Can you solve for the shaded areas? 3 great puzzles
ฝัง
- เผยแพร่เมื่อ 2 มิ.ย. 2024
- These are 3 interesting questions. How many can you solve?
0:00 problems
2:35 solution 1
4:03 solution 2
6:15 solution 3
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Problem 2 AoPS
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I almost feel offended whenever I can't solve a problem and he comes out with "This question was given to 15-16 year olds"
I wouldn't be surprised if it was the very hardest problem on the PSAT.
I get over that by remembering that those kids are in school and still have the tools for solving these problems fresh in their minds. My tools have been rusting in the attic. 😅
Helps when you realize this type of question, like most math questions, has extremely limited relevance to the real world, and even the "smart" people who manage to secure jobs that practically use core principles applied here rarely see problems like this if ever. It's more mathturbation than anything else and not much different to sudoku or other puzzles.
As a 15 year old, I also feel offended
We're you able to solve this one?
Plot twist: the white regions are the shaded parts
Wait,
Their areas would depend on their thicknesses, which are not given
The solution for problem 1 was elegantly done. Great job!
For problem 1, _any_ curve going from the center to the circle and rotating around the center will sweep the same area for the same rotation angle, regardless of its shape (this gets trickier to define for certain shapes where you have to contend with negative areas, but it holds). Therefore the sector (due to straight line rotation) and the claw (due to circular curve rotation) must have the same area. But a 15 year old would not know that (it probably requires calculus), so I like your explanation a lot.
Good point, similar principle to stacking infinitely thin squares to build a pyramid but the apex can be moved and the shape even warped and still work as long as the slices are the same. Gets kind of funny in this problem and is most likely polar in its proof though.
This is equivalent to the following statement: Any curve from the center to the circle and rotating around the center sweeps the same area as the sector with the same angle.
Let the center be C. Any point on the curve A would be taken to another point A’ as if A has moved along the circle centered at C and with radius AC=A’C for the same angle the entire curve has rotated. And the same applies for the straight line segment of the sector as well, for a sector can be seen as the area that a radius of the circle swept by rotating around the center for some angle. In other words, there is some B and B’ on the circle with radius AC such that the sector CBB’ has a central angle as the curve has swept. Then the arcs AA’ and BB’ are equal and thus the whole curve sweeps the same area as the sector.
This is very much not rigorous, but if you were to make it rigorous it’s pretty much the same as calculus.
idk I’m sixteen and I figured that out. It’s basically the same visual logic as proving that the area of a parallelogram is equal to the base times the height, regardless of the angles & other side lengths of the parallelogram
In terms of question 1, I visually just thought if you kept rotating the offset circle by 20 degrees, you would get a partition of a whole circle into 18 equal 'claws' because there are 18 20s in 360. So if the half circle is 72, a whole circle is 144, and an 18th of that is 8. This doesnt require addition and subtraction of overlapping areas.
exactly the method i used (and how i realized that The Claw must have an area equal to The Wedge)
That's a brilliant approach 👏 👌
That's how I solved the problem also. But before doing so I came up for a geometric argument for why it works: the tangent lines to the two arcs at point Q are both perpendicular to their respective semicircles. So like the semicircles, they'll be at an angle of 20°. The two arcs are also equivalent, since you can just rotate one to the other, which also hints towards the process of rotating the claw in increments of 20°.
3rd problem is easiest if you color the middle part Green. So Yellow = 49-8pi-Green and Blue = 49-9pi-Green . Therefore Yellow - Blue = pi
Perfect 🍦
Nice!
In problem 3 you don’t need the 7 (as long as it is > 6; even if it is > 10, b = 0 but the answer still holds). Great visual presentation as always!
BINGO!!! I got all three correct!!! A huge sense of satisfaction!!!
These were fun! The solution you did for 3 is so satisfying I can't believe I didn't do it that way!
I stared at problem 1 for a minute and saw the really neat method prior to watching the video. It's the same area as the 20 degree sector! The problem in general is a very cool idea.
As for problem 3, I immediately thought about canceling areas since there are some overlaps. It basically boils down to a radius 3 circle minus two radius 2 circles since each overlap is uniquely determined by a quarter circle and semicircle. You overthought that one, eh? We all do that sometimes.
Broo that's such a great observation!!
But I wanted to know how to write something that counterintuitive as a proof.
Also, are you preparing for the IMO (International Mathematics Olympiad)?
Those were some of the best area puzzles you’ve shown us in a while! They were challenging, but fair; and I like that I didn’t need advanced math to solve them.
As I did each problem I learned how to think about the next one differently.
I did the first one in a really time consuming way where I found the radius and the area of the bulging triangle (with good ol pythag cause no trig allowed) and eventually came to 8 for the shaded region, after which I noticed 8 was the same thing I got for the 20 degree sliver and then all the dots connected in my head and I realised this problem could’ve been done in seconds.
Then for the next one I was more on my guard, I started thinking about different ways intersections could cancel and I started writing algebraic word problems with things like semi, extra, main area, mini triangle, big triangle. Eventually I figured out you could divide stuff into mini triangles then from there it was just working stuff out albeit still in a more convoluted way then shown in the video.
Then for the last one seeing it was a subtraction and having done the other 2 I figured you would be able to do this with some huge cancellation at the end so I pretty much immediately wrote out the solution shown and ended up solving it much faster than the others.
Cool video
The hint in the first one that this was given to pre-trig teenagers was (for me) pretty helpful; I'm not the best at these puzzles, but once you're looking for tricky bits of equal area or sneaky symmetries, then the knowledge that there has to be a straightforward one helps finding it. For me the third one was the easiest; and the second the hardest - the explicit subtraction in the third one seems like a good hint that it's worth considering parts that are equal between the two operands of the subtraction - hey, those black bits! I gave up on number two after a minute or so.
I just used Calculus and threw everything into a calculator to get the answer. I know this isn't the smart way of doing it, but I wouldn't be able to solve it otherwise.
Hi Presh,
the 3rd problem is much easier to solve. Since the center section stays the same for all considerations, I can ignore it. The black overlapping areas are completely symmetrical and account for equal area deductions and can therefore also be ignored.
All you need to do is to assemble the complete circles, which are ONE 3^2Pi (from 4 quarters in the corners) and 4 half-circles, being 2^2Pi (sitting on the center edges), and compare their respective area difference.
3^2Pi - 2*2^2Pi = 9Pi - 8Pi = Pi. Done.
neat and clever!
Had a real hard time with understanding # 3 as I only saw the thumbnail and thought it was looking for the sum of orange and blue areas. Then I went back and actually listened to the problem...
Problem 2:
Draw lines from each vertex of the bestagon to the opposite vertex. This will separate the hexagon into six equilateral triangles, each with side length 2. Now draw two more equilateral triangles, each of side length 3, connecting the centers of each semicircle with the two on either side of the semicircle opposite (you'll get a six-pointed star). This will further break up the hexagon into 24 equilateral triangles, each with side length 1.
The area we need is the six central triangles (A), minus the six circular segments corresponding with those triangles. The circular segments are each composed of a 60° sector (B) minus a triangle (A). So the formula is 6A - 6(B-A), or 12A - 6B. 6B is equal to one full circle, and 12A is equal to half the area of the hexagon.
A = (3√3/2)(s²/2) - πr²
A = (3√3/4)2² - π1²
A = 3√3 - π
Problem 3:
This one is actually much simpler than it initially appears. You can see the solution if you imagine orange as positive and blue as negative. The black areas in between are where the positives and negatives have cancelled out.
The area we are looking for is 4 of the corner quarter circles (which equals to one circle of radius 3), minus 4 of the side semicircles (which is 2 circles of radius 2).
A = πR² - 2πr²
A = π3² - 2π2²
A = 9π - 8π = π
My solution for 3 was, I think, a little simpler. I just rephrased Orange and Blue in terms of each other -- in other words, let QC = (the area of the quarter-circle segments) and SC = (the area of the semicircle segments), and
Orange = QC - SC + Blue
Blue = SC - QC + Orange
In order to find Orange - Blue, we can simply substitute one of our expressions, so
Orange - Blue = (QC - SC + Blue) - Blue = QC - SC + Blue - Blue = QC - SC
Since we can easily find the area of the quarter-circle segments (9pi) and the semicircle segments (4pi x 2, or 8pi), we can see that Orange - Blue = 9pi - 8pi, or simply pi.
The solution of problem 3 was very elegant!
I rarely comment, However when I saw how elegantly you solved these problems which to me were so intimidating in appearance. It made me it made me audibly go "huh that's really fascinating", anyways just wanted to say I enjoyed the unique (to me) perspective.
I did all the calculus for part 3, to come up with the area of the overlap is about 3.126. It was at that point I realized this number had to be completely meaningless, and went back and solved it the way you did. Nice problems!
Sounds like your numerical integration wasn't that precise, but that's close enough to pi that I might be thinking about it since the problem is all about circles.
i did it without calc and i got 1.789
Glad to see I was able to solve all 3. Sometimes I worry I can't think through stuff like this anymore since I finished college.
The solution to no 3 is genius!
For problem 3:
Orenge - Blue
= (Orenge + intersected) - (Blue + intersected)
= Corner ¼circles - Middle semicircles
= π3² - 2π2²
= π
I did it this way too. :)
Congratulations Presh for that you are going to have 3 million subscribers , we ( the audience) demand a story of yours , about how u started the channel and all that stuff.
It’s interesting just how helpful it was to look at the big picture with these problems, thinking less about the calculations and more about the big picture. For the third problem, I embraced the symbols and looked at the area of the whole square minus the 4 middle circles. I saw it was equal to the area of the 4 orange plus that weird middle square. Doing the same with the square minus the corner circles, I saw it equaled the blue plus the same weird square! Realizing the square cancelled simplified the problem instantly.
I just love how it shows the value of the big picture and encouraged me to think in new ways
PSAT is Pre-SAT (Scholastic AptitudeTest). SAT used by colleges (universities) as part of admissions consideration.
OMG! I happen to get the correct answer to the 3rd one in my first try! without looking on your solving. I got pi and checked the final answer and voilà! I'm sooo happy!
from Morocco...all my respects mr genious
You are a real professional in Math
These videos make me realize that I'm just not as good at math as I used to think I was.
We need more problem solving methods like 1 and 3, solving problems intelligently instead of relying on high level techniques
I couldn’t solve part 3 at first as I decided to start with that one. But I couldn’t do it so I did part 1 first and that somehow made me able to do part 3.
I thought "man that's really hard". Then I had the answer in 15 seconds.
Ok, quick intuition on the first problem.
We have a circle(by extending the semicircle) around the point O.
Then we have a semicircle (equal to the first on by having the same area) attached to O at its coroner, free to rotate.
If we rotate that semicircle around the O joint it sweeps out an area.
The trick here is to notice that if it rotates the whole way around the circle, the arc edge between O the border must sweep out the entire area. And by symmetry pieces sweeping out equal angles must be the same regardless of the starting angle.
If we sweep out a fraction of the angle around the circle the swept are must be proportional to that fraction.
The fraction is 20/360=1/18.
The area of the entire disc would be 2*72.
Hence the blue area should be 72/9=8.
as an eighth grader who learned this type of math quite literally yesterday, i feel angered that problem three was simply pi. i’ve gotten used to such complicated equations leading to complicated answers that such a simple answer shocked me.
i am in 8th and i finished most of calculus other than vectors.
Problem 2 visual solution is nice but it relies on the fact that the semicircles’ intersection point aligns with the triangular grid, and no proof of that was given.
For problem 1 I used a rotation matrix to get the points of intersection. Then drop a perpendicular from the first point of intersection to the third semicircle breaking the area up into two sections. Then just add two integrals.
I liked the first one: a full circle consists of 18 "claws" (
You can use a similar overlapping areas argument as you did for problem 1 to solve problem 3 very quickly. The overlapping area is being subtracted from both the orange and blue areas, so it has no impact on their difference. Thus, we can simply take the total area of the quarter circles and subtract the total area of the semicircles. The only tricky bit is that the square has to be large enough for none of the semicircles to intersect each other (so it has to have a side length greater than or equal to sqrt(2)*semicircle radius), and similarly for none of the quarter circles to intersect each other (so side length greater than or equal to 2*quarter circle radius). Cases where you do have such self-intersections (eg. if you changed the side length to 5) are much more complicated, depending on how the new overlap regions are colored.
Really great problems, all of which hinge on how you plan the approach. Nice to link them together in one video.
I have NEVER felt this hype about solving these without pen and paper. Every time I had an insight I was fistpumping. Every time I managed to do all of the calculations and get to the right answer at the end I got a "Booya!" out. Very satisfying!
Intuitively, it's a 20° rotation of the same figure (relative to the center) on the half circle so 1/9 of the angle, hence 1/9 of the surface area of the half circle.
re prob 1, there is a simple intuitive solution, as follows:
the "claw" results from rotating semicircle R by 20°. if you rotate R' by 20°, you get another claw with an identical area. if you do this 16 _more_ times, the claws perfectly fill _circle_ Q, which has an area of 144. 144/18 = 8
At this point forming these shapes on a polar coordinate system and finding the area by integration is taking the easy path.
The 1st & 3rd problems are very easy. I had somewhat more trouble with the 2nd problem.
yay i figured out the 6:25 part (now ive learned to not be so greedy with calculating stuff bc it might just cancel out, but it still was insightful to solve)
the area of one of those weird shapes is ≈ 1.789
2nd question is impossible to solve without the help of equilateral triangle.
Le me, who can never think of a triangle in this problem. Now I am aware about eq.. tri.. in hexa.. problems. Thank you
i earlier knew that
that was an AMC 10 previous year question
4:24 Indeed this did create a ... mess!
Hmmm mmm , this question was on easy side 👁️👁️ . I would say , it is doable in 3-4 minutes by an average student .
Wow, algebra works almost everywhere
simple proof that the two prove the two shapes in common have the same area:
draw triangles inside the shape, two triangles with the exact same side lengths will be congruent, in this case we have two equilateral triangles, each side of the triangle is a sine of the same length arc (because of it is an equilateral triangle)... hence one shape is really a 20 degree rotation of the other
First I thought it is hard problems. Then I tried to solve them and found an easy solutions to all three.
The 3rd puzzle is deceptively simple. The overlapped areas are shared, so they don’t factor into the difference. For this reason, they should be ignored. The four quarter circles make up a single circle with radius 3, so the area is 9pi. The four semicircles make up two circles with radius 2, so the area is 8pi. The difference is 9pi - 8pi = pi.
My jaw dropped on the last question, what a simple approach
For the third problem I solved it by doing (49-8pi)-(49-9pi) as the middle are of the square cancels out as its in both shapes
just to straighten out the thumbnail: the SAT is used in college admissions. The PSAT is a pretest that is administered several years prior and is used for preperation for the SAT and as a placement barometer for High School.
Problema 1.- (2*72)/(360/20)=72/9=8
Problema 2.- Área de un hexágono de lado =1 → p=6*(√ 3/2)/2 =6√3/4 → Área círculo que lo circunscribe =⊓*1²=⊓ =c → Área azul =p-(c-p) =2p-c =3√3 -⊓
Problema 3.- Naranja - Azul =(4*⊓3²/4) - (4*⊓2²/2) =9⊓ - 8⊓ =⊓ .El exceso de naranja y azul considerado corresponde a los solapos entre semicírculos y cuartos de círculo; es decir, es el mismo para ambos colores y no influye en la diferencia entre ambos.
Curiosos problemas. Gracias.
👌
problem 1: repeating the tilt will show that 18 copies of the shaded area would cover a full disk, so the area is 1/9 of the half disk, so the shaded area is 72/9=8.
problem 2: drawing central circle and hexagon would reveal that the hexagon cuts the lens shaped areas exactly in half. So L=2(C-H) and A=C-L = 2H-C =12T-C = (12sqrt(3)/4 - π)r^2, which for r=1 equals about 3×1.732 -3.142 = 2.054
problem 3. the black areas overlap, so the area difference is one large minus two smaller circles: π(3^2-2×2^2)=π
I wrote this at 2:16 from top of head, without pen paper or calculator, so I may have made mistakes.
For question 2, I thought:
A = Hexagon with side = 1 - ( Circle with r = 1 - Hexagon with side = 1)
= 2 Hexagon - 1 Circle
= 12 Triangles - Circle
= 12 (1/2 * b * h) - Circle
= 12 (1/2 * 1/2 * Sqrt(3)) - Circle
= 3 * Sqrt(3) - Pi
For question 3, I thought:
Blue = Square - 4 Quarter Circles with r=3 - Middle Bit
Orange = Square - 4 Semi Circles with r=2 - Middle Bit
So difference = 1 (4 quarter) * 9 Pi - 2 (4 halves) * 4 Pi = 9 Pi - 8 Pi = Pi
Solution 2 was visually interesting practically I don't think so anyone can think about it but I have another way to solve:
Area of hexagon - 3 × area of semicircle = x (assume)
2x = hexagon - eye shaped area + shaded area.
Now I am stuck and with the help of definite integral and trigonometry I can find the eye shaped area but it take time.
They're symmetric lens' which have a nice formula: A = r^2 (theta - sin(theta)), where theta will be the angle in the equilateral triangle.
I had the similar idea for it but did: hex - 6semi + 6lens, since the lens were over subtracted by the semis.
Hi, youtube doesnt always inform the authors whenever there is a reply to a comment. Atleast its a preference setting. Anyway i have answered your comment. Hope u find the answer satisfying. Its essentially group theory.
20/360 fraction of circle's area. Problem 1 in under 1 minute.
3rd problem is the most beatiful
For problem 3 you don't even need algebra. You can just think of the areas where semicircles and quarter-circles overlap as orange and blue areas canceling each other-after all, the areas are subtracted in the end, so adding an equal amount to both sides of the subtraction sign doesn't change the result. So the solution is just the areas of the quarter circles minus the areas of the semicircles.
I solved problem 3 as follows:
Notice that the areas of square - 4 quarter circles = blue area + shape in the centre
square - 4 semicircles = orange area + shape in the centre
Subtract the first equation from the second you get:
(square - 4 quarter circles) - (square - 4 semicircles) = (orange area + shape) - (blue area + shape)
Open the parenthesis and simplify you get 4 semicircles - 4 quarter circles = orange area - blue area
My solution for #3 is to find the difference between the square less the 4 quarter circles and the square less the 4 half circles. Visionally this is just the orange with the center squarish area minus the blue with the same squarish area. Mathematically this is trivial: (49 - 9π) - (49 - 2*4π).
Yup, that’s how I got it too, congrats!
Interesting approach, but since 9π > 2*4π, then (49 - 9π) < (49 - 2*4π) and (49 - 9π) - (49 - 2*4π) < 0
So you should be subtracting the other way: (49 - 2*4π) - (49 - 9π) = π
May I ask which tool / software you used to enable you to apply color to a shaded region bounded by different curves. Many thanks!
the 3rd part was also calculated using the trapezoid formula:
10 vdu5:for a=0 to 15:gcol a:print a:next a:c1=4:c2=11:gosub 330:rem farbtest
20 l1=7:r1=3:r2=2:xm1=-l1/2:ym1=-l1/2:xm2=0:ym2=-l1/2:rem den schnittpunkt berechnen
30 sw=(r1+r2)/100:xs=xm1:nu=100:xc=0:yc=0:wu=pi/2:@zoom%=@zoom%*1.4:goto 80
40 ys=ym1+sqr(r1^2-(xs-xm1)^2):dgu1=(xs-xm2)^2/r2^2:dgu2=(ys-ym2)^2/r2^2
50 dg=dgu1+dgu2-1:return
60 xvx=(xu-xc)*cos(wu):xvy=(yu-yc)*sin(wu):xv=xvx-xvy+xc
70 yvx=(xu-xc)*sin(wu):yvy=(yu-yc)*cos(wu):yv=yvx+yvy+yc:return
80 mass=1E3/l1:gosub 40
90 dg1=dg:xs1=xs:xs=xs+sw:xs2=xs:gosub 40:if dg1*dg>0 then 90
100 xs=(xs1+xs2)/2:gosub 40:if dg1*dg>0 then xs1=xs else xs2=xs
110 if abs(dg)>1E-10 then 100 else print xs,ys:ar=0:goto 130
120 xbu=(x+l1/2)*mass:ybu=(y+l1/2)*mass:return
130 yul=-l1/2:yol=yul:xul=-l1/2:dx=(xs+l1/2)/nu:xol=xul:gcol c1:for a=1 to nu:dxa=a*dx
140 x=-l1/2+dxa:if x
That's very generous to call the second one "a great puzzle". First and third were nice though
Answer 20/360 × 144 = 8 . the two semicircles bounding the area can rotate through 360 degrees seprated by 20 degree
For qno. 3 I have joined the four quarter circles of radius 3cm and calculated its area pi.3^2 then I subtracted the four semi circle of radius 2cm (1/2.pi.2^2)*4
Re solution for problem 1: it's a bit of big jump from "all the semicircles have the same area/are congruent" to "the intersection between two semicircles has the same area". In this particular configuration perhaps, but I am missing the explanation as to why.
Consider 2 circles(full circles) intersecting ach other's radius they have a constant common area regardless of how both are rotated. The problem is the same thing with just the upper half.
In the PSAT problem, you don't have adequate time to do calculations...have to just eyeball it. Half of a semicircle is 36, the shaded part is about a quarter of that, so 8-9. Pick the multiple choice answer closest to that and move on.
idk why, but problem 3 was the easiest for me. Probably because it's more connectet to algebra (i instantly thought of making a system of equations).
While I wanted to do the trick at 4:22, it wasn't completely obvious that the intersections of the red lines had to fall at the intersections of the semi-circles. I felt like I needed to prove that that worked. It does, but is this a known fact about circles inscribed in hexagons?
I don't think so, but it was such an intuitive assumption that I almost didn't spot that it needed proving. Just something that seems natural when you've done a lot of triangle and circle puzzles, I guess. Pretty obvious once you lay the mesh of smaller triangles over the hexagon, though.
That’s an elegant solution for the 3rd problem, but you seem to have skipped a step going from 4(3²pi÷4)-4(2²pi÷2) to pi. My middle school math teacher would have (and did) dock points for not showing all of your work.
3:25= you could've used this reasonment to deduce that the shaded area is exactly the area of the circle : 18 since 360: 20 = 18
Problem 1.
Realize that if the construct was rotated 18 of then would completely fill a circle. Hence, 9 would fill the area of a semi-circle. Hence answer is 72/9 = 8
In problem 2, shouldn't we prove that the points where semicircles cross lie on the lines that connect side centers? It looks right but doesn't seem obvious to me.
I had pretty much the same thought process for the last one but for some reason I assigned b to blue, a to orange, and p to the overlap
I don’t remember why I used p and I’m a little confused thinking back on it
It’s funny that the side length being 7 is mentioned tho bc it’s not needed
Bro this is Krishna from India. Here is a problem from my side. Find the volume of Intersection between two spheres of equal radii, with their centers lying on the surface of each other?
I derived a direct formula for this problem in my 11th standard using Trigonometry and Solid angle concept, And verified my formula using integration in my 12th standard. I want to share with you. Because I love you videos.
Could you explain to me how you used the concept of the solid angle to find the volume?
The formula for circles is (pi-1)r^2/2. What is it for spheres?
Easy questions except 3rd it was moderate.
I solved all questions in 15 min😂
I solved the first one and second one but stuck on the third one, but it turns out it was easiest of them all
6:38 while i love math for easy and elegant ways for solving task like this one unfortunately not every task have it =(
problem 1: the way i thought thru this was that the claw/rotated curved lines is the same as drawing curved lines like a hypnosis circle. if u draw lines splitting it into 18 equal parts (360/20 for the 20deg rotation), the circle is in 18 equal parts regardless of the shape of those parts. or for the half circle, 9 equal parts or 1/9, 72/9 being 8
might not be the "right" way of getting there but its the way i got it 😂
The first one, I noticed that the angle for the 'claw' was the same as the small segment of 20 degrees. I can't readily prove it, but it seems that the 'curved sides' of the claw, having the same radii, simply 'cancel' and you have the same area as the circular segment. From there it's easy to see 1/9 of the semicircle area.
Same. I thought of using calculus to sum the radial arcs over the size of the left circle, but then figured I’d just estimate it based on a straight wedge in case this was a mutiple-choice test because that should be close enough. Nice to know it was exactly right. :)
Why do you need to multiply everything by 4 in the last problem? Wouldn't it be easier to leave it as is and continue the same?
3:07 why exactly do these two have the same area? What if you tilt the titled one even more? Now it's intersecting more of the left and less of the right semicircle.
3:14 thats the most unusual way I've ever heard someone pronounce congruent
All three problems have a much simpler solution than the one shown here.
After a few sub-par videos recently, this one is _Sheer Poetry_ ..!
on the psat i had a question about cosine so there is trig stuff on it
Even being reasonably good in Math, I find these type of problems difficult. My mind find it arduous to parse the spatial information given into meaningful patterms to solve. Also, if I've ever solved these- usually end up taking the long route which involves more brute method and less elegance. 😮💨
Imagine a problem with circles without even the mention of pi...
for the second part, i have written this code with variable number
of polygon points (set nz=6 and rkg=2 in line 10):
10 nz=6:sw=.001:lg=2:dr=0:wi=2*pi/nz:rkg=lg/2/sin(wi/2):wk=asn(lg/sqr(4*dr^2+lg^2))
20 @zoom%=@zoom%*1.4:rkk=sqr(lg^2/4+dr^2)
30 nu=50:dim xs(nz-1),ys(nz-1),x(nz-1),y(nz-1)
40 xm=0:ym=-rkg*cos(wi/2)-dr:print ym:xc=0:yc=0:wu=wi
50 xu=xm:yu=ym:goto 80
60 xvx=(xu-xc)*cos(wu):xvy=(yu-yc)*sin(wu):xv=xvx-xvy+xc
70 yvx=(xu-xc)*sin(wu):yvy=(yu-yc)*cos(wu):yv=yvx+yvy+yc:return
80 gosub 60:xm1=xv:ym1=yv:print xm1,ym1:xs=sw:goto 110
90 ys=ym+sqr(rkk^2-(xs-xm)^2):dgu1=(xs-xm1)^2/rkk^2:dgu2=(ys-ym1)^2/rkk^2
100 dg=dgu1+dgu2-1:return
110 gosub 90
120 xs1=xs:dg1=dg:xs=xs+sw:xs2=xs:gosub 90:if dg1*dg>0 then 120
130 xs=(xs2+xs1)/2:gosub 90:if dg1*dg>0 then xs1=xs else xs2=xs
140 if abs(dg)>1E-10 then 130 else print xs,ys
150 xs(0)=xs:ys(0)=ys:for a=1 to nz-1:xu=xs(a-1):yu=ys(a-1):gosub 60:xs(a)=xv:ys(a)=yv:next a
160 xmin=0:ymin=0:xmax=xmin:ymax=ymin
170 for a=0 to nz-1:if xmin>xs(a) then xmin=xs(a)
180 if ymin>ys(a) then ymin=ys(a)
190 if xmaxx(a) then xmin=x(a)
500 if xmaxy(a) then ymin=y(a)
520 if ymax
run in bbc basic sdl and hit ctrl tab to copy from the results window
important info: set nz=6 instead of 5 in line 10.
I didnt understand in problem 2 how you know that the lines of the mid points cross the point of intersection of the circles
@4:20 (answer 2), it is not obvious to me that line segments connecting the midpoints of sides two sides apart will go through the intersections of adjacent semi-circles. How do we know that is true?
Because each small "semicircle" is one third the circumference of the six semicircles, and you can draw triangles to verify this.
The edge of the circles are the center of the other circles? this would be easy
the curved triangle. well its area can be calculated but not needed. The important factor is the blue area+ middle area= middle area+ bottom curved triangle therefore blue area= bottom triangle
And the bottom triangle is a shard of the semicircle. Since it's 20 degrees it's 1/9 the area of the semicircle. Solved.
Well since you asked for the size it's 72/9=8 in area.
i have written a code using the trapezoid rule that uses the trapezoid data for the graphics for the first part:
10 r=1:dim xm(2),ym(2):xm(0)=0:ym(0)=0:xm(1)=0:ym(1)=-r:nu=200:w=20
20 xu=xm(1):yu=ym(1):xc=0:yc=0:@zoom%=@zoom%*1.4
30 xvx=(xu-xc)*cos(rad(w)):xvy=(yu-yc)*sin(rad(w)):xv=xvx-xvy+xc
40 yvx=(xu-xc)*sin(rad(w)):yvy=(yu-yc)*cos(rad(w)):yv=yvx+yvy+yc
50 xm(2)=xv:ym(2)=yv:sw=.01:w=sw:i1=0:i2=1:w=3/2*pi:goto 130
60 dx=r*cos(w):x=xm(i1)+dx:dy=r*sin(w):y=ym(i1)+dy
70 dgu1=(x-xm(i2))^2/r^2:dgu2=(y-ym(i2))^2/r^2:dg=dgu1+dgu2-1:return
80 gosub 60
90 w1=w:dg1=dg:w=w+sw:if w>2*pi then stop
100 w2=w:gosub 60:if dg1*dg>0 then 90
110 w=(w1+w2)/2:gosub 60:if dg1*dg>0 then w1=w else w2=w
120 if abs(dg)>1E-10 then 110 else return
130 gosub 80:xs1=x:ys1=y
140 i2=2:w=3/2*pi:gosub 80:xs2=x:ys2=y
150 print xs1,ys1:print xs2,ys2:mass=5E2/r:goto 170
160 xbu=x*mass:ybu=(y+r)*mass:return
170 for a=0 to 2:x=r+xm(a):y=ym(a):gosub 160:xba=xbu:yba=ybu:for b=1 to nu:wb=2*pi*b/nu
180 dx=r*cos(wb):x=dx+xm(a):dy=r*sin(wb):y=ym(a)+dy:gosub 160:xbn=xbu:ybn=ybu:goto 200
190 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return
200 gosub 190:next b:next a:gcol5
210 da=0:ar=0:xa=0:y=r+ym(1):yah=y:yal=y:dx=xs2/nu:for a=1 to nu
220 xn=xs2*a/nu:ynh=ym(2)+sqr(r*r-(xn-xm(2))^2): if x>xs1 then 240
230 ynl=ym(1)+sqr(r*r-xn^2):goto 250:rem print ynl,ynh
240 ynl=-sqr(r*r-xn^2)
250 da=dx*(ynh-ynl+yah-yal)/2:ar=ar+da
260 x=xa:y=yal:gosub 160:xb1=xbu:yb1=ybu:x=xn:y=ynl:gosub 160:xb2=xbu:yb2=ybu
270 y=ynh:gosub 160:xb3=xbu:yb3=ybu:x=xa:y=yah:gosub 160:xb4=xbu:yb4=ybu
280 move xb1,yb1:move xb2,yb2:plot85, xb3,yb3: move xb1,yb1:move xb3,yb3:plot85, xb4,yb4
290 xa=xn:yal=ynl:yah=ynh
300 next a:print ar
310
0.866025404 -0.5
0.984807753 -0.173648178
0.179551341
>
run using bbc basic sdl and hit ctrl tab to copy from the results window
Problem 1 and 3 were easy but 2 i couldn't do it was hard to visualize wow.
There are 18 of the warped sections in a full circle. So 2*72/18=8; no need to concern ourselves with shapes.
How to find number of digits in n factorial where n is a whole no.