@@mathevengers1131 Assuming you’re talking about BPRP challenge, I think it’s about you. You have to choose the best video, aka the video you’re the most proud of. I can give you my opinion but maybe the video I consider the best on your channel might be one you dislike lol
If we start from drawing the pink equilateral triangle, we'll be able to count the area of the quarter as Area of 120 degree sector + Area of the triangle + Area of 30 degree sector, which mean the total Area = 4 * ( π r^2/2 * (120 + 30)/180 + sqrt(3)/4 ) = 4 * π *5/12 + sqrt(3) = 5π/3 + sqrt(3). It will be faster )
Draw a radius from the centre of each circle to each of its closest intersecting points with adjacent circles, and draw a vertical line from the higher intersecting point on each side to the lower one. The areas above the top radii and below the bottom ones are each 2/3 pi (two-thirds of a circle). The middle area consists of six triangles and the two side arced areas. Two of each of these account for 1/3 pi, the remaining four triangles sum to 4 * sqrt(3) / 4 = sqrt(3). Total area therefore 5/3 pi + sqrt(3)
Thanks Michael, nice geometry. I arrived at same solution by adding the two small 'axe-head' areas to the area of two circles. I based everything on the inscribed hexagon (6 equilateral triangles, side lengths r) inside the centre circle. The 'axe head' area is area of 1 triangle - 1/6 of the difference of area between the main circle and hexahgon (minus 2 of these + 1 = minus 1, since two of the -sectors- segments are internal to the axe-head triangle and one is needed in the final area).
When you drew everything in 1st quadrant I was like "Let's see if I remember integrals", went on writing out the integral and realised it is never meant to be solved like that :'D
theres no problem per say but using calculus for these geometry problem always felt like brute forcing so not interesting to me. you usually dont get any new insights.
I took the dirty calculus route! Area, A = 3pi - 8 Integral ( sin^2 t dt ) from t = 0 to t = pi/3 sin^2 (t) + cos^2 (t) = 1 cos (2t) = cos^2 (t) - sin^2 (t) Using these two identities, you get sin^2 (t) = ( 1 - cos (2t) ) / 2 A = 3pi - [ 2 (2t - sin 2t) ] with the bit in square-brackets evaluated at pi/3 3pi - 2 ( 2pi/3 - sin (2pi/3) ) = 5pi/3 + root3
While you were dividing your problem into four pieces I divided it into two and found the answer while you were reiterating and scratching your head, much the way Kevin Martin comments. Oh dear. Please keep trying with geometrical questions though, it is very entertaining. (You are very good a number problems but I like geometry ones more! I half expected you to start using a coordinate system and double integrals everywhere.)
I don't think the symmetry/split into 4 did anything very useful. The final area is 3 times the area of any of the circles less 4 times the area of any of the segments bounded by one of the circles and the chord connecting the intersection points with another circle. In turn the area of any of these segments is the area of the corresponding sector minus the area of the triangle formed by the chord and two radii. Assuming you don't just want to treat this as the well-known geometric arrangement that it is, you show (in a manner similar to in the video) that the triangle has an angle of 120 degrees (pi/3), a height of 1/2, and a base (the chord) of √3, and you fall into the same calculations. By not splitting the figure into 4 you avoid the notational problems of that "strange shape", and instead are working with well-known (and with well-known names) geometric shapes: Circles, segments, sectors, radii, chords, and triangles.
Used integral to find an area of that lens-shaped thing (actually, I found an area of ¼ of it first). It happened to be Scut = 2π/3-√3/2. The answer then will be 3π - 2 Scut.
What about the area of n chained circles, (denoted below as C)? If we take a sum of areas of these n circles, (which is n * PI), there will be (n-1) double counted segments, which area A is this: A = (2/3) * ( Area of the circle - Area of isosceles triangle inscribed in the circle with radius 1) = = (2/3) * ( PI - (1/2) * (3^0.5)/2 * 3 ) = (2/3) * PI - (3^0.5)/2 In general we have: C = n * PI - (n-1) * A = n * PI - (n-1) * ((2/3) * PI - (3^0.5)/2) = ((n+2)/3) * PI - ((n-1)/2) * (3^0.5) So, for our shape (in this case n=3) we will have indeed: C = ((3+2)/3) * PI - ((3-1)/2) * (3^0.5) = (5/3) * PI - (3^0.5) For Audi shape (4 circles) we wil have: C = ((4+2)/3) * PI - ((4-1)/2) * (3^0.5) = 2 * PI - (3/2) * (3^0.5) For comparison, when n = 100, C = ((100+2)/3) * PI - ((100-1)/2) * (3^0.5) = 34 * PI - (99/2) * (3^0.5) Sometimes the right approach to a problem might be pointed out by good old generalisation technique, which here, in my opinion, trumps the "simplification by symmetry" technique. After all, by using generalisation we derive solution for a class of problems, not just for a one particular case.
at 2:30: draw the equilateral triangle inside the overlap area. The area above it is 2pi/3. The area to the right is pi/6. The area of the triangle is sqrt(3)/4. The area of the quadrant is thus 5pi/6+sqrt(3)/4. Thus the final area is 10pi/3+sqrt(3).
The 1/4 circle comprises a "half axehead" (to use BatchRocketProject's terminology) and a convex region, whose 3 vertices form an equilateral triangle. The convex region, in turn, comprises a 1/6 circle, area π/6, and a segment bounded by one side of the equilateral triangle, area (π/6 - √3/4). So, the area of the half axehead is π/4 - π/6 - (π/6 - √3/4) = √3/4 - π/12 and the total area of the 3 circles is 4 times this plus the two outer circles.
I solved It drawing the line that pass in the intersection points of the two top circles and then taking One of the 4 triangles of the rhombus iscribed you can see that Is 30-60-90 degrees. I wasn't very clear but i'm not very good with english
I tried setting up an integral to find 1/4 of the leftover area from the middle circle and got the integral from 0->1/2 of sqrt(1-x^2)-sqrt(1-(x-1)^2) before remembering that my integration skills are rusty at best and I didn't know how to solve it. I imagine you could do some trig substitution to find this integral and then the area would just be 2pi + 4 (integral value).
Think of dual basis as a set of functions {f1,f2,....,fn} from a Vector Space V over some Field F -> F, where each f_i associates each vector a€V, with it's i^th coordinate of it's coordinate matrix WRT some basis, this is a very nice intuition to have. Feel free to ask me if you have not understood what I said.
today i learned how to calculate the size of that cats-eye shape. thank you. this makes me wonder though: if the circles were not overlapping exactly at their centers but instead were further distanced by some value D, does this make it significantly harder to calculate the size of their overlap? is there a more general solution to this using integrals?
Far too much faffing about at the start. If you first consider two overlapping circles radius r, whose centres line on the circumference of the other circle, then draw a chord joining the points of intersection of the circles. Next join the centres of the two circles by a line which has to be perpendicular to that chord by symmetry. Draw the 4 radii from the centres to the points of intersection of the circles. It then becomes obvious that the chord subtends an angle of 120 degrees or 2.pi/3 radians at the centre of either circle, and that the chord has a length of r.sqrt(3). By symmetry, the overlap is cut in half by the chord. That half-overlap is the difference between a sector of one of the circles (area = pi.r^2/3) and the 120-30-30 isosceles triangle made up of the radii and the chord whose sides are r, r, r.sqrt(3), (area = 1/2 . r/2 . sqrt(3).r = sqrt(3).r^2/4). So the area of overlap for two circles is twice r^2(pi/3 - sqrt(3)/4) = r^2.(4.pi - 3.sqrt(3))/6. Now to find the total area of N overlapping circles, we just take the area of N circles = N.pr.r^2 and subtract the area of the overlaps = (N-1).r^2.(4.pi - 3.sqrt(3))/6. For 3 circles with radius = 1, that equates to 3.pi - (4.pi - 3.sqrt(3))/3 = (9.pi - 4.pi + 3.sqrt(3))/3 = 5.pi/3 + sqrt(3).
if the middle circle doesn't exist and r → 0 and number of circles approached to infinity, could we say the area of circles are equals to length of a line starting from the lowest point of the lowest circle to the highest point of the highest circle? and if no, why not?
Let's do it the physics way: the answer is 3*pi - (> 2/3 but < 1)*pi, so take the average and we get (3 - 5/6)*pi ~= 6.81. The exact answer is ~= 6.97, so we're within 2.5%. Not bad, let's go get lunch. :)
Yes, if you take the y axis to be the horizontal line and the X axis to be the vertical line, one quarter of the figure's perimeter can be described with the following piece-wise function: f(x) = sqrt(1-x^2) for x between 0 and 1/2, f(x) = sqrt(2x - x^2) for x larger than 1/2 Then if you integrate this over the interval 0 to 2 you will find the answer. Note that it isn't easy to integrate.
You are... Amazing and too much funny you take minutes to explain us something we easy understand which it was π/3 the angle of the triangle; but at the same time you immediately come up with the equation more developed. Hahahaha you dude 😉
@Ricardo Fodra Assuming the centres are co-linear, you work out the overlap between two circles first, and you'll find it's r^2.(2.pi/3 - sqrt(3)/2). Then for n circles you have the area of n circles less the area of (n-1) overlaps. Total area = n.pi.r^2 - (n-1).r^2.(2.pi/3 - sqrt(3)/2)
Thank you for explaining this geometrical problem. I always have to think too long about these kind of problems, and even for me difficult to solve. But your explanations make it easier to understand, Professor Penn!
I divided the quarter-area (upper-right quadrant) to be calculated into: (1) a quarter circle of radius 1 on top, (2) an equilateral triangle of side-length 1, with vertices at the center of the top center, the center of the middle circle, and at the intersection of the two circles, and (3) two 30 degree sectors of a unit circle, or equivalently 1 60 degree sector of a unit circle. (1) + (2) + (3) = A/4
For anyone wondering about using integrals: The are of the topmost and bottom most circles are trivial, so for the leftover area in the 1st quadrant, you can use this: the integral from 0 to sqrt3/2 of [-sqrt(1-x^2)+1]dx, and add the integral from sqrt3/2 to 1 of [sqrt(1-x^2)]dx.
I cut the circles on their intersection points. This way I receive 3 full circles minus 4 times a bow segment that can be calced via either using the formula or integration. Was nice to do this integration stuff again because I didn't had the bow calc formula at hand. Nice problem, like to solve this kind of stuff.
![Twenty One Pilots - “The Line” (from Arcane Season 2) [Official Music Video]](http://i.ytimg.com/vi/E2Rj2gQAyPA/mqdefault.jpg)
6:00 🍕
8:36 🟡🟢🔵
Hey can you help me find my best video?
@@mathevengers1131 Assuming you’re talking about BPRP challenge, I think it’s about you. You have to choose the best video, aka the video you’re the most proud of.
I can give you my opinion but maybe the video I consider the best on your channel might be one you dislike lol
@@goodplacetostop2973 ok. Thanks for your opinion.
😩+🍕=😋-->🤗
If we start from drawing the pink equilateral triangle, we'll be able to count the area of the quarter as Area of 120 degree sector + Area of the triangle + Area of 30 degree sector, which mean the total Area = 4 * ( π r^2/2 * (120 + 30)/180 + sqrt(3)/4 ) = 4 * π *5/12 + sqrt(3) = 5π/3 + sqrt(3). It will be faster )
Came here to say just that.
This looks simpler (& faster) way.. hmm.
Draw a radius from the centre of each circle to each of its closest intersecting points with adjacent circles, and draw a vertical line from the higher intersecting point on each side to the lower one. The areas above the top radii and below the bottom ones are each 2/3 pi (two-thirds of a circle). The middle area consists of six triangles and the two side arced areas. Two of each of these account for 1/3 pi, the remaining four triangles sum to 4 * sqrt(3) / 4 = sqrt(3). Total area therefore 5/3 pi + sqrt(3)
Thanks Michael, nice geometry. I arrived at same solution by adding the two small 'axe-head' areas to the area of two circles. I based everything on the inscribed hexagon (6 equilateral triangles, side lengths r) inside the centre circle. The 'axe head' area is area of 1 triangle - 1/6 of the difference of area between the main circle and hexahgon (minus 2 of these + 1 = minus 1, since two of the -sectors- segments are internal to the axe-head triangle and one is needed in the final area).
Nice and clear, just getting back into maths, easy to understand
Very clear. Thank you for coming up with this problem.
When you drew everything in 1st quadrant I was like "Let's see if I remember integrals", went on writing out the integral and realised it is never meant to be solved like that :'D
You can use integrals to find the third area you see in the main formula! Who cares if you're not supposed to!
What's the fun in solving a problem in the way it's "intended" to be solved, anyway 😉
Why couldn't one use integrals? Are the Geometry Cops going to get you? If so we can get Real and Complex Analyses to testify on your behalf.
You can use whatever you want to and get the same answer (if you do it well)
theres no problem per say but using calculus for these geometry problem always felt like brute forcing so not interesting to me. you usually dont get any new insights.
I took the dirty calculus route!
Area, A = 3pi - 8 Integral ( sin^2 t dt ) from t = 0 to t = pi/3
sin^2 (t) + cos^2 (t) = 1
cos (2t) = cos^2 (t) - sin^2 (t)
Using these two identities, you get
sin^2 (t) = ( 1 - cos (2t) ) / 2
A = 3pi - [ 2 (2t - sin 2t) ]
with the bit in square-brackets evaluated at pi/3
3pi - 2 ( 2pi/3 - sin (2pi/3) )
=
5pi/3 + root3
3*pi - 2*(2 circles overlap) = 3*pi - 2*2*(pi/3 - (area of unit equilateral triangle)) = 3*pi - 4*(pi/3 - sqrt(3)/4) = 5*pi/3 + sqrt(3)
While you were dividing your problem into four pieces I divided it into two and found the answer while you were reiterating and scratching your head, much the way Kevin Martin comments. Oh dear. Please keep trying with geometrical questions though, it is very entertaining. (You are very good a number problems but I like geometry ones more! I half expected you to start using a coordinate system and double integrals everywhere.)
I don't think the symmetry/split into 4 did anything very useful. The final area is 3 times the area of any of the circles less 4 times the area of any of the segments bounded by one of the circles and the chord connecting the intersection points with another circle. In turn the area of any of these segments is the area of the corresponding sector minus the area of the triangle formed by the chord and two radii. Assuming you don't just want to treat this as the well-known geometric arrangement that it is, you show (in a manner similar to in the video) that the triangle has an angle of 120 degrees (pi/3), a height of 1/2, and a base (the chord) of √3, and you fall into the same calculations.
By not splitting the figure into 4 you avoid the notational problems of that "strange shape", and instead are working with well-known (and with well-known names) geometric shapes: Circles, segments, sectors, radii, chords, and triangles.
Pretty close to what I did. I took the 3 circles, subtracted the 4 sectors, then added the 4 triangles.
Yeah I thought he was doing that to make it easier to use integrals. Did it anyone solve it that way?
Used integral to find an area of that lens-shaped thing (actually, I found an area of ¼ of it first). It happened to be Scut = 2π/3-√3/2. The answer then will be 3π - 2 Scut.
What about the area of n chained circles, (denoted below as C)?
If we take a sum of areas of these n circles, (which is n * PI), there will be (n-1) double counted segments, which area A is this:
A = (2/3) * ( Area of the circle - Area of isosceles triangle inscribed in the circle with radius 1) =
= (2/3) * ( PI - (1/2) * (3^0.5)/2 * 3 ) = (2/3) * PI - (3^0.5)/2
In general we have: C = n * PI - (n-1) * A = n * PI - (n-1) * ((2/3) * PI - (3^0.5)/2) = ((n+2)/3) * PI - ((n-1)/2) * (3^0.5)
So, for our shape (in this case n=3) we will have indeed: C = ((3+2)/3) * PI - ((3-1)/2) * (3^0.5) = (5/3) * PI - (3^0.5)
For Audi shape (4 circles) we wil have: C = ((4+2)/3) * PI - ((4-1)/2) * (3^0.5) = 2 * PI - (3/2) * (3^0.5)
For comparison, when n = 100, C = ((100+2)/3) * PI - ((100-1)/2) * (3^0.5) = 34 * PI - (99/2) * (3^0.5)
Sometimes the right approach to a problem might be pointed out by good old generalisation technique, which here, in my opinion, trumps the "simplification by symmetry" technique. After all, by using generalisation we derive solution for a class of problems, not just for a one particular case.
and that's how you can draw an equilateral triangle using only a compass and straight edge!
at 2:30: draw the equilateral triangle inside the overlap area. The area above it is 2pi/3. The area to the right is pi/6. The area of the triangle is sqrt(3)/4. The area of the quadrant is thus 5pi/6+sqrt(3)/4. Thus the final area is 10pi/3+sqrt(3).
off by a factor of 2 in the triangle :)
and apparently in the angle...
The 1/4 circle comprises a "half axehead" (to use BatchRocketProject's terminology) and a convex region, whose 3 vertices form an equilateral triangle.
The convex region, in turn,
comprises a 1/6 circle, area π/6, and a segment bounded by one side of the equilateral triangle, area (π/6 - √3/4).
So, the area of the half axehead is π/4 - π/6 - (π/6 - √3/4) = √3/4 - π/12 and the total area of the 3 circles is 4 times this plus the two outer circles.
I solved It drawing the line that pass in the intersection points of the two top circles and then taking One of the 4 triangles of the rhombus iscribed you can see that Is 30-60-90 degrees. I wasn't very clear but i'm not very good with english
I tried setting up an integral to find 1/4 of the leftover area from the middle circle and got the integral from 0->1/2 of sqrt(1-x^2)-sqrt(1-(x-1)^2) before remembering that my integration skills are rusty at best and I didn't know how to solve it. I imagine you could do some trig substitution to find this integral and then the area would just be 2pi + 4 (integral value).
Imagine a vertical line at x=sqrt3/2 (where the circles intersect) and add two separate integrals.
Can you go through dual spaces and dual basis? I really dont understand why they are useful, thanks
Think of dual basis as a set of functions {f1,f2,....,fn} from a Vector Space V over some Field F -> F, where each f_i associates each vector a€V, with it's i^th coordinate of it's coordinate matrix WRT some basis, this is a very nice intuition to have. Feel free to ask me if you have not understood what I said.
This video has been ¾ sponsored by Audi.
And japan
I have a question: What happens if the radius of each circle is R (or an unknown non-negative, and non-zero constant)
You have to multiply everything by R²
since it's about area, you multiply the result by R^2
That's like stretching both x and y axes by R. So you have to multiply the area by R² (as the small pixels each increase in area by R² times).
Quite funny problem. Thanks for the video!
today i learned how to calculate the size of that cats-eye shape. thank you. this makes me wonder though: if the circles were not overlapping exactly at their centers but instead were further distanced by some value D, does this make it significantly harder to calculate the size of their overlap? is there a more general solution to this using integrals?
What a fantastic problem. I would have drowned myself in integrals.
Thank you, professor!
Great job
Hi,
For fun:
5:31 : "ok, great".
Far too much faffing about at the start.
If you first consider two overlapping circles radius r, whose centres line on the circumference of the other circle, then draw a chord joining the points of intersection of the circles. Next join the centres of the two circles by a line which has to be perpendicular to that chord by symmetry. Draw the 4 radii from the centres to the points of intersection of the circles. It then becomes obvious that the chord subtends an angle of 120 degrees or 2.pi/3 radians at the centre of either circle, and that the chord has a length of r.sqrt(3).
By symmetry, the overlap is cut in half by the chord. That half-overlap is the difference between a sector of one of the circles (area = pi.r^2/3) and the 120-30-30 isosceles triangle made up of the radii and the chord whose sides are r, r, r.sqrt(3), (area = 1/2 . r/2 . sqrt(3).r = sqrt(3).r^2/4). So the area of overlap for two circles is twice r^2(pi/3 - sqrt(3)/4) = r^2.(4.pi - 3.sqrt(3))/6.
Now to find the total area of N overlapping circles, we just take the area of N circles = N.pr.r^2 and subtract the area of the overlaps = (N-1).r^2.(4.pi - 3.sqrt(3))/6.
For 3 circles with radius = 1, that equates to 3.pi - (4.pi - 3.sqrt(3))/3 = (9.pi - 4.pi + 3.sqrt(3))/3 = 5.pi/3 + sqrt(3).
Homework : expand the answer to work for N overlapping circles
if the middle circle doesn't exist and r → 0 and number of circles approached to infinity, could we say the area of circles are equals to length of a line starting from the lowest point of the lowest circle to the highest point of the highest circle? and if no, why not?
That's a good placed circle
Looking for more geometry questions
nice shot
Л ж
love the jingle
Let's do it the physics way: the answer is 3*pi - (> 2/3 but < 1)*pi, so take the average and we get (3 - 5/6)*pi ~= 6.81. The exact answer is ~= 6.97, so we're within 2.5%. Not bad, let's go get lunch. :)
That's a good observation, but I don't get the physics reference.
@@skeptic1000 Just a joke about the stereotype that physicists usually do non-rigorous, hand-wavy mathematics.
@@leif_p I would have gone with economists
Great.It is pretty idea
Can we solve this with integrals?
Yes, if you take the y axis to be the horizontal line and the X axis to be the vertical line, one quarter of the figure's perimeter can be described with the following piece-wise function:
f(x) = sqrt(1-x^2) for x between 0 and 1/2, f(x) = sqrt(2x - x^2) for x larger than 1/2
Then if you integrate this over the interval 0 to 2 you will find the answer. Note that it isn't easy to integrate.
Fun problem
One could parlay that demonstration to 3 overlapping spheres. No?
We know the formula for the 1/4 and 1/2 circle, apply integral to figure out the area
Nice
here we stacked 3 circles on top of each other. What if we stack n circles like that?
You are... Amazing and too much funny you take minutes to explain us something we easy understand which it was π/3 the angle of the triangle; but at the same time you immediately come up with the equation more developed. Hahahaha you dude 😉
What if there where n circles like that?
@Ricardo Fodra Assuming the centres are co-linear, you work out the overlap between two circles first, and you'll find it's r^2.(2.pi/3 - sqrt(3)/2). Then for n circles you have the area of n circles less the area of (n-1) overlaps.
Total area = n.pi.r^2 - (n-1).r^2.(2.pi/3 - sqrt(3)/2)
Very nice problem with a lovely explanation; very patient and thorough
8π/3 + sqrt(3)? Typing down to check if my intuition is right before watching the video
almost
Thank you for explaining this geometrical problem. I always have to think too long about these kind of problems, and even for me difficult to solve. But your explanations make it easier to understand, Professor Penn!
I divided the quarter-area (upper-right quadrant) to be calculated into: (1) a quarter circle of radius 1 on top, (2) an equilateral triangle of side-length 1, with vertices at the center of the top center, the center of the middle circle, and at the intersection of the two circles, and (3) two 30 degree sectors of a unit circle, or equivalently 1 60 degree sector of a unit circle. (1) + (2) + (3) = A/4
For anyone wondering about using integrals:
The are of the topmost and bottom most circles are trivial, so for the leftover area in the 1st quadrant, you can use this: the integral from 0 to sqrt3/2 of [-sqrt(1-x^2)+1]dx, and add the integral from sqrt3/2 to 1 of [sqrt(1-x^2)]dx.
The are is about 74% of the area of the 3 circles if they were separated (no overlaps). I.e. the overlap is about 26% .
I cut the circles on their intersection points. This way I receive 3 full circles minus 4 times a bow segment that can be calced via either using the formula or integration. Was nice to do this integration stuff again because I didn't had the bow calc formula at hand.
Nice problem, like to solve this kind of stuff.
You should have a "And this is a good place to stop" T-shirt made!
Sir i want to know that, what do i need to compete math olympiad. I read in class 10 from Bangladesh.
Long explanation but the final result is false? Embarrassing to me..
A me viene 2pi-2+2rad2, quasi uguale