How To Solve This Viral Math Problem From China
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- čas přidán 15. 06. 2024
- This is harder than it looks!
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There's another, possibly much easier, way to solve this problem! I am impressed many people emailed me this: czcams.com/video/_2lzv41RksA/video.html
Yes, and there are another ways to solve this too. Don't even using sin cos or any tan.
my way can go without integral.
it's only 4 short steps
Which is larger 2^100! or (2^100)! ?
czcams.com/video/OGkUIiVIXMw/video.html
Of course. :)
WAOT A MINUTE PRESH THERES ANOTHER FASTER WAY YOU DIDNT SHOW..TAKE THE AREA OF THE SEMICIRCLE AND JJST SUBTRACT AREA OF THE OTHER CIRCULAR ARC 1/2 TIMES R SWUARED TIMES THETA WITH THETA BEING THE INVERSE TANGENT OF 1/2.. AND YOU GET THE PURPLE AREA MUCH QUICKER..DIDNT ANYONE ELSE DO IT THIS WAY??
The first half was ok, until the purple part came
Yeah, same😂 stopped watching after that because it was more complicated than I thought...😂
Thanos of this math
thanos really fcked u up
As you can see the isosceles triangle, just draw its height lenght which it would divide isosceles triangle into two right triangle and do a simmiliar triangle,
Because as you can see, the right triangle(with hypotnuse 4) and the bigger right triangle(with based 8 and height 4) both are simmiliar cuz the angles are same
It hit me right before bed and I started writing it down like a mad man. It's actually pretty simple.
China is the only country where this could go viral
@@mailasun really china is overall good at mathematics
US kids are smoking pot. So math is too hard for them
Asian math skill is op
Lmao
Try Indian's in mathematics and JEE advanced second paper than will seeeeee
I mean, I understand it
I just can't pull these equations out of my head
ur pfp XD
Then you don't understand it
@@officersmiles9114 I said I understand what he needs to subtract from the original areas to find the shades area.
I'll never be able to remember those equations for those particular shapes.
math is pure logic ... end of story
@@bobbiusshadow6985 thanks for the fuckin irrelevant comment, pythagoras
why the hell is this in my recommendation, im not even smart
@Devesh Gupta what is brain?
Thats the point
it is there to make you smart I guess
@@realiti773 Had the exact same thought
I’m*
This is a math problem from middle school in china, using integration is not allowed, so the first way is the best way for a student in middle school
Y.Z they are Chinese though...
Middle school 😂 I have a bachelors degree working on masters and I could never do this 😂
I think I did a similar question when I was in grade 5-6 ish. Lol
Nope. In HK, integration can be used in middle school. But Integration is taught in extended math modules in high school level
@@jerryqi2494 dont make make stuff up man. there is no way u learned this in America until high school. i just graduated high school and this kind of problem is at least math2level.
Looked away for one second and got lost
🤣🤣🤣
Trust me I watched the whole thing
The time he messes up with the sin and the tan I completely get lost
🤣🤣🤣
Expecting simple algebra. Got a warp theory equation
It's geometry
exactly i saw the thumbnail and thought this is really easy...then i saw the video😑 really different way of explaining
@@nimverxza2485 Usually, stuff simplifies (Algebra wise) in geometry . You rarely found problems with a tan inverse that does not go away like this.
And he didn´t use coordinate systems. Once you learn them, finding areas and volumes will never be the same
@@xtheslipknotmaggotx Then it would cease to be simple euclidean geometry and be straight out real analysis (yikes).
Studied mathematical equations
Solved complex problems, used calculus during my course in Mechanical engineering.
Only to find myself not being able to use that knowledge,by trying to answer my wifes question.....
Who's Trixy?
why is trixy?
I used integration to solve it. Took a bit for me to think about it
@骑士黑暗 What?
@骑士黑暗 🖕
@骑士黑暗 mama mo go to jail hahahah xd
And here I was hoping it would simplify at some point.
Yeah, problems like these are just tedious. Obvious brute force solutions leading to ugly answers. Meh.
It wasn't that complicated, for me atleast
Butt Sez you are very smart
@@stephenjiang8099 i am high school pass out , maybe that's why
Butt Sez
That is cheating.
I'm 16 and I'm a math passionate since an young age and I've been the best in maths at my grade and I thought I was really good at it, but then I see you and I noticed that what I really wanted is to have this problem solving hability/creativity, I just want to be that good in math outside the school and solve problems that easily. Respect to you Presh, you are really an inspiration.
EXACTLY BRO! People say I am good at math, but I can't even participate in a maths competition because of my inability to solve such problems.
Same. People say I am so smart and good at math a d I am just thinking to myself, I can't seem to display sharp problem solving skills though. Bery frustrating.
same, brother, math is such a beauty
Don't get discouraged,the answers don't come right away,but they eventually will...oh,and the satisfaction when they come!
Yeah, The first step is to keep trying and then gain an intuttion for math problems.
1:00 ok, easy peasy
1:30 hey, i think i can actually do this one!
2:30 ...ok , little bit of a challenge
3:30 wha-
4:00 what the cluck is going on?!?
5:30 stahp it! Enough!
6:30 waaaaaaaaaah!
don't worry, calculus isn't hard, it's not high level math. I know your comment is supposed to be a joke but I just wanted to say that if you start learning something with the idea that it is hard, it will be actually be harder for you to understand. So, be optimistic, you're too smart for this!
@@beautifullife1259, i know exactly what you mean. I'm actually quite far in algebra (not very, but quite), but i don't feel the need to learn about calculus. But i don't doubt that i could learn it if i tried. The only thing i always fail to get into my head is reading sheet music. Now, in any ordinary case, this wouldn't appear to be considered as an issue, but in my particular situation, it can be a rather mentionable obstacle, becuse i'm a hobby pianist. But i just can't learn sheet music, no matter how long i try. So i said "cluck it", and i'm using synthesia now.
@@IssaMovie you SHOULD be able to understand it at 16
@@justaregulartoaster I play the piano too. I'm gonna be annoying here(because you already know what I'm gonna say) and say that it's actually really beneficial for you to learn sheet music in the long run. Music theory is helpful for musicians, and everything becomes clearer with sheet music.
I mean how can you blame me
czcams.com/video/-3WuQxnA7Hg/video.html
China : Releases a viral math problem
China after a year : Let's make something even more viral
And they took it literally
i.e. Corona 😭
USA ： hold my Fort Detrick
Dark
This is chinas equivalent to America’s viral pemdas questions except this a bit harder than some subtraction
Pemdas is much harder
I think you mean BODMAS
I'm confuse on what the video just taught me
I solved it using calculus, turns out you did the same,
Also you can find the Green Area by Subtracting the area of semi circle from that of whole rectangle then simply divide the whole by 2.
Good question BTW.
I used y=mx+c and trig. I hate integration lol
Hasibun Nisha wait- huh?
Integration can give you the area under the curve shape in almost one step as the equation, if the origin is set at the bottom edge of the rectangle halfway between the corners you would get (x^2)/4
This is a question from primary school graduation and junior middle school selection exam. calculus is not taught in primary school yet.
5:05 tbh this is what I immediately came to, because let's face it, questions about "calculating areas between curves, lines or otherwise graphs" just beg to be solved by calculus.
Solutions by geometry are often more elegant tho.
What an incredibly unsatisfying solution.
Marvin Candal Why ? it’s not like it ended up being some long expression that couldn’t be simplified down. This is as satisfying as it gets.
@@MrTaylork1 rounding it off at 3 decimal places is not simplifying. Its unsatisfying because there was no concise way to express the exact answer like pi or e or something.
@@chronyx685 well you sure can express it in terms of inverse trig functions; that's a valid form of answer.
An integral would have been way easier
Sure?
The integral was much easier tbh when shown in the video
He showed it after. You obviously didn't watch the whole video.
Forget doing stuff with angles and just set it up as the double integral.
Int[0,0.8](int[2y,-sqrt(16-(y-4)^2)+4](1dxdy))
@this that I didn't. I don't do any dividing at all.
I did it another way, similar to the first method Presh described but I think easier to follow:
Area of the rectangle = 8 x 4 = 32
Area of the semi-circle = pi x radius squared divided by 2 = 8 pi
Area of rectangle - semi-circle = 32 - 8 pi = area of both shapes between the rectangle and the semi-circle.
Two of these shapes (one on each side), so each side shape = 16 - 4 pi (As Presh found in his solution as well @2:05).
Split rectangle in half vertically, forming two squares of side length 4, and so area of each square = 16
Right easy so far, the next part is still easy but it is easier to draw on your own diagram:
Each of these squares, when looked at with the original diagonal line included, are formed from a rhombus stacked on top of a triangle.
This triangle has an area of 1/2 base x height = 4/2 x 2 = 4.
Therefore, the rhombus has an area of 12 (as each square's area is 16). This can be backed up by the area of a rhombus being 1/2 the sum of the parallel sides x the perpendicular distance between the sides = (4 + 2)/2 x 4 = 12.
Would really recommend drawing this bit, much easier to visualise:
Looking at the left square we made only, this rhombus area is made up of a circular segment from the original semi-circle from the the top left corner of the square to where the diagonal line cuts the circle edge, we'll call this area A.
As well as this, it is made up of a triangle with the three points being the top-centre of the rectangle (the top right corner of the square), the centre point of the rectangle (the centre of the right edge of the square) and the point where the diagonal cuts the circle edge. i.e it shares an edge with the circular segment. We will call this area B.
The final part of the rhombus is made up from the top part of the side shape with the area of 16 - 4 pi. We will call this area C.
Basically if we can find the area of C, we can find the area of the shaded part we need to find!
Now you really need to draw to visualise:
Give each of these points on the rhombus a letter. The top left point will be D, the top right will be E, the bottom right F and the bottom left G. Give the point where the diagonal line cuts the circle edge K. If you're following along correctly, as you go clockwise around the rhombus, the order of points should be DEFKG. Hope this makes some sort of sense!
Now we need some angles. Similar to Presh calculating theta @3:57, we need the angle of the bottom left corner, DGF. This is calculated from trigonometry with tan = opposite/adjacent. Therefore, tan (DGF) = 8/4 = 2, and so the angle DGF is = tan^-1 (2) which is approximately 63.4 degrees.
Then this angle forms a 'C - angle' with angle EFG, meaning combined they add up to 180 degrees, meaning angle EFG is approximately 116.6 degrees.
To find the angle FKD, we do some more trig, since we know that the length of EF is 2 (half the rectangle width) and the length KE is 4 (as it is a radius of the original semi-circle).
We can use the sine rule (can't do simple trig as the triangle is not right-angled) which is when sin(a)/A = sin(b)/B (google it) to find the angle FKD.
Therefore, sin(EFK)/4 = sin(FKD)/2 ===> 2 x sin(EFK)/4 = sin(FKD) ===> sin(FKD) = 2 x sin(116.6)/4 ===> sin(FKD) = ~0.447 ===> Angle FKD = sin^-1(0.447) which is approximately 26.6 degrees (coincidentally this is identical to theta, the angle that Presh calculates in the video).
Now, we can calculate the angle KEF, which is easy as it is the final angle of the triangle EFK, and so is 180 - the sum of the other two angles.
Therefore, angle KEF = 180 - (26.6 + 116.6) ===> 180 - 143.2 = approximately 36.8 degrees.
The final angle we need before we can put this all together is the angle DEK, and this is simply 90 - angle KEF.
Therefore, angle DEK = 90 - 36.8 = approximately 53.2 degrees.
Now we can find the area of A and B.
Area A (the circular segment):
We know the angle DEK, and since this is from the centre of the semi-circle, and is bound by the circle edge, we can find the exact area of this shape.
The total semi-circle area is 8 pi, and this corresponds to 180 degrees. Therefore, 1 degree = 0.0444 pi. So, 53.2 degrees = ~2.36 pi. This is the area of the circular segment, A.
Area B (the triangle EFK):
Normally the area of a triangle is half of the base x the height. However, in this case this is not obvious so we can use the alternative formula for the area of a triangle, 1/2 a x b x sin C (please google if you don't know this formula).
Therefore, the area of the triangle EFK is 1/2 x 2 x 4 x sin (36.8) = 2.4 (exactly).
Now we know the area of A and B, we can find the area of the unknown part of the side shape, area C from before. We can do this as the sum of all the areas, A, B and C, in this rhombus must be equal to 12, as we calculated before.
Therefore, A + B + C = 12 ===> 2.36 pi + 2.4 + C = 12 ===> 9.81 + C = 12 ===> C = ~2.182
Finally, we can use this value to find the answer to the problem.
We know the area of each side shape is 16 - 4 pi, therefore 2.182 + the answer = 16 - 4 pi.
Therefore 2.182 + ANSWER = ~3.434.
And so the answer is = ~1.252, just as Presh found @7:04.
Now if you draw this out, it is quite a nice solution which is different to Presh's method and in my opinion is easier to follow as it uses middle-school maths which all makes sense in each step.
However, reading over this solution I wrote it actually sounds just as complicated! I really would recommend drawing it if you actually want to understand what I'm saying.
Either way hope this alternative was interesting to at least a couple of you who made it this far - I'll make a video of it on my channel if the demand is there.
I really need to find something better to do with my time...
lmao dude 😂
STOP IT
Even my thesis was less complicated than this.
AEA
i just read your first sentence. and i'm done.
For the purple area, I imagined a scaled down version of the isosceles triangle inside the unit circle, fit perfectly so that it's circular sector is part of the unit circle. For there I took the arctangent of 0.5 for the left facing angle on the border, and used the inscribed angle theorem to find the center's version of that same angle, which I then subtracted from angle π to find the angle of the sector, which I then plugged into the isosceles triangle, where I divided the isosceles triangle into 2 right triangle, found the sine and cosine of half the angle, times 4 (because that is what the radius is supposed to be), divided by 2 because it's a triangle and multiplied by 2 because there are 2 of them(ultimately changing nothing), to find the area of the isosceles triangle, which I then subtracted from the sector of the circle(16π * angle/(2π)), plus the spike(4(4-π)), subtracted from the overall triangle in the original shape(8*4/2), to get the same 1.25199... answer.
This method requires more steps, but because it's much more visual and understandable, mainly people are pretty comfortable with the unit circle anyways.
Yes. You’ve done this before didn’t you?
Framed slightly differently but yes. It includes the same solution
(spoiler) czcams.com/video/xnE_sO7PbBs/video.html
Just goes to show we do learn something watching these videos :-)
I'm sure he did because I remember solving it and discussing it with a friend at highschool like three years ago, perhaps he shoves it differently this time idk
I'd take the integrals and subtract
He probably wanted to show us the solution using integrals.
This is the first problem I attempted after my break of 1.5 years , I m glad I solved it and grateful to you for this video! Keep up the good work!
simple,but the more important thing is how many ways we have to solve it
and one of the way, is by asking our Chinese friend
You had to ve Chinese
I thinked that too
@Devesh Gupta the solution is only one, but the number of paths are infinite.
I used a graph😀😀
You failed the test!
Correct answer:6.4 - 8 sin^-1(0.6)
Your answer: 12
@traversing cloud r/swoosh
traversing cloud it’s not an accomplishment lol.
More like
your answer: WW2
This is an absolutely correct incredibly beautiful but actually confusing question
As expected from a Indian.
@@Zhov96 actually i think indians are quit smart in mathematics i swear dude this problem is not even a bit closer to the toughness of maths questions asked in JEE advanced and later GATE exams. I once looked at one of them and i was like heck how can an undergrad student think of solving them and i was like yo screw this man i am lucky that i have'nt born there
@@sumanrai8921 I think you misinterpreted what I said :D ai totally agree with you cause I have Indian friends who are really good in math.
Hey! There was a question similar already on this channel! Literally the only differences are the values and that it's only part of the original problem.
I knew it
Yep, that's why I remembered the geometrical solution to this. Calculus still jumped to me first though.
I thought that to
Even easier by integration on y-axis:
Integral (from 4/5 to 0) {4 - sqrt(16 - (y-4)^2) - 2y} dy = 6.4 + 8* arcsin(4/5) - 4*pi ~= 1.252....
You are right. Well, I like pure geometry solutions. Please see mine:
czcams.com/video/vXG-XSquhXA/video.html
It would take me much longer to think of the first solution than just solve it with integrals, that was the first solution that came to my mind and I'm somehow impressed you thought of the first solution in the video. Shows how at first, calculus is intimidating, but once you get the hang of it, many otherwise complex problems that require creative thinking can be done in a very straight-forward matter.
Please see my solution: czcams.com/video/vXG-XSquhXA/video.html
I lost you at the part where rectangle is 8×4
XD
Using Calculus:
Area = 32/5 - 8*arcsin(3/5) units^2 ~= 1.252 units^2
Heart filled with joy . Very good explanation . Keep it up!
Using an integral is the right idea. But you can make your integral easier by using radial coordinates and doing a double integral over the angle and the radius. For the limits of integration for the angle, you can use two constants: zero and inverse tan half. For the limits of integration on the radius, you use the trick of going from zero to the radius expressed in term of the angle (and think about coming up for a formula for the circle in terms of the angle). While this sounds like a more complicated approach, this approach usually simplifies more easily.
"Solve for the shaded area using ALL of highschool math"
Yeah I think there's a video on bilibili that solves a similar problem using pure middle school mathematics
i solved using 3rd prep. math in egypt :D
Idk why I can't see euclidean theory in his vids, its wonderful
@@joshuamason2227 it's not possible coz value of theta is unknown!
assuming 45° or 30° is completely wrong, it's nearly 23 or 25 something (you could check by drawing it on paper)
@@webflyer035 yeah bruh but trigonometry is covered in middle school maths
WebFlyer0 don’t you use soh cah toa to find theta
draw it in a CAD program, hatch the area, look at the properties -> area, ;)
good idea
Hey I did that to pass calculus lmfao
Crying rn cuz that’s exactly what I did LMAO
that s the easy way to solve, even no need to hatch...
You're a wizard harry!
Congratulations!! I watch you from Fortaleza, Brazil. You help us very much!!! Please, go on!
I like you channel, had a lot of fun doing all kinds of math problems these days. just one VERY important thing for the future: explain which tools we can use in the beginning. e.g. this one here definitely needed a calculator.
Every body : easy
Me : crying on the corner "oh God this is so hard*
2:28 From here things are started going above my head
I was going to comment so !
Same ugh
The moment you completed grade 9 in China:)
The way In which this problem is solved is far more complicated than need be but I still loved the video! Good job, your making mathematics more enjoyable.
For physicists, they would assume the area of the purple part to be half of the semicircle. And the whole question is hence simplified. But in the end an adjusting factor n is needed for the answer.
Is that even physically possible I mean the radius of that "semi circle" is a constant already. And so is the length of the arc. the length of the arc wouldn't match up with the radius because the length of the arc is related to the radius in the equation theta(r) but the actual arc of the purple sector is longer so in the end. Wont it contradict itself? And that being said you would also need to find the length of the radius to find the area of that purple sector. Which is a lot more steps than the one he showed, because that's a chord and you don't know how long the chord is until you mess around with the angles
@@cockernadan5283 He is trying to mock the physics world.
not physicists hun, engineers
I'm going to go ahead and outsource the work for this one
How I wish that you have trigonometric identity/substitution tutorials
a^2-x^2 similar to 1-sin^2 x = cos^2 x
a^2+x^2 similar to 1+ tan^2 x =sec^2 x
x^2-a^2 similar to sec^2 x -1 =tan^2 x
just compare the constant and varieable parts for both 3 identities and you can easily figure it out
not to mention you should use tri sub for sqrt case simply beacuse only constant terms and power of x term is easily calculation by those law
I was already mindblown when you cut the rectangle in half. All that inverse tan thing practically killed me.
Agradezco que haya canales como este que en realidad sirven a los niños, jóvenes y adultos
Excelente contenido
It hurts.. STOP HURTING MY LITTLE BRAIN
Tmjon great u still have brain
*my* *brain.exe* *not* *responding*
The circle sector you chose was very clever because the sector I chose when solving by myself was only through the curve of the red part, so it took me longer but I still got the same answer.
For some reason, I find your explanation kind of hard and confusing lol. Even though I solved it myself, I caught myself becoming lost while your trigonometric explanation. I don't know why lol
I was trying to sleep when this pops up on my recommendation. What a great find.
Flip the axys so that x is y and viceversa. Double integral with polar coordinates. This way is a simple double integration, you can do it under 2 min.
Luis Breva Can you please explain it in detail?
I've solved it in another way
I am not English speaker and I am study at the middle school.
I don't understand anything, but it seems interesting.
Solved it myself, and your explanation was over the top, but we got the same answer so 👌.
0:28 intersect line from center with diagonal and calculate the red triangle.
From the coordinates of the intersection calculate the angle of circular part
whole circle=r*r*pi
part of circle=r*r*w/2-r*cos(w/2)*r*sin(w/2)
substract the sector from 2nd triangle
A=triangle1+triangle2-(r*r*w/2-r*cos(w/2)*r*sin(w/2)
install bbc basic and hit f9(run)!
10 print "area of part of rectangle mit abweichung von einer geraden"
20 xm=4:ym=4:xn=0:yn=0:xg2=8:yg2=4:rk=4:sw=.1
30 x1=xn:y1=yn:x2=8:y2=4
40 xu=4:yu=0:nf=sqr((xm-xn)^2+(ym-yn)^2):goto 90
50 dx=rk*cos(w):dy=rk*sin(w):rem abweichung von einer geraden
60 yi=ym-dy:xi=xm-dx
70 df1=(y1-yi)*(x2-x1):df2=(y2-y1)*(xi-x1):df=(df1+df2)/nf
80 return
90 w=0:gosub 50
100 dfs=df:w1=w:w=w+sw:if w>2*pi then stop
110 w2=w:gosub 50:if dfs*df>0 then 100
120 w=(w1+w2)/2:gosub 50:if dfs*df>0 then w1=w else w2=w
130 if abs(df)>1E-10 then 120
140 sc=sqr((xu-xi)^2+(yu-yi)^2):wc=2*asn(sc/2/rk)
150 at1=(xu-xn)*(yi-yu)/2:at2=(xu-xi)*(yi-yu)/2
160 asector=wc/2*rk^2-rk^2*cos(wc/2)*sin(wc/2)
170 ages=at1+at2-asector
180 print ages: ar=(x2-xn)*(y2-yn)
190 print "Das sind";ages/ar*100;"% der gesamtfläche"
it's about 6.9% of total area
Basic School Math is almost impossible to pass in China.Calculus already started in Grade 2 !
Calculus is not that hard
CZcams ask me to study again....
Me: No Korean beauty teacher, bad class!!!!!
whachusay u should learn Chinese first
Jumbomuffin13 whatever u said I don’t care English, either... u don’t care Chinese because u r not the man of far-sight.😂
@whachusay ......All u learn is a dirty word, ”smart“ guy😂
Hi Sir. Thank you very much for giving mind blowing questions.
You already have a definite triangle on one side, as for the other just deduct by height and length plus angle. or you can calculate by dimensional capabilities, directly by the angle for the first half. and plus pie square roots for the second.
Please see my solution: czcams.com/video/vXG-XSquhXA/video.html
FYI, this question comes from a primary school in China and that's the reason why it becomes famous.
Bloody china
Damn chinese
I think this question should be at least for middle school students. The regular primary school in China won't teach sin/cos. Unless it is for math competition.
@@wkfYT i agree. Where I first see this question is in a topic "is it possible that a primary student can solve this question?" Many people just think this is a very easy question and answer "yes". However, this is a difficult question in fact.
I got sin cos tan in high school.
6:49 - "16×(theta/2 + (2 sin theta cos theta)/4 *)* from arcsin(-0.6) to 0" - Missing right parenthesis between right end of expression and 'interval' vertical bar. Why?
Really nice video. Thanks!
I used pretty much the same method as in this video, although I split the shapes up slightly differently.
Roses are red
Violates are blue
There's always one Asian
Who do better than you
*I am already lost at triangle...*
Thank you sir. What a brilliant solution 👍👍👍
I did it a bit differently. I assumed the origin was at the middle of the top of the rectangle and used algebra and the quadratic formula to find the intersection between the circle and the line. -2.4, -3.2 then used arcsin to find the angle 36.87 of the sector that sweeps from the bottom middle to the line-circle intersection. Once you know that angle you know that it has 36.87/360 times the area of the whole circle. From there it's just some simple arithmetic dealing with the area of rectangles and right triangles of known dimensions. It was pretty satisfying when I saw I got the right answer.
It was okay not until the solution of thag violet semicircle's (not actually a semicircle showed
5:30 - how do you know that x = 1.6?
respectively, you get its coordinates when you cross the two functions.
How do you get the function of the circle?
I'm having the same doubt
I remember there was a similar multiple choice question in middle school public examination. Tutor taught us how to guess the correct answer within one minute
Note that the area of the four sided polygon made from the centre of the semi-circle and the vertices of the region in red is the same as the light blue triangle in Presh’s answer. Calculate the area of the light blue triangle and subtract the area of the circle sector made from the two vertices of the region shaded in red.
I think in the next patch they’re gonna nerf Asian math skills because it’s too OP
They need to buff our social skills though, because they are UP.
lol, does anybody here plays world of warships?
I solved it another way: if we'll imagine that the bottom left corner of a rectangle is a dot (0; 0), we'll get the two functions: y = x/2; y = 4 - sqrt(16 - (x-4)^2);
So the square of a red figure will be an integral [1.6 is a solution for a system of equations above]: (from 0 to 1.6)int: x/2 dx + (from 1.6 to 4)int:4-sqrt(16-(x-4)^2) dx; so the answer is the same.
how did he determine that section is at 1.6 and 0.8?
That's exactly what he did duh.
@@korencek solving this y = x/2; y = 4 - sqrt(16 - (x-4)^2)
@@ddm1912 sry, I guess I've watched only the first half =|
@@dimakoss5142 and how do you get y = 4 - sqrt(16 - (x-4)^2) ?
Decompose the large triangle in the upper left corner into an irregular area, a sector and an isosceles triangle. The isosceles triangle can be decomposed into two congruent right triangles, and the degree of one angle of the right triangle is obtained by the value of sin (4/4√5) (arcsin 4/4√5 = sin⁻¹ 4/4√5 = Aº ), then you can calculate the area of this large isosceles triangle ((4√5/5)/tanA*4√5/5*1/2*2=B) and the degree of the sector (180-2*(90-A)=Cº), calculate the sector area (C/180*4*4*3.14*1/2=D). Furthermore, by subtracting the two parts from the area of the upper left corner of the triangle, the area of the irregular area is obtained (4*8*1/2-B-D=E), the semicircle is subtracted from the rectangle, and the area of the irregular area is subtracted by two. (4*8-4*4*3.14*1/2-E=answer).
I am confusion
I studied all this stuff back in school. Now I’m running a company and thankfully, a phone basic calculator is enough to solve our daily work. I can’t even remember now how to use the sin cos tan in scientific calc.
Very quick solve with an Integral and using the functions :
x^2 + y^2 =16
y = 1/2(x)
a) your semi circle function is incorrect, it's not even a semi circle and the center is not 0,0
b) it's not as easy as you think to solve with integration, give it an actual shot
@@petrosathanasiou1480 it is quite easy if you know how to integrate
Ok this is not for me.
I'll go back watching Thom Yorke dancing.
To simplify one can remember that area is homogenious of second degree. so instead of (4,8) one can take (1,2) and than do at the end a Dilatation with factor 4.
Hi Presh,
1.252 . No need for angles and subtracting areas. First find the coordinates for the intersection of circle and diagonal in the left half of the rectangle, here (1.6|0.8), with respect to origin at the bottom left corner. That also gives me the little triangle forming part of the solution, which has the area of 0.64. Add the Integral of the little "ramp" next to it, being:
4 - Sqrt(4^2-x^2), latter resembling the area under the circle, while using boundaries 0 and 2.4 for the Integral (as if you were integrating the mirror image in the 1st quardrant from left to right. Gives me an area of 0.612. Adding both values yields the solution: 1.252
Got same result with another formula
area = 4*pi + 6.4 - 16*arcsin((4/5)^0.5)
eh bro i used tht too
I got the triangle part
Me too bruh , stuck after that
This can also be done with the Al-kashi formula to find an angle and determine the portion of the cercle area, it's a bit more straightforward. Red area is 1.25199... with this method, so it's just as good.
Calculate area under the diagonal till x=4 then subtract the area intersected by the semicircle with the triangle till 4. Which can be calculated by area under the diagonal - area under the semicircle with limits from the point of intersection of diagonal and semicircle to 4. And you get your answer, simple.
The true fact is in most cases, this question is for primary school or middle school students, so calculus is banned here. There are other ways to do it.
I should not say! but u made the solution more complicated.
I think it should be quite easy with integration .
It can be solved by simple geometry by drawing perpendicular bisectors
absolutely too complicated, just use the upper left 4 ; 8 triangle, 1/4 circle, substract a bit and you are there.
@John Wick he's also Indian...his last name is talwalkar
That's quite simple as well 🤔 although he didn't use predefined formula as we all did.
Excelente vídeo, socio. De manera recreativa resolví el problema con geometría analítica y puede resultar interesante para otro vídeo en el futuro.
You can also solve the green area by subtracting the full circle from the full square and dividing by 4.
Dude this is like 7th grade math in China...
I’m really glad I’m not in China right now just saying 🙂
can you explain? why not 8 X 8 square with inscribed circle 8 in. diameter?
Problem will be simple u don't need trigo : area (square - circle ) / 8 = 1.72
8 is due to the symmetry
The first one is like our normal addmath examination question would pop out
that can also be solved by integration and the limits can be determined by solving the intersections of the two curves
I am a Chinese from Shanghai. Weibo is something like Twitter. It‘s amazing to see this familiar picture again. hahahah. Actually it should be a Math Problem for Chinese student in Secondary School, around 15 years old.
The life of a Chinese student:
-get 99% on a test
-get scolded by daddy
I was surprised yet impressed that I was able to solve this!
Nicely done.
I used the correct steps to solve in the first one but I got my math wrong on paper 😭
*brain.exe has stopped working
Overhaul unfortunately
Pretty cool use of integrals
For second step just use similar triangle to get the height then calculate the angle theta.
got slightly lost at the (1.6,0.8) coordinates can anyone elaborate that for me thanks :)
if you consider bottom-left corner as origin, then equation of the diagonal of rectangle is y=(1/2)x.
and equation of the semi-circle is (x-4)^2 +(y-4)^2=16. Now you have to find the intersection points of both.
Put y=(1/2)x in equation of circle, you will get x=8 or 1.6 and after that you can find their corresponding value of y.
ohh makes total sense!..I got it thanks!..
@@06nishant how about integrating (x-4)^2 +(y-4)^2=16 with respect to x. I think it's not possible. do you break some bodmas rules or what? y = ?? Like how does it end up y = 4 - sqrt(16-(x-4)^2)
@@davemwangi05 It's a simple transformation from the semi circle function around the origin y = -sqrt(r^2 - x^2) (negative because it's the bottom semi circle). The whole thing plus 4 moves it up by 4, and x - 4 moves it to the right by 4.
Why are u guys making it so complicated lol
That point is the point of intersection
How did he get the area of the isosceles triangle?
(product of the two sides times)*sin(angle between them)/2
(same side = 4^2)*sin(π - 2θ)/2
@@023Clifford Okay, so basically a fancy way of doing "base times height"
I got a simpler solution.
Equatoion of circle=x2+y2=R2.
Y axis pass through the center of rectangle and x axis from top of rectangle. Equation of diagonal intersecting the arc is x=2y+4
R=4
Substituting in the equation for x we get y=-3.2 x=-2.4
Hence the diagonal intersects the arc at a point 1.6 along the length of rectangle and vertical distance is 0.8.
The length of chord =8-1.6x2=4.8
The angle between the two radius joining the centre and end of chord =2xsin inverse 2.4/4=145.55 deg
Area of sector =(22/7)x4x4x145.55/360=10.30
Area of triangle =4.8x3.2/2=7.68
Curved area in trapezoid =10.3-7.68=2.62
Area of trapezoid =(8+4.8)/2)x.8=5.12
Area of curved surface in trapezoid =5.12-2.62=2.5
The asked shaded portion =2.5/2=1.25
That is the answer.
No need of integration and higher degree of proof. Please let me know hows that. I am waiting for ur response.
well, or you can put the whole thing into coordinates, and then integrate twice: first we shold get the equalations of the line and the semicircle. thoose are:y= x/2 and (y-4)^2+(x-4)^2=16.
then we are looking for the point where they meet.
then integrate the line from 0 to the meetingpont then integrate the circle from the meetingpoint to 4 ant there you go there is the shape