Its crazy watching these as someone who so far only knows middle school math and seeing that I *could* actually figure it out with my current information
It's why I like to make the distinction between complex and complicated/hard. Most problems are complex, in that they seem daunting but are trivially decomposable like this one. Some problems however are daunting because they are actually hard and can't be decomposed, at least not without great effort. In that case clever simplifications grounded on good assumptions usually go a long way into turning the problem into a complex one.
what a crazy solution to a problem like this. as someone who doesn't know the formulas to solve these off the top of my head, i see things like this at work all the time (im a contractor and volume and area come up constantly) and i always just end up estimating. but to be able to crank out a real solution would be so satisfying
Man, your channel shows exactly what all my math teacher once told me: math is not hard, in fact, it is easy, you just need to decompose the complex processes into simpler ones until you solve everything Really nice content, keep on the great work!
I am a student in the second year of secondary school, and I solved this problem correctly, but in a different way, and it took me an hour and 45 minutes of my time.🥲
I did it a bit simpler: cut diagonally the shape is made from two white circle segments cut from two red circle segments. The two radius 8 segments are equal, so we can fit them together and the total area is a radius 12 segment, minus a radius 4 segment, or (36pi-72)-(4pi-8)=32pi-64.
You seem pretty good at geometry. I wanted to ask for a few tips. I’m in 7th grade, and I’m taking a geometry class. I’m struggling a lot and so is everyone else. Our last test was somewhat in the middle of difficulty, and it was a derivation of Heron’s formula with no prep, and a few problems from the harder half of AMC10. I just don’t know how I can minimize time wasted. I don’t even erase anymore.
the curves must be circular if they have constant curvature and intersect the squares at exactly their vertices. Neither of which is specified in the problem. So you're right it's a poorly conditioned problem.
Then you estimate and get as close as is reasonable. Or, you use calculus. You measure the rate of curvature for each section of the curve, represent that with a function, and integrate over the range.
This is truly beautiful. Like these kinds of problems, you mostly just need to break it up into nice pieces and then it all comes together (pun intended) in such a beautiful way. Love your explanation too.
Thanks so much for this stepped solution. I struggled with algebra and have not done it for nearly 20 years yet watching this really made my memory trigger with how do all that - i understood it and feel like i could apply those principles in other circumstances.
You can also cut the claw along the diagonal of the bottom left square. Then shift the smaller part to the top right corner of the box and turn it by 180 degrees. Then x is just (1-1/3^2)*A, where A is given by the quarter of the area of a disc with radius 12 minus the area of a square with diagonal 24.
This problem is extremely easy with some basic calculus. Calculating area by integration is one of the most common problems in early calculus courses. For this problem all you need besides integration is the equation for a circle.
I mean, that's essentially what he's doing, he's just using the final value of the integral of the equation for a circle from 0 to r, that being pi*r²/4. And I'm not sure this would actually be easier by explicitly defining a bunch of piecewise functions and doing one big integration; that feels like introducing unnecessary complication when you already know the formula for the area under each individual quarter-circle in the grid.
@@Idran They both accomplish the same thing in nearly the same way. Which you find easier probably depends on what you are more familiar with. Fewer fundamental equations are required with calculus, but as you said, more piecewise functions and you need to know how to do basic integration.
What a great video! I had no clue how the process for those inner sections would be calculated when I started and was shocked at how intuitive it was that making them quarter circles and removing the triangle came right as you started saying the solution! Very well structured and paced!
If you add and subtract the obvious quarter circles, and use a pen to keep track of double counting for each little region, you find that you simply overcount by exactly 4 of the squares :) Quarter circles: 36pi + 16pi - 16pi - 4pi = 32pi Subtract the squares: 32pi - 64
You can also factor out the 16 and radii, seeing that each quarter has a same-size right triangle subtracted, and the r=2 quarters cancel each other out, thus may be omitted. Then you simply evaluate 16*(3*3-1*1)*(π/4-1/2) = 128*(π/4-1/2) = 32π-64, which is the correct answer I got just by looking at the figure.
I watched the other video in the description and I have another solution as well: 1/ calculate the A of the right claw by substracting the larger 1/4 circle (r=12) from the smaller 1/4 circle (r=8) and 2 squares on the left. 2/ calculate the A of the left claw by substracting the larger 1/4 circle (r=8) from the smaller 1/4 circle (r=4) and 1 square in the bottom mid. 3/ Add the A of the 2 claw and substract 1 extra left corner square. This is fun! Thanks for the vid!
Recently found your channel and i am absolutely in love with your content. As someone who always used to fear math, i recently began on a journey to like math and i cannot tell you how awesome your content has been in guiding me along that journey. As of this peoblem i came close but i coulsnt find out rhe area pf the segment because i didn't visualise it in that way
Idk why my mind went straight to putting this image on a graph, splitting the curves up into different functions, and finding the area under the curve with integrals and adding them together.
@AjayKumar-mg3xc Well, it's a 12 by 12 grid that you can put in the first quadrant. You can separate and label each line as a quadratic. For example, that curve that goes from the bottom left corner to the upper right corner can be labeled y=(x^2)/12.
I tried to to do this but trust me.. it is NOT easy, you will have to find a lot of intersection points and figure what all areas to subtract. Wouldn't recommend.
tl;dr: You can shift the white area above into four white squares and a white quarter circle, turning this problem into something elementary. It can be done even quicker and in a much simpler way (even simpler than the trick in the other video). Since you have two congruent quarter circles, a lot of symmetry can be used. Start with the area of the biggest quarter circle π*6^2, the white area outside it is not necessary. If you look at the two quarter circles with radius 8 you can actually find two full white squares with them, fit the white area in cell 3 and 6 above the red into the white area above the red in cell 4 and 7. Together with the two cells in the top row you have 4 white squares, so you can subtract 4*4^2 from the total. Now the very middle cell is left. In that cell, if you see that the red part in the bottom right has the same area as the white in the top left (again due to symmetry of the two congruent circles) then it suffices to just subtract the area of the smallest quarter circle for the solution. So π*6^2-8^2-π*2^2= 32π - 64. I think this would be the method with the least amount of calculation.
im a decade past recalling exact formulas for things like area of a quarter circle or circle segment, so the explicit numbers didn't come to me, but i still got some decent problem solving, worked out the clever shortcut you mentioned from the other guys vid on my own (asking myself "why wouldn't you simplify and do it like this" and felt very vindicated when you pointed to that video and the guy presented the same alternate solution) super neat stuff!
I don’t even watch math videos but for some reason this popped up. I watched the whole thing through, super entertaining which I would have never guessed beforehand.
Yeayyy got it right on my first try! Your questions remind me of the math olympiads I took part in when I was in elementary school anywayyy. More challenging questions please, I'm so curious!
Nice! I managed to do it, similar basic idea of subtracting a group of smaller shapes from a square, but didn't choose such clever ones, hence resorted to using an integral.
I am obsessed with this sort of stuff. I knew all of those equations but I just did not see the broken down shapes. This is soooo easy once you break it down
If you cut the area along the diagonal from the top right to the bottom left, and rotate that piece on the left around the center of the large square by 180 degree, you can simply your calculation. The area would be a large circular segment minus a small circular segment, which is 36pi-72-(4pi-8)
A polar planimeter! I have a K&E device - of course I bought mine in 1967 as an engineering student. But, today there are some interesting computer programs to integrate the thing! Love the computer!
I just love when you see a strange shape in nature, abstract as it may be. Encase it in a symmetrical construction, and calculate the difference. Symmetry, what a wonderful word!
The formula I used involved cutting x down the two bottom left corners. Then if I match the r=8 circle edges together, I get that the area x is: ( a quarter-circle r=12 minus a triangle b=12 h=12 ) minus ( a quarter-circle r=4 minus a triangle b=4 h=4 ) Since it’s trivial that the two parts are similar, I can just simplify it to (3×3 - 1×1) = 8x the hole. 8 ( a quarter-circle r=4 minus a triangle b=4 h=4 ) = 8 [ (π×4×4/4) - (4×4/2) ] = 8 [ (4π) - (8) ] = 32π - 64 (units²)
Cool solution, but if you split the shape down the diagonal, you can solve it much easier, because then you can subtract whole quarter circles (plus a small rectangle+triangle shape) from the two larger quarter circles that make up to two arcs we see in the shaded area.
It gets even better when you see that the two intermediate quarter circles have the same area, so in the end you just need to subtract the tiniest segment of a circle from the biggest one
@LukesVGArea it gets even even better if you print out the problem, cut out the area you're trying to find and then weigh the section and compare it to the total weight of the entire square. You have now calculated area as a measure of weight and then take that ratio and match it against the total area of the shape and itl give you your total area
The assumption that the two areas are segments of a circle could be wrong. He's assuming by inspection that the curves have an eccentricity of zero and are, hence, part of the circumference of circles, but this might not be the case. Other types of curves can also connect the two endpoints. So, interesting solution but based on what could be two faulty assumptions.
This is a crazy problem to solve. I thought by clicking on the video I was going to see this being solved with calculus. However, your problem solving solving method stunned me as you were able to make complete sense out such a odd (and complex) question. This is definitely one of the coolest videos I’ve seen lately.
@@TheSpacePlaceYT That kind of information always has to be given in these kinds of problems. The hardest part of the problem shouldn't be you sitting there wondering if a shape is actually what it looks like. So if you find yourself having to think about that, always go back and check to see if it was already clarified.
I dont know why im addicted to watching these. Like, I know the possible ways of figuring it out, but I dont know any of the formulas. Its like watching a friend play a PlayStation game and you know what to do, but you dont know how to handle the controls hahaha
I got the same result but with a different way of calculating. I calculated the area of the big quarter circle and subtracted the triangle part and the round part of the smaller quarter circle and so on.
I went by a different route (also I used X for the length instead of 4). I cut the red shape along the diagonal, which meant: Large Shape area = Quarter circle of 3X radius - Quarter circle of 2X radius - 2 squares of length X - 1/2 squares of length X = (Pi*(3X)^2)/4 - (Pi*(2X)^2)/4 - 2(X^2) - (X^2)/2 Small Shape area = Quarter circle of 2X radius - Quarter circle of X radius - square of length X - 1/2 squares of length X = (Pi*(2X)^2)/4 - (Pi*X^2)/4 - (X^2) - (X^2)/2 Add the two together: Total_area = (Pi*(3X)^2)/4 - (Pi*(2X)^2)/4 - 2(X^2) - (X^2)/2 + (Pi*(2X)^2)/4 - (Pi*X^2)/4 - (X^2) - (X^2)/2 Multiply everything by 4 to get rid of the divisors: 4 * Total_area = Pi*(3X)^2 - Pi*(2X)^2 - 8(X^2) - 2(X^2) + Pi*(2X)^2 - Pi(X^2) - 4(X^2) - 2(X^2) Open the squared parentheses 4 * Total_area = 9Pi(X^2) - 4Pi(X^2) - 8(X^2) - 2(X^2) + 4Pi(X^2) - Pi(X^2) - 4(X^2) - 2(X^2) Add up 4 * Total_area = 8Pi(X^2) - 16(X^2) 4 * Total_area = (8Pi-16) X^2 sq. units Divide both sides by 4 Total_area = (2Pi-4) X^2 sq. units Same solution, just plug in whatever value you want for X.
I agree, having to calculate only 4 areas (of the same shape) is much faster (and better) solution. Saving x as symbol to plug it in later is cherry on top
I estimated that there is about 2.25 equivalent of red squares total. Multiplied that by 16 and got pretty close (36). Good way to get a quick idea of what the answer should be close to on a test (or at least eliminate obviously wrong answers).
I did this a little differently, took the upside down quarters and subtracted their area which left me with the larger part of the claw and the first column with a partial left bottom corner. Similarly i took the smaller claw from the upright quarter circles and got the smaller claw and part of the first and complete second square from bottom row. Now i subtracted the whole boxes which were not part of the claws and the bottom left corner once to get the desired area
And here i am, randomly getting this video on my feed, looking more dam stupid. Even though i used to go to maths faculty in high school, i feel i know nothing but the basics.. also never have thought maths could be this interesting. great channel fam, best wishes to you!
Hey, good solution, I think I got to a similar solution by applying "boolean" additions and substractions. I guess it is just a different solution but it involves, i think, less calculations. I don't know how to explain in text but if we name vertices of the matrix as A,B,C,D vertically and 1,2,3,4 horizontally (so bottom left is A1, bottom right is D1, top right is D4). Also when I refer to quarter circles i will indicate as follows (End#1, Centre, End#2), While Squares or rectangles will be notated diagonally like this [Vertex 1, Vertex 2 (oposing diagonal)] then it would work as follows: (A1, A4, D4) + (A1, A3, C3) - (B2, B4, D4) - (B2, C2, C3) - [A4, B1] - [B1, C2] Which translates into, generalizing for any distance between vertices d: (9𝜋d^2)/4 + 𝜋d^2 - 𝜋d^2 - (𝜋d^2)/4 - 3d^2 - d^2 = 2𝜋d^2 - 4d^2 = (2𝜋-4) d^2 And the result is exactly the same. Fascinating how one can come up with different solutions to the same problem.
I thought it was gonna need some advanced formulas that I don't know, but everything used here was things I learned in 5th or 6th grade. Just used very cleverly. Cool.
There's a most elegant solution of only sums and subtractions of segments of quarter circle if you look at this problem diagonally (literally) Hence only one area formula is necessary, applied to 3 different radii
An even faster way to solve this, is to rearrange the picture by nestling the little red horn into the curve of the bigger red horn. that way you notice that the shape is actually one big segment with r =3*4=12 minus one small segment with r=4. this gives us a pretty short equation of x=1/4*π(12)^2-1/2*(12)^2-(1/4*π4^2-1/2*4^2)=144π/4-72-16π/4+8=128π/4-64=32π-64 which is the result you ended up with. When there are multiple ways to cut up a shape like this the first step should be to find the best way to describe your area with the least amount of composited shapes in order to avoid doing as much work as possible.
I find that for shapes like this, it's usually just easier to write the curves as semicricle equations (y = sqrt(radius^2 - x^2)) and then use integrals to find the volumes of solids. however, the way that you did it is super cool!
1:11 I think these are called spandrels. At least that's what we called them when dealing with the centroid and the rational inertia about these shapes.
You can rotate the piece above the (bottom-left to top-right) diagonal line around the center of the image by 180 degrees. Then the problem becomes very simple.
Theres another way of solving this. You can place another 4*4 in the middle making total area 160. You can then take out the four round areas without using any diagonal sections. The areas you would eliminate would be taken out twice in the middle hence the extra 4*4 only leaving you the required area of shape. Maths works out the same way but it skips the step of requiring the additional area divisions.
