Viral question from China
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- Опубликовано: 30 сен 2024
- This was shared as a primary school question in China, which I gather is at most year 6 and students aged 11-12 years old. Can you solve it? Thanks Blake for the suggestion!
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I’ve taught primary school in China. This is not the kind of question they have to deal with. Circles and geometry, yes; but not this.
This is not for typical primary school students but for competitions
@@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 is very common in Singapore primary school
@@vooeyhooey9165I was a primary school student a few years ago, and I can say that we, in no way, learned theta or arctan.
@@vooeyhooey9165I have a friend from singapore that went to NUS or something and no apparently it's not "very common" in primary school there.
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
This is a classic trick question on social media that's faked as a "primary school question" because it looks simple at first. But no even grade 6 math competitions in China will not have arc tangent. Plus that using a calculator is not allowed for primary school exams anyway. However, apart from the calculating arctan(2) part, the rest is indeed primary school level, but as a bonus question that's only meant to be solved by top students.
Me in college, unable to resolve this: 😢
@@lucascamelo3079 Me too. I would just drop this into xy coordinates and calculate the integration. Never saw sth like this until grade 10.
@edkk2010 Well, have you? At least I have been through a few and I got prizes to prove it. It is not going to be normal for primary school kids to know arctan. we learnt to solve problems that were hard, but not this hard.
I took grade 6 math in China. I still have my text book. This is obviously in the grade 6 math book of China. It is in an exercise section of a chapter, a type of questions called 思考题. So yes, the publisher did not make anything up. The majority Chinese here have long forgotten their elementary school.
@@EuroTravChannel You seriously have arctan in your textbook?
Since I have relatives from China, I asked them: They think it is highly unlikely that this is a questions asked in primary school. They confirmed that primary school is up to the 5th or 6th grade ... depending on where you go to primary school in china, but even for 6th-graders they said this question is to hard. Calculating with squares and circles to a certain degree yes, but not to this level of complexity. - They think it is more likely that the phrase "Primary school in China" was written beside it, to cater to prejudice that all Chinese people are top notch when it comes to math, which they are not.
so true
Tbf, some prestigious schools definitely could be teaching this to primary school students, but it is definitely not normal.
@@NaThingSerious
Same in USA, many prep kids are doing problems like these at ages like 10. We got grade skippers too who go to college when they're only like 12
FWIW, i have a Soviet (as in USSR) education, we were studying trigonometry in classes 9 and 10, which is equivalent to Western high school. i think 14 years old is the minimum. of course you could teach that to a smart 10 year old, but what would be the point of it, just for raising someone's eyebrows?
You don't need this kind of math for a general kid. Maybe for that one genius kid out of 1,000 kids, who can actually use it one day. For everyone else it's a waste of time.
we dont all have the same primary school 😂
The hardest question at my primary school was √225=15
My high school question's was Quadratics not calculus
For Real 😂
Not sure this is funny though
🤣
Tbh i think this is why Chinese people were so smart.
I wasted 3 hours trying to solve this problem from the perspective of a primary school student (by not using calculus), believing that a simple solution exists. 😢
..i doubt it would make much sense for any1 in primary school..
firt u need the intersection point between the two cirles
C1: (x)² + (y - 4)²=(4)² and C2: (x - 2)² + (y)²=(2)² ->
(x)² + (y - 4)² - (4)² = (x - 2)² + (y)² - (2)² = 0 ->
..wich gives x=0, 16/5 ;Y=0, 8/5 [point 1: 0, 0; point 2: 16/5, 8/5]
integrate with respect to x between 0 to 16/5
∫f(x) = ( (x - 2)² + (y)² - (2)² ) - ( (x)² + (y-4)² - (4)² )dx ->
∫f(x)dx=−2x²+y²x−(y−4)²x+16x+C
or ∫f(x)dx=−4(x−2y) ; Simplified ->
∫f(x)dx=∫( (y²−4)x+( (x−2)³)/3+C ) - ∫( x³/3+(y−8)yx+C ) [C=0, Integral by parts]
Solution: ( (200y²−3584)/375 ) - ( (1200y²−9600y+4096)/375 ) ->
(640y−512)/25 ->
16 ( (40y−32) / 25 ) = exacly 16 * 3.84 ...
using arctan u get 3.847 (according to the video)
..it may be more precise, but is it worth it?...
..not at all an easier solution...
I teach physics in Brazil's public schooling and students often have a hard time with simple trigonometry problems by the age of 15-16. I have a hard time believing a problem like this would even be proposed as a university entrance exam
brazil
Trigonometry is the neglected child of math. Students have no way to visualize how it works when it's taught to them in a classroom, and over time they form mental blocks that stop them from ever figuring it out.
My university entrance exam had a decently difficult double integral that ended in arctan(1/x) + arctan(x). I know many people that solved the double integral and got stuck at the end.
Acho que o senhor está enganado... como ex-vestibulando deste ano, questões semelhantes a essa não são incomuns em universidades federais. Não que isso signifique que a maioria dos vestibulandos consiga acertá-las, mas já vi questões quase idênticas a essa.
Here in France, this type of problem is chinese for highschoolers, and still a pain in the ass for university students. I'm a private teacher myself for highschoolers, I understand well the solution of course, but was unable to figure it out alone.
As a Chinese primary student back 20 years ago, I can comfirm i have had done some like this one before. Reason for young students doing this was bcs they need to attend after-school maths competiton, to get some awards for better chances to enter better high school after graduation from primary school.
I’m surprised that Presh didn’t solve it using multiple techniques (like he often does), and then show how to do it using calculus…
@@kaithedoge5861integral of the small sphere minus the integral of the the curve of the big sphere ending at the intersection point
didnt wanna make the video too long prolly. and there was no need to, the calculus maybe wasnt that interesting
There's a couple different ways to set up this as a double integral, which anyone with calc 3 knowledge should be able to do
I tried the calculus solution and it requires solving an integral with u-sub. You need to guess that's in the form x=k*sin(u), which isn't obvious if you have never solved similar integrals. This method is much simpler and intuitive.
The calculus approach is a rabbit hole you don't want to go down. Trust me.
Arc tangent being used in Primary school...this question is tough even for high school students 😢
Well trigonometry used to be in primary school 😊
@@kwestionariusz1 in some kindergartens even.
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
@@jamesfeng5374The angles of each circle sector, not to find the area of a triangle
@@excentrisitet7922In some daycares even 😁
Guys I'm from China and I've never seen such hell in primary school. The hardest stuff i got was probably solving system of equations in 5/6th grade. So i assume this question is designed for teenage math olympiad?
china never fails to show themelves as some extraterrestrial intelligence society🤣
In indian board, system of equations come for the first time in Class 10 (last year of secondary). We are basically 15 years old when we learn it for the first time 😂
@@gaminghellfire it comes first time in class 7 with linear equations idk you are talking
@@atopgaming9000 You are from ICSE right? Im talking about CBSE and state board
@@gaminghellfire you studied 7-8 online right? linear equation in one variable and polynomials are literally chapters of class 7-8 both in cbse and state board and algebra is taught in ICSE board at class 6.
This may not be the best example of a tricky primary school question due to the involvement of trig functions, as others pointed out. However, I did see a friend asking for help on social media about a similar problem that was supposed to be their 4th-grader's homework. It features cleverly constructed geometric shapes with multiple overlapping areas and, like this question, essentially requires the student to uncover hidden quantitative relationships among the various parts. A smart student would be able to visually or verbally articulate the relationships and solve the problem using basic algebra (she may not have learned to formally use variables and will be doing it implicitly using words/drawings). Calculating the result was trivial, so long as the student knows area formulas for basic geometric shapes. That was a much better question, in my opinion, if you want to figure out who the smart cookies are.
Yes, I have seen a similar solution that uses no trig functions. They presented the solution on RUclips because it was one of the math problems given on an entrance exam to a middle school in Japan. That means that 6th graders needed to be able to solve this problem.
Can you describe it?
It can be solved without the use of trig functions.
This is not a primary school question anywhere in the world
problems like you describe are current; still they are way easier ones
I would consider taking two integrals and calculating the difference between them, rather than trying to find the solution with basic geometry
I've just done it. Certainly, it can be done, but it is really messy. Maybe it's on me, but without my calculator, I wouldn't have done it. Maybe there is a simpler method, but one of the functions describing a circle has extra minus where the other one hasn't. It took me a lot of time to figure it out. I just wanted to solve it with the calculus so bad.😂
Well, if you want to do it in polar, you'd need to do that. In cartesian coordinates this could be set up in one double integral, although it wouldn't be too pleasant to solve
Solving with polar results in -8 + 2pi + 12arctan(1/2) which is equivalent to the calculated answer
would be messy but probably easier TBH i would just plot the circles in auto cad and have it calculate for me....
@johnbruner5820 I've done it as a check, which helped me find an error in my integral equation.
Doing it using integral is skill issue.
1) I recall solving a similar problem in the High School exam. However, we were just requested to discover areas that can be calculated (in circle or triangular parts). The expected answer was like:
MIN < AREA < MAX
2) That was the era when logarithm, sine, cosine, etc., were looked up from a table or a sliding rule.
Hey! I really like your videos, but I want to solve more problems on a regular basis to keep in touch with all the concepts, and so can you suggest some resources for the same? Like they must have problems from random topics, and the hardness level should be customizable according to our needs. Thanks :)
Presh’s videos used to feature ads about several books he’s written. Since he’s no longer including those ads, I’m not sure if the books remain available for sale-but if they are, that might be a good place to start. If you look through some of his older videos, they still include those ads and would likely help you locate the books.
Premath@ RUclips
CAN U DO IT USING CALCULAS
I did it using calculus:
The Funktion of the quarter circle in the square is:
f(x) = 4 - sqrt(4^2 - x^2)
The function of the semicircle is:
g(x) = sqrt(2^2 - (x - 2)^2)
The searched area can be calculated with Integral of g - f in the boundaries of 0 and the intersection point.
The intersection point is calculated with f(x) = g(x) and it's value is 3.2
I calculated this point.
I didn't calculate the integral per hand, but the value is correct.
For calculating the integral you might need some substitutions.
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.
😂😂😂
Draw it in solidworks and get the area.
Loved this!
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?)
If this is indeed a primary school question, then I'd think the students were learning about approximations rather than trying to give an exact result. Figure out that the square has an area of 16, note that the blue area is somewhere around 1/4 of the area of the full square, and give an approximation of 4 or so for the final answer.
No, Chinese schools never ask that
I‘ve started with a probability calculation too.
I used calculus to solve and it seemed quite smooth
So the equations of the circles are x^2 + (y-4)^2 = 16 and (x-2)^2 + y^2 = 4
Then find the intersections by isolating y and making x the variables, and you get x = 0 and x = 3.2
As you transformed the circles to functions, the equations are y = sqrt(16 - x^2) - 4 and y = sqrt(4 - (x - 2)^2)
Then integrate from 3.2 to 0 using the following equation,
A = sqrt(4 - (x - 2)^2) + sqrt(16 - x^2) - 4
And you get 3.8469......
There was no way I could've done it the trigonometric way, that was just. Insane
Well done. I tried doing it that way but mucked up my algebra somewhere. I have not done calculus in about 20 years lol, just glad to know my method was correct!
I did it the exact same way. Also for others, you can change the equations by changing the coordinates. This question is kind of coordinate bashing and use of calculus.
@eyeofregret4362 You should have said integrate from 0 to 3.2 rather 3.2 to 0. From your perspective it’s still negative
very good but I get the solution to the integral to involve arcsin (Dwight 350.01) and then the values are outside of the domain of the function. (I can try to calculate in excel)
get 3.8418 with dx=0.05
Are we SURE this was primary school?!? I didn't use arc tan until high School. Is there a more simple solution?
It's not from a regular school.
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.
@abrvalg321 ah, yeah, we have 9 years in czechia, and we start trigonometry late 9th grade
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 aced this very problem when I was in kindergarten, the answer is "the area of the overlap of the circles is the blue one".
As someone who actually went to primary school in china, I can confirm that although I've never had to do this type of question at school, many parents send their kids to special math academies called "Math Olympiad" in order to gain an advantage to their peers. These type of math training can start as early as 2nd grade. I have done a number of questions similar to the one in this video during my primary school years in china, so in that sense, although it is not taught in school, many chinese elementary schoolers are expected to be able to do this problem.
> advanced trigonometry
> primary school
pick one
It isn't an "advanced trigonometry" until Euler's formula is involved.
Primary 🎉
This was actually one of the coolest problems you've presented in some time, Presh. Well done! ^^
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?
@@danquaylesitsspeltpotatoe8307 ha that’s kinda funny.
@@iqrainstitution4365 Funny cause its true!
I first started down an integral path, but they became intractable. I figured there was a tricky solution, so played around and did the exact thing he does! Was pretty surprised I got this. One difference was that I just did 16 arctan(1/2) for the second sector.
It works fine with integration, not intractable. You do have to find the intersection of the 2 curves. I plotted the curves, solving for y, as f(x), g(x) and noting the intersection at y = 2.4 in the coordinate system with the origin at upper left hand corner of square.
@@johnpinckney7269 I don't know how you'd do the integration without the intersection point, and doing it by hand was intractable for me 😂. I guess it depends on your definition of intractable, but just getting a computer to do it, doesn't reduce its complexity, and to me kinda defeats the purpose.
Dude just distribute the integral and do a trig substitution
@@theboss5929 could you provide a little more detail?
@@atrus3823 The curves can be expressed as f(x)=(x-2)^2+y^2=4 and g(x)=x^2+(y-4)^2=16 corresponding to the corner of the square. Rewrite them in terms of x giving f(x)=sqrt(4-(x-2)^2) and g(x)=-sqrt(16-x^2)+4. They intersect at x=3.2, so the bounds of integration are from 0 to 3.2. Write the integral as f(x)-g(x) to find the area between curves. For each function, you can just do a trig substitution where for f(x), x=2sintheta+2 and for g(x), x=4sintheta. The differential for each respectively is dx=2costheta dtheta and dx=4costheta dtheta. Rewrite both using pythagorean identities to multiply costheta by itself, resulting in 4cos^2theta and 16cos^2theta. Take out the constant and apply the reduction formula. Integrate with u sub, rewrite back in terms of x and do the bounds. It's a lot of work though
this is so simple, we can imagine 2 circles one with radius 4 and center (0,4) and another circle with radius 2 and center (2,0) now we just have to find the point 2 of contact of the circles, point A will be at origin, and other point can be found by assuming the circle equation. C1:- x^2+(y-4)^2 =16 and C2:- (x-2)^2 + y^2 = 4, point A is (0,0) let the other point be E, point E is going to be (3.2,1.6) this can be found by solving C1 and C2. now with integration we can find the area under the curve, integral of C1-C2. with area element as dx or dy according the limit changes too.
with this we can find the area of the intersection. this is another approach.
what i did was form two equations of the circle, equated them to find a line that goes through both points of intersection. used that line a a chord for both sides and used (1/2r^2θ)-(base x height/2) to find the area of the segment for both sides of the line and add them
if its 'so simple' you would be able to devise a method without writing out a whole paragraph lol
@@godofjimjams1611 hmm, i think its was not clear for you to understand LOL
find equation of 2 circles and now find the area of the intersection of those 2 circles using calculus. now simple?
In fact, in China, every newborn child is born knowing calculus from its mother's womb. At the age of 5, they could automatically solve the Riemann Hypothesis in their head and explain it in two minutes. Accurate information, let's spread it.
The first thought that came to mind was a subtraction of integrals, double integral, but that point where they meet gives you a lot of power to slice things up and just use 2 regular integrals and area of a circle.
I also thought of that and was wondering if double integration could solve it (I'm positive it does)
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 solved this problem through integrals and I am so glad that the results matched
This is the correct way to do this question
How may I ask like why was the element that you considered @@chuashanganluciennhps9992
i got 2 pi.
x = Square area - ((1/4 circle #1 area -x) + (1/2 circle #2 area -x) + ( Square area - ((1/4 circle #1 area -x) + (1/2 circle #2 area -x) + x))
x = (4*4) - ( ((pi *(4^2)/4) -x ) + ((pi *(2^2)/2) -x)) + (16 - (((pi *(4^2)/4) -x ) + ((pi *(2^2)/2) -x) + x))
x = 16 - (4pi -x +2pi -x +16 -4pi +x +2pi +x)
x = 16 -4pi + x - 16 + 4pi -x - 2pi + x + x
x - 3x = -4pi
2x = 4pi
x = 2pi
prolly got it wrong, but hey...
We can reduce the problem to find the overlapping area of two circles with radius 4 and 2, the centres are sqrt(20) apart. The solutions are well known. However, the primary students in my home country are not allowed to use a calculator during homework or exams. I wonder how a student can work out this problem without a calculator. 😂
that actually. doesn't work,i thought at first too
Looks like somebody is just over bragging and boasting about themselves.
This is not that difficult. You know the total area of the square. You know the total area of the quarter circle, the difference is the negative space. Similarly you know the total area of the other half-circle. That leaves some thought as to the negative space of the right side (along CD) which is just the negative space of the two circles minus that overlap.
yeah but subtracting the overlap is very difficult. To calculate the overlap you add the two negative spaces together and then subtract the total space of the negative spaces. The total space of the two negative spaces is just the negative shape of the area we're trying to find in the first place. With simple algebra I don't think It's readily solvable and you need the advanced equations that he outlines
I tried that route, but unfortunately you have three equations and four unknowns. There are four regions. The three equations are: 1) area of the "big" quarter circle 2) area of "the small" semi-circle, 3) sum of the four regions = area of square. Without an equation for the area of no overlap, you can't find the area of overlap.
Instead, you could also convert the equations to polar, so you have
r=4cos(θ) for 0 θ=arctan(1/2)
Then use the polar integration formula integrating 1/2 * r^2 using arctan(1/2) as a bound
For 0
It's a straight-forward calculus problem, though it comes in four parts:
1. find the intersection of the circles at about (3,1.5)
2. find the area of the large circle under the curve from 0 to 3-ish (edit - and subtract it from the enclosing rectangle (4*3-ish) to find the crescent area at the bottom.)
3. find the area of the small circle under the curve from 0 to 3-ish
4. subtract those values to find the answer
I can believe a handful of 6th graders could do this - China has a HUGE population and there are exceptional people in abundance for this reason. I do not believe their bog-standard 6th grader would even know the words.
I looked at the integrands and thought, "Nope, not going to do that." 🤣
In order to take an integral (whether you're taking it in terms of x or y) you need to know the top and bottom functions. You can't do this with circles in the traditional sense because they follow a parametric coordinate system as a conic. Without having rectangular coordinates, we can't qualify the given the equations into a quadratic or exponential equation without a little more work. So, no, you can't calculate the cross section with just a four step calculus procedure until you determine what functions coincide with the 2 circles.
@@factsthatyouneverthoughtyo9188 I see. So I think I can get the area under the semi-circle but you're telling me I can't get the area under the large circle because what I want isn't within the circle #notamathemetician . Since I know where the x-axis is wrt the large circle's center, I ought to be able to get the curve under the circle just the same. Maybe it is 5 or 6 steps then?
For the issue @facts brought up, I just calculate the area of the large circle from 0 to 3-ish (it's 3.2, really) and subtract this from the enclosing 4*3.2 to find the crescent area at the bottom... so it isn't a lot harder. That being said, I'm not a mathematician and I can tell solving this problem would take me all day. So maybe straight-forward for a someone who does this sort of thing more often than I do...
@@factsthatyouneverthoughtyo9188 Umm, dude.. the formula for a circle in rectangular coordinates is one of the most fundamental geometric formulas in existence. It comes from basic geometric identities and has been well known for literally centuries:
(x - h)² + (y - k)² = r² ( (h,k)=center, r=radius )
Therefore, in this case, over the portions of the circles that we care about (0 ≤ x ≤ 4, 0 ≤ y ≤ 4):
Large circle: f(x) = 4 - sqrt(16 - x²)
Small circle: f(x) = sqrt(4x - x²)
It's basic algebra..
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?
How is this primary school???
Teacher: It's a fun and easy homework! It is only equivalent to elementary school kids' homework
Also teacher: *have arctan in the solution*
I'm the test paper in china these types of geometry questions are not written on me for primary school students
Edit:- Thanks for 1 like
How do you know that the two triangles that you've filled in purple are congruent? I think we've assumed that the angle made at the intersection point between the quarter circle and the semicircle by the radii we've drawn is a right angle, but how do we prove that?
We know that the corresponding corner of the other triangle is a right angle because it is also a corner of the square. The side lengths of the two triangles are also the same, so the triangles are congruent.
@@dancledan We don't know that the second triangle has a right angle, because we simply connected two points, we didn't draw a tangent or construct a right angle. The proof is: these triangles have sided 2 and 4, and they share the third side, so their sides are equal (pair by pair), therefore the triangles are equal.
@@vsm1456 This is exactly what I said. The two triangles are congruent because their side lengths are the same. We know that the leftmost triangle has a right angle because it is also the corner of the square. Therefore, since the two triangles are congruent, the other constructed triangle is also a right triangle.
@@dancledan that's not what you said, as far as I see. you can't use angles as a proof that these triangle are congruent because we don't know the angles. we prove they are congruent by looking only at their sides. and only after that we can talk about angles
@@vsm1456 It is exactly what I said. The angle of the leftmost triangle is a right angle regardless of whether the triangles are congruent. Because the triangles are congruent, the rightmost triangle also has a right angle. There are two separate parts: proving one triangle is a right triangle and proving both triangles are congruent. Put both together and you prove both triangles are right triangles.
In China, there is a kind of course/program /class named "奧數"(Olympic math), it's just like honor program, it's much difficult and more deeper than normal math class,for 6th grade students.
Olympiad*
How are 12 y.o. Kids solving this?
I call BS on Primary School.
Wow, that’s an incredibly difficult math problem for primary school children! That’s more like a high school or early college problem!
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 triangles with a 90° angle and the adjacent leg and opposite leg have the same lenght.
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).
He didn't stated they are equal, neither used this fact in any point.
They could not even be equal (they are though), the proof doesn't depend on that
@@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 was like I could now get the angle by trig but theres surely simpler way, I gave up after 10 minutes
God bless those super IQ pupils in primary school who able to solve it. Many of still struggling to remember basic tables once taught in primary school.
its olympiad question, chinese students learn trigonometry at 9-10 grade and they dont even have arc tan, arc cos and many other stuff so i believe this is just some internet thing
Area of quater - area of hemisphere = area we need +( area of square - area of quater) easy method i solved
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.
So did I
Total area of the cube is 4²=16
Now the radius of quarter circle is 4, so the area is [(π×4²)÷4]=4π and the rest area is (16-4π).
The radius of the semicircle is 2, so the area is [(π×2²)÷2]=2π and the rest area is (16-2π).
Now let's take the area of the coloured place is X.
So,
X+(16-4π)+(16-2π)=16
=> X= 2.89
So where is my mistake..?? 😗
You counted the area twice (triangle shape)
Are you sure it's not a translation error? I'm pretty good at English, and I think everyone in Czechia would call grades 1-9 primary school , which is up to age 15... I dont know if there is any other name but it's a single school huilding, students ages from 6 to 15...
In China and HK, schools are separated into 小学 (primary Y1-6), 中学 (secondary Y7-12) and 大学 (university); since the characters in front of 学 mean small, middle and big respectively, it’s probably not a translation error 😅
More likely, it's just some troll making up BS on the internet again. Unless somebody can quote me a source, I'm assuming this problem was probably never actually asked of any school children at any level in China to begin with, and somebody just stuck on that bogus claim just to make people share the meme around more...
Here in England (except in Northumberland which has first, middle and high schools like the US) primary school is ages 5-11 and secondary school is ages 11-18.
Completely bogus. The way to treat this problem is to apply the probability intersection formula. Treat the quarter circle as P(A) and the semicircle as P(B). Then the intersection of the two independent areas is merely the product of P(A) and P(B).
P(A) = ((1/4)*(pi*4^2))/16, and P(B) = ((1/2)*(pi*2^2))/16
Multiply the two and you get the area.
That doesn’t take into account the relationship of these two specific areas. For example an area equivalent to the semicircle exists such that there is no overlap, and yet your method would treat it the same.
My solution literally came to me in a fever dream. Basically, you separate the lines into three individual shapes then overlay them:
Area of Square - (Area Small Circle / 2) = Area of Cut Square
Area of Square + Area of Cut Square / (Area of Large Circle / 4) + Area of Cut Square + (Area Small Circle / 2) = Ratio of DC wedge to Square.
(Area of Large Circle / 4) - (Area of Cut Square * Ratio of DC wedge to Square) = Blue Highlighted Area = ~3.85
Dont ask me where the order came from, I have a high fever right now and things are fuzzy.
Hello! Your solution is basically the best (simple but tricky) I've found here. I wanted to repeat the solution for myself, but it didn't work. - Possibilities for misunderstandings: the DC-Wedge is the small area at the right side outside "the large circle / 4" and outside the aerea of "small circle / 2" (?) What is your result of the "ratio of DC-Wedge" to the whole square (?) - Please explain it to a newbe. Kind regards R
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.
but actually it's much easier if I use calculus.
you can solve it in a much easier way, by finding the area of the small space outside of the 1/2 circle and the 1/4 circle, then the rest is just finding the other areas outside of the overlapping space, which is easy once you have the area of the space in the right side
how would you get the area of that space?
thought that at first too but nope
Can't we just use integration...like area under graph concept?
You could have also take 2 circular segments. It is basically the same but simpler.
Conceptually a bit simpler, but the math is actually a fair bit more annoying, IMHO. I kinda like the elegance of this method, to be honest.
You should not use trigonometry to solve, as it is for primary school kids.
It's cool that this equation is actually an equation for sum of two sets.
A∪B=A+B-A∩B
so
A∩B=A+B-A∪B
Where A∪B is the sum of areas of two shapes, A is first shape, B second and A∩B is the overlap.
i dont understand? they are not the same space? in fact you say A+B=A∪B+A∩B
Yeah, it is cool..., except that finding AUB is of the same level of difficulty as the original problem!
@@MrTiti yep, they are the same.
When you're doing A+B you're counting the overlap twice. And when you're doing A∪B you're counting it only once so you have to add the overlap, hence this A∪B+A∩B
thanks a lot! @@cyraxoo2791
Lmao i think this is too but i know that not gonna work here 🤣
He could have just used the sixty thirty ninety triangle method
Nice solution! I used circle equations to find coordinations of second intersection. Then I found length of secant of the circles. And then I calculated and added two circular segments.
"Some of his students in America are having troubles solving this problem."
Dude Even I am having trouble solving this (I am a software engineer)
How?
I believe a faster way to solve this is to model two functions for the circles, find the intersect and integrate.
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.
@@xqandstuff well I suppose you're right; only problem for me is that I suck at geometry and all :(
I solved it using Calculus.
First, I figured out what the two relevant functions are...
𝑓₁(𝑥) = √(4 - (𝑥 - 2)²)
𝑓₂(𝑥) = 4 - √(16 - 𝑥²) .
Next, I figured out that, as you move along the 𝑥 axis, the two functions eventually intersect at the point where 𝑥 = π .
Next, I calculated the *definite integral* of 𝑓₂(𝑥) from 0 to π .
Next, I calculated the *definite integral* of 𝑓₁(𝑥) from 0 to π .
I then performed the subtraction...
∫𝑓₂(𝑥)𝑑𝑥 - ∫𝑓₁(𝑥)𝑑𝑥
Never this is from chinese primary. It‘s from german kindergarden.
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 *100% CERTAIN* that there's a *_SIGNIFICANTLY_** EASIER way of solution* out there... :P
.
This way IS to be the easiest. I had almost the same solution as the Author had.
I solved it using calculus... I used the equations of the two circles
x²+y²-8y=0 and
x²+y²-4x=0...
Solving these equations, point of intersection was (16/5, 8/5)
Then I just solved the integrals to find the required area...
Integral(√(4x-x²)) -Integral(4-√(16-x²))
This came out to be 3.84695
Edit: I am 15 years old
Edit 2: It is my great pride that I solved the integrals on my own without using WolframAlpha
I tried this, but ran into the problem of getting a linear equation y = x/2, which didn't even seem to pass through the intersection point. Then I tried solving each circle equation for y first and got x = {0,6}, neither of which can be the x-value for a point in this square. How were you able to put these two circle equations together to get an actual solution for the intersection point (x,y)?
@@skoosharama You were proceeding correctly... Y=x/2...
Now just put it in any of the equations of the, circles and you will get the intersection point... Hope it helps...
We had integrals at the age of 18-19 in Switzerland. And we had a lot of math. Something went wrong with our education. 🤔
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.
1:45 It feels like you kind of hand-waved that the radius of the quarter circle that you drew is tangent to the semicircle. (It is tangent, but it seems like something you need to prove before assuming it.) Likewise at 2:05 where you assume the radius you drew of the semicircle is tangent to the quarter circle.
it was defined as being so
@@chimkinovania5237 What was defined as "being so"? One was defined as being the line segment from the center of the quarter circle to the intersection point of the two circles, for example. It's not "defined" to be tangent, it has to be proven it's tangent.
@@tBagley43 Thanks, good proof, that definitely should have been in the video.
It happens to be tangent-but the solution doesn’t rely on it being tangent. So it was irrelevant to the problem, and therefore wasn’t necessary to state or prove. It also wasn’t assumed to be tangent-as I said, it was irrelevant
@@verkuilb Actually the lines do need to be tangent for the proof in the video because the proof implicitly assumed there was no overhanging area outside the drawn regions (i.e. any tiny circular sections along the edges of the triangles where parts of the circles might have jutted out. In other words, if they weren’t tangent, the “sums of the areas” would have had to include some additional tiny areas as well that weren’t shown on the diagram because the diagram assumed tangency.
How dare you round of 3.8469 to 3.847 , the beauty, nooooo
I am Chinese and this is definitely NOT a question you are required to solve at the time when I was little. Though some top middle schools' entrance exam could have something of similar complexity, but I am not sure if that's still true to this date.
yeah there is so much weird thinking about china. I think it just feeds and is fed by some stereotypes that are positive towards chinese people, but that don't make sense when looking at chinese people who have not emigrated - it is westerners looking only at immigrants and making generalisations. In the past it was always the best students and/or the richest chinese people moving abroad. Nowadays, the level of student is lower due to foreign universities just wanting money. Also, it filters a lot because those who have good english and enough money are from families that are doing well - people don't realise china is just like anywhere else with a bell curve and a lot of kids struggling and a lot of families struggling. It is rough out here for kids who arent super bright or who fell behind early.
@@doyourownresearch7297 I feel ya, though the stereotyping is everywhere including China. When I was young I used to think all foreigners were blond with blue eyes lol. I was just as ignorant thinking about it now.
Agree with chaken6187. No kid learns arctangents in any primary school on planet Earth...
As a British person I was really confused with arctan but it turns out in the UK it's just tan^-1 haha it all makes sense now
I got this question in class 10th here in india it seems hard but its easy
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.)
So basic geometry which isn't this.
i think i found a super very easy way to solve it,
lets take take the 1/4 of circle ABC, the area is π(4)^2/4=4π,
and the small circle (AD) the one with diameter of 4,the area is π2^2/2 (divided by 2 because its half of circunference), so his area is 2π
and the last area is the ADC wich is the square minus the area of the bigger circle with we already know (4π)
so his area is 4^2-4π=16-4π
now we have all informations to solve, the blue area is the area of the semicircle minus ADC
so 2π(area of small circle) - (16-4π) (area of ADC) 2π-(16-4π)=2π-16+4π=6π-16
lets take pi as 3,141592653... So 18,849555918-16≈2,849555918, and this is the final result
sorry for my english
The primary school part is silly. I don’t think trigonometry is taught in primary schools in any country. Not that it couldn’t be, but it just isn’t.
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.
Why would you learn other bases? It’s completely pointless apart from binary and hexadecimal
@@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.
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.
Here's what I started with when I tried to solve it (from the thumbnail alone!)
I considered B to be the origin: quarter circle centered at (0, 0) and semicircle centered at (2, -4).
I recalled the implicit circle equation:
(x-h)^2+(y-k)^2=r^2.
That means the quarter circle has the simple x^2+y^2=16 and the semicircle
(x-2)^2+(y+4)^2=4.
After expanding the second, I set everything equal to 0. Now that this was the case, I set the sides with the variables equal. Once the dust had settled, I determined (subtracting the big circle from the small) the line going *through* the crossing points was, in slope/intercept form, y= 0.5x-4. I plugged the RHS into the big circle (x^2+y^2=16) and got a quadratic in x- but after that, I was stumped!
Yes, solving that integral is quite difficult if you haven't encountered that form before. Try u-sub with x=k*sin(u). You get an integral with arcsin.
You're still trying to find the coordinates of the intersection of the two circles?
It's actually much easier if you'd define the bottom left vertex of the square as (0,0) . But let's continue your way (you were already quite close):
You found the equation for the line connecting the intersection points of the circles, as y = 0.5x - 4 . Enter this result into the equation of the large circle (x² + y² = 4²):
x² + (0.5x - 4)² = 16
x² + (0.5x)² - 4x + 16 = 16
x² + (1/4)x² = 4x
(5/4)x² - 4x = 0
x * ((5/4)x - 4) = 0
x = 0 OR ((5/4)x - 4) = 0
x = 0 OR x = 16/5
... remember y = 0.5x - 4 ==> if x = 16/5 , then y = 0.5(16/5) - 4 = 16/10 - 4 = -24/10 = -12/5 ...
(x,y) = (0,0) OR (x,y) = (16/5 , -12/5)
Now that you know that the distance between the intersection point and the left side of the square is 16/5 , determine and calculate the proper integrals, and find the area of the blue region!
:-)
EDIT: Oops, I've let a mistake slip in: the last line of the calculation should be
(x,y) = (0,-4) OR (x,y) = (16/5 , -12/5)
(I still had the calculation with point A as the origin in mind, hence I wrote (x,y) = (0,0) instead of (x,y) = (0, -4) .)
@@yurenchu I had back-substituted because only two cases of x=2y+8 were actually intersection points: (0, -4), which the diagram gives right away, and (3.2, -2.4) which I used the resulting quadratic to determine.
@@wyattstevens8574 I see now that I made a mistake in the last line of the calculation because I still had origin (0,0) = vertex A (instead of vertex B) in mind. I've added a rectification in my previous reply now.
But now that you've determined the intersection points, you can calculate the appropriate integrals and determine the requested area!
:-)
I wrote two functions: y1 = sqrt(4 - (x-2)^2) and y2= x^2 /4. The intersection of y1 and y2 is 2.7293. At end, I finally apply the integral of sqrt(4-(x-2)^2) - x^2 /4 , 0 to 2.7293. And I got 2.8729.
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.
Great job making simple problems 100x harder! 👍
how would you do it?
4.56, got it in 5 minutes with mental math 😭😭😭😭😭😭 I'm too Asian
u dont need any trigonometry or additional figures to solve this, its just a combinatorics question where u put a bunch of initial areas into a system of 3 linear equations (i found the odd area at the right and the answer was free from there)
How did u find the area of the right triangle ?
You've clearly never been to China if you think this is what they are working on - this is graduate level in China. Chinese schools and education are really bad, it's a myth that the education there is good.
I doubt this is a primary school question in China. Maybe in some advanced mathematics schools. Took me a few hours to solve.
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!
How far a person will go not to do calculus
why dont we just subtract area 1/2 circle with the square substract with 1/4 circle
I got it but i am ashamed as indian that i cannot solve this mentally
This problem can be solved using another method and the exact answer is: 3.4016 square inches.
primary school dont learn trignometry in china
Hey, presh! I’d like to know why I got a different answer if I approximate this problem using integral. I wrote two functions: y1 = sqrt(4 - (x-2)^2) and y2= x^2 /4. The intersection of y1 and y2 is 2.7293. At end, I finally apply the integral of sqrt(4-(x-2)^2) - x^2 /4 , 0 to 2.7293. And I got 2.8729. Thanks!
Your formula for y1 describes the small (semi-)circle, in a coordinate system with the origin (x,y) = (0,0) located at A (= bottom left vertex of the square). So this formula is appropriate here.
Your formula for y2 doesn't describe a circle arc, but a parabola; so this formula doesn't appear correct. Instead, it should have been
y2 = 4 - sqrt(16 - x^2)
I got the same answer as you by using trigs
Take 2 equation considering a coordinate plane
A circle with centre (0,4) radius is 4
And a circle with centre (2,0) radius 2
Use the standard eqn of circle
(X-h)² +(Y-k)²=r²
Find point of intersection
1st point is (0,0)
As seen in figure
Now we use a concept of calculating area under curve using calculus(integration)
Did I miss something? How did you prove that the second triangle was right and also congruent to the first triangle?
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.
With the video titled “viral question from China”, I’m surprised the comments aren’t yet filled by Covid conspiracy theorists… 😂
Imagine learning trig in primary..
Lmao, my solution was to use f(x)=sqrt(4-(x-2)^2) and g(x)=-sqrt(16-x^2)+4, calculate their intersects at 0 and 3.2 and subtract their definite integrals over that interval, this is how uni drains your creativity. The solution is no less correct tho.
You haven’t Soviet kindergarten!
Before I saw the final answer, I had speculated that there might be multiple ways to solve it in addition to using trigonometry, and, if so, clearly some basic geometry taught in primary school would lead to the solution. However, since the answer has a trigonometric element, using trigonometry turns out to be a must even though there may still be many ways to solve it. In China in my generation (1980s), we learned trigonometry in so-called senior high school (students aged between 16 and 18), which is right before going to college. So I guarantee you by no means would this question attest to the math competence of primary school students, even for math competition purposes. This is unbelievably exaggerated.