'Impossible' question stumps students

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The recent GCSE exam left many students baffled. Here's how to solve one of more challenging questions. Thanks to Joe and Rob for the suggestion!

The 'impossible' question on GCSE exam that had 16-year-olds across the country stumped
Thousands of kids left baffled by ‘impossible’ GCSE question they claim can’t be solved – do you know the answer?
Science Filtered solution on Twitter
HarrySurplus solution on Twitter

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i actually managed to figure out how to solve this during the exam because of you! i remembered a video about a lens and that got me the method

deekoei
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Another way to do it is to recognize that each area that you’re not trying to solve for (I’ll call that the negative area) is symmetrical across the axis of the intersection points (DE and FG). Based on the triangles drawn out, the angle creating that segment is 120° because it’s two equilateral triangles. You can then solve for one sector, then multiply it by 4. At that point, you just subtract this negative area from the area of the circle (16pi), and you have the positive area

Jaymac
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I started by finding the area of the circular segment, then just subtracted two of those from two sectors. I also kept it in terms of r until the end for easier bookkeeping.

eroraf
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Without knowing the formulas for the equilateral triangle and the circle segment, but knowing pythagoras and area of circle, you can still get there. You can half the triangle to a Pythagorean triangle with hypotenuse 4 and short side 2. And you can easily see that 6 segments fit in a circle, so area of the circle, minus 6x surface of triangle, div by 6 is the circle segment. I think that showing how to solve with minimal formula knowledge is often best (you can still show the formulas for the more advanced viewers).

Misteribel
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That feeling when you used a completely different approach and still got the very same result

iknet
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First time I actually solved one of your puzzles without seeing the answer first; thanks for all the videos, man, you rock.

RefluxCitadelRevelations
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Just imagine the centre of the middle circle as (0, 0). Then find all the coordinates required and then use integration to find the area under the curve

shreyaskarchakraborty
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I feel great again. It took me more than several minutes to look at it and tried to work out the solution without watching your answer. In the end I could solve it and am so happy. Keep up the great work please. I used to love geometry as a kid and I go back to my childhood thanks to you:)

canbalcioglu
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It's basically two equilateral triangles, each with one circular segment added and two dropped. So we have to subtract one circular segment from each of the two equilateral triangles:

Area of equilateral triangle = r²√3/4

Area of circular sector = r²π/6

Area of circular segment = r²π/6 - r²√3/4 = r²(π/6 - √3/4)

A = 2 * (r²√3/4 - r²(π/6 - √3/4)) = 2r²(√3/4 - π/6 + √3/4) = 2r²(2√3/4 - π/6) = 2r²(√3/2 - π/6) = r²(√3 - π/3)

A = 4²(√3 - π/3) = 16(√3 - π/3) = 10.958

Waldlaeufer
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i got this in my gcse, one of my friends actually got it which is mad.

also, the paper (specifically this question actually) was leaked on a discord server about 3 hours before the exam was sat but only a few people saw it.

gadgetlab
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This was the final question on the British GCSE Mathematics rexam on May 20th. The students always complain about the final question and it usually a tough geometry one.

EnemyOfEldar
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Managed to figure it out during the exam but I'm incredibly happy to see this question show up here. Props to Edexcel - it's a really well made and well crafted question that's not incredibly overcomplicated or wordy, just heavily problem solving. Was fun to find out in the exam.

tobysuren
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Another useful strategy when dealing with problems like this is "if you have symmetries, use them" - I personally used symmetries to divide the thing to be solved into 4 pieces that I called X, then I defined a piece Y, which is half of the intersection of two circles bit, and X and Y together form a quarter of the central circle. That's a lot easier to draw and keep track of. Just don't forget that your final answer is 4X at the end.

Pretty symmetries usually lead to pretty solutions.

georgiishmakov
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I solved it by putting the circles in a grid, while ignoring the left most circle. This gives you two circles with the formulae: x² + y² = 4² and (x - 4)² + y² = 4². Substitute one in another and you get x = 2, which is the x-cordinate of the intersection. Now use integration of the middle circle with 2 and 4 as limits (or use the other circle with 0 and 2 as limits, but this formula is easier to integrate) this comes out to be 2⅔π - 2√3. Multiply this by 8 because there are 8 of these in the middle circle and then get the surface of the circle, which is 4²π, and subtract 8 × (2⅔π - 2√3) from that and you get -16/3π + 16√3

ragingmajesty
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this is one of the few ones in here I could actually do! I just split the parts where both circles overlap in half by drawing a line, then found the area of one of the segments, multiplied it by 4, then subtracted that from the total area of the circle

Trep
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I know a few others may have already stated this solution or a similar solution, but I want to type it out here anyway:
using the diagram at 2:17 as a reference (note: [ABCD] means the area of shape ABCD):
Connect D to E and F to G instead of what was shown in the video. This creates 4 "caps" which are congruent by symmetry which can be solved for quite easily.
To find the area of one "cap": we will subtract triangle ADE from circular sector ADBE; for the same reason the equilateral triangles can be drawn in the video, angle DAE is 120 degrees, so the [ADBE] = 1/3(circle A) = 16pi/3
next, we find [ADE], which can be done in a multitiude of ways like [triangle] = ab sin(c)/2, special right triangles, etc; so [ADE] = 4sqrt(3)
from this, the area of a "cap" (such as DBE) is 16pi/3 - 4sqrt(3)
lastly, we subtract 4 of these from one circle, so we get 16pi - 4(16pi/3 - 4sqrt(3)) = 16sqrt(3) - 16pi/3, just as the video said.

cybr._.
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Imagine calculus doesn't people lived like that.

jesselaw
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People studying for Colégio Naval and Epcar watching this: easy
However, it's a good question! Awesome solution too!

*For people who don't know what's CN and Epcar: both are military schools for students between 15 and 18 years old in Brazil.

vicenzo.grazinoli
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explanation and visual was beyond perfect. truly appreciate the effort Presh 👏👏👏

-AnweshaDas-
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It's an interesting solution. I thought of a very different one though. We can break the blue area into 4 equal parts, due to symmetry, and find one part's area as a difference between two finite integrals.

BeyondKawaii