@@aporifera I was once a math competition student until middle school. No there's no arctan for primary school math. Unless it is somehow hinted in the question, like if a right triangle has sides equal to 1 and 2 the angle can be assumed to equal to blabla...
This question can be done without using arctangent. The two triangles are all right angled triangle, so just use 0.5*base*height will give you the answer
How to solve this problem WITHOUT trig or calculus: Cut a square piece of plywood. Weigh it. Use a jigsaw to cut the two arcs. Weigh the resulting shape. Finishing weight divided by starting weight is equal to the answer divided by 16.
as late as the turn of the century, my bff who works in pharmacology would calculate integrals using essentially this method. they used to manufacture graph paper with an extremely uniform "density" (mass/area) for this exact purpose.
I don’t think there is time for that in an exam. And no student brings such things in anyways unless they knew this was coming out (but then wouldnt they have just learnt the method instead?)
Other questions from china primary school Why do we love our beloved leader xi jinping? Why is our country the best in the world? Why is china the best place to live under the rule of xi jingping? Why is china the most fair and democratic country in the world?
I cheated a bit here to find the relevant integral and ended up asking wolfram alpha for "integral of sqrt(4 x - x^2) - (4 - sqrt(16 - x^2)) from 0 to 3.2", and this produced the correct answer.
Maybe a translation error, im from czechia and im pretty good at English, and I think everyone would call grades 1-9 primary school, school everyone goes to , ages 6 to 15. I was supposed to learn cos/sin/tan in 9th grade too but due to covid we skipped it and our high school teacher was annoyed we skipped it due to covid... So my guess is just translation error. Not every country in the world has different school names, and primary school
@@petrkdn8224 primary means mandatory or everything before high school. idk exact curriculum around the world but trigonometry should start in 8th grade in public schools. Should be much earlier in specialized schools.
My guess is that the part about Chinese primary school students supposedly being able to solve this is so that some gullible viewer (i.e. American) will look at this and wonder why the kids in their country are so far behind. It's purely a means of getting attention by making a claim that doesn't hold up to close scrutiny.
I'm actually comforted by many of the other posters here, some of whom attended primary school in China, who all said that they never received such complex math problems. On the other hand, I do wonder about the complexities that they CAN handle, when compared to American students. I am under the impression that they have a "no nonsense" education. There's no focus on gender, sexuality, or "feelings", but actual hard-core learning, especially math and science. It's where we Americans need to be, but we've gone far, far, far off that path.
It wasn’t immediately obvious to me that the line from B to the center of AD splits each circular sector in half. It deserves a quick proof in my opinion
You have 2 triangle, with 2 sides with same length, and a shared third side.(meaning all 3 sides have the same length) That means they are identical(and in this case mirror) of each other. the known length sides, are radius of each circles, that both start at the center point, which was what you linked at the end. It also make the second circle right angled.(as they are identical).
@@migssdz7287 Actually, the proof does depend on that, because it relies on the fact that the angles on either side of the dividing line are equal. If they weren't, it would have taken more work to figure out what the sector angles ("2t" and "2s" in this case) actually were (still doable, I think, but you would have needed to use a different method). However, as others have pointed out, it's pretty easy to prove this (and that this would always be the case), because the triangles have all three sides the same length, and therefore they must be the same triangle (just mirrored).
@@foogod4237the proof depends on the two triangles being equal, and their angles being equal-but that’s not what this comment is discussing. The comment is stating that the shared side of those two triangles bisects the two ARC SEGMENTS which form the boundary of the area being calculated. THAT is not stated in the video, nor is it necessary to know in order to solve the problem.
I don't know if this is originally from china but defo not a primary school question, I've seen it in Professor Povey's Perplexing Problems, however which already makes it at least a high school level problem.
You could also calculate it by using trigonometry to construct functions, calculating the intersection point, and then using the integrals to calculate the area
I'm a highschool math teacher and I didn't solve this... I rarely do geometry stuff and my problem actually was the construction at the beginning of the solution. When I saw those two lines at 2:13 I was like: well, NOW this is easy... but seeing the right starting point actually gave me quite a bit of trouble.
this is exactly what i thought of in the first place, and i even have the working for it, but the video's method is much simpler as you skip all the algebra to lead you to finding the intersection points and also an integratable form of the 2 circles, plus by integrating you require integration by substitution, which i do not want to delve deeper into that.
I'm going to venture that this is not a primary school question in China or anywhere else in the world for that matter. I have worked this out, but it's a rather messy answer involving trigonometry as well as geometry unless I've missed something. Deriving the length of the shared chord isn't too difficult, and the rest isn't too bad, but I can't see a nice, clean answer.
such geometry questions are common in Chinese elementary school math competition they don't use arctan in elementary school, so the question is often designed to use some special angle (30, 45, 60 degrees), students are expected to know the ratio of edges (1, 2, sqrt(2), sqft(3) etc.)
It took me 10 minutes and calculated the upper part of the shaded region has an area of 1.02 sq units. The lower part has an area of 2.83 sq units. Total shaded area therefore is 1.02+2.83 = 3.85 sq units.
it looked easy, but it had been quite a while since i last did any calculus or geometry-related math, so this ended up being much harder than expected.
It's 12.55 am now in India. I've thought on this question and figured my process to arrive at the answer in the following way : Solve for the equations of quarter & semicircle to find the coordinates of intersection. And then solve for each segment areas. I'm starting to solve now. Let's see how much time it takes ☺️ EDIT 1 : Solved it same above way comfortably & in disinterested way by leisurely watching an entertaining series for half an hour or so in the meantime. Otherwise this question shouldn't even take 30 minutes to solve fully, it's that simple if solved this way. Coordinates of intersection : (16/5, 8/5) Answer : 7.999999 = 8 sq. units FINAL EDIT : I checked the video answer and found my answer wrong, and then realized that I'd assumed wrong formula to find segment areas (I used triangle area trigo formula by mistake 😂) My final answer : 3.846968 = 3.847 sq. units ✌️☺️
I tried a simpler method but I'm surprised I didn't get the same answer. I thought that if you got the area of the quarter and semi-circle and then subtract what isn't in either circles from the total area then you would get the highlighted area. But I got 2.8525. Here are my steps. I first got the areas of both of the circle portions: Quarter = 12.5675 Semi = 6.285 And then I subtracted them separately from the total area in the square to get what isn't in the circles: Quarter = 3.4325 Semi = 9.715 Ah wait, I think I found out what the problem was. Doing this in this way caused me to subtract areas that weren't in either circles twice. At least I think that's what the problem is.
yeah, i got this problem too, which sent me down a whole rabit hole of how to NOT do that. it turned out to be fruitless, as the only way to do this would be a simultaneous equation, but there were simply too many unknown variables; 4 in total.
I'm a secondary 2 student in china, that question is even a challenging question in our grade, I think that question isn't for primary school students. Anyway, I used 30 minutes to solve it by myself.
Of course, you could also write the answer as 16*arctan(3) + 8*arctan(2 + sqrt(5)) - 8 - 6*pi That's what you get if you integrate sqrt(4 - (x - 2)^2) - (4 - sqrt(16 - x^2)) between 0 and 16/5. Easy!
I found an alternative method: solve it using circle equations and graphically let the origin be the bottom left corner of the square the circles have equation: x^2 + (y-4)^2 = 16 and (x-2)^2 + y^2 = 4 make x^2 the subject of the first equation, and put it into the second equation and simplify to get : 2y - x = 0 and then make x the subject of the first equation by square rooting both sides and then put it into the current equation to get: 8y - 4(sqr root(16 - (y - 4)^2)) = 0 then make the y outside the square root the subject of the equation, and use iteration starting with y(zeroth iteration) = 3 because looking at the diagram, 3 is around the point of intersection for the y coordinate. after using iteration, you get y = 1.6. use this value to get the x coord of the point by putting y = 1.6 into the equation for one of the circles, you get x = 3.2. once you have the coordinates of intersection, find the length of the line from the origin to the point of intersection using pythagorous. Now you can use the cosine rule to find the angle of each sector in the two circles, since you have all the sides of the triangles you need, and then use that to find the shaded region which is made up of two segments of each circle, by finding the area of the sectors and minusing the triangles area using 0.5absinC. Now you got the answer.
The question ceases to be interesting when it requires the computation of a trig function. When looking at these questions, the main assumption is that they are solvable without requiring a calculator. At most, we are left with a dangling Pi. Anything beyond fails as a pure geometry question.
I got the same answer, but did it by finding the x values where they intersect (in my case 0 and 16/5 because I chose point A as the origin) then subtracting the integrals of each on that domain with respect to x. Here is the wolfram alpha code: Integrate[Sqrt[4 - (x - 2)^2], {x, 0, 16/5}] - Integrate[4 - Sqrt[16 - x^2], {x, 0, 16/5}]
I think I have found a solution, can anyone say if this solution works? If we imagine the curves in the square as curves in a coordinate system, then one curve would be an exponential growth curve and the other curve looks a bit like a sine curve. I'm not a mathematician, but it could be a Fourier series semicircle. So if you set up a function for both curves, you can calculate the point of intersection. Then you use a derivative to calculate the area of the two curves between 0 and the x-value of the point of intersection and subtract them from each other. Then you have found the area, haven't you?
Probably it should be at least a question that only appears in primary school mathematics Olympic competition which ordinary students should not deal with this kind of question.
The best way to do this is by using calculus and transformation of axes. The method shown here is quite involved and is not a generalized method. No one in exam hall would be able to adopt the method explained in this video. The best way is to do as follows: 1. First set up the equations of the 2 circles using coordinate geometry and considering A as the origin. 2. Solve the 2 equations thus established to get the points of intersection as (16/5, 8/5). 3. Next use calculus to find the required area by integration. There will be 2 integrals - one for the arc AC and the other for the arc AD. The difference of the integrals will give the required answer. The limits of integral will be from 0 to 16/5 (for x). 4. While carrying out the integration consider the standard formula of integral of sqrt(a^2 - x^2). 5. To simplify integration for arc AD, you may have to use transformation of axes and change of limits. It is an absolute crap that this question was given to primary school students in China. Even I can say that this question was given in nursery level in India. Just saying such things and attracting viewers attention is not good. One should post the question paper of the exam at primary school level in China as a proof to make the whole thing credible. RU-vid must take a note of this.
Back during the second half of 70s and the first half of 80s in Ljubljana (capital of Slovenia, then part of Yugoslavia) we were learning *binary, ternary, quaternary, quinary, senary, septenary, octal, nonary* and of course decimal number systems during the *fourth year of primary school when most of us were 10 years old* and some were 9 years old. We had to convert numbers from any base to any base. On the other hand, we didn't learn hexadecimal and when I one year later arrived in Zagreb (capital of Croatia, then too part of Yugoslavia) I was surprised they didn't learn number systems other than decimal.
@@NaThingSerious I can't say for others but for me learning more than just bases related to programming and digital circuits design (binary, octal and hexadecimal) was *very* helpful in developing various visualisations related to numbers and being able to "see" the numbers directly without converting them. For example, if I see some single digit number where the digit is the highest one and if there is a square root of a base which is integer - then I "see" the same number as "two highest digits of the base which is a square root of the original base", e.g. if I see 8 in base 9 then I immediatelly "see" that is 22 in base 3 too. Similarly to octal is useful when looking at groups of three binary digits (or hexadecimal being useful when looking at groups of four binary digits), quaternary is useful when looking at groups of two binary digits. The most important thing is to learn those things at very young age, exactly like learning many languages which in that case all become native. If you learn a language when you are older you can never learn it like it was your native language.
I still can’t see where this would be helpful or worth teaching, ig it could be interesting and possibly help with coding, but it still just seems like a waste of time
@@NaThingSerious If you can't see where this would be helpful doesn't mean it isn't. I've described just a few aspects where it was *very* helpful for me. Here is another example - when I was about 11 I got ZX Spectrum and very soon I was able to write short machine code routines directly in hex without looking at Z80 op codes or loading the assembler. Yes, one could learn the opcodes but the code I was writing involved bit masking, OR-ing, XOR-ing, shifting to program multiplication and division "by hand" etc. and learning numerical systems helped me a lot because of learning to visualize the numbers in various systems and because I didn't have to convert them as I could "see" the values. Depends of what you are doing if something you learnt you would find useful. Someone could say learning mental math is just wasting time because we have computers but there is more in using mental math than just calculating the numbers - it's about influencing brain development. It isn't just learning about number systems - it's about learning it at very young age in which case numeric systems become as natural as reading letters which makes a huge difference. As you have noticed when you said that could "possibly" help when coding (which itself would be enough reason for having that in the schools), I can assure you it is of crucial help when developing embedded systems hardware and firmware which is what I've been doing. Someone could say 99% of what they learnt at the school was waste of time, for someone else waste of time was maybe 80% of what they learnt, for someone it was 40%, for someone 20% etc.
It is very likely a multiple-choice question. It can be estimated at the primary level. The simplest way of finding a solution is to draw a scale diagram on a grid paper and count the number of squares.
Much easier to see the angle S as arctan(1/2) rather than 2pi - arctan(2), since you're going to need a calculator to obtain any arctan value anyways...
Gentlemen, the task is greatly simplified if you draw a diagonal between the vertices of the right angles of the quadrilateral. All resulting right triangles are similar. to find the answer you just need to calculate all the sides of all the triangles. I solved the problem with the side of the square equal to 2, not 4. I found the area Pi-2.
You must first prove that the two lines from the centers of the circles to the arc intersection point are both tangent to the other arc, which is definitely not a primary school question.
Why is it complicated? With the right triangle, there are 4 and 2, it is half of an equilateral triangle, so t = 60 => A=r^2t = 2^2*60 = 480 Same with s
We know that the triangle on the left is a right triangle due to the definition of a square. We also know it has sides length 2, 4, and the hypotenuse. We also know the triangle on the right has sides length 2 (semicircle radius), 4 (quarter circle radius), and shares the same hypotenuse. Therefore, the triangles are congruent and share the same measures for angles and sides. Therefore the right triangle is also a... right triangle.
Solution using calculus. The equation of y = f(x) for the top/bottom half of a circle centered at (x0,y0) with radius R is f(x) = y_0 +/- sqrt{R^2 - (x-x0)^2}. Put the origin at the bottom left corner of your diagram. The equation of the larger circle segment with radius R =4 and (x0,y0) = (0,4) is f1(x) = 4 - sqrt{16- x^2}. The equation of the smaller circle with radius R = 2 centered on (x0,y0) = (2,0) is f2(x) = sqrt{4- (x -2)^2}. The limits of integration are x = 0 to x = x* , where x* is the non-zero solution of when the two curves meet, f1(x*) = f2(x*), x* = 16/5. So integrate f2(x) - f1(x) from x = 0 to x = 16/5, and we recover your answer
Simple. Lets make another big quarter of a circle that will cover that area on the right. Then we have two big quarters (lets call them 2q) and one small half of a circle (1s) They overlap but cover the whole square. We can calculate their total area and compare it to the area of the square. it will be bigger than a square by the two equal overlapping parts, one of which is the red one
Calling it “TRIG” does not make the matter any easier. The problem is just in those four letters: it demonstrates the intention of making things less complex than what they really are, lowering the ability to think. It is “TRIGONOMETRY” and it can be complex.
My approach was way off 😆 First worked out the area of the square (16), the bottom right arc ADC ( 16-((pi*4^2)/4) ), then the the smaller half circle AD ( (pi*2^2)/2 ) and then thought I could do some manipulation with adding or subtracting combinations of them (I couldn't !)
So this is the "new math" I heard parents complain about. I didn't know how to solve it but the trig stuff I understood. I was pretty close with my estimate of 4
There is a trend in China that encourage student in primary school to take Mathematical Olympiad class which will have such a problem shown in this video. The goal of such a class is to use basic algebra and trigonometry skills to solve harder questions that are easier to use calculus or differental equation to solve in college.
Nice one, I also spent some time trying to figure it out without use of arctan, but I think its not possible. A side not, I think you might have missed a step (or i missed it). At some point you have to argue that 2 purple triangles are the same (i.e. same angles) in order to use "times 2" for the area. For this you have to use a theorem (i.e. 2 triangles that have 3 sides of same length are identical). Otherwise you cannot argue that second one is orthogonal.
If we note big sector without overlap with A, overlap with B and small sector without overlap with C we will have: A+B=4pi C+B=2pi => B is 1pi Please tel me know where I am wrong
Let's keep it real ... in China every 5 year old can solve this! That's why every nobel price goes to China and why every new technology is invented in China. So impressive!
Well, this can't be real since you have to use a scientific calculator to calculate the answer and I know for a fact that Chinese students (preschool through university) are not even allowed to use double-zero calculators.
This can actually be explained with some probability properties, whereas (AUB)=A+B-(A(intersection)B) So what you did was rework this and switch the intersection of both with the reunion of both.
Hold your horses! I don’t believe even Chinese children would be expected to solve the problem by that method. Here is briefly what I did: There are 3 known and 4 unknown areas. Label the unknown areas W, X, Y, and Z. Depending on how you labeled the unknown areas, you should get something like: X + Y = 2Pi, X + Z = 4Pi, W + Y = 16 - 4Pi, W + Z = 16 - 2Pi, and W + X + Y + Z = 16. Solve by substitution. MYD got 3.847. I got 3.44 (with lots of rounding along the way).
If in the problem statement we had the angle ABF = β, so that β is the "known" angle, then it would be THE kind of question Chinese are able to resolve even in primary school (the answer would be S = 12β - 8 + 2π). My solution is the same as the author's. ----------------------------------- Let's put F as a centre of the AD segment and E as the intersection of both archs (at not 0). Then obviously BE=4, EF=2. The quad is symmetric as to BF, so S(ABEF)=2*(4*2/2) = 8. Then, if ABF angle is β=arctan(1/2) => ABE angle is 2arctan(1/2), and AFE angle is π-2arctan(1/2). Therefore, if x is a sought-for square, we have S(ABEF quad) = 8 = S(ABE sector) + S(AFE sector) - x = = 4^2 *2arctan(1/2) /2 + 2^2 *(π-2arctan(1/2) / 2 - x = = 16arctan(1/2) - 4arctan(1/2) + 2π - x = 12arctan(1/2) + 2π - x, and x = 12arctan(1/2) - 8 + 2π ~ 3.847.
The approach I tried was: The quarter circle is 1/4 the area of a circle with radius 4. 16pi/4 = 4pi = Area of the whole quarter circle. Are of the half circle = 1/2 the area of a circle with radius 2. 4pi/2 = 2pi 4pi - 2pi = 2pi I'm guessing what's missing is I didn't find the area of the shape on the right that's between the two circles, which feels really obvious in hindsight. Unfortunately, it's probably impossible to do without weird tangent shenanigans that I've forgotten and can't be bothered to retain for long enough to know what they're useful for. :/ Hyper geometry sucks and I pity anyone who has to do it.
i would do an approximation, not try to solve it exactly. i'd start with a bounding box, and chip away at it with triangles with lines tangent to the curves. i would never have come up with that proof.
You got a intersection of a circle with radius of 8 and a circle with a radius of 4, just calculate the angle and make a comparation of the areas and the intersection, and done 👍
I'm not a fan of the approach to this problem in particular. 1. Making the triangles creates geometric assumptions that I can't prove outright and justify. 2. The use of arctan in primary school is kind of an egregious statement to make. The problem can actually be solved using simple area calculations that are indeed within primary school capacity.
I am a tenth grader in India who is currently giving board exams and to be honest different countries have different types of questions with varying difficulty levels. But this question seems to be more of a HOTS questions aka Higher Order Thinking Skills. What I could do (atmost) is calculate the area of the quadrant = (pi R²)/4 = 4pi sq. units and the area of the semicirle = (pi r²)/2 = 2pi sq. units and nonetheless, all my efforts except this are all in vain. But still this is a worthy question lol.
Dear All I got 4π-8. Ι think its elegant,since the angle on top left is 2*π/6 with radius 4 and in the lower section the angle is 2*π/3 with radius 2. So every surface can be calculated between the two orthogonal triangles with bases 4 and 2. Please let me know if you agree. Thanks
I used integration after writing formulae for the curves enclosing the region. Not very imaginative, but that's what mathematical tools are there for I guess. Primary school? I. Don't. Think. So.
The proficiency of our species appears to be in a state of considerable decline when a significant number of individuals deem the presented problem to be challenging. One would expect a standard child to effortlessly resolve this matter mentally. One must question the educational institutions within their nation if such a basic matter eludes comprehension. The expectation that individuals are unable to surmount this challenge by the age of 12 is disconcerting, as this pertains to rudimentary concepts.
Solution is not true. You can not assume that the extra radius of length 2 that you drew just smoothly intersects at the top right corner of intersecting circles (blue area) without crossing into the blue area.
hmm…. I found a different way: 1. Square area is 4x4=16 2. Difference between square and first segment is 16-4PI (Segment area is PI X r square / 4) 3. Difference between square and lower segment is 16 - 2PI 4. Overlap area is square mine (2) minus (3) = 6PI - 16 My result is 2.849 which differs from his result so I might make a mistake here. Maybe someone could point it out.
I tried to do 4 simultanous equation system for the 4 areas inside the big square. but in the end, my system of 4 equations didn't have a discrete solution ;(
Because of the pi and arctan both appearing, there can't be any elegant solution. Mine was similar to 3:33 except that I summed 2 areas and subtracted 2 other areas.
