The notorious three-circle problem from the GCSE
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- Äas pĆidĂĄn 25. 04. 2024
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Here's the last question from the 2022 GCSE maths paper that made the news. We have three circles as shown and each radius is 4 cm. We have to find the area of the shaded region in the middle. I made a horrible mistake last time when I said the area of a circular sector is r*theta. The correct formula should be A=1/2*r^2*theta and theta has to be in radians.
My first time doing a British GCSE maths paper: âą First time solving a G...
Reddit: / self_last_years_gcse_m...
The Sun News: www.thesun.co.uk/news/1864787...
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and you'll get a lot of views as well
If you didn't make (and show) your mistakes, people would respect you less. Well done for showing not only your fault, but the solution.
Bruh so many people here saying how clever they are and that they have no idea how this is "notorious". Its a GCSE question done by all 16 yo a lot of whom have no interest in maths, just because youre doing complex analasis or linear algebra at college or uni doesnt mean that a 16 yo kid cant struggle without you saying how much better you are, why cant we appreciate that some people struggled and be happy that this vid might help
they can't get the appreciation they want from their teachers or parents, so they resort to youtube comments to boast about their math skills
i too think the question isn't hard at all tho, all you need is practical knowledge
I am 16, turning 17, and I have never heard of this shit. But that is because our school system here is dog crap. I would have failed this very hard.
I had this question during my gcse, was the only one in my class who had got it right on the test and ended up with 235/240 marks out of all 3 papers
yall are learning linear algebra at college or uni?????? wtf?
@@activatewindows7415pretty standard for STEM majors
You wouldnât be expected to know about radians at GCSE. The area of the sector is one sixth of the area of a circle.
That looks like a solvable question. You don't need to make a fancy construction once you realize the broken arches can be made from by adding 2 circle sectors and subtracting the extra triangle.
Seeing the previous problem (13) with proving the semicircles grown on the right angle triangle reminded me of the last problem in my maths textbook for circular arcs (romanian school). We had to find our the area of the lunules of Hippocrates. You get shaded area of arches drawn between semicircles drawn on cathetes and semicircle of hypotenuse passing by the right angle point, which turns out to be exactly the area of the right triangle.
Seeing the similarity of previous problem and having this come up leads one to just consider addition and subtraction of surface areas, knowing this exam is precalculus.
You are expected to know
@@anghme28ang11nope
That is a UK problem then.
Because average Italian, French and German 16yo students know about trigonometry and radians.
GCSE are joke exams. đ đ»
Your second sentence is weird.
Yes the area of THIS sector is a sixth, but you have to show that it's a sixth in the working out, you can't say "cause I see it's a sixth" đ
Great video. The only comment I will make as someone who teaches maths here in the UK, is radians are not taught until A level (beyond GCSE). GCSE students are taught area of a sector by considering it a fraction of the area of the whole circle using degrees: (theta/360) multiplied by pi.radius^2
That is a UK problem then.
Because average Italian, French and German 16yo students know about trigonometry and radians.
GCSE's are very weak exams.
đ đ»đ đ»đ đ»đ đ»đ đ»
Continental kids are better that Brits up until uni level.
Then in uni, with the relationship you have with the work environment, UK is the very best.
So yeah... if you have a kid in England, make him study abroad from 11 to 18 and when he comes back he can kick arse to his peers đ
@@MartiniComedian you sound like an incredibly grating person to be around, i hope only the worst for you, much love from bob 10
â @@MartiniComedianwhat ? Different countries learn different syllables. Other countries might learn about radians sooner in the uk but the uk would learn other things sooner as well. Weird comment.
@@MartiniComedian mate people just have different curriculums. We were taught trigonometry at 13 years old. And while one GCSE is not that bad, the fact that you have to do 10 subjects, with 2 papers for each, which will go on your job applications for the rest of your life, they are fucking brutal.
"Great video. The only comment I will make as someone who teaches maths here in the UK, is radians are not taught until A level (beyond GCSE)."
I didn't see the point of using radians, I'd not have done it that way, but I have to say the 1968 SCE "O" grade paper included radians, but even that was beyond GCSE... I'd have calculated the area of one 60 degree sector, (1/6th of a circle's area) then the area of an equilateral triangle of side 4, subtract that from the sector area, giving the segment area, and subtract 4 x one segment's area from 2 x a sector's area, we have the area of the shaded portion. But I have to express that in terms of pi? I'd have divided my answer by pi, giving a neat answer (I think) And I'd hope not to be penalised for not following a prescribed path...
In my day, pi was expressed as 22/7 (no calculators in 1968...) and circle radii/diameters expressed as multiples of 7... (they weren't that cruel). Maybe we were permitted/obliged to use slide rules and log tables?
I didn't advance much beyond "O" grade maths as a discrete subject, only as part of machine shop and sheet metal calculations... (where everything's pared down to its simplest, since complexity brings with it opportunities for error)
GCSE maths will only be in degrees no angles in radians. Radians are only introduced to maths students doing A level in the UK.
Yeah, I was able to understand this because I was in Year 12 when the question came out, but then I realised I was using A-level formulas to do it, idk how the kids in the year below me were coping with it, most of whom weren't even going to continue maths to A-level!
They use an alternate formula, pi*r^2*(theta/360). It simplifies to the same formula but is a lot more digestible for them.
The "rule of thumb" I give my students is spend on average 1 minute per mark. This is based on a 90 minute exam, for 80 marks, with 10 minutes for checking work.
Of course there are some 1 mark questions at the start of the paper that can be answered in seconds, and I think you really do need to work quickly on those basic questions to build up a time cushion because in my view some of these later questions can be quite involved, and will take longer (in minutes) than the number of marks being awarded.
This was a video of over 8 minutes and yes, there was some time taken for explanation etc but as a calculus professor BPRP knows exactly where to start and which direction to take. I think most 15-16 year olds would in any event take longer than 5 minutes to answer it, especially given the number of steps and the need to methodically set out working.
I do think that's a bit of a shame as it over-rewards simple questions and under-rewards the difficult ones. If I were designing the paper I would increase the number of marks and keep the very basic questions that can be answered in 30 seconds or less at 1 mark, whilst weighting the difficult questions (normally up to 5/6 marks) more and giving them say 10 marks.
Exam boards should be able to work out what is the required amount of time to spend on a question, and allocate marks accordingly, but until they do that, I'll keep advising (a) pick up the easy marks quickly (b) don't dwell on a question too long, move on to the next and (c) come back later if you have time". That's general exam strategy, not specific to maths, of course.
half the people never get to the end of the paper, there is no point weighting questions more if people aren't going to end up there, thats just lowering the threshold and making it harder to classify into students that at different grades.
All it does is makes it better for students that are good at pattern recognition and have done many past papers to get the answer right than for students who take their time to think through a solution.
@@rbanerjee605 I'd have to see some evidence for the claim that "half the people never get to the end of the paper" but leaving that aside, my point is that all marks should be worth the same.
If the problem is (as you state) that too many students are not getting to the end of the paper, then that is a solvable problem - you either (i) have fewer questions or (ii) increase the time allowed to complete the paper or (iii) mix up the order of the questions more between the "simple" and "complex" ones.
If, say, the order of the questions were mixed up, (some) more difficult questions would come before (some) easier questions, mitigating the issue somewhat. Assuming there will be a number of students who do not read ahead, and who will simply deal with questions sequentially, they may miss out on some simple marks altogether because they are further in, whilst possibly struggling with the earlier, harder questions.
Therefore, as long as the structure of the paper is such that you get basic, 1 mark questions at the beginning, with progressively harder questions, I think my argument holds, which is that a reasonably good measure of the difficulty of a question and the number of marks it should be awarded is the average time it should take to answer it.
Otherwise, not all marks are created equal and I can't see any logic for that approach.
@@tanelkagan The purpose of examinations is to distinguish different skill levels. Adding some disproportionately difficult questions helps distinguish the highest skill levels. If they're kept to the end of the exam then they won't impact less-skilled students who will attempt the questions in order.
@@4hodmt Of course, I don't think it's ever been disputed that exams are designed to distinguish between students of different levels.
Mixing the order of questions up was not something I was pushing - I was merely exploring the possible implications of such an approach given the alleged problem of "half of students not finishing the paper".
I don't believe that anyone sits down and specifically tries to design an exam paper so that half the students will not finish it. If we are assessing different skill levels, it is far better and far more meaningful to do that based on questions that *are* answered, rather than questions that are not.
This goes back to my original point - if the order of questions is to be kept as it is, we should still reward questions as finely as possible based on complexity, with the time taken to answer the question being a measure of that complexity.
In simple terms, it should take on average 5 times as long to answer a 5 mark question than it does to answer a 1 mark question. That's all. Where you have 1 mark questions that you can answer in 20 seconds and 5 mark questions that take 8-10 minutes, something isn't quite right and that's why I would prefer to see better time/mark weighting. I can't see how anyone could logically argue against that.
Really hoping for questions on Thursday to be beauties like that. I would run out the exam hall in glee if it was simply just trig lol
This was in the non-calculator exam and it brought back some funny memories of the exam, it's nice you're covering it nice video!
Seriously?
@@lawdohio Yes, I sat my GCSEs and now Iâm doing A Level Maths
@@sampanchung1234 Brutal. It's not that bad but under exam conditions... I think anyone could justifiably blank out.
@@lawdohio yeah I think I got only 2 marks in that question, I was pretty exhausted as it was the last question on the exam but I did learn a lot from that experience
It was so weird lol everyone was so confused after the exam.
An alternative approach is to find a segment of either of the outer circles (radius 4cm, angle of 120 degrees (since radians aren't technically in the GCSE syllabus)) and then multiply that by 4 and subtract from 16pi.
That was my approach yeah
The only thing that gave me a pause was proving that the intersection points (vertically) marked 1/3 of the circumferences. Because of the radii of the center and one side circle, you wind up with triangles of identical side lengths, which implies 60-degree angles, etc.
@@josepherhardt164 saying in the q A, B and C have the same r, so its kind of already implied and ofc easy to prove... its also easy prove CC A-C is a straight line... after that its becomes rather simple
Shaded area = rÂČ(2Ï/3) - 4[ ( rÂČÏ/6) - ( r(râ3)/4 ) ]
Clarifications
sectors containing shaded regions = 2 * rÂČ(Ï/3) = rÂČ(2Ï/3)
sector segment = ( rÂČÏ/3)/2 = ( rÂČÏ/6)
triangle = ( r( (r/2)â3)/2 ) = ( r(râ3)/4 ) [ Base=r, Height=(r/2)â3 ]
@@josepherhardt164 I confess I over complicated that part when I did it in my head. I did inverse cos(1/2) (requiring me to further use the result that triangles from the diameter of a circle to the circumference are right-angled) instead of just observing the perfectly obvious equilateral triangle.
Yeah, radians superflous as the intersections are all based on perfect hexagons thus equilateral triangle based, and thus the circle is cut in sixths (or thirds, depending which approach you go)
I did it slightly differently with a mix of geometry and basic algebra: if you draw some additional circles of radius 4 centered on intersections, you can partition the central circles into 6... let's call them concave triangles (CT) and 12 ... almonds (A). The grey area is then equal to 2CT + 2A. Since the circles have radius 4, we know 6CT + 12A = 16Pi and we also know that the area of an equilateral triangle of side 4 can be described as follow CT + 3A/2 = 4 Sqrt(3). From there you manipulate these two equations to reach 2CT + 2A = 16 Sqrt(3) - 16 Pi / 3.
I looked at this paper as a final year engineering student and it stumped me for a few minutes too
A less cumbersome method:
Call the upper intersection of the circles A and B, Y, and that of B and C, X.
BX=BC=XC=4. BCX is therefore equilateral, so 6 such triangles will form a hexagon in Circle C. Call each triangle area A(T)
Equally, Circle C comprises 6 pie slices of equal size - the triangles plus a curved portion. Call each slice area A(P). It's just 1/6 of the area of the circle.
Next, the part in each slice but not in each triangle is a sliver. Call the sliver area A(S)
But also BXY is another congruent triangle, and half the area we want, with 2 slivers extra.
So for the whole of the desired area, we need 2A(P)- 4(A(P)-A(T)) = 4A(T)-2A(P)
A(P) = 1/6(pi*4^2)
A(T) = 1/2(4*(2sqrt(3))
4A(T)-2A(P) = 2(4*(2sqrt(3))-1/3(pi*4^2) = 16sqrt(3)-16pi/3
I was feeling so confident going into this exam, i finished the rest of the paper with half an hour left and i thought it was going so well, and then the last question hit me đit took me all of that remaining half hour to solve this!!
I just did integration, I converted it into geometric coordinates considering B as origin, found each coordinate, found equations of the circles, found points of intersection of the circles the integrate from 0 to 2 for 1 part,, then 16pi-8(segment area) = ans, should mention each segment is that area from 0-->2 that's part of the A and C circles that are 8 in total
Pretty interesting way to do it damn, but yh this is done by 16 year olds in the UK đ, 80% which dgaf
samee I too did it with integration
but tbh its a lengthy way.. (especially if you dont know remember the formula)
Not sure you would get the marks as integration isn't on the GCSE syllabus.
I got
|(-16Ï - 48 â3) Ă· 3|
without using the formula.
I divided B into 6 equal pieces, then made a triangle for every piece.
I calculated the area of the triangle (At).
Then I calculated the area of the small thing left,
As = (â circle B - At).
Finally, it will be
Shaded Area = 16Ï - ((As Ă8) + AtĂ4).
I simplified and got my answer.
For the people in the comments ! Yes this question isnât incredibly hard to solve if you know your stuff, but remember that most of the students that did this question were 16 year olds that dgaf with probably around 5~10 minutes left if lucky đ
And as such this is probably the differentiator question for those that get absolutely top grades vs just very high grades. You're not going to fail GCSE because you can't do the one hardest question on the paper.
A task in my maths book when I was 15 was a lot like this. It was my favorite one I had ever done at that point and it really further solidified my love to maths. I think I may still remember the exact page and task number of it.
Alternatively you could draw the 6 small equilateral triangle created by the intersections of the circle and conisder the top, bottem and both on the right ones. After that move the gray circular segment such that the gray areas look like a equiliteral triangle with 1 circular segment missing and move the white area of the top and bottom equiliteral triangle to the right side of the white triangles on the right. Then the gray area is:
A_gray = 4*A_triangle - 1/3*A_circle = 4*(1/2*4*sqrt(3)/2*4) - 1/3*(PI*4^2) = 16sqrt(3) - 16PI/3
This one was easy thanks to your explanation
Here is a different way to do it: Let AB = 1, then let D be the upper point of intersection between circles A and B. Then find angle, E, between segments AB and BD. Drop vertically down from D, point F, at intersection of segment AB. Point F must be mid-point of AB. Thus, from right-triangle, Z, formed by FB, BD, DF, angle E can be solved by arccos(BF / BD) = arccos(1/2) = pi/3. Area of corresponding sector is then, Y = pi/3 * 1/2 * r^2. Radius is same as segment AB, so sector area becomes, Y = pi / 6. Area of triangle Z is 1/2 * BF * DF. Because Z is equilateral, it height, segment DF, is srq(3)/2. Area of Z is then 1/2 * 1/2 * sqr(3)/2 = sqr(3)/8. The area of one full circle, W, is pi*r^2. Because r = 1, W = pi. Then, to get the area in question, V, that's the circle, W's, area minus 8 slivers; each sliver is sector, Y, minus triangle, Z. Or, W = V - 8 * (Y - Z) = pi - 8 * ( pi/6 - sqr(3)/8 ) = pi - (8/6)*pi + sqr(3) = sqr(3) - (2/6)pi. To get the final answer, V, requires scaling AB back up to 4, which results in 16 times the original calculated area, or 16 * V = 16*sqr(3) - (16/3)*pi. I actually paused the video this time! :D It's basically the same way blackPenRedPen solved it, but using the full circle, instead of a quarter-circle.
I'm an International GCSE student. And I'm glad to see that you made a video about solving a GCSE math question. Please upload videos like this regarding the hard GCSE maths questions.
The point here is to not freak out and think you have to subtract stuff from a square. My approach: Complete the red circle, draw the vertical chord lines where the circles intersect, and then draw the radii from the three centers to the intersection points. After that, it becomes clear.
Even quicker: use your sector formula to get the address of the 30deg sector in one step, instead of taking the 90deg quarter sector and subtracting the 60deg one.
Or: having got the quarter sector, the 30deg sector must be 1/3rd of it.
Since 1 circle is made up of 6 equilateral triangles and 6 minor segments:
Let O = area of a circle = ÏrÂČ
let â = area of equilateral triangle = ÂŒrÂČâ3
let â = area of minor segment =?
let â = shape internal to area of â such that, â= â +2â
hence we can form a system of simultaneous equations;
A = area required = 2â + 2â
O = 6â + 6â
â = â +2â
Solving for unknown â by seeing that
2â = â O -2â
2â = 6â - â O
adding these two eqn. gives solution,
A = (â O -2â) + (6â - â O)
= 4â - â O
= 4(ÂŒrÂČâ3) -â (ÏrÂČ)
sub in r= 4cm
gives area A = 16(â3 - â Ï) cmÂČ ăĄăĄ
Finally some real math notation
@@TheSourovAqib I struggle with "real math notation" But I know how to calculate the area of a circle, and I know that, as a proportion of the circle that circumscribes it, a regular hexagon has an area cos 60 / 1 of the circle's area... from that, I can calculate the six segments' total area, and thence the area of four of them, the area of two sectors, and subtract the smaller figure from the larger. that gives the area of the shaded portion,
It's not expressed in terms of pi, but, divide by pi, you get there... in a rough and ready way.
@@robertlawson8572 indeed
@@TheSourovAqib A somewhat brief response... Having worked in engineering for most of my life, I found practical maths needed to be cut down to its bare bones to be performed quickly and (sufficiently) accurately on the shop floor. Esoteric stuff doesn't fly well in the workshop.
Nice job!
How about this:
Area of the Square suroundin middle circle is 4R^2
Eliminate area covered by left and right half-circles = subtract pi R^2
Now you have to get rid of the wavy bit along the top. This is a curve x^2 + y^2 = R^2 that goes from 0 to R/2 , four times across the top, and four times underneath. You need to subtract this too.
The area of this is 8 { area under the curve from x=0 to x=R/2 ; curve is y^2 = R^2 - x2...
Yeah we get to do some integration
I = integral(0 to R/2) of sqrt(R^2-x2) dx
The overall result is 4R^2 - pi(R^2) - 8 I
The integral I comes out to be R^2/2 ( arcsin(X/R) + (X/R) sqrt(1-(X/R)^2) )
Which evaluates to zero at x=0 and to
I = R^2 /2 arcsin( 1/2) +(1/2)sqrt(1-1/4)
I = R^2 /2 arcsin( 1/2) +(1/4)sqrt(3)
arcsin(1/2) is pi / 6
4R^2 - piR^2 - 8 I = ... = R^2 ( 4 - sqrt(3) - Pi / 3)
Which I think is what you got.
As a check, you can see by inspection that the shaded area is a little less than 1/3 of the area of one of the circles, i.e,
The worst thing is my math teacher put a somewhat similar question for me on the board, just because I do calculus in my free time, and judging by his face, he wanted me to find it's area without a limit. Solved it like you did. After class he told me, that he wanted it to be somehow get solved with the golden ration, like wtf. anyway, got an A :)
LOL. Your teacher has conflated sqrt(3) with sqrt(5).
How do you solve an area with a limit?
Limits apply to functions with a range of values that converge to a number. This area function simplifies to A(r) = (r^2)[ â3 - (pi)/3 ]. Thus, lim(r-->n) A(r) for some value "n" converges to an area A(n).
If you want an area equal to the golden ratio, then solve for "r" in, (phi) = (r^2)â3 - (pi*r^2)/3 ----> (phi) = (r^2)[ â3 - (pi)/3 ] ----> r = â [ (phi) / (â3 - (pi)/3) ] = 1.537074995 .
If you want a radius equal to the golden ratio, then substitute "r =phi" into A(phi) = 1.792969103
By some fluke got this in about 12 minutes first time. I worked out the cord length (4 * sqrt(3)) and the angle subtended (120 degrees). Then worked out the segment area with that cord length wand subtended angle. That is the sector area - area triangle of height 2 and base 4 * sqrt(3). There are obviously 4 of those sectors where the two outer circles intersect the first one so just subtracted 4 of those segments from the area of a circle with radius 4.
In any event, a tough question and a lot of time for just 5 points.
Apparently the term for the shape bound by a chord and the circumference is a *circular segment.*
explain how it is bounded by a chord
@@savitatawade2403 It is bounded by a chord *and the circumference.* In other words, that third shape he found the area for.
@@ZipplyZane oh I thought you were referring to the overall area
For the area of equilateral triangle, wouldn't it be easier if we just use "Area = 1/2 ab sin C"?
You could have gone the calculus way by integrating (r^2 - x^2)^{1/2} - r/2 from x = - r*sqrt(3)/2 and + r*sqrt(3)/2. After that you just need to substract 4 times the result to the area of the circle. Not sure if it is faster, though.
Well you see, this was a non-calculator exam and integration is not taught at GCSE level so you cannot do that as you wonât get marks for it
@@Colea1010 They don't give you the mark if you use a technique that they didn't teach you directly? I didn't know that. Also, you don't need a calculator.
@@jeremielhomme8572 yeah because you get a mark for the correct answer, but most the marks come from working out. Theyâll have some methods for working out listed and if your method isnât there - like if you use integration, then you donât get the marks for it. Itâs happened to me a few times doing past papers - for example 1 Q asked 27^x = 3, and doing log_27(3) to get the answer wouldnât award more than 1 mark (3 mark question btw), and you instead had to recognize that cbrt(27) = 3 and cbrt(27) = 27^(1/3) and get x = 1/3 that way
@@Colea1010 ah ok. Thank you for the context. I am not familiar with that exam. It seems pretty hard!
@@Colea1010 That's not true. You get marks for any mathematically valid method (unless they explicitly state a method, which they don't here.) The examiner may not understand it, but they would escalate it to a chief examiner as it's not on the mark scheme.
I used slightly different shapes, and thus had different interim values, but I did still come to the same result.
how about drawing the three circles on the coordinate axis(take point B as origin for convenience) so you can get the equation of the three circles and then find the area using integration
This is an exam for 15/16 year olds. Integration is not on the syllabus at that age
That 'wedge' is just a segment, or a minor segment.
Integrate to find area under curve of root(16-xÂČ) from 2 to 4, let this be = k
Therefore, the required answer is 16pi - 8k
And k comes out to be = 8pi/3 - 2(root3)
Answer : 16pi - 64pi/3 + 16(root 3)
= 16(root 3) - 16pi/3
Value of k :
Put x = 4 sin y
dx = 4 cos y dy
When x = 2 then y = pi/6
When x = 4 then y = pi/2
Therefore, now we have to integrate 16cosÂČy from pi/6 to pi/2
Put cosÂČy = 1/2 + cos 2y/2
Therefore after integrate it will be 16(y/2 + sin 2y/4)
Put the limits
16(pi/6 - (root3)/8)
= 8pi/3 - 2(root3)
Hope u like this đđ
Itâs a nice idea but sadly would get no marks in the exam as integration is not on the curriculum for maths GCSEs đą
@@Daniel-yc2ur Ohh
Remember these are 15 year olds taking this exam
The mark scheme for this question did give a mark for the correct answer (or equivalent), so they'd likely get 1 of the 5.
The GSCE is similar to a GED in the U.S., and it would be crazy if this appeared on the GED.
This was the last question on the higher paper, so only the candidates who were getting the highest of grades would be expected to get it right.
Not quite, a student will take GCSEs when they are 15 or 16.
I think it's easier to find the shaded area by thinking of it as the whole middle circle (centred on B) minus the overlapping (white) regions from the other circles.
If you draw a chord line between the points where two of the circles intersect (say the circles centred on A and B) that will split one of those white regions directly in half. The region to the right of the chord is a segment of the circle centred on A.
You can find then the area of that segment (which is a skill a GCSE student would be expected to know), multiply it by 4 to get all of the white then minus that from the area of the circle to get your final answer of the area of the shaded region.
This was my approach as well.
If you have a function f(x)=x+inf. What would be its derivative? On one hand infinity is like a constant so its derivative is 0 and that of x is 1 so f'(x)=1. But on the other hand the function is always inf. and, again, infinity is like a constant so f'(x)=0. So is there a definite answer?
"inf" isn't a number so you can't really use it in equations like that - the only time bprp uses it is to represent limits like when x -> inf of 0.5^x: you can sort of write that as 0.5^inf but that's only really a representation - you can't actually take 0.5 to the power of infinity. Your question is a bit like asking what inf/inf equals: on the one hand infinity is infinity no matter how much you divide it but on the other hand inf*1 = inf, divide by inf on both sides you get 1=inf/inf so inf/inf must be 1 but then you'd also get inf/inf = 2 with the equation inf*2=inf. In short, equations with infinity don't make sense because infinity isn't a number
This was very easy! You observe that the wanted area is the area of the central circle subtracted by the areas of the two "US football" shapes that happen when you intersect the central and the side circles. Meanwhile, half of each that shape can be seen as the area of two overlapping thus summed circle sectors of 1/6th of a circle but since overlap subtracted by one area of the equilateral triangle (since in the two sectors added there is one triangle that is added twice, thus we need to subtract it). And there are 4 of these half-shapes.
So, the whole equation goes:
A = r^2*pi - 4*(1/6*r^2*pi + 1/6*r^2*pi - r^2*srqrt(3)/4) = r^2*(srqrt(3) - pi/3)
If you have to write a thesis to explain it then no, it's not "very easy", especially for 15/16 year olds, I got a 9 which is the best grade for GCSE maths and I would have had an aneurysm if this was the question they gave for my year, thank fuck it wasn't
@@mwgaming5167 â OK then, for 8th grade (with geometrical awareness):
a) 2 circles (centers A and B, same for B and C) intersecting form equilateral triangles between their centers and intersection points (call those E, F, G, H). 5th grade math
b) connect the dots between 3 circles centers (A, B, C) and all intersection points (E, F, G, H). You get 6 equilateral triangles (also making up a hexagon AEFCGH, not important but so you can visualize easier)). 6th grade math
c) see circle segment ACE? it's 60 degrees cause its equilateral. Its sector area is full circle x 1/6 or 60/360, as you wish. 7th grade math
d) get 2x that sector area and deduce one triangle area, gives you exactly half of a "crescent" area. That one has to be seen, I agree not for everybody, some people can never see it. Here is where geometry awareness is needed.
e) there's 4 of these areas in the 2 "crescents", so you multiply by 4. 2nd grade math
f) deduce that product from area of one full circle (the middle one, with center B). 1st grade math
@@z000ey I mean when you break it down anything is easy, but well done you you're so cool for flexing your maths knowledge on the world
@@mwgaming5167 not my intention. Just wanted to point out we don't need radians (which might make it uncomfortable to some), just stay in plain 1/6th of the circle, keeps it simpler for pupils.
i got a whooping 1/5 marks on this question while doing it as practice for my gcses this year lol
good luck!
Don't dismiss a single mark. They all count. Do as much as you can on any question.
@@ellentronicmistress4969 yea a single mark can be a big difference and thankfully I learnt that before the real thing, thanks
Hey, can you solve imo 2007 p6, its very intuitive
please find a formula for summation of n^p from 1 to a for variable n
I liked the video however theres one thing to point out.
I sat this exact paper and this was the one question I couldn's solve. I still have nightmares.
Anyway, the method you used utilised radians, which is very convenient for finding areas. But in our GCSE specification we do not learn about radians so I don't think this method is fair for GCSE students.
I mean, it's still pretty simple. A 60° sector is just 1/6 the area of the circle.
Solved it faster than the duration of the video and correctly, so I call it a win.
Slightly simpler way: Consider the top half of the diagram. The bulge out at the top is equal to the bulge in the either side of it. So replace the top and 1 side with straight lines (let's use the side near A). Draw straight lines from B to C and from C to the upper right corner of the gray. Now you have a parallelogram. Figure it's area. Subtract 1/6 of a circle (the sector from C along the 2 lines drawn previously). Double the result, to add the bottom half of the diagram.
What
Maybe a little easier: Let's call the equilateral triangle area T, and the little crescent area C. The top shaded area is just a copy of the equilateral triangle with 2 crescents deleted from the bottom and 1 crescent added to the top. So its area is T - 2C + C or simply, T - C. The video shows that T = 4â3, and C = 8Ï/3 - 4â3. So the top shaded area is T - C = 4â3 - (8Ï/3 - 4â3) which simplifies to 8â3 - 8Ï/3. Double that to get both shaded areas and we get 16â3 - 16Ï/3.
Also, you don't need to know about radians to solve the area of the pie-shaped piece. It's simply 1/6 of a full circle, so its area is ÏrÂČ/6 which is 16Ï/6 or 8Ï/3.
Absolutely no radians needed! It's a piece of a perfect hexagon, either you are looking at the whole crescent and thus 1/3 of it, or you're looking at the half crescent in which case it's a 1/6th of it, either way, radians superflous
I guess we can solve it easily using the areas of ellipses which is Ïab
You can make it easier if you use the formula for the area of a circular segment.
How many people have that memorized?
ermmm i dont remember GCSES being like this lol. although once you notice that the outer circles share an intersection with the centre circle, youre basically done.
yeap i gave igcse exam on that year and that was the paper got memories back
great that you got it right but its a GCSE maths question and the techniques that you used i dont even think theyd count it as a marking point. thats like doing integration and differentiation in a gcse higher maths question. not necessarily incorrect but doesnt follow the spec
can you tell me what's the app you're using please?
Can You make a video on Gudermannian functionn Calculus? And also Diffrential Equations of 2nd order?
I can't wait for that day many years in my future when I'll be sifting through the fruit section of a mostly empty supermarket late one evening and suddenly this problem appears out of a portal from another dimension and demands I solve it... Thanks to this video, I'll be ready.
4 * integral sqrt(16-x^2) - sqrt(16 - (x-4)^2) from 0 to 2 â 10.957652, which is the same as BPRP's "exact" answer
I dont think they do integrals in gcse...
gcse is done by15/16 year olds, no integrals in gcse mathematics
I was gonna say integration but remembered it's GCSE
My advice if you get a question like this is at least have a crack at it and show working. Even if you don't get the solution, doing this can still net you a good portion of the marks. Also consider if your time is better spent trying to get the whole solution to this or doing what you can on it efficiently and taking a second look at any other (perhaps easier) questions where you can still improve your mark.
I'm a physicist by training, 1st class Masters from a uni generally considered in the top 10 globally, and the solution wasn't immediately obvious to me - so if you're one of those who didn't get the answer under time constrained exam conditions, please don't feel bad about it. This is the hardest GCSE question I've seen. Of course taken step by step the maths is not difficult, but the devil is in figuring how to break it down. These kind of questions invariably boil down to figuring out that you can get the area of an odd shape by subtracting the area of two simple shapes.
@blackpenredpen Dear Blackpenredpen ,by the way, I want to ask a question. Is the arbitrary constant of indefinite integral is written in capital 'C' or small 'c' ?
Either, doesn't really matter as long as you're clear and consistent. We always used a capital "C" perhaps to distinguish from variables which are typically lower case? đ€) but I don't think it matters as long as the meaning is evident from the context.
We always used "y = mx + c" for the slope-intercept form of the equation of a straight line and thought nothing of it. Many countries use "y = mx + b" instead and I don't like that quite as much since it makes sense to use "c" for a constant there also. Once we got to calculus however our teacher always capitalised "C" for the constant of integration.
I suppose if you were in the unusual position of solving a problem that dealt with straight lines (which typically don't require calculus) *and* anti-derivatives then using c/C *might" get confusing and b/C or b/c would be better. As always, just be clear on your definitions and notation and all will be well!
@@tanelkagan thank you very much, sir!!
@@HeinThantAung-xs2lo You're welcome - you didn't get a reply from BPRP but I hope my reply was a good consolation prize!
think only 1 person in my year managed to figure out how to solve the question in the exam although they didn't have time to finish it
I remember having this in my paper. It was horrible but managed to get the 5 marks. Thereâs another horrible question in 2023 papers near end of paper 2 I think but might not be avliable yet.
what was the question again? i remember everyone near me getting paper 1s last question wrong
@@DotDotEight it was the hexagon one
I'd call it fiddly and tedious, but horrible? It's about picking the problem to pieces then assembling the pieces.
â@@3TAN12Eoh I remember that one when I sat the paper. But I'm so grateful for 2022's circle question because they were pretty similar so I understood what to do
@@3TAN12E oh that one? i coulda sworn i did something similar in class but maybe im wrong seeing as i sat this same paper. i remember doing this question but i forgot if i got to the correct conclusion
Clearly the application of this problem is to calculate how much fabric is needed to make a thong. đâș
bruhh
This pretty kaizo for gcses
Nice one!
But why don't you solve all equations for r as a variable, rather than a constant 4?
So, at the end we can have a general type also, you never know, may be a nice formula will emerge...
Thank you!
I shouldâve got this question right. I scribbled out my answer because it looked dumb and did some stupid assumption that gave me a âbetter lookingâ but *incorrect* answer, still got a grade 9 overall tho (highest grade possible)
I did my gcse in maths more than a decade ago. They didn't teach us about radians, sectors or segments. Things have changed a bit apparently.
radians aren't in gcse and aren't needed to answer the question, it will also work in degrees.
We still donât do radians but the rest is included đ
So it would have been pi*r^2Ăž/360
@@powerhouseplays that's good. Hopefully theyll continue to expand the topics they teach
To solve this question you don't need to use radians and they still aren't at GCSE level, sectors and segments have been part of Maths in the UK since forever, and you actually do them in geometry in year 7 at age 11/12. In this question they wanted students to notice the equilateral triangle in the segments and go from there to solve the problem, which can be done by knowing the basics of circles and triangles.
Glad I wasn't the only one
Yeah I got all of them other than the last one, probably since im in 9th grade but thx i wanted to know how to do it
I was one of the ones who did this paper
I'm doing A-level maths now and at first glance I still didnt know how to solve it lol
Even as a grade 9 maths student, if I got this, I would be very unhappy.
Very good
I could show you how to solve this problem using Polar Equations and Polar Coordinates. From that perspective, the problem solves readily.
This exam is taken by 15 year olds
yep iâm not doing all that thank god i know integrals
Just a doubt can't we use calculus.
This is my least favorite type of problem. Each time I see a problem where I donât even need to think to find a way to do it and then itâs all boring computation⊠I swear they just have to put those types of questions in every single test. If there were some clever way of doing it fast I wonât be complaining but there often isnât one that is accessible to people who donât know absurdly complicated active research stuff.
lol i sat this one, spent 20 minutes on that question, got halfway through and ran out of time
-8pi/3 - 8pi/3 is not 0, it is -16pi/3. What kind of a chancer is this guy.
lacrimosa must have started playing in these students heads once they turned the page to this horrorđđ
I believe the area is wrong. You missed the small wedge in the triangle.i will post a solution to it tomorrow.Great job.đ
No, the result is correct.
@@mcichael9661 Sorry. THE RESULT IS CORRECT. I HAD PROvEN IT USING CALCULUS .IT IS UP IN MY CHANNEL. THANK YOU.
I got the same result as the video using a different technique. The answer given is correct.
Please try to integrate 1/(x^5+1)
(4*ln(abs(x+1))+(sqrt(5)-1)*ln(x*(2*x+sqrt(5)-1)+2)+(-sqrt(5)-1)*ln(x*(2*x-sqrt(5)-1)+2)+2^(3/2)*sqrt(sqrt(5)+5)*arctan((4*x+sqrt(5)-1)/(sqrt(2)*sqrt(sqrt(5)+5))))/20-((sqrt(5)-5)*arctan((4*x-sqrt(5)-1)/(sqrt(2)*sqrt(5-sqrt(5)))))/(5*sqrt(2)*sqrt(5-sqrt(5))) + C
this is a hard problem? WHAT???
I was part of the year this question was on đ
I am from Ethiopia I like your teaching videos and I masterd at mathematics especially calculus and I have a dream to join mathematics or physics but I joined medical school because of our country give minor attention to maths and physics field
What is 0+0+0+0+0.... â equal to?
Technically it should be 0.
But when you treat this series as a Geometric Progression and apply the formula for sum of infinite terms of G.P., the result comes out to be 1/(1-(0/0)).
How is this possible?
(0/0) is undefined; so technically your question is meaningless. BUT, if you take (0/0) to be infinity, then the fraction evaluates to 0, since 1/infinity is 0. One reason 0/0 is undefined is that logically it can be shown to be equal to any number you like.
The formula for a geometric progression formula is:
a/(1-r), where a is the first term, r is the ratio, |r| < 1.
If we take a=0 and r to be whatever we ratio like where |r| < 1, we get our series (as 0*r = 0):
0+0+0+0... = a/(1-r) = 0/(1-r) = 0.
So no problems.
No, your first term is 0, so the sum is 0/(1 - 0), which obviously does not equal 1.
ahaha i remember turning the page, seeing that monstrosity, and turning back to the rest of the paper immediately
we used to do questions like this in 10th grade in india cbse
thats the same age as this question was built for
@@gabriql yeah, so i don't see why it's considered hard; it's quite a basic question for 16 year olds
@@dingus42 it really isnt that paper is for EVERY 16 year old in the country, even if they dont choose it; its not like only for gifted students. I don't think you understand also that they had about 5 mins for that length question, and also have around 9 other subjects to revise for other than maths. It is an easy question for someone in hindsight maybe, but dont say its easy
@@gabriql idk, it is genuinely very very basic geometry, literally just area of a circle and area of a triangle are needed. concepts which smart upper primary-lower secondary school kids can solve quite easily. Even if you use trig it's quite simple trig, so by 16 i would not say it's unfair or particularly hard at all.
Only a real man can admit mistakes in public
edexcel really just decided nah screw the 2005-06 kids (this was the only question i struggled on)
It's not that hard, it's just a sector of a circle = (1/3)piĂ4^2
MINUS
Area of triangle= (1/2)absinC = (1/2)Ă4^2Ăsin120 = 8Ăsqrt(3)/2 = 4sqrt(3)
Then just a full circle minus 4 of those D shapes, easy.
Alt way: Take B as origin and convert the whole system into coordinates and use integration....
I REMEMBER THIS IN 2022 IT DESTROYED MEEEE đđđ i ended up just putting 16pi as my answer đ
Please try the 2016 JEE Advance paper. It's a tough pill you will like it
I was able to solve this as a 10th grader CBSE student in India without radians
thank heavens I did iGCSE đđ
NotoriousâŠ? I tried it and it wasnât hard at all⊠You can do it without radians, too.
if you got it correct but forgot the 4x or forgot cm2
Here I was using integrals and trig substitutions.
That will work but it will take very long
@@Samir-zb3xkand also a 15-16 year old wonât be able to do that
â@@Samir-zb3xk Yeah, that was kind of my tone, it was more like a "woe is me, what am I doing when it's not even that complicated" type of comment.
@@evanpoole7829 I'm not too sure I agree with that sentiment, or that it should even be a consideration when constructing tests like the GCSE. If you want to find exceptionally talented people, you have to construct exceptionally challenging tests. I hear the counterarguments that "well my child isn't a mathematically-inclined person, but they're better at X," and use that as a reason for why the testing shouldn't be difficult. And I find that patently absurd because they're not concerned with finding exceptional people within the youth of our society, they're concerned with their own child's marks and dumbing the rest of society down so that their non-maths-inclined child still has opportunities abounding without the necessary talent.
@@vilelive I get the desire for challenging tests. They're much more fun than the boring stuff you do at GCSE. But that's what the further maths GCSE is for. There are also Olympiads available to year 11s, and if you really want a challenge you can sit the BMO and try and sit the IMO, which is very rare but some people do manage it even in year 11. You also get to sit entrance exams like STEP III which are far harder than anything on the A level once you get to sixth form.
GCSE maths has to be sat by nearly every kid in the country, there's not much point including integrals in a normal maths GCSE question. You can still get marks for it, so if you want to go that way then do, but questions shouldn't include them. If you want integration, do the further maths GCSE or sit A level maths early.
nice
blue pen yay
Wait...we start learning radians in sixth form college not high school, how were they be able to solve it?
Because it is not really relevant.
they used a formula that used degrees instead probably
this question was supposed to test students on their exact trig value knowledge so to get the area of the triangles theyd have to use 1/2 ab sin(angle) or 8sin(60) and stuff, it would be much quicker with radians but as you said thats only taught at A level +
@@harley_2305 "much quicker". *roll eyes*
@@samueldeandrade8535 dont get what youre trying to say, it is much easier and more efficient to use radians
Draw a vertical line where the circles intersect so there are four identical segments. Just find area of one segment and multiply by four. Then area of a circle minus that area.
That's the way I did it.