How To Solve This Viral Math Problem From China

This is harder than it looks!
Many people found a possibly easier way too! Here's a video I made about that method: • Viral Math Problem Fro...
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[@MindYourDecisions]

There's another, possibly much easier, way to solve this problem! I am impressed many people emailed me this: kzbin.info/www/bejne/lWPPq6lqZreeqaM

[@jananonlertvilai9212]

Yes, and there are another ways to solve this too. Don't even using sin cos or any tan.

[@FATFATHER506]

my way can go without integral. it's only 4 short steps

[@PuzzleAdda]

Which is larger 2^100! or (2^100)! ? kzbin.info/www/bejne/hXjOhnyfi66Lg9k

[@kamiladamczyk4710]

Of course. :)

[@leif1075]

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??

[@qzepio5222]

The first half was ok, until the purple part came

[@eddiebert6648]

Yeah, same😂 stopped watching after that because it was more complicated than I thought...😂

[@aakashsahani2991]

Thanos of this math

[@mrpepekroni]

thanos really fcked u up

[@murrayfranklin8390]

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

[@EthnHDmlle]

It hit me right before bed and I started writing it down like a mad man. It's actually pretty simple.

[@Oscar-zp6io]

China is the only country where this could go viral

[@black_jack_meghav]

@@mailasun really china is overall good at mathematics

[@MarcoAshford]

US kids are smoking pot. So math is too hard for them

[@dreammaker3037]

Asian math skill is op

[@josep9016]

Lmao

[@redhair1401]

Try Indian's in mathematics and JEE advanced second paper than will seeeeee

[@Broomie63]

I mean, I understand it I just can't pull these equations out of my head

[@orf2072]

ur pfp XD

[@officersmiles9114]

Then you don't understand it

[@Broomie63]

@@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.

[@bobbiusshadow6985]

math is pure logic ... end of story

[@officersmiles9114]

@@bobbiusshadow6985 thanks for the fuckin irrelevant comment, pythagoras

[@Jvx_M]

why the hell is this in my recommendation, im not even smart

[@Jvx_M]

@Devesh Gupta what is brain?

[@HumbleMicheal以仁慈]

Thats the point

[@antoniusjuan466]

it is there to make you smart I guess

[@ndiyabucaphukelaubusobakho1732]

@@HumbleMicheal以仁慈 Had the exact same thought

[@ouwebok3356]

I’m*

[@yz6445]

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

[@evanning9803]

Y.Z they are Chinese though...

[@85yr]

Middle school 😂 I have a bachelors degree working on masters and I could never do this 😂

[@jerryqi2494]

I think I did a similar question when I was in grade 5-6 ish. Lol

[@allen2759]

Nope. In HK, integration can be used in middle school. But Integration is taught in extended math modules in high school level

[@richarddu3797]

@@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.

[@alexandregoncalves7482]

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.

[@krishnannarayanan8819]

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.

[@aidan-ator7844]

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.

[@tsunakbayev]

same, brother, math is such a beauty

[@cristyoutube212]

Don't get discouraged,the answers don't come right away,but they eventually will...oh,and the satisfaction when they come!

[@RR-bs9mr]

Yeah, The first step is to keep trying and then gain an intuttion for math problems.

[@lincolnross9000]

Expecting simple algebra. Got a warp theory equation

[@nimverxza2485]

It's geometry

[@sanjitharajesh635]

exactly i saw the thumbnail and thought this is really easy...then i saw the video😑 really different way of explaining

[@everlastingideas8625]

@@nimverxza2485 Usually, stuff simplifies (Algebra wise) in geometry . You rarely found problems with a tan inverse that does not go away like this.

[@xtheslipknotmaggotx]

And he didn´t use coordinate systems. Once you learn them, finding areas and volumes will never be the same

[@everlastingideas8625]

@@xtheslipknotmaggotx Then it would cease to be simple euclidean geometry and be straight out real analysis (yikes).

[@justaregulartoaster]

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!

[@beautifullife1259]

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!

[@justaregulartoaster]

@@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.

[@beautifullife1259]

@@IssaMovie you SHOULD be able to understand it at 16

[@beautifullife1259]

@@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.

[@justaregulartoaster]

I mean how can you blame me kzbin.info/www/bejne/Y2S6poSuo6Zqfsk

[@antaresmaelstrom5365]

And here I was hoping it would simplify at some point.

[@akaRicoSanchez]

Yeah, problems like these are just tedious. Obvious brute force solutions leading to ugly answers. Meh.

[@buttsez4419]

It wasn't that complicated, for me atleast

[@stephenjiang8099]

Butt Sez you are very smart

[@buttsez4419]

@@stephenjiang8099 i am high school pass out , maybe that's why

[@heyhey97777]

Butt Sez That is cheating.

[@humpfzzz7962]

China : Releases a viral math problem China after a year : Let's make something even more viral

[@arvindiyer1649]

And they took it literally

[@chitranshbindal1874]

i.e. Corona 😭

[@raymondzhao9557]

USA ： hold my Fort Detrick

[@krish1349]

Dark

[@LucasKingster]

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...

[@franciscacindy9202]

lmao dude 😂

[@stephenlee3911]

STOP IT

[@kashfzahra1897]

Even my thesis was less complicated than this.

[@diegovasquez7610]

AEA

[@afifassihab7953]

i just read your first sentence. and i'm done.

[@sheauiwne5294]

Looked away for one second and got lost

[@AAAAAA-gj2di]

🤣🤣🤣

[@forg7864]

Trust me I watched the whole thing The time he messes up with the sin and the tan I completely get lost

[@ulialiuleo]

🤣🤣🤣

[@chr50n]

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?

[@qwertylink9066]

why is trixy?

[@AnhLe-zl4ws]

I used integration to solve it. Took a bit for me to think about it

[@tyronebiggums2710]

@骑士黑暗 What?

[@somethingalongthelinesof7946]

@骑士黑暗 🖕

[@valdrain6246]

@骑士黑暗 mama mo go to jail hahahah xd

[@hasibunnisha2612]

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.

[@dotdotdot1113]

I used y=mx+c and trig. I hate integration lol

[@yerro504]

Hasibun Nisha wait- huh?

[@LeMuDX]

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

[@briansiev15]

This is a question from primary school graduation and junior middle school selection exam. calculus is not taught in primary school yet.

[@ethanbates7734]

This is chinas equivalent to America’s viral pemdas questions except this a bit harder than some subtraction

[@abhinavpy2748]

Pemdas is much harder

[@joeywild2011]

I think you mean BODMAS

[@ninjapirate123]

I'm confuse on what the video just taught me

[@tosyl_chloride]

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.

[@ARBB1]

Solutions by geometry are often more elegant tho.

[@everlastingideas8625]

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!

[@Nacho_Meter_Stick]

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.

[@yuluoxianjun]

simple,but the more important thing is how many ways we have to solve it

[@rifqiaufan3994]

and one of the way, is by asking our Chinese friend

[@hell2920]

You had to ve Chinese

[@rafagm1384]

I thinked that too

[@Thatguy-mz6uk]

@Devesh Gupta the solution is only one, but the number of paths are infinite.

[@sarthakdas6059]

I used a graph😀😀

[@stevenmellemans7215]

Yes. You’ve done this before didn’t you?

[@peetiegonzalez1845]

Framed slightly differently but yes. It includes the same solution (spoiler) kzbin.info/www/bejne/rp-okKaFbLWVeNU

[@stevenmellemans7215]

Just goes to show we do learn something watching these videos :-)

[@gnikola2013]

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

[@jamesandonian7829]

I'd take the integrals and subtract

[@JakubMikes23]

He probably wanted to show us the solution using integrals.

[@cakeandicecream1582]

What an incredibly unsatisfying solution.

[@MrTaylork1]

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.

[@chronyx685]

@@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.

[@nuklearboysymbiote]

@@chronyx685 well you sure can express it in terms of inverse trig functions; that's a valid form of answer.

[@madhavkarnani7131]

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....

[@spacescopex]

You are right. Well, I like pure geometry solutions. Please see mine: kzbin.info/www/bejne/rImqXouJptqbjqM

[@joeldick6871]

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.

[@hello_there2458]

This is an absolutely correct incredibly beautiful but actually confusing question

[@Zhov96]

As expected from a Indian.

[@sumanrai8921]

@@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

[@Zhov96]

@@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.

[@InDstructR]

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.

[@Artaxerxes.]

I knew it

[@zecuse]

Yep, that's why I remembered the geometrical solution to this. Calculus still jumped to me first though.

[@nahbro7282]

I thought that to

[@thelickpolice1210]

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.

[@spacescopex]

Please see my solution: kzbin.info/www/bejne/rImqXouJptqbjqM

[@artmiraa706]

Every body : easy Me : crying on the corner "oh God this is so hard*

[@RobertLock1978]

Using Calculus: Area = 32/5 - 8*arcsin(3/5) units^2 ~= 1.252 units^2

[@Zir2904]

I lost you at the part where rectangle is 8×4

[@ddm1912]

XD

[@kjhin6431]

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.

[@crw_5319]

You failed the test! Correct answer:6.4 - 8 sin^-1(0.6) Your answer: 12

[@jemlap]

@traversing cloud r/swoosh

[@crw_5319]

traversing cloud it’s not an accomplishment lol.

[@da-lx4gz]

More like your answer: WW2

[@faiz2320]

*my* *brain.exe* *not* *responding*

[@Diegoscomeback]

draw it in a CAD program, hatch the area, look at the properties -> area, ;)

[@dbsse7503]

good idea

[@Montana270]

Hey I did that to pass calculus lmfao

[@ggdatboi]

Crying rn cuz that’s exactly what I did LMAO

[@homer123452]

that s the easy way to solve, even no need to hatch...

[@dewittdaniel87]

You're a wizard harry!

[@satyapalsingh4429]

Heart filled with joy . Very good explanation . Keep it up!

[@andersondavi561]

Congratulations!! I watch you from Fortaleza, Brazil. You help us very much!!! Please, go on!

[@luisbreva6122]

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.

[@krishnachaitanya8353]

Luis Breva Can you please explain it in detail?

[@krishnachaitanya8353]

I've solved it in another way

[@bonkdragon5504]

How I wish that you have trigonometric identity/substitution tutorials

[@samwong9489]

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

[@danielchipper6781]

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.

[@zdrastvutye]

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

[@coolmaster3190]

2:28 From here things are started going above my head

[@aira321tob737]

I was going to comment so !

[@felixdemosquito1126]

Same ugh

[@kagura-423]

The moment you completed grade 9 in China:)

[@hikohiko5377]

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.

[@cockernadan5283]

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

[@KarthikHebbar96]

@@cockernadan5283 He is trying to mock the physics world.

[@user-oi1rx4hh7s]

not physicists hun, engineers

[@angelpaz4720]

Agradezco que haya canales como este que en realidad sirven a los niños, jóvenes y adultos Excelente contenido

[@jamesconduit333]

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.

[@fburton8]

"Solve for the shaded area using ALL of highschool math"

[@joshuamason2227]

Yeah I think there's a video on bilibili that solves a similar problem using pure middle school mathematics

[@user-so3hp4xv5t]

i solved using 3rd prep. math in egypt :D Idk why I can't see euclidean theory in his vids, its wonderful

[@webflyer035]

@@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)

[@joshuamason2227]

@@webflyer035 yeah bruh but trigonometry is covered in middle school maths

[@catherinesc88]

WebFlyer0 don’t you use soh cah toa to find theta

[@rush2225]

KZbin ask me to study again.... Me: No Korean beauty teacher, bad class!!!!!

[@rush2225]

whachusay u should learn Chinese first

[@rush2225]

Jumbomuffin13 whatever u said I don’t care English, either... u don’t care Chinese because u r not the man of far-sight.😂

[@rush2225]

@whachusay ......All u learn is a dirty word, ”smart“ guy😂

[@samer5751]

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

[@PuffleBuns]

I was already mindblown when you cut the rectangle in half. All that inverse tan thing practically killed me.

[@ynahbanguilan5918]

It was okay not until the solution of thag violet semicircle's (not actually a semicircle showed

[@dimakoss5142]

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.

[@korencek]

how did he determine that section is at 1.6 and 0.8?

[@ddm1912]

That's exactly what he did duh.

[@dimakoss5142]

@@korencek solving this y = x/2; y = 4 - sqrt(16 - (x-4)^2)

[@dimakoss5142]

@@ddm1912 sry, I guess I've watched only the first half =|

[@korencek]

@@dimakoss5142 and how do you get y = 4 - sqrt(16 - (x-4)^2) ?

[@TheEternalHermit]

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.

[@TheArtPerspective]

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.

[@spacescopex]

Please see my solution: kzbin.info/www/bejne/rImqXouJptqbjqM

[@forg7864]

It hurts.. STOP HURTING MY LITTLE BRAIN

[@gorgeousg7296]

Tmjon great u still have brain

[@James_Moton]

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?

[@pisces52able]

Area of shaded part = Area of triangle - Area of Circular segment - Area of circular sector = (4x8/2 ) - [4^2/2 (126.88 x 2pi/360 - Sine 126.88 )] - [ 16 - pi/4(16)] = 16 - 11.312 - 3.4336 = 1.2544

[@rainsworth-]

I was trying to sleep when this pops up on my recommendation. What a great find.

[@BL-gt1hn]

I'm going to go ahead and outsource the work for this one

[@canman5060]

Basic School Math is almost impossible to pass in China.Calculus already started in Grade 2 !

[@metakatana]

Calculus is not that hard

[@WoodyC-fv9hz]

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

[@user-bd2wu3wz2t]

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).

[@allytartsnarts9373]

I am confusion

[@stevenbanton5073]

Got same result with another formula area = 4*pi + 6.4 - 16*arcsin((4/5)^0.5)

[@general_prodigy]

eh bro i used tht too

[@Desk45Wiv12Line]

I am not English speaker and I am study at the middle school. I don't understand anything, but it seems interesting.

[@Peter_1986]

I used pretty much the same method as in this video, although I split the shapes up slightly differently.

[@Caturiya]

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.

[@MrExtr1234]

Very quick solve with an Integral and using the functions : x^2 + y^2 =16 y = 1/2(x)

[@petrosathanasiou1480]

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

[@Akash231196]

@@petrosathanasiou1480 it is quite easy if you know how to integrate

[@weitaotang5702]

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.

[@MagicBot.1570]

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.

[@adityanadgir3769]

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.

[@Vitalstatistix]

*I am already lost at triangle...*

[@ghimbos]

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?

[@ADVENTURESofPUP]

I'm having the same doubt

[@MrCostandi]

My procedure before watching: 1. Solve simultaneously between the equations of the diagonal and that of the circle to get the point of intersections. The distance between the points yield the chord length from which the area of the segment can be gotten (using the radius and sector angle). NB: Quite more difficult to calculate than expressed here. 2. The other unknown part beneath the diagonal is gotten by subtracting 1/4 of the area of the circle from half the rectangle. 3. Sum the obtained areas. 4. The required area is the difference between half the area and the sum in step 3.

[@worldagency]

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.

[@reymisteryo9163]

Ok this is not for me. I'll go back watching Thom Yorke dancing.

[@dawnheroes7574]

The solution is coronavirus ?

[@jothegamingpro3439]

Solved it myself, and your explanation was over the top, but we got the same answer so 👌.

[@bryanfuentes1452]

that can also be solved by integration and the limits can be determined by solving the intersections of the two curves

[@YuriDokiDoki]

The life of a Chinese student: -get 99% on a test -get scolded by daddy

[@kitlovecat]

FYI, this question comes from a primary school in China and that's the reason why it becomes famous.

[@flanbenflen9069]

Bloody china

[@jikaikas]

Damn chinese

[@wkfYT]

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.

[@kitlovecat]

@@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.

[@calyco2381]

I got sin cos tan in high school.

[@allen2759]

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

[@9itowl]

You can also solve the green area by subtracting the full circle from the full square and dividing by 4.

[@dawgsout4free]

I think in the next patch they’re gonna nerf Asian math skills because it’s too OP

[@nishit7147]

They need to buff our social skills though, because they are UP.

[@siralanturing9103]

lol, does anybody here plays world of warships?

[@cindypeng9657]

Dude this is like 7th grade math in China... I’m really glad I’m not in China right now just saying 🙂

[@yeleelim]

(8파이+32)/5 오분의 팔파이 플러스 삼십이 (여러가지 이유로 수식을 쓸 수 없어 읽는 소리 대로 표기) 적분 없이 풀이 가능합니다. 과정이 워낙 복잡해서 중간에 계산 실수가 있었을 수도 있는데 계산에서 실수만 없었다면 맞는 답입니다.

[@smaug9833]

Just find the intersection of the line with the circle. X=2y and (x-4)^2+(y-4)^2=16. Area of red= that of triangle+ integrate circle curve limits from X intersection point to X=4 . Done

[@ToZik]

Roses are red Violates are blue There's always one Asian Who do better than you

[@abrahamdinian1363]

I got the triangle part

[@rawvid9065]

Me too bruh , stuck after that

[@Bustical]

Solved this in a much easier manner (apart from integration which is also easy): Mirror the triangle on the other side, find area of the new sub triangles, the original triangles and the semicircle and subtract to find area of any portion you want. *Symmetry*

[@vilmosszerecz7819]

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

[@Frie_Jemi]

can you explain? why not 8 X 8 square with inscribed circle 8 in. diameter?

[@johnfu9073]

Problem will be simple u don't need trigo : area (square - circle ) / 8 = 1.72

[@johnfu9073]

8 is due to the symmetry

[@enjoy7869]

I should not say! but u made the solution more complicated.

[@rahulmalik1083]

I think it should be quite easy with integration .

[@revat8294]

It can be solved by simple geometry by drawing perpendicular bisectors

[@earlhickey1557]

absolutely too complicated, just use the upper left 4 ; 8 triangle, 1/4 circle, substract a bit and you are there.

[@piyush6494]

@John Wick he's also Indian...his last name is talwalkar

[@tvclipshollybolly2014]

That's quite simple as well 🤔 although he didn't use predefined formula as we all did.

[@adnaannamazee9763]

You could have used cosine rule to figure out length of diameter of semi circle. This way you could have worked out area and took all took areas away from triangle

[@dragondadvoodoo]

For second step just use similar triangle to get the height then calculate the angle theta.

[@paraleluisjoaquing.1007]

*brain.exe has stopped working

[@Nahrvwi]

Overhaul unfortunately

[@bannrl4310]

Using integrals is like cheating.. I solved the problem just with euclidean geometry and trigonometrics functions.

[@mikeyhoban7762]

i got a similar answer about 1.266 (probably rounding) by finding the area of the triangle, finding the area of the semicircle, and the area of the integral between the two lines, then subtracted the integral from the triangle and subtracted half of the area left after subtracting the semicircle from the whole rectangle giving me the answer since there were two equal corners left

[@smbhquasar1527]

It took me around 3 hours to merely rely on geometry to find it, but after being limited to time, I gave up and plugged in the law of sine to solve it. I was able to find it in a few mins after using law of sines. I'd say that a basic understanding of geometry and trigonometry is sufficient enough, yet it's really how you'd use them to find it.

[@rickybloss8537]

I used the correct steps to solve in the first one but I got my math wrong on paper 😭

[@david21686]

How did he get the area of the isosceles triangle?

[@023Clifford]

(product of the two sides times)*sin(angle between them)/2 (same side = 4^2)*sin(π - 2θ)/2

[@david21686]

@@023Clifford Okay, so basically a fancy way of doing "base times height"

[@vibhutisaha2294]

Instead assume the origin of the axes of the Cartesian system at the top, left hand corner of the rectangle and use integral calculus to find the required once you gain the equations of the semi circle and the straight line ( which won't too much of a pain to derive once the coordinates― under our assumption of the placement of the origin at the corner as mentioned above― of the top, right hand corner and bottom, left hand corner turn out to be (8,0) and (0,-4) respectively.)

[@robertlynch7520]

There is a pretty good "third way" to solve this, too. Taking the center point of the half-circle as (0, 0), then it has an equation № 1.1: 𝒚 = -√( 4² - 𝒙² ); Likewise, the diagonal line has equation № 1.2: 𝒚 = ½𝒙 - 2; Setting them equal delivers a reducible equation: √( 16 - 𝒙² )² = (2 - ½𝒙)² … expanding, combining 12 = -2𝒙 + ⁵⁄₄𝒙² … rearranges to ⁵⁄₄𝒙² - 2𝒙 - 12 = 0 … which is quadratic, so 𝒙 = (4 or -2.4) Which is exactly helpful; sure enough the '4' case is at the top right corner where the diagonal and the circle meet. Feeding -2.4 back into № 1.2, we get 𝒚 = -3.2 Now, taking the lower left hand corner as point (0, 0), we find that the intercept is at (1.6, 0.8); the whole triangle is thus 𝒂 = ½ 𝒃𝒉 𝒂 = ½ (4)(0.8) 𝒂 = 1.6 units. Finally, we need to remove the 'lens' as cut into the triangle by the circular pie section defined by θ pie = ½ θ𝒓² … where 𝒓 = 4; But theta? Well, its not that hard inverse-sine of (opposite-over-hypotenuse). The opposite has to be (4 - 1.6 = 2.4), so θ = asin( 2.4 ÷ 4 ); θ = 0.643501; … thus pie = ½ 0.643501 4² pie = 5.148009; To find the lens, now subtract out the triangle between the centerpoint, the bottom of the semicircle, and the intersect: lens = pie - ½( 2.4 × 4 ) lens = 0.348009 The final area then is obvious: solution = 1.6 - lens solution = 1.251991 Which thankfully is the same solution as our Fearless Leader found by the other methods. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅