A very challenging question

[@geoninja8971]

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

[@goldenanime2088]

3 days ago?

[@kishandebnath9228]

@@goldenanime2088 yeah man how is it possible

[@Jopppz]

@@kishandebnath9228 a site with unlisted youtube videos

[@kishandebnath9228]

@@Jopppz oh understood a bit ......btw can i get some details about it

[@leif1075]

How are decimals nkt exact? It almost sounds like you contradicted yourself?

[@wwoods66]

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.

[@ArteePhotography]

Came here to say the same thing.

[@Ksraghu1c]

Same here

[@gblargg]

Wouldn't you really just need to take area of enclosing circle, subtract 2 grey circles, and the triangle that has the central shape within it? 2 grey circles because you subtract 3, then the triangle (0.5 of a circle), then need to add the three wedges twice (totaling one circle).

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@cobokobo2115]

amazing .....kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@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

[@lorenzcangiuz1519]

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

[@Ndiedddd]

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

[@rohangeorge712]

fax

[@kaosarsufian1119]

Same here bro

[@kaosarsufian1119]

Same here bro

[@zona5332]

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

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@cobokobo2115]

amazing .....kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@Felix-ge4sd]

Good job young bro

[@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)

[@ericwilliams1832]

Yep! That’s what I did

[@GeeeDann]

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

[@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

[@MichaelBuck]

Bravo! Thats what I did, lol. Was way easier than the solution proffered in the video. I always try to keep things simple, no need to make it more complicated than necessary!

[@luiscarlosvieira3966]

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

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@SensSword]

I spent all my childhood tessellating the bathroom tiles during difficult poops. I wish I'd be able to watch this and similar videos instead of wondering when I'd liberated from the toilet. C'est la vie...

[@markstahl1464]

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

[@holybasilisk2248]

Oooo👏👏👏👏

[@Matthew-fb5mo]

daaamn you right, that would have made it so much easier

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@cobokobo2115]

amazing .....kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@cobokobo2115]

@@Matthew-fb5mo amazing .....kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@kacubemember3077]

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 :))))))))

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@cobokobo2115]

amazing .....kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@mustafizrahman2822]

This is one of the simplest questions. Very easy.

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@tariqislam4915]

❤️❤️❤️

[@mustafizrahman2822]

@@tariqislam4915 Oh! How are you, brother? I am fine. Amio tumake dekhi na.

[@mustafizrahman2822]

@@tariqislam4915 Kotahi hoi na. Nobel purosker paya udhao hoia gela ga naki?

[@mustafizrahman2822]

@@tariqislam4915 Bolo ki help?

[@gm_matthew]

At 4:45 I realised that I had made one slight mistake and miscalculated the area of the central triangle as 8*sqrt(3); oops! After fixing that mistake I was able to figure out the correct answer.

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@mosab1340]

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.

[@Unidentifying]

solved it in 5 minutes :) thank you for these videos bro !!!

[@atingupta77]

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

[@eomoran]

Yeah, I was surprised he bothered to vermiculite it

[@anandk9220]

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. 👍👍😘🤗😊😊

[@lw4ve1488]

You can also build a three radiuses of big circle that intersect sides of a triangle in the middle of each side respectively and easily prove that they’re perpendicular to the sides

[@anandk9220]

@@lw4ve1488 That's almost what I've done, dear friend. Just wrote it in greater detail. I got your point that 3 radii of largest circle in 120° manner will exactly meet at the center of the circle. But proving that center of largest circle is exactly same as circumcenter is not clear and straightforward. 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. However for proving center to be circumcenter, I can think of only my way. Kindly suggest and explain other better ways if any. Thank you so much. 😘🤗😊

[@Grizzly01]

No-one, including Presh, bothered to prove it, as it is evident by just applying some very simple circle theorems. "3 circles pair-wise tangent", "large circle circumscribed", "3 circular sectors... between the 3 centres of the small circles" give you all you need to know.

[@anandk9220]

@@Grizzly01 Nothing is evident, dear. He would've realized that it's a huge long thing to prove and explain that part. So perhaps he'd have ignored that. Ideally, that proof is absolutely necessary to determine further calculations.

[@Grizzly01]

@@anandk9220 Absolutely disagree. It certainly isn't a huge long thing to prove anyway, and need not be, just as we don't need to prove Pythagoras every time we use it.

[@-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°

[@cobokobo2115]

amazing .....kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@muneebhaider104]

I don't know how but I am able to solve most of the questions on this channel now after watching it for a year, maybe he is the best teacher after all.

[@kurzackd]

03:14 - "which means that the other leg would be equal to 2/sqrt(3)" aaaand... how exactly do we know that?

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@Grizzly01]

Look up the properties of a 30-60-90 triangle

[@TheApplesauce1992]

I just realized how cool making your own math problem is

[@2mbrothers547]

I actually found it !! What an amazing effort, keep going

[@cobokobo2115]

amazing .....kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@schmidt3110]

The fact the triangle edges and the lines from its center to a corner are in a ratio of root 3 is luckyly well known in the electrical engineering world due to the Y-Δ transformation for elctrical circuits. This fact makes the task quite enjoyable, but I can see why it might be tricky to get this ratio.

[@joaovmlg]

I hated doing that Y-Delta transformation, luckily my physics professor teached a easier way to do the transformation, not sure it works for extremely complex circuits you might see on college, since I was at high school.

[@vincenttavani6380]

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.

[@judah738]

At roughly 3:09 in the video, how do you know that the constructed triangle is 30, 60, 90? I get that you constructed a right triangle, hence the 90, but where do the 30 and 60 come from?

[@satoutatsuhiro866]

the triangle is constructed by taking the 3 middlepoints of the circles as corners. since those points all have the same distance to each, the triangle can only have all sites equal length and all angles equal degrees. (triangle 180 degrees and 3 corners so 180/3=60degree each) then since the line that indicates the radius of the big circle is yet again dividing the triangle into 2 parts 60/2 =30 degree on both sites of the line. Questions?

[@judah738]

@@satoutatsuhiro866 Thank you! That was a great explanation! Appreciate it.

[@justablank1184]

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.

[@siamsadman1950]

I am a 9th grade student of Bangladesh and I was actually able to solve this question. Thank you for this amazing problem.

[@crocophobia]

I used the fact that the centres of an equilateral triangle are all the same and so the centroid of the triangle must be the centre of the large circle. I used a property of the centroid which is that it divides the line that passes through the centre in the ratio of 2:1 and then i found the height of the triangle and multiplied it by 2/3 and then added it to the radius of the smaller circle to find the radius of the larger circle and it was pretty easy from there

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@ibroeswfm414]

The center of an equilateral triangle is located 1/3 from the base, which makes it a little easier.

[@AlchemistOfNirnroot]

You could also use the area of a sector formula for the sectors. A=1/2r^2*angle in radians. Although it is equivalent.

[@omrgdf3611]

In the small right triangle, how did we get the 2 divided by square root of 3?

[@Grizzly01]

You either look at it as a 30-60-90 triangle (as Presh explains), so therefore the sides are in ratio 1, 2 and √3 or use the fact that cos30° = (√3)/2. The hypotenuse is therefore 2/cos30° = 2/(√3/2) = 4/√3

[@mandeltownthekillerfrombab5202]

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.

[@anant5977]

Circum-radius of an equilateral triangle is side/root 3 . Radius of large circle = 2 + side/root 3

[@canbalcioglu4229]

This one was easier for me. Great job!

[@michaellautermilch9185]

I found the big circle radius a bit differently. Big circle radius = small circle radius + distance from the corners of the triangle to the center of the triangle. Small circle radius = 2 Distance to the center of the triangle was calculated by recognizing that it is equilateral, and can be subdivided into 4 similar equilateral triangles of equal size by connecting the midpoints of the sides. Then, repeat this process on the inner triangle, subdividing it into 4 smaller triangles. Repeat to infinity, always subdividing the innermost triangle into 4 smaller ones. The center of the triangle will always be contained in the center triangle no matter how many times you divide it, therefore the cumulative heights of every other triangle in this series can be added together to converge on the center. Alternating triangle heights are: sqrt(3) + sqrt(3)/4 + sqrt(3)/16 + ...

[@shaunbrowne9870]

From looking at the thumbnail: Assuming the triangle is tangential to the small circles as it appears to be, it must then be an equilateral triangle, therefore the three shaded sections of it sum to a half circle. Therefore, we are looking for the area of the large circle minus 2.5 small circles. The radius of a small circle is given at two, so figuring that area out later will be trivial. If we allow the radius of the large circle to be 4+X (4 being the diameter of the small circles) and the distance from the point of tangency between the triangles and the centerpoint of glyph to be Y, we get a right triangle with angles of 30 degrees, 50 degrees, and 90 degrees and the side lengths of Y, 2, and 2+X. I believe the fact that this is specifically a 30/60/90 triangle is of some special significance which ought to allow the formally educated to solve this in their heads, but I am not formally educated. In any case we've reached the point where we can solve this by just plugging values into a calculator, and that's good enough for me.

[@sarthaksiddhpura4538]

I solved it just by staring at it for the area in the order : First the bigger circle Then 3 small circles Then equilateral triangle Then the 3 sectors

[@mr.scream2934]

your math teacher would be proud

[@sampeter1322]

I think this problem is from math book by R.D. Sharma Class 10.

[@sarthaksiddhpura4538]

@@sampeter1322 might be true There are many similar problems in that book

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@anandk9220]

@@sarthaksiddhpura4538 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) !!! 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. 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. 👍👍😘🤗😊😊

[@nimishagandhi7626]

I'm a 10th grade student and this is a pretty standard question that we get in SOF olympiads so I found it pretty simple

[@Cheerwine091]

This isn’t challenging so much as it is expansive. It covers a lot of area in fine detail, and one dropped number sets you off by more than you’d expect. But, if you keep track of everything correctly, you’ll eventually pull through! That’s the hard part of it, making sure you know how each part works in tandem. The other hard part (for me) was figuring out the area between the circles… didn’t realize it was just the triangle - 3 wedges!

[@darshamuralidharan2516]

Is 2 the radius of the small circle? I assumed it to be the length of the sides of the equilateral triangle. And therefore the radius of the small circles to be 1. Request for clarification

[@sidrockx6399]

2 is radius

[@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)

[@ravitejakakarala7858]

Radius of large circle is 2 + radius of circum circle. R = abc/4A. it is equilateral triangle so A = root (3) / 4 * a^2 . So R = a/ root(3) . a = 4.

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@mayankraj9249]

I can't lie, this question is the easiest out of all the presh's video. I'm satisfied I solved correctly 👍

[@mahmoudalbahar1641]

Many thanks for your great videos... and I suggest you make video about: non-integer base of numeration.

[@tosuchino6465]

The blue area = the big circle - 3 of the small circles - the equilateral triangle + 6 of the blue wedges = the big circle - 3 of the small circles - the equilateral triangle + 1 of the small circle = the big circle - 2 of the small circles - the equilateral triangle

[@Qsalis]

Thank you for going for black backgrounds. It allows for nicer nighttime watching

[@dudono1744]

When you draw tangents it's ez, I got : big circle's radius : r = 2 + 2/cos(30°) white area (in the circles) : 10π white area (center) 4 sqrt(3) π - 2 π total area : π * (r^2 - 8 - 4 sqrt(3))

[@ivanadriazola1991]

I could solve this, but it would take me a while. Big circle of unknown radius Minus, 3 circles of radius one Plus, the triangle with easy enough to get height ( split in half then Pitágoras) Minus, (same triangle minus half a circle of radius one) The hardest bit would be getting the radius of the bigger circle, and I'm pretty sure it's not that hard, there must be a formula or something, but even if not I can probably get it through geometry, worst case scenario I would recreate with equations an find an intersection, but I doubt it would get to that.

[@ivanadriazola1991]

I considered 2 as the side of the triangle.

[@ivanadriazola1991]

And about the radius, yeah it was easy enough, but I hadn't thought how to do it yet.

[@satoutatsuhiro866]

I recommend actually calculating the answere to see if and where mistakes occure. If you only go through it in theory your learning will be lessend I believe.

[@EngLhag]

Nice one. I love when I get the correct answer :)

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@fguo79]

One doesn't need to calculate the center small area (4*sqrt(3)-2*pi)) specifically.

[@wombat4191]

Wow, I actually got it, and with pretty much the same methods as Presh. I didn't bother calculating out the decimal answer, I was happy with (4/3 + 16/sqrt(3))pi - 4sqrt(3)

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@yassineattia6679]

To know the radius of the large circle, I used the propertie of "gravity point" stating that it is at 2/3 of the median of a triangle. The median is also the bissectrice and u can calculate it thanks Pythagore

[@adrien5568]

I did the same

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@Forever._.curious..]

Plsss start other topics of mathematics too 😊🤗

[@trickyepithet9122]

That satisfaction you get when you solve a hard problem

[@mathmusicandlooks]

A fantastic and fun problem, but one thing kept me cringing throughout the video: how frequently you substitute numbers back in. It’s so easy to lose track of the meaning of various numbers that way. Just leave everything in variable format until you have an answer, THEN plug in the numbers. It makes the whole process so much easier to keep track of. It may not be a familiar practice to many of your viewers. That’s okay. This is a beautiful opportunity to teach them this simple, useful, helpful practice.

[@ayanahmedkhan2580]

Your Video helps me to improve my math , Thanks 🤩

[@matthewbell4273]

It took me a little over an hour, but I am very happy to say that I did it, I got the right answer (approx 26.281376802258367420012324600624 units-squared). But: (1) I had done it using decimal numbers on the calculator (and so I could not follow all the fractions when I watched the remainder of the video afterwards); and (2) When it came to the small right-angled triangle, I knew the length of only one side (the radius of the small circle), not of two sides; and so I could not Pythagorise; instead I need to use Cosine 30 (Adjacent / Hypotenuse) to let me calculate the Hypotenuse (for addition to the small circle radius in order to find the big circle radius).

[@agentdai]

How did you calculate 2/root 3 and 4/root 3? Please answer me.

[@spreekcreations]

Got the radius of the larger circle wrong but still ended um with 26.281... Hmm...

[@Aerythraea]

This process can be more easily simplified actually. The shaded area equals the large circle minus the unshaded area. The unshaded area equals 5/6ths of 3 smaller circles, and a small middle portion. That's 2 and 1/2 smaller circles and the middle. The middle portion is equal to the triangle in the middle minus 3/6ths of a small circle. That means the middle portion plus 1/2 of a smaller circle equals the larger triangle. Therefore, the shaded area equals the area large circle minus the area of 2 of the smaller circles minus the full middle triangle. Let 'r' be the area of the large circle, 'h' be the height of the triangle in the middle, and A be the total shaded area. A = πr²-2π2²-2h Lets draw a right triangle with the following vertices: one bisecting the base of the triangle (F), one at the bottom right vertex of the triangle (G), and one at the very center of the large circle (H). The hypotenuse of this triangle (FG) plus the radius of the small circle (2) equals the radius of the small circle. The hypotenuse of said triangle can be found with trigonometry. Angle FGH is 30° bisecting an equilateral triangle, which is 60°. The adjacent side is known and has a length of 2. To find the hypotenuse relative to the side adjacent side is found using sine. cos(30) = 2/FG Divide both sides by 2 cos(30)/2 = 1/FG Exponentiate both sides by -1 (which takes the reciprocal) 2/cos(30) = FG r = FG+2 Therefore r = 2/cos(30)+2 cos30 = √(3)/2 r=4/√(3)+2 Lets draw another right triangle with almost the same vertices: F, G, and one at the top of the original triangle (J). Line FJ is equal to 'h', and the hypotenuse is known to have a length of 4. Angle FGJ is 60° because it's part of an equilateral triangle. To find the opposite side relative to the hypotenuse is found using sine. sin(60) = h/4 Multiply both sides by 4 4sin(60) = h sin(60) = √(3)/2 h = 2√(3) Insert the values of 'r' and 'h' into the original equation. A = π[4/√(3) + 2]² - 2π2² - 2[2√(3)] Solve for A A = π[4/√(3) + 2]² - 2π2² - 2[2√(3)] π[4/√(3) + 2]×[4/√(3) + 2] - 2π4 - 2[2√(3)] π[4/√(3)×4/√(3) + 4/√(3)×2 + 2×4/√(3) + 2×2] - 2π4 - 4√(3) π[16/3 + 16/√(3) + 4] - 8π - 4/√(3) 16/3π + 16π/√(3) + 4π - 8π - 4√(3) 16π/√(3) + 16/3π - 12/3π - 4√(3) 16π/√(3) + 4π/3 - 4√(3) Calculate ~29.02 + ~4.19 - ~6.93 ~26.28 A = ~26.28

[@unknown_url]

Hi, is there a relation between the perimeter of the three inside circles and the one on the exterior? like the one of only two circles, which, as long as they're touching themselves, their perimeter will be the same as the circle on the exterior. Sorry if these is not the best grammar, my first language isn't English.

[@HerrLavett]

I like the dark background very much better than the white

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@trollar8810]

I recently discovered your channel and started binging your videos this is the 3rd one that i could solve (after the x x x=6 and the oxford shape thing)

[@cringeconnoisseur6037]

Most of this can surprisingly be done with the knowledge of 9th grade math. I'd know because I just learned this recently. This looks like a regents question from hell.

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@andre.santus]

Thanks for the dark color background!

[@mehulpunia6174]

I use trigonometry to find the radius of the large circle.

[@michaellarson2184]

This might seem cheap but instead of using pure geometry I found graphical equations for the three inner circles. Then I used them to work out the radius of the larger circle. Everything after that was easy.

[@asheeesh_09]

can you give some step by step way how you solved it, your method seems fun

[@EvenHeisenbergIsCertainAboutMe]

I got the answer by solving my own😊✅ btw thanks❤️

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@wvvwwwvvw]

3:18 sorry, but why the other leg(lag?) is equal to 2sqrt(3)?

[@sudhirjoshi7782]

Your videos are excellent. This one is easy.

[@animezoneamv9116]

But I actually got an answer approximately 32.048 cm². Where can I prove my method? My interpretation is just right as yours sir. How can I reach you?

[@Grizzly01]

You went wrong somewhere. Show your method here if you think it is correct.

[@joelsmith3473]

Biggest hurdle here was self doubt over how messy the final answer looks and how the terms didn't nicely cancel out like many problems. I even self doubted enough to find the height of the central triangle two different ways and get two different answers until I realized 2√(3) = 6/√(3) = √(12) and they weren't different answers after all.

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@vidyadharjoshi5714]

The centre of big circle is 1/3 height from base of triangle. Height of triangle = sqrt(4sq - 2sq) = 3.46, (2/3)*3.46 = 2.31. So radius of big circle = 2+2.31 = 4.31. Area of big circle = 4.31sq*pi = 58.34. Area of each the smaller circle = 4pi = 12.56, Area of triangle = 0.5*4*3.46 = 6.92, Area of the blue sector inside the triangle = 3*(1/6)*4pi = 2pi = 6.28. Grey area inside the triangle = 6.92 - 6.28 = 0.64. Grey area of smaller circles = 3*(5/6)4pi = 31.416. Total grey area = 31.416 + 0.64 = 32.056. Blue Area = 58.34 - 32.056 = 26.284

[@mauryagoel4857]

My calculation - (The answer is slightly off because I took pi as (22/7) [((2(2+√3))/√3)^2*(22/7)]-[(12*(22/7))+(4√3-(44/7))]+[44/7] It's quite funny, whenever you present a simple problem, I find it tough, and whenever you present a challenging problem, I find it simple. This one was quite easy.

[@FireyDeath4]

Right, sooooooo...(SPOILERS, NOT STOOPING TO YOUR LEVEL YOU SMART ALECKS) ...the area of a circle is, I think...πr². So we have 4π for each circle with a radius of 2 units. Now...we got three circles together intersecting with a triangle along each radius, so they're all 60°. Sooooooooo: I guess we have 12π/6, which is probably 2π. I remember π to be...3·141592 and that's all I can remember (it's transcendental). So I'll just multiply all that to get: 6·283184. Definitely floored, unsure if rounded. -That's the best answer you're getting from me- (I don't vigorously double-check during math tests) so I hope that's right Did think I'd have to think about radians at the angle-checking point but I guess not Aaaaaaaaaaand I just looked back at the video and I have to check the big circle as well. Okay. Alright. Okay...alright Well...alright so what's the radius of it? Umm...the radius of the small circles are 2. So I need to know how much more it is to get to the middle of the intersecting triangle. Wellllllllll...the, um...0·5bh is the area, how does that help me? ... well...the triangle has a side length of 4, soooooooooooooooooooooo urgh I'm tempted to give up already. How am I gonna know the height? Probably heard about it a few years ago while passing by a formula in a book or an obscure math course online, but I only really remember things I have a use for. I'm: gonna be generous and assume, that: the bottom of the triangle precisely intersects the horizontal axis on the lower two circles' centres. Good, now we can just go ahead and assume the triangle height is 4. But wait, that makes no sense! ...Does it? damn you Presh Talwalkar for ruining my life (partly) Okay well the three circles have their centres connected while touching each other, so...well the two radii are 4 so the height...is...ah I see where I was wrong. I was thinking in squares, but the bottom two circles are moved up a bit because they're off to the side otherwise not touching. Right. Wellllllllll...I'm tired of being oppressed by you and your decent commentators. I'm tired of feeling like dying because of you and you know what? I'm not gonna do this. Instead, I'm going to start visibly employing all of my psychological defence mechanisms at once, rather than doing math that will potentially save an entire building of people who are in danger when I'm the one in control. So long, suckers!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! EDIT: This video made me feel genuinely upset for some reason

[@vladimirrainish841]

It will became much simpler, even trivial, task if to reverse blue and white colors in each of the 3 small circle. Then, the total area of 3 original outer blue shapes in the initial picture will be the area of the entire large circle minus area of small trianle in the center ( which will be all in white after such reversal) and minus 5/6 of area of the small circle multiplied by 3.Radius of the small circle and side of the small triangle are known (2 and 4), Radius of the large circle can be easily computed. The areas of each of 3 sectors in small circles which are in blue in the original picture is 1/6 of the area of the small circle. This reversal is the the only challenging(mildly) part, the rest is straightforward computations.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@davidmurphy563]

Do the corners of the triangle extend to the centre of the circles?

[@markstahl1464]

Yes. The vertices of the triangle are the centers of the small circles.

[@knowone3529]

What is % area of mandorla? Little help be very much appreciated..

[@markstahl1464]

First time through, I accidentally lost a pi, but my answer was negative. Second time through I got the same answer as you but didn’t factor out the (4*pi)/3 so it looked a little different. Had a hell of a time putting this into my smartphone calculator at the end. I didn’t round early so as not to throw off the final decimal.

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@tianqilong8366]

great video as always

[@akashraj6391]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@Quintanedu]

It's interesting to think it as big circle minus two small circles and minus the triangle

[@lamborghinitestarossa2133]

This was so simple. Just throw a couple of things in cad and ask it what the area is

[@D3ltus]

I know what I must do but I don't know how to get the radius of the big circle

[@EK_FAKIRI]

O is the centre of the circle and ab is the diameter if oda = 90 and OB = BC find angle 1..?!

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@udhaya_shankar_V]

I got 26.262, getting larger circle radius is key with 2/3 from Apex of triangle.

[@oloyt6844]

when you get everything right except forgot to divide the triangles area by 2 in A = bh/2 and have to spend 20 mins trying to find you mistake

[@nvztsnl]

i saw that when the 3 slices of little triangle equals to half of little circle then i realized (area of the big circle)-2,5x(area of the little circle)=(shaded area) and i think this is easier

[@cobokobo2115]

amazing .....kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@bader2347]

Guys, in 3:55, how the equation became like this? I mean from (28/3+16/sqrt3) to (16/sqrt3 - 8/3) please i need a specific explanation.

[@isaakyhsialf4369]

i remembers when in maths test there was a quest like this Demontrez que x= √(π/2)-ln(1/4) This implies that result is corrects But I never got that result After trying for 30 minutes to find what I did wrong I got enlightenment My answer was x=√π/2-ln(1/2/2/1) So as a greats mathematician I simply said x =/= ✓(π/2)-ln(1/4) therefore the question is wrong And still got marks after recieving tests

[@peteiiseasy3217]

Ahaha😃

[@yelloweater5506]

How can I send you my own questions to you? I found your email and texted you but got no answer , so what can I do?

[@topatosalt]

lol not me totally ignoring that there is blue anywhere in the circle except the ones inside the triangle and immediately thinking the area is 2pi because the 3 sectors make a semicircle

[@rileyjeffries1282]

People who have taken Statics know the drill, figured the centroid of the equilateral triangle was the center of the circle, solved in about 20 min.

[@nicherix]

Get it as rule please: if we have a fraction that has square root in denominator, we shall multiply both numerator and denominator with denominator for convenience: a/√b → a√b/b.

[@piolix0004]

Have you ever thought of using 3b1b style?

[@manofpowers25]

What is it

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@osamadandy7553]

Wonderful solution!! can anyone explain how the triangle is thirty-sixty-ninety in 3:03?

[@vincentjoseph8438]

You know an equilateral triangle angles are 60°, you take the half of it and it's 30° Then you just add 60° to make the sum of the angles 180°

[@osamadandy7553]

@@vincentjoseph8438 Yes I know, but how can your prove that the line is a bisector for the sixty-degree angle?

[@vincentjoseph8438]

If you proved that you have an equilateral triangle I think you can say that since it passes through the middle of this equilateral triangle it is a bissector of its angles

[@anandk9220]

@@osamadandy7553 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. NOTE : 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. 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. 👍👍😘🤗😊😊

[@Grizzly01]

@@osamadandy7553 Simple circle theorems will show it to be the case

[@arnoahmed9269]

what's the song name used at the end?

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/j3TPdKV_rK6Apas

[@AMANSHARMA-ci2yh]

You also use this concept in one of the previous question

[@lensenstark9819]

I calculated the outer circle by turning the outer Diameter and the radius into a linear function