This popular problem is stumping everyone, and many, many people emailed it to me. I give extra thanks Chaitanya from India who was the first person who sent it to me. I also thank David K, Achille Hui, and Seyed who found geometric solutions to this problem when I posted it to Math StackExchange. Watch the video for 3 ways to solve this problem!
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The first comment is from 1 month ago because I posted it to Math StackExchange where I received great help: math.stackexchange.com/questions/3016718/a-way-to-find-this-area-without-calculus Early comments also come from Patreon, as I do my best to give early access to supporters on Patreon: www.patreon.com/mindyourdecisions And finally, thanks for 1 million subscribers! We are building one of the best mathematical communities in the world. Thank you.
Hi Mind your decisions, isn't the solution just a specially sheared version of a slice of the upper semicircle limited by points P and Q?
Hey, I solved it in a different way that does not require as much knowledge of trigonometry!
Now noone has an excuse for saying "first"
Hi mind your decisions I have a problem for you: In a trapeze abcd (ab||dc) The ac and bd diagrams are cut at point e the ratio between the bases is 4:9 and the area of the triangular bec is 108 What is the area of the whole trapeze?
Matheus Mendes yeah I also did it , well I used basic geometry and one equation system.
Just check the resolution of picture and count the red pixels. It will be surprisingly accurate.
Yeah this is big brain time.
That is actually a legit answer. The method is Monte Carlo method.
Tell me this Sfhnxdkmvddghcssasfhkkmnjggcghjhy cnbvyul iyvvjygvbjuydfvjjnnb hgjjbffhjgtynoolmnbc hhyrszxvnklopplkkjhgh
That is the basis for calculus.
@@technosiam4480 I see what you mean.
Another solution: print it on a paper, cut out the red part, weight it, divide by the total weight.
this may sound funny but it's how they did it before calculus/geometry was common, in fact no so long ago, math became more common in 18th century or so
@ then print a modified picture of it without the red part colored in
Lame
@Aryan Sharma you think from perspective of today. Note that only few in only few nations knew this in ancient times. Rest of nation was not able to read or write. What more geometry was expressed primarily graphically, there wasn't kind of todays' equations.
Engineers vs mathematicians
-"the first method is simple" me: "alright, this might be the first time where i can solve a hard Math problem on my ow..." -"shows space rocket schematic" me: "okay, next video"
I'm just watching it to see what I DON'T understand.
@@schwartzseymour357 Just watch the first method, the other ones are waaaay too abstract and you're needlessly complicating things. With some knowledge of geometry and testing your 3D logical insight it's fairly doable. If you lack these concepts, yeah this video is as good as impossible for any newbie.
@@janjan55555 I disagree. If you know integral calculus, the third method is VERY straight forward, and is definitely the way I would have went before he even mentioned the geometric solutions.
@@williamhexberg6748 I do and I still disagree with your method
@@janjan55555 To me the integral method feels brute force and it’s the way I went about solving it before watching the explanation. I was expecting some clever/creative geometry trick and it turned out to be really simple. I just couldn’t see it.
For those who are wondering where π came for θ, it is the radian system or circular system for angles. 180 degree = π radian. And 1 radian = 57 degree (approx). *And 1 radian can also be represented by the angle formed at the centre of a circle by 2 radii separated by an arc of 1 unit length (on the circumference of the circle)*
this is a 7th grade problem
@@Mike-rw1jw akshully I could do it at 1st grade 🤡
@@Mike-rw1jwno. We get this in 11th grade
Next video: Put many circles inside a square; make most of them pass from intersections of other circles. Find the most absurd looking piece of area. And then solve for it.
@Tokisaki Kurumi using only a ouija board
And then realise a 7 year old Korean kid solved it quicker than you did.
U don't have any other work or what?🤔😑
haha, this one is so complicated i don't want to watch through it.
AhmedHan that actually sounds fun
I’m majoring in Mechanical Engineering and the calculus method seemed to be the easiest method. 😂
It included the most important step I remember from my calculus class - "annnd... skip some steps". Though in my class this was the "and from there... it's obvious" step :P
I dont think so, its easier to error with calculus, and more work demanding in my opnion. If you can work with basic algebra or geometry and have a good trig perception you should use it. Calculus is too powerfull, like trying to catch a fly with a gun when you have a fly swatter.
1! +2!+ 3! +...........25! divided by 13 then what will be the remainder left?
@@theunique140 Using python. x = 0 for i in range(1,26): x += factorial(i) x%13 = 9
@@theunique140 If you're still looking for an answer, you can do it like that 1! + 2! + 3! + ... + 12! + 0 (x!%13 = 0 for x>=13) 1(1 + 2(1 + 3(1 + 4(1+5(1+6(1+7(1+8(1+9(1+10(1+11(1+12))))))))))) % 13 11*(12+1) = 11*0 10*1 9*11 = 99, 99%13 = 8 8*9 = 72, 72%13 = 7 7*8 = 56, 56%13 = 4 6*5 = 30, 30%13 = 4 5*5 = 25, 25%13 = 12 4*0 = 0 3*1 = 3 2*4 = 8 1*9 = 9
I'm proud of myself for this one. It is the first time I solved one of your math puzzles. It took me like an hour but I got the right result of 9.77 I'm only 14 years old and I've never had anything like this in school before.
Ayo good job!!
@@Soefae lol, you answer on my 11 month old comment. But anyways, thank you
@@PerspectiveInsightReviews out of curiosity, if you didn’t learn it in school then where did you learn it? And which method did you use?
@@commonlogicart1604 I found some method to calculating the different circle pieces area, and subtracting them. But it's almost a year ago, so I don't remember exactly what I did.
I'm 14 too and I get 21.4
I paused the video and solved the problem on my own before watching the full video. I approached it geometrically. I did it in my head without even using paper and pencil. I thought it was quite simple. I continued to watched the full video just to make sure I got the same answer. I was surprised that all the 3 methods in the videos are so long. Below is the outline of my method. Using the same labelling as 00:34 of the video, Area of shape DPQ = Area of shape DPQC - Area of shape DQC. It is easy to see that the shape DPQC repeated 4 time will fill the square. So the area of DPQC is just 25. Next step is to solve for area of shape DQC. Let midpoint of DC be E. Area of DQC = Area of sector QEC + Area of quadrilateral DEQA - Area of sector DAQ. It can be easily checked that Angle QEC = 2 * Angle QDC = Angle QAD = 2 * tan^-1(0.5). Therefore, Area of shape DQC = pi * 5^2 * (2 * tan^-1(0.5)/2pi) + 2 * 1/2 * 5 * 10 - pi * 10^2 * (2 * tan^-1(0.5)/2pi) =50 - 75*tan^-1(0.5) Finally, Area of DPQ = Area of shape DPQC - Area of shape DQC = 25 - (50 - 75*tan^-1(0.5)) = 75*tan^-1(0.5) -25 This is the same answer. I didn't finish reading all the 1.5k comments. But so far among the comments I've read, I haven't seen a similar method. This is the first time I see this problem. I'm not sure whether or not this method above has been posted before elsewhere on the internet.
Finally an elegant solution without huge calculations and variables. Bravo!
@@GGReaver Thanks for taking notice of my solution. I thought no-one would bother reading it 😆
you know i've been watching a couple of your vids and there's usually a common theme in finding the correct answer. Either you use coordinates or geometry. When using geometry there's always one rule... find triangle and you'll find the answer. Just throwing that out. It's amazing to how much extent you can use this simple concepts to solve something like this
Solve this problem guys. This is a challenge for you. Problem given in the link below. kzhead.info/sun/o8uQqJt7oKJ8n6c/bejne.html
Yup. This is exactly why trigonometry is taught as a standalone course and not as the subset of geometry that it is.
i have enough problems in my life
😂😂😂😂😂😂😂
This stuff is precisely to distract you from problems in life. XD
😂😂😂😂... Swears
LMAO relatable😂😂😂
Maths book: "Cute"
Any mathematics enthusiast will be blessed by finding this channel...Hats off to the creator❤️🎉🙏👍
I Just loooooove the feeling when you can use so many different solutions to find the same answers. Keep up the good work!
I once solved this really hard Q in class which no one could solve and then the teacher told us how to do it and his method was completely different than mine but we had the same answers.
I used the 4 option.... I watched the video
Me too 😅
Sven Minoptra , 😂😂😂😂 I did the same
If I only had 5min to solve, I would answer "definitely less than 100" & walk away
exactly
"definitely less than 50" to be more accurate
@TheMutedGamer I would bet it was less than 18. (haven't watched yet)
Well definetely more than zero
@Yash Hriddha I bet it's 9.77
The amount of work that goes into your videos is truly phenomenal 🙌
9:00 instead of using trig functions, we can also apply the similar triangles rule. Find the ratio of the kite triangle and smaller triangle hypotenuse, then their areas have ratio to the square
Option 4 - Depending on how many sig figs you need. Step 1: Find a sheet of graphing paper. Step 2: Sse a compass and rulers to draw out the figure. Step 3: Spend the last 3 hours of the exam to count the number of little squares.
This is essentially how some computer programs compute areas. They just make the squares small enough so that they get the desired precision and of course they can count the squares much faster :)
마인드 유어 디시젼
@@user-mx4sx1ul8h dak chua!!
I work at a railway testing lab. There's a test that involves pressing a rubber pad in between a mechanical actuator and a cast iron mold with small high relief squares carved on it, over which a powder is spread. You place the rubber pad, press it using a specific pressure, and leave it there for a while. Afterwards you remove the pressure, shoots a photo of the rubber side with the square marks the powder made, transfer that photo to a computer, and have it calculate the ratio powdered area vs. non-powered area by literally counting how many pixels in the photo got this or that color and dividing one number by the other.
That's what I would do in exam..
Your first "little lens" can be calculated much simpler, namely as the 5x5 square without the two other shapes, so that's 25 - (25 - quarter circle) - (25 - quarter circle), so that is a half circle of radius 5, minus 25 all in all, making it 12.5pi - 25. That's a whole lot faster than your "lens" calculation.
I've had a similar reasoning to do that part (square - 2 whole circles and then divide by 4). I wish there is a close way (substract areas) to solve the second part... i'm pretty sure there is one but i can't find it
Isn't this basically a way he went in the second option?
His first “little lens” could be calculated in the way you showed, but what about the second “little lens”? You probably will need a different method to solve it, on the other hand, his method of the first “little lens” was applicable to the second lens, thus, saving him time and making the overall solution easier, at least in my opinion. How would you calculate the second “little lens”?
I did the same.
@@--Za did the exact same thing and the second part is bothering, you need to get the area of about 6.47 by using the circles but I cant seem to figure out how
Me: I know calculus well Me after watching this: I know nothing about calculus
Facts. I haven't even begun learning integral calculus, only differential, so I found the geometry a million times easier to understand.
This was basic calculus actually, you do have some problems
@@DanTheManTerritorial lmao, you have the reduction formulas memorized?
Don't feel bad, unfortunately this problem calculus wise is very reliant on formulas from table of integrals. The solving was largely done by the reduction formulas, which I don't think any reasonable calc teacher would expect u to have memorized because they don't pop up that often, outside of situations like this where the curves happen to be parts of circles.
@@kameronrogers4000 no it would if you have enough practice of indefinite integration during 12th std, or even if you face such difficulty later you could always look on internet, these are basic reduction
As an alternative, one might use symmetry and scale to simplify the solution. Steps: 1) 'Symmetrize' the picture by drawing another quarter circle radius 10, centered at the TOP, RIGHT vertex of the square. This divides the square into 8 regions, of which 3 are pairwise identical, thereby resulting in 5 distinct shapes ( with areas x1 thru x5). Our target region (shown in red in video) is one of the paired regions (say x2). 2) Express the areas of the small semicircle, the large quarter circle, and the square as linear equations in the variables x1 thru x5. 3) Thus far we have 3 linear equations, (and 5 unknowns) so we need 2 more. These are a) The area of the small 'lens' region formed by the intersection of the two semicircles, which can be easily computed as described in the video, or otherwise. and b) The area of the larger 'lens' formed by the intersection of the two large quarter circle. Since the larger lens has the same shape and twice the linear dimensions as the smaller lens, its area is 4 times as large. 4) Now we have a complete set of equations, from which x2 can be easily obtained.
Thought that would be the "best" solution as well when trying the problem on my own, but sadly the equation system is underdetermined and you can't solve for x2 (the red "wanted" region) with these equations alone. You somehow need to calculate the area of one of the "paired" regions on their own and for that you always need the right most corner of the red region (or its symmetrical counterpart). And at that point you can just use the calculus solution from the get go...
I just recently discovered your channel. I did the problem with both methods, the area subtraction using geometry (since that stood out to me) and the calculus way (because that just made sense integrals for the win). I set up the problems in my head and didn't particularly solve for an exact solution. Once that was done I finished watching the video. I m glad my physics/math brain is always on the ball great problem. Math is just awesome!
I put this in AutoCAD 😎😎😎 Ans - 9.7736
you are THE MAN
9.773570675
@@ahmedmonjid5696 Thanks man
@@mayanktomar5022 exactly
How did you measure the area of the shape? (I created the same things as shown in the video but I can't measure it)
The method by the integrales is great !!! And it's a good mean to understand how powerful are the integral concept and tool !!! Very nice video !!!
I did't solve it but just watched the solutions. Very interesting, and of the three I think the second solution is simpler and better. The Calculus solution is very drawn out and would need a person who is good at it - we all can be good at Calculus, just keep practicing and memorize first pricipals. Thanks for posting and really enjoy your maths videos.
I literally looked at the thumbnail and tried to solve it for 10 mins. My friend said “are you looking for something on the screen?”
1! +2!+ 3! +...........25! divided by 13 then what will be the remainder left?
I graphically solved it without paper after looking for hour and a half HAHAHAHA ocd
Thanks, dude you made me realized I need to retake Calculus and Integration.
Solve this problem guys. This is a challenge for you. Problem given in my channel.
Wow, this was so brilliant! I've got stuck with the area of lenses. I tried to get the area of circle fractions, then I tried to subtract them, but it needed the lens part. Thank you so much for teaching me how to calculate new shape!
Thank you for your helpful lecture. The method of subtracting area to calculate the area of the figure to find.
It's funny how calculus is 10 time easier than geometriy
that's because the calculus way is the robot way, the geometric way is the intelligent way.
avhuf why is it robot
No it's not
@@SticklyDaKing because u don't need to use a lot of ur brain , even a 10 th grade student can solve it that way given hr know all the formulas
Calculus is mostly used to program computers and calculators which uses the same definite way to solve one type of questions. Intelligent ways are much more diverse and allows shortcuts that cannot be discovered by machines. It also varies from one person to another, signifies creativity. Unlike calculus, geometry defines intelligences that a machine will never have. (Unless somebody invented artificial intelligence that will eventually cause global extinction.)
And i was there thinking i can do it in my head with my calc 🤣
Solve this problem guys. This is a challenge for you. Problem given in my channel.
Hello! I am from Brazil and I love your channel. Thanks for everything
I used the 2nd method with some modifications. When calculating the blue shape I first calculated both of the sectors, and then I took the total "kite" area (= 50) and substracted one of the sectors from it to get the part of the other sector that is outside the blue lens. And then I simply subtracted this area from that other sector to calculate the blue shape. Seems to be easier.
One of the best math channels of KZhead, greetings from Brazil!
yes really
Papa Flamme ficou triste agora
Super exatas>>>
Xande god
Yeah This and 3Blue1Brown
I solved this problem using calculus, it was the fastest and easiest way, in my opinion. I really liked the other two geometric forms of solving it!
Area of intersection of two semi circles can be calculated by drawing all four semi circles inside a square. Suppose area of semi circle is A, area of intersection of two semi circles is B and area of the square is C than 4*A - 4*B = C. Because when you add areas of four semi circles together, each intersection will be counted twice so you have to subtract four times of the area of intersection in order to make it equal to the area of squire. For a square of side 10, the value of B is comes out to be 14.26. It is first half of the solution.
Wow! What an amazing problem. I used something close to the long geometric method. However, I wouldn't call this 'insanely hard' compared to some of the other maths problems on your channel haha. Really interesting to see the different methods of solution
Wow! You really had a lot of work with this one! Great job! I used a mix of methods. For the 5x5 lens, I used geometry b/c its very easy to calculate: Area of 1/2 circle with radius 5 - 1/4 Square area. For the 5x10 lens I used calculus, but I turned the original pic upside down. So then it became integral of smaller circle from x=0 to 8 minus the integral of the larger circle from x=0 to 8. To solve the "sqrt's" integrals I had to look it up in a table of common indefinite integrals. PS: I also did it in the geometrical way, but my final solution uses arcCos function b/c I used the cosine law to determine the angle of the circumference arcs.
Love your vids! Please don't stop!
okay, at the end of the video, you asked if i got this question and what method I used, but I did Not in fact get an answer, but I had the thought process of what methods I would have used which were all correct. I haven't been in math for two going onto three years, but I knew and followed everything you were explaining. Made me smile
I ended up a lot closer than I expected. I've been challenging myself recently to not use a calculator while solving problems, so when I tried using the geometric solution mentioned here, I ran into a major snag when having to calculate the inverse tangent. I ended up taking a best guess on it based on what looked right for the problem and ended up with a final answer of "9.9 or a little less". I'm surprised that I ended up using the correct method though. Usually geometry is my Achilles Heel for mathematics. I didn't even consider using calculus to solve for the answer, which surprises me given how much I tend to default to it.
I am still struggling with this problem so I haven‘t watched the solution- it is driving me nuts! I am not yet ready to give up. Wish me luck. 🤔🙄😳😠😩
u done yet?
Affinizationz lmaoo 1 year later
I bet he stills solving it
Updates ?
I don't know of a way to do it without calculus. This isn't one of those "clever" problems that has a trick solution once you see it. Instead, it's a lot of grinding calculation.
I literally wrote a python code to do this computationally: start with an nxn grid. Imagine a circle at point (0,0) of radius n. (circle 1) Imagine a circle at point (0,n/2) of radius n/2. (circle 2) Imagine a circle at point (n/2,0) of radius n/2. (circle 3) Loop through literally every point in the nxn grid. Add to variable A if the point is: inside circle 1 AND outside circle 2 AND inside circle 3. The ratio A/n^2 is the percentage of the square containing the area we want. Multiply by 100 square units and you have the answer to the question. I blew n up to 5000 and waited about 45 seconds for my code to run and I got 9.77, so there is a "Red Green" solution lol.
share it please
What you did is known as monte carlo integration
GG
nice
that seems unnecessarily brute force...
I'm proud to report I sat through the entire video without skipping. That alone is an accomplishment given how little I understand anything that was said.
LETS GOOOO I ACTUALLY GOT ONE OF YOUR VIDEOS RIIGHTT ✊🏼😤 Damn I've never been so proud of myself lmao
I did not calculate it, I guesstimated 10%, being a building contractor has it's perks ! ;)
Lol
"its" used as a genitive gets no apostrophe.
@@cubeeater6458 dude, get a life.
whoa, that's great!
@@cubeeater6458 Ha ha. I couldn't do the maths problem, but like you I can see all the grammatical errors in the comments :-)
Thanks for making me feel like I learned both something and nothing in this video
🤯 I get the second solution best out of all of them but I wouldn't have got there in a million years and I wouldn't be able to do it even now without watching this again!
Great explanations! Thank you for this very detailed solutions.
can u please explain to me how did he get those curve equations?
I just got it he transformed the origin .
I literally just finished calc 3 this morning so it was still very fresh in my head. I was thinking "oh, you could do polar coordinates and use single or double integrals (whatever you fancy), broken up by the intersection point, with the origin being in the top-right corner" I actually hate that these solutions were significantly easier and used a much more basic version of math
Yea exactly, I was counting it by transforming into polar coordinates and in my head it was much easier with a double integral, than three basic integrals in cartesian
I liked it at the end when he said "did you figured out this problem ?" XD Please have mercy
For the calculus solution, I used polar coordinates. Rotated the figure and shifted the origin to the point where the left, top point (0,5) is now located at. Doing so, all the curves pass through the origin and r-theta functions are easy to set up. Used A= 0.5 x integral of r^2 d theta. All the cartesian equations of circles can be converted easily to polar x = r* cos(theta) etc. This is a shorter and less painful method than using cartesian system shown here. All the angular limits are easy to find using regular trig/geometric formulae.
Same with me. As soon as I saw the picture I thought this is to complicated to think about. Let's just integrate and get it over with. But I knew for sure there must be a geometric solution.
As an observation, the formula found for the area of a lens in the video is for the special case of a lens where the two circles meet at right angles - in general, you won't have right-angled triangles, and end up with a longer general formula (the standard version is given in terms of r and R, the two radii, and d, the distance between the two centers)
Can you state the longer one?
Absolutely. It's only going to be a right angle if you have a tangent which, as you say, is only in one special case. I'd stick with the calculus !
@@we-are-electric1445 But being bent at an angle other than 90 degrees would complicate the Cartesian plane needed for the integration
@@goyonman9655 It is in two dimensions so the integration should be straightforward. I haven't checked the proof. The narrator seemed to be skipping bits from the part I listened to and his calculus explanation didn't seem clear. The key point is equating two of the curves to determine the x,y limit points through which you need to integrate. When people skip bits of an explanation most usually it's because they are so clever they expect everyone else sees the missing bits as obvious or they don't fully understand the proof. If he comes back and corrects me then fine but at least i hope he comes back with more information.
@@we-are-electric1445 Remember the dimensions of the square gave him values of points on the axes and how to pinpoint their values on the plane
Something tells me we may be able to solve this with a system of equations. I named each section with constants a --> f (with b being the objective), and got to b + e = pi*(10)^2/8. Only worked on it for a couple of minutes but I'm sure I can represent e (which would the the lower right section of the quarter circle) in terms of the total area of the square then I'm good! Thanks for sharing this Presh!
Did you manage to do it that way, I'm really curious because I have been trying that myself but no results yet
I did the same and got to a matrix 6x6. I’ve tried solve it using Cramer’s rule but didn’t work. So I had to appel to the integral method and found 11,0743. I still didn’t watch the video.
I’m just gonna pretend I understand what he said
worked for me!
hahahahahaha so so good to know its not just me
i could solve the yellow one with an easier formula but i will have to pretend to know about the blue one
What his teacher hears: So if 3 - 2 = 1 it means that 1 is the answer What I hear: So if the power of xy squared divided by 5.73x10 to the power of -82 you can calculate the coefficients of angles bdgx2 and x over y to the power of 68 in an a standard form and that’s how you calculate the efficiency of booster 2 in a 2500 litre rocket...
Love it! I actually tried to solve it for about an hour. Then I gave up. I used a smilar methode like your first one, I even used triangles but only once instead of twice and could have hit myself as you explained it lol
This is solvable using geometry and algebra only. Algebra can find the coordinates of the right-most circular intersection using the general equation for circles and the known centers. Knowing that, a sufficient system of triangles can be constructed such that the desired area is obtained purely by addition / subtraction of triangular areas and arc segment areas. I will confess it took 2 hrs to come to this conclusion.
Exactly, i did it. And i got 9,78%
It took me 3 hours using the same method!
I did integrals but I’m finding a common theme. Whenever looking for an area of a circular shape, find a right angle triangle that splits the needed area into a right triangle and a proportion of the circle
I love this channel ,it is great!!
Intersection between two equal circles in that position will always be (pi/2 - 1)*rˆ2. You can easily deduce it drawing the combination of the two circles into a square of its diameter.
My ignorant younger self took AP Stats instead of AP Cal..so I did this the long geometric way, and essentially just nodded my head as you narrated the on-screen calc operations, like, "Yes, the audio appears to be in sync with the video here," so, long story short... 🙏 I appreciate your videos, which make me wish I'd taken more maths courses at university, because I always enjoyed the subject and I ought to have continued to be good at it (Well, that wasn't very short, was it?) ✌️
As soon as I saw this picture, the word "calculus " came to my mind, even though I had no idea how exactly...
Solve this problem guys. This is a challenge for you. Problem given in my channel.
I used method 1, except: first, I was too lazy to do the math for the larger lens, so used an online calculator for that (area=24.043), and second, I found the area of the smaller lens by a simpler method. Not having the graphics available, I describe it thus: the smaller lens is found by calculating 1/2(area of left-side semicircle) - 1/2( 1/2 the box area - area of upper semicircle), or 1/2(39.270) - 1/2(50-39.270) = 14.270. So large lens less small lens is 24.043 - 14.270 = 9.773.
1. 3D print the shape but give every edge the same height wall 2. Measure how much water it hold and how much red area holds 3. Red water divide by all water X100 4. Area of Square X Step 3 = Area of red
When you showed that the length of the side of square is 10, my first guess was "the area is somewhere around 10". And then you showed that it is around 9.77. :D
Presh, love your vids, very helpful and fun
That was fun! Reminds me of being in university long ago, thank you.
The way I did it was drawing the diagram and putting A, B, C, D, E and F into each of six parts of the jigsaw puzzle, with B being the shaded bit. If we take half of one side of the square as r, the area of the square is 4r2 then express B = 4r2 - A - C - D - E - F and you can find other relations where two of the letters add together to make half an area of a circle etc etc and eventually all other variables cancel out to give the answer.
i just drew this up in Fusion360 and had it measure the area for me.
I used hyperoperated calculus with stable equality with a trig failsafe, also including operation successorship. This method is unbeatable.
Nice. Before watching the video I managed to do it with method 2, but my calculus is way too rusty (engineering degree >30 years ago!) to have done method 3. But I must admit that back in those days that would have been the quickest go to solution.
I solved this when I was a fetus in India.
Could you work out the area of the semi circles and the quarter circle, then add them up, then take away the area of the square, then since you have two layers of the required area then dive it by 2?
I've graduated high school 2 years ago and I almost forgot those math elements.. Now I remember that I was genius in math when I was a student and now I realized that my brain is rotten since graduation... Better study again. I've been wasting my time since graduation playing games and internet surfing... Thank you so much.
Thanks for the video!
The second method seemed to me to just be the first method worked out twice with specific values for a and b. In both cases you find all the triangles etc.
Solve this problem guys. This is a challenge for you. Problem given in the link below. kzhead.info/sun/o8uQqJt7oKJ8n6c/bejne.html
0:53 yeah
THe calculus method is very elegant. Congratulations. I def lack the skills but I get some of the geometrical solutions thou.
Calculus part is nice. Verifications and comparisons of the three answers are interesting.
I used up all of my brain cells just for this
So was it worth the try?
@@wolframalpha8634 He can't comment to react. He used all of his brain cells, so he's officially braindead.
After watching the solutions, I realize that I'm simply mathematically undereducated, I didn't have the tools to solve it with what little trigonometry I remember from school. I guess, I'll have to start from scratch with basic maths if I want to be able to tackle problems like these... Oh well, it was fun trying to derive what I could with what little I could remember.
My hearty thanks:from India
Thank you for doing this… all solutions are good… I liked calculus one…
Hi Presh, many thanks for this gem. It's a lot more simple though. Focus on area PDA (via inner arc). I call it the battle axe. It's area is obviously 25. You instantly also know the area of PD, the double-sided symmetrical konvex lens which I call the fish. (done via computing semi-circle minus 25). Now work out area BCD (via point Q), I call it the "kite", like a hangglider (done by the "square" minus "quartercircle"). Thereafter you work out area QBC using year 10 high school maths (drawing a line from Q orthogonal to line BC) followed by integrating separately on both sides of this line "under" (here to the right of) the two arcs of the respective circles. Take QBC away from the "kite" and you get QCD, known as the cleaver. Take away cleaver and fish from already known semi-circle and yield PQD being 9.7735... Woody
wha...
Exactly how I did but got stuck at the QBC part..
@@ShittingStar07 . QBC looks like a tail fin of a fish or a double-sided ramp. I come across this shape quite often . Even though calculus feels like cheating, the simplicity to work out the area is too quick and easy. I draw a line from Q orthogonal to BC and integrate both sides and add them together. It's merely 2 simple circles, easy. Two things to watch out for. We integrate above the circle not under it. Second thing, since there are two different radii, 3 component's change when writing down the integral, not two.
@@WoodyC-fv9hz can it not be done without using calculus or trigonometry?
2:02 Solution using geometry and trigonometry 6:49 Longer solution using geometry and trigonometry 12:06 Solution using calculus (integrals)
My first thought was to use calculus as well. It's just such a powerful tool.
I enjoy solving your questions. I think questions apart from geometry would be better
Here is another way of showing the geometrical method: For the semi-circles, we have the circle area formula of πr², so: half circles area = 25π/2 = 12.5π quarter circle area = 100π/4 = 25π Considering that the small lens goes between two identical circles, we can claulcat it as: smalllens = semi_circle1 + semi_circle2 - kite smalllens = 12.5π + 12.5π - 25 smalllens = 12.5 x (π-2) or smalllens = (π-2) x side² / 8 And for the biglens we need the lens formula: biglens = side² x (6 x arctan(0,5) - 4 + pi) / 8 f = side² - 12.5π - 25π + biglens Then we can solve: Answer = biglens - smalllens
Nice and clear presentation. Great Job!
How can your comment be one month old...🤔
@@shyamsharma1746 About 1+ months ago, I found the problem on MSE and there was a comment by Presh pointing to this vid. The vid already exist then because I indeed watched it and commented. I don't know why the published date now was changed to Jan 10, 2019. There must be a mess-up by YT or some other reasons I don't know. Today is the 11th, it is unlikely that video of this nature could receive so many comments in just one day.
The first yellow lens shape is the intersection of two semi-circle (with the diameter of the side of the square). So you can draw four of this kind of semi-circle in the square, and every two of them has an intersection with the shape of this lens. Now you have four of them in the square. The total area of these four lens is actually the area of two full circles subtracting the square (when you take out the square from the circles, what left is the intersection). Therefore, the lens area is (2x(5^2*pi) - (10^2))/4 = 12.5pi - 25.
I remember a few years ago I wasn't able to solve this. I was able to solve it today having learned calculus. The geometric way requires less calculations, but the critical thinking involved in that method would have taken me longer to be honest. So the calculus way was the fastest way to solve it for me.
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I also did this with calculus, but I'll keep kites in mind from now on. On another note, your trig relation between arcsin and arctan is interesting (arcsin(opp/hyp) = 2arctan(opp/(adj+hyp)) -> arcsin(4/5) = 2arctan(4/(3+5)) = 2arctan(1/2), wow!), but if you glue a 3-4-5 triangle to an appropriate second triangle, the relation between arcsin(4/5) and arctan(1/2) pops out nicely. I leave the details to you though ;)
Thank you very much for this video
A simpler expression for arcsin(x) in terms of arctan(x) can easily be derived from a right triangle in which sin(y) = x (take opposite site to have length x and hypothenuse to have length 1, e.g.). Then the adjacent side has length sqrt(1 - x*x). Now the angle can be expressed as: - y = arcsin(x) - y = arctan(x / sqrt(1 - x*x) Since both expressions must yield the same angle, we deduce that arcsin(x) = arctan(x / sqrt(1 - x*x)). (this also holds in case x < 0, since both arcsin and arctan then flip signs; but we have to take x -1 and x 1 - although 'limit-wise' it still holds)
I looked at that and thought: It is pretty straightforward with calculus, though it would take a while with multiple integrals. With geometry, that's tough. EDIT: I wrote that before watching. That makes me feel better.
3 definite integrals if i count correctly that is if you chose 0 for y at conviniant position, not even watching :D
Sam two double integrals, the region can be broken down into two type 1 or two type 2 regions.
1! +2!+ 3! +...........25! divided by 13 then what will be the remainder left?
@@theunique140 0. 13 is a factor.
@@theunique140 Slightly more interesting variant I just thought of: Divide by 13^13. Same answer.
Wait a second, this video was a month ago. Was it only for patreons or something?
You posted before my comment appeared...The first comment is from 1 month ago because I posted it to Math StackExchange where I received great help: math.stackexchange.com/questions/3016718/a-way-to-find-this-area-without-calculus Early comments also come from Patreon, as I do my best to give early access to supporters on Patreon: www.patreon.com/mindyourdecisions And finally, thanks for 1 million subscribers! We are building one of the best mathematical communities in the world. Thank you.
Solve this problem guys. This is a challenge for you. Problem given in my channel.
why this have 69 likes ?
A math professor in my college once introduced this problem in a calculus class. He use this as an example of the power of calculus, cuz you just take a few mins to write down a function and put it inside the integral, hence the geometry solutions takes a lot of steps and assumption, which potentially can cause a lot of mistakes.
If you draw a diagonal from the lower left to the upper right, then compare the area of the upper left triangle formed to the combined areas of the two semicircles, the over-count in area could be thought of as 4 half lenses. This seems like a way easier method of calculating the lens area than the method described.