How To Solve This Viral Math Problem From China

This is harder than it looks!

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[Oscar-zpio]

China is the only country where this could go viral

[qzepio]

The first half was ok, until the purple part came

[yz]

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

[Broomie]

I mean, I understand it

I just can't pull these equations out of my head

[Jvx_M]

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

[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

[alexandregoncalves]

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.

[antaresmaelstrom]

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

[chrn]

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
Who's Trixy?

[lincolnross]

Expecting simple algebra. Got a warp theory equation

[humpfzzz]

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

[sheauiwne]

Looked away for one second and got lost

[hasibunnisha]

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.

[ethanbates]

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

[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...

[stevenmellemans]

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

[yuluoxianjun]

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

[cakeandicecream]

What an incredibly unsatisfying solution.

[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.

[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.