Viral question from China
ฝัง
- เผยแพร่เมื่อ 8 ธ.ค. 2023
- 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!
Subscribe: th-cam.com/users/MindYour...
Send me suggestions by email (address at end of many videos). I may not reply but I do consider all ideas!
If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.
If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.
Book ratings are from January 2023.
My Books (worldwide links)
mindyourdecisions.com/blog/my...
My Books (US links)
Mind Your Decisions: Five Book Compilation
amzn.to/2pbJ4wR
A collection of 5 books:
"The Joy of Game Theory" rated 4.3/5 stars on 290 reviews
amzn.to/1uQvA20
"The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" rated 4.1/5 stars on 33 reviews
amzn.to/1o3FaAg
"40 Paradoxes in Logic, Probability, and Game Theory" rated 4.2/5 stars on 54 reviews
amzn.to/1LOCI4U
"The Best Mental Math Tricks" rated 4.3/5 stars on 116 reviews
amzn.to/18maAdo
"Multiply Numbers By Drawing Lines" rated 4.4/5 stars on 37 reviews
amzn.to/XRm7M4
Mind Your Puzzles: Collection Of Volumes 1 To 3
amzn.to/2mMdrJr
A collection of 3 books:
"Math Puzzles Volume 1" rated 4.4/5 stars on 112 reviews
amzn.to/1GhUUSH
"Math Puzzles Volume 2" rated 4.2/5 stars on 33 reviews
amzn.to/1NKbyCs
"Math Puzzles Volume 3" rated 4.2/5 stars on 29 reviews
amzn.to/1NKbGlp
2017 Shorty Awards Nominee. Mind Your Decisions was nominated in the STEM category (Science, Technology, Engineering, and Math) along with eventual winner Bill Nye; finalists Adam Savage, Dr. Sandra Lee, Simone Giertz, Tim Peake, Unbox Therapy; and other nominees Elon Musk, Gizmoslip, Hope Jahren, Life Noggin, and Nerdwriter.
My Blog
mindyourdecisions.com/blog/
Twitter
/ preshtalwalkar
Instagram
/ preshtalwalkar
Merch
teespring.com/stores/mind-you...
Patreon
/ mindyourdecisions
Press
mindyourdecisions.com/blog/press - วิทยาศาสตร์และเทคโนโลยี
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.
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.
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.
Twang twang twang twang
Me in college, unable to resolve this: 😢
Such a long rant from someone has zero experience in Chinese math competition. I am not surprised to see this type of question in Chinese primary school competitions.
@@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 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 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.
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’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.
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.
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 😁
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?)
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 😂
It's pro-CCP shills. They post this stuff to "assume dominance" when it's so obviously a bluff
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.
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
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".
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.
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 TH-cam 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 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
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
Wow, that’s an incredibly difficult math problem for primary school children! That’s more like a high school or early college problem!
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
> advanced trigonometry
> primary school
pick one
It isn't an "advanced trigonometry" until Euler's formula is involved.
Primary 🎉
This question can also be solved by two easier ways,
1- using area of lens formula
2- using coordinate geometry, area under the curve
I solved this problem through integrals and I am so glad that the results matched
This is the correct way to do this question
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@ TH-cam
You could also calculate it by using trigonometry to construct functions, calculating the intersection point, and then using the integrals to calculate the area
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)
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.
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)
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.
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.
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 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
I’d solve it using probability rules: the probability of something being in a quarter circle is its surface divided by the surface of the square (pi*16/4/16=pi/4). probability of something being in half circle is its area divided by the surface of the large square (pi*4/2/16=pi/8).The probability of something being both in semi and quarter circle is the product of these two probabilities =pi^2/4/8. Sir fade if this area is probability times surface if the square=pi^2/2=appx 5
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
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.
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
This problem can be solved using another method and the exact answer is: 3.4016 square inches.
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!
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*
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
Teacher: It's a fun and easy homework! It is only equivalent to elementary school kids' homework
Also teacher: *have arctan in the solution*
criticize please:
total area = 16 ("2d aquarium")
big circle quater area = 3,14*4^2/4 = 12,56 (so I imagined that it's a big stone)
small circle half area = 3,14*2^2/2 = 6,28 (a jelley)
total area - big circle quater area = 3,44 (space left after we immersed our "stone" into "aquarium")
So jelley will fill this remain space and the rest will be pushed beyond our aquarium. The rest is the red area, so 3,44 - 6,28 = -2,84 (is our red part of jelley that was pushed out of aquarium).
Hmm I guess I can see a problem here, a part of the squeezed "jelley" will move to this lil triangle on the right side, so we need to know its area too...
How beautiful mathematics is when everything fits together
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.
I call BS on Primary School.
If it was multiple choice I would use the areas of the square not covered by the quadrant and the semicircle and guess at the overlap, then subtract it from the total area of the square and pick whatever answer is closest. This is how I do a lot of things, which is why I am a great chef and a lousy carpenter.
The problem is a particular case of a more general situation, viz., the intersection of two circles, where two "lunes" and a "lens" are formed. In your video, it is required to find the area of the "lens". The more general solution gives the areas of the two "lunes" as well, and of course it requires a little trigonometry which, as it may be supposed, is way above the possibilities of a normal child in primary school (even in China).
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 🤣
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.
you can simply solve it by writing their equations in x-y plane and then doing integration of f[x] - g[x]
but your approach was also very simple.
I'm a student from China and to the people that says this is fake, they are wrong and when I was in primary school, I did encounter this type of question. I can't tell you the name because of well, security reasons, but yeah, this is real
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.
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 remember this question, and yes, actually I saw this in my olympic exercise when I was at my primary school. The only one solved it was the "genius girl" on my class. When she talked about the method, I totally had no idea.
You can just use coordinate geometry to find the equation for both circles, set equal and integrate to find the area between.
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
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.
Three years ago I could have solved that. But after a cerebral hemorrhage and a second ischemic stroke I'm not as sharp anymore. But I still love this channel.
This is the kind of problem we used to see in the "exams" we would have for our high school math team. Completely forgot what it was called and I'm not sure if this was nationwide, but essentially our school in New York would register a 5 (6?) person team that would take these tests every month or so, and our scores would be combined and then ranked against other high schools who have taken the same tests with their teams. Funnily enough, I think our year was a bit weak as most of our team could really only manage 4 out of 6 with two of us occasionally getting the 5/6 or 6/6, and this would be the sort of question that would pop up in the latter questions of the exam and most of us would get wrong. This was also still freshman/sophomore year of high school so we haven't even started learning pre-calc yet, much less calc, so using calc to solve this kind of problem was not possible, unless you already knew calc in freshman year of high school, which in that case, all the power to you. But they would often include these types of tricky problems that wouldn't require you to have a higher knowledge of math concepts, but still would demand you to be sharp enough to find alternate solutions, or in this case, geometrically form familiar shapes by simply drawing lines inside or around the original shape to help deduce the final answer.
Interestingly enough, I never saw these types of problems outside of those exams even during the AMC. I was never able to qualify for the AIME and had to quit math team after sophomore year for external reasons, so I can't verify if these types of questions would be seen in the AMC 12, but it was always funny to us how these random exams simply used to rank our high schools for seemingly innocuous reasons was much harder than the national math competition itself, albeit just an introductory competition.
my school that was opt-in, and basically everyone who liked math took the exams, because it was a way to get out of random other lessons for a fun math puzzle once a month. only the team people's scores were used, but the rest of us were basically trying out to be on next year's team and didn't know it
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
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.
@@danielcpeters 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.
@@danielcpeters 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.
you can also do it like this
sqrt(4-x^2) = (-sqrt(16-(x+2)^2)+4) => x= 6/5 to get the point where the lines meet
integrate (sqrt(4-x^2) - (-sqrt(16-(x+2)^2)+4)) from -2 to 6/5 dx = 3.847
As comments said, it's definitely not regular even in competition level. Trigonometry was taught to me in Taiwanese middle school(a top class), so I can see this be a middle school problem for advanced students.
Great job making simple problems 100x harder! 👍
how would you do it?
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'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.
I am 14 and from India and we usually deal with this kind of questions so it was easy for me
I added the two segments that I could form in that shape to find the area. It was a relatively easy approach and less complex.
Sure, but you also got a Masters in Tea by 16 and dance ballet by 18 and performed brain surgery after that until you discovered programming. We had a student from India in a statistics class, he could do math in his head, he was wrong 100% of the time, what are the odds?
lier
How is this primary school???
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 moment I realized we'd need areas of sectors, I knew I couldn't do it without looking things up. That's a formula I never remember.
Isn't it just the area of a circle multiplied by the ratio of the angle to a full turn?
can be easily done within minute using area under the curve integration method ,considering left bottom corner of the square as origin
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.
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.
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.
For this kind of question, which is likely for student maths competitions, calculators are not allowed, and the students would not have learnt radiant angle or arctan in primary school. So in the actual question, it will ask you to recall the angles of a 1:2 right triangle. The skill being tested is that you can identify these triangles and recall the ratio between fan areas equal to that of the angles.
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.
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.
I've used brute force and calculated the area using an integral of a difference of functions representing the circle curves. Zero creativity, but it works
I mean i'm in highschool and i could see this being a question the teacher would put as a challenge for us, but for primary grade?
My solution: Find the equation for both circunferences and match them, this way you will find the 2 points of intersection. Next use the vectors that go from the centers of their circunferences and go to the point of intersection, then use the dot product to find the angle that separate them, then calculate aproximately the area of the circles sectors, and subtract the triangles, and voilá, there you have it. I know it's not clear, i put the solution in my community page, bu if it's still not clear i can solve it in my channel. Let me know.
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 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. 🤔
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.
I thought a bit and got to conclusion that we need to solve the following situation. One circle "stays" on a horizontal line touching it with its one bottom point. Its radius is 4. Second circle stays at the same line at the distance of that radius touching the line also with one bottom point. Its radius is 2. So between the two bottom touching points of those circles the distance is 4. R1=4, R2=2. We need to find the crossing point of the circles. Eventually we need to find the area between the horizontal line and the circles. Once we find it we can solve the problem.
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 ?
Not a primary school question. Not right. No click baiting plz.
Great! What an elegant way to CONSTRUCT an answer (like literally). We could've also used properties of triangles (cosine rule) to figure out what the angle is(in terms of pi radians), rather than using inverses.
But these geometry problems are seriously obsessive!!
Nope. Spend 40 years creatively avoiding maths (I'm a science batchelors graduate). Maths at school was a continuum of frustration and despair.
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.
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
I think this could be a great lesson on how calculus can allow you to shortcut a lot of very difficult geometrical thinking at the expense of an immediately intuitive explanation
By "primary school math", you may be referring to Math Olympiad students, a subgroup of primary school students selected to participate in Olympiad Math education and competitions (almost all primary schools have them). I took part in such a group back in primary school. They rely heavily on memorizing typical auxiliary lines, and yes, complex trigonometry and geometry are on the table, just like sequences and series. One of my worst childhood memories
Great combo of trig and geometry. Maybe quicker to look for intersection of the two circles using calculus?
I doubt this is a primary school question in China. Maybe in some advanced mathematics schools. Took me a few hours to solve.
If a student solves it while he is in primary school, he must be appointed as Minister of Education
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}]
same! I am glad I am not the only one.
can you teach me how you computed their functions to find where they intersect?
I haven't bothered to perform the calcs but it looks easy enough to calculate as the sum of a segment of a 2 unit radius circle, plus one of a 4 unit radius circle, the angles their chords / arcs subtend not being terribly difficult to calculate...
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.
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.
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
✌️☺️