAT 'O' IS BOTH THE CIRCUMCENTER OF TRIANGLE AND THE CENTER OF THE LARGEST CIRCLE.


This was somewhat crazy to think. But I really enjoyed this and solved the problem in roughly 40 minutes.

Do like and comment on your thoughts and ideas.

Thank you and love you all, dear friends and readers.

👍👍😘🤗😊😊

[vincenttavani]

There is another equilateral triangle sharing a side with the centered triangle, with its vertex on the large circle. Therefore the diameter of the large circle is twice the height of the triangle plus the radius of a small circle.

[zecuse]

Simpler approach:
Big circle - 3 small circles - equilateral triangle + 6 sectors (because we over-subtract the sector portion of the triangle with the 3 small circles)
Which simplifies to:
Big circle - 2 small circles - equilateral triangle (because the 6 sectors = 1 small circle)

[mandeltownthekillerfrombab]

2 is the radius of the circle. The area of the small circle is 4π. Equilateral triangle area is sqrt(3). Like the angle of equilateral triangle is 180. This area will be multiplied by 1/2 (for the 1/2 circle’s area : 2π. The small circle inside a big circle is sqrt(3) small than than the circle. The area for a small circle is 4π then the triple is 12π. Knowing the small circle is in area 3 time smaller than the big circle and the radius from the big circle is 2sqrt(3) (or sqrt(12). The area from a big circle is 12π . This area vaults 10π.