A nice quick geometry problem.
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- Опубліковано 7 вер 2020
- For some "lighter" fare today, we look at a nice quick geometry problem.
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This has some massive recommended in 8 years energy.
I actually expected it to be published 6 years ago.
Hello people from the future!
Fr tho
@@israelserrano1271 same. I thought it was on old video that got recommended
Agreed lol
UA-cam:
Yo! Here is some geometric problem and how to solve it!
Me sitting on toilet:
*interesting*
Hey its been a day have you gotten off the toilet
@@randallmokjialung3592 yeah, but I was sitting a little too long.
@@remigiuszdarmach4233
*interesting*
me rn
Shit, you spying on me?🙄
Oh, that's a square.
Its a square till you see the circle
Yeah I thought it was a rectangle. AB looks clearly larger than AD, and I thought he would find a formula that would apply for any AE / DE length.
Americans are to picky, where i live a circle looks like a squares
An elusive Parker square?
Kaisar Sihaloho I don’t think so, you can prove __AM__ =/= __MB__ for midpoint M, by proving R is either half the inscribed horizontal line of symmetry, or half the inscribed diagonal line of symmetry, but not both, which is trivial since reflecting R over the vertical axis would create a contradiction r=/=sqrt(2)r. Ergo, the shapes are not concentric.
Now, set up a Cartesian plane with the origin on M. Complete the diagonal diameter by rotating line segment BM 180deg. Label this diameter’s bottom-left circle intersection F. By symmetry across x-axis, we can see that F is on the square, at coordinates (-x,-y); so, it must be on their intersection. But if it is on their intersection, then its distance from the origin is greater than r, else the shapes would be concentric.
That’s no square, which you can tell precisely by the circle
Dunno how the heck I landed here (probably missed a stop on the UA-cam bus or something), but I'm glad I arrived. :-) cheers!
This is a good place to stop
You cheated your way here cheaterman
@@vikranttyagiRN For what it's worth, I suspect this is part of the "Stand-up Maths audience also watches this channel" recommendation series! If you all know Matt Parker, then this explains that!
@@cheaterman49 Or BlackPenRedPen, Dr. Peyam, or Prof. Omar.
1:15 me in a presentation I have done no work on. "and then furthermore we know some stuff"
Summarizing other group members' information
I should use that sometime!
Lmak
Fun fact: I paused the video right at the start and tried to solve it on my own. I did it but it took several hours and involved trig functions and the inscribed angle theorem. Needless to say I was pretty shocked that it actually had such a simple solution.
It's kinda sad it took you several hours ? What theorem appealed to you at first ?
Plus the answer here is pretty useless, assumes it's a square while misleading with the rectangle drawing and isn't even about the general case for any square
@@Revengeur72 how long did you take mr. Ramanujan?
@@Revengeur72 What do you mean it isnt about the general case for any square? Didnt he solve it for general r and y?
Also if you cant work with a hand drawn sketch like that, youre probably not a geometer xD
Hahahaha lok
Yeah i used an angle theorem, it’s easy enough if you remember your sin double angle formula!
Neat, it’s good that the perimeters turn out to be really similar as it makes the question hard to guess. The division evaluates to about 0.98.
now imagine your are on test with no calculator
@@Ivan-wb4id it isn't that hard, you can calculate the π needs to be 3.2 to be equal, since π is smaller, the square is bigger
As a mathematician he wasn't at all interested in how close they were to each each, only that one was greater than the other!
This needed to be highlighted in the video. It's not obvious based on mental math.
"And furthermore we know some stuff" I couldn't tell if he was trolling or not
When ur the impostor and trying to convince others ur not....
...sorry I just had to make an among us refference
It's a joke that only actually smart people who realize the depth of their ignorance about the world can pull off
Based on the comments, I think a good lesson is to always preface any drawings with the fact they aren't always drawn to scale.
Anything drawn in chalk (or ink or....) will _never_ be an exact circle or square. If I had done this in class, I would have drawn the circle without a compass just to illustrate that point.
I mean, you people had math classes at school right?
It should definately be stated somewhere, I find that sort of thing very confusing. Why make a drawing at all if it´s nowhere close to what it´s supposed to represent? On the other hand the circle appears to be perfect, so one would assume the squaery thing to be as it appears too.
@@teppo9585 Diagrams are supposed to help you gauge relationships between shapes, points, lines, angles, etc. They're not meant to be perfect representations of those things. That's why when we want perfect representations, we add labels or descriptions.
Is it even possible that there could be a rectangle at its place???
U cant place a rectangle there....
Don't know why it was recommended and why I watched it, but I liked it!
Very nice job of analyzing the problem!
Another way I found to do this which avoids the Pythagorean calculation is to extend the line through EO (where O is the center of the circle) until it meets the circle again at G. Let the midpoint of BC (which lies on EG) be labeled F. Then we have EF x FG = BF x FC (using a well-known theorem about intersecting chords in a circle). Using s for the side length of the square this reduces to s x FG = (s/2) x (s/2) which implies that FG = s/4. Then we have 2r = EG = EF + FG = s + s/4 = 5s/4. The rest then follows as in the video.
I think yours is one of the best method. Once you find the relationship 2r = 5s/4, i find it helpful to think of the perimeters using a clock analogy. It will take 360 deg for the circle to complete a full rotation. If you then stretch out the square so that you fits its perimeter (4s) to an arc of radius r. You find that taking 4s = r theta and plugging in the above yields 4s = 5s/8 theta. Then cancelling on both sides simplifys to theta = 32/5 rad = 366.7 deg. So the square will move a greater distance in the same amount of time (like that of a clock).
Absolutely brilliant videos. Gets my brain working in ways it hasn't for a very long time. Thank you!
I really enjoy your videos, Michael. I wish your channel grows a lot.
@@mr.knight8967 Really does not need to be spammed here.
Unique Math^ 👍
Perimeters: Square Wins = 32/10π.
Areas: Circle Wins = 25π/64
Oof perimeter of square almost had it .6 if I’m correct
After all, a circle has the highest ratio of area to perimeter of any shape
@@Piper_____ yeah
Um, you might want to indicate what the ratios represent.
32/(10π) ≈ 1.02 = Perimeter of square / Circumference of circle
25π/64 ≈ 1.23 = Area of circle / Area of square
But your ratios are inconsistent. Both ratios should be circle/square or both should be square/circle.
@@Piper_____ This is true, and would be simple if both the square and circle had equal perimeter. But the perimeter of the square is greater than perimeter of circle. So we can't just by the fact that a circle has greater area for the same perimeter. In this case, the slightly longer perimeter of square is not enough for the square to have greater area.
Outstanding Michael. You’re awesome.
And here I was using the inscribed circle theorem to get to the radius.
Great vid
I solved it using similar triangles and didn't need to use the Pythagorean theorem.
I'd like to see you challenge Presh to an arm wrestle and if you win, he has to call it the Pythagorean Theorem.
lol
I think he called it gougu theorem
Haha,same
What I did was to form lines EB.and also EF(which passes through E and centre)
I found EB with pythogoras theorem,found it to be (√5/4)a.
Now when you bisect any chord in a circle,and draw a perpendicular to it(GH,H is centre),it will reach the centre.
Now the triangles EGH and EFB are similar. Therefore EB/EF=EG/EH.
(EH is the radius)
Now EH being (5/8)a, the permiter of the square is 4a,the permiter of the circle is 6.28*5/8 which comes out to be 31.4/8 which is a slight less than 4. Therefore, the permiter of square is bigger. :)
(Idk how I managed to pull this off but I'm reall happy)
Prof. Penn has been climbing as a hobby for a while.
I doubt Presh stands a chance.
@@karenamma7716
eb/ef = eh/eg
Do you think i’m around today because someone said “That’s a good place to stop”? I want more!!
Excellent presentation
It is also good to throw in some of these easy ones. Helps shore up self-confidence.
Amazing video my man, keep it up
I'm happy I watched this video, the solution using ratio is really helpful. Thank you so much!
Nice solution! I tried coming up with my own and took a different but still algebraic route.
Without loss of generality, assume the radius of the circle is 1.
We can represent the distance AB with respect to AE using the coordinate formula for a unit circle, shifted up by 1: AB = 1+sqrt(1-AE^2)
Also, because we're working with a square, AB = 2AE.
We can substitute the first equation to get:
2AE = 1+sqrt(1-AE^2),
(2AE-1)^2 = 1-AE^2,
4AE^2-4AE+1 = 1-AE^2,
5AE^2-4AE = 0,
AE(5AE-4) = 0,
the positive solution is AE = 4/5.
The perimeter is 8AE = 32/5, which is greater than the circumference of the circle with radius 1, which is 2pi. This can be verified by hand by halving both values, and noting that pi = 3.141... < 32/10 = 3.2
Here's an alternative: Using his sketch, move line AED to the right such that A and D now sits on the circumference. Let these now locations be denoted by A' and D'. You now have a inscribed rectangle of sidelengths 2x and 2y. Let the centre be denoted by 'O'.
Let theta = 0.5(angle D'OC). Now x = rsin(theta) and y = rcos(theta).
From the original sketch we have DC = r + x and DC =BC = 2y. Now equate them: r + x = 2y r(1 + sin(theta)) = 2rcos(theta).
Divide by r, substitute cos(theta) = sqrt(1-sin^2(theta)) then square both sides: 4(1-sin^2(theta)) = 1 + 2sin(theta) + sin^2(theta).
This is a quadratic equation: 0 = sin^2(theta) + 0.4sin(theta) -0.6. This solves to sin(theta) = 0.6.
The perimeter of the square is 4DC = 4(r+x) = 4( r(1 + sin(theta))) = 4r(1+0.6) = 32r/5. Area of circle = r2pi. The ration as given in the video is: 2rpi/(32r/5) = 5pi/16
Assigning a value to the circle's radius it can easily be calculated with functions as well. For example, the equation telated to the function of a circle with r= 3 is x^2+y^2=9. The "ceiling" of the square necessarily needs to fulfill the equation x+3=2y. Solving the system of equations we get a side for the square of 4.8. The perimeter of the square is thus 19.2 whereas the perimeter of the circle is 18.8
Loved this one, really nice. Easy to understand and a nice solution.
That's a really surprising answer, just based on the picture it looks like the parts of the circle that lay outside the square vastly exceed the parts of the square that lay outside the circle
That is because the drawing is way off. The "square" is far from square. As drawn, the circle definitely has a larger perimeter than the rectangle.
love the way this guy solves the problems
The chord joining the intersections between the circumference and the square divides it into two rectangles, one of which is entirely internal to the circumference, and in the other it is easy to see that the part outside the circumference has a smaller area than that in common.
Perfect timing for some carpentry work I need to do today!
This is a great problem. I am going to use this problem to introduce some properties of circles to a Cambridge IGCSE class. Let's see how that goes!
I haven't done math in earnest in AGES. Stumbled across the 'three squares sides a, 2a,3a inscribed in a triangle' problem and this one and it's been fun! This one I solved in a much different way (more complicated :/ used inscribed angle theorem and some double angle identities ) but came to the same answer, so yay!
Good job mate
Nice shirt! Discovery is amazing. Happy Star Trek Day!
Let's take a unit square, so r + x = 1. Since y = 0.5, we have that the most right point of the circle is 0.25 away from the square (Thales and uniformity). Hence the circle has diameter 1.25, and a perimeter of pi*1.25 which is smaller than 3.2*1.25 = 4, which is the perimeter of the square (4 times 1).
Nice puzzle! I solved it differently - I set the side length of the square to 2, then AE=ED=1. Setting the angle BEC=\theta, the angle EBA=ECD=\theta/2 and tan \theta/2 = 1/2 so tan \theta = 2(1/2)/(1-(1/2)^2) = 4/3 - and by the central angle theorem, the angle BOC at the circle center is 2\theta - and taking F as the midpoint of BC, we can see that the triangle OFB is a 3-4-5 triangle with opposite length 1, since tan \theta = 4/3 - so the radius is 5/4, giving the same result that the perimiter of the square is larger.
Nice little peace of geometry, thanks a lot
Nicely explained...
I love these type of intelligent teachers
Travelling round the square or the circle, he asks “Which takes longer?” I knew right away : the square. Because you have to slow down at the corners.
How can you kept your head upright when you got such a big brain 👏👏👏
I love this guy and his videos.
Note that right from the beginning you can treat the side length AD of the square as 1 unit. Thus y is always 1/2 and you can skip the ratio/Substitution of y/r and go right to 5pi and 16.
Thank you!
When Conan and Neill Patrick Harris have a math teaching child
Not watching the video yet:
Draw three lines connecting the center of the circle (call it point R) to points E, B, and C. These lines each have length = radius r.
R, B, and C are corners of a triangle with base BC and height h.
r = BC - h
r^2 = (BC - h)^2 = BC^2 - 2BCh + h^2
by Pythagoras, r^2 also = (0.5 BC)^2 + h^2
BC^2 - 2BCh + h^2 = 1/4 BC^2 + h^2
BC^2 - 2BCh = 1/4 BC^2
3/4 BC^2 - 2BCh = 0
3/4 BC^2 = 2BCh
3/8 BC^2 = BCh
3/8 BC = h
r^2 = 1/4 BC^2 + h^2
r^2 = 1/4 BC^2 + (3/8 BC)^2
r^2 = 25/64 BC^2
r = 5/8 BC
square perimeter = 4 BC
circle perimeter = pi (5/4) BC = pi (5/16) * square perimeter
16/5 = 3.2 < pi
5 pi < 16
pi (5/16) * square perimeter < square perimeter
circle perimeter < square perimeter
wow you have to be really smart to come up with that on your own
@@jonasmach5728 Not really. It's a good problem but not a very difficult one. Just takes a few minutes to think about all the relations you can come up with between the different variables.
Same,
What I did was to form lines EB.and also EF(which passes through E and centre)
I found EB with pythogoras theorem,found it to be (√5/4)a.
Now when you bisect any chord in a circle,and draw a perpendicular to it(GH,H is centre),it will reach the centre.
Now the triangles EGH and EFB are similar. Therefore EB/EF=EG/EH.
(EH is the radius)
Now EH being (5/8)a, the permiter of the square is 4a,the permiter of the circle is 6.28*5/8 which comes out to be 31.4/8 which is a slight less than 4. Therefore, the permiter of square is bigger. :)
(Idk how I managed to pull this off but I'm reall happy)
Thanks for the hard work, can u take a look at 1985 Putnam B6 and hope u make a video solving it .
I don't wanna be "that guy" since it would be cool to see him solve it, but isn't it not that hard? Since G={A1,A2,A3,...,Am} is closed under matrix multiplication (as it is a group) then we know that AiAj=Ak. So if we have a matrix, S, such that S is the sum of all the terms of G, then we know that Ai*S=Ai*(A1+A2+...+Am)=Ai*A1+Ai*A2+...+Ai*Am=some other ordering of S=S. This means that we can write S*S=(sum [m≥i≥1] Ai)*S=sum [m≥i≥1] Ai*S=sum [m≥i≥1] S=m*S. If S²=mS, that gives two possibilities for S, either it is the zero matrix or it is m times the identity matrix. It can't be m*I, unless m=0, as the trace of m*I is 2m, which can't be 0 unless m=0, which if it is then that would mean G is the empty set and S is an empty sum and thus, the zero matrix. This means S must be the zero matrix.
@@jadegrace1312 it doesn't follow a priori that S is either the identity or the zero map, it could be any diagonal matrix with entries 0 and m. I think it does indeed follow that all Eigenvalues are 0 or m, and then the trace condition kicks in, but there's definitely a bit more to say.
@@achimkrause7971 That's probably true. My solution may be incomplete so it would interesting to see him solve it.
I wish to have such chalk powers someday
Thank you
The trick is to always make up variables for the things you don't know!
that's like, what variable are
How can I show that the points B and C are on the cercle ?
UA-cam: just a nice geometry problem
Everyone: yeah we like math
*clicks in even though they hate math*
Nice one! You Rock!
Wow, I massively overcomplicated my answer by finding the chord EC and forming a triangle with the radii to E and to C. I worked out it had sides (r, r, sqrt(5) y) and angles (atan(1/2), atan(1/2), pi - 2 atan(1/2)). I ended up with the ratio between y and r as cos(atan(1/2)) / sqrt(5/4), which as it turns out is exactly the 4/5 required, though I did require wolfram alpha to tell me that!
I spent hours trying to do that by figuring out the equations of each line of the square and taking the hypotheses that the left one is tangent to the circle, that both right ones intersect exactly at the circle, that the distance (computed with pythagore theorem) between the horizontal segments adn the vertical ones should be the same... All of that to lead to an equation 5x^2 - 4x = 0, leading to the same result :-(
Lot of time taken but not really lost after all and it's nice to see that there are different ways to reach the same result !
Easy and neat question.... Ratio of perimeters of circle to square is 5π : 16
@@mr.knight8967 yeah one time saw.... Easy question!
4:52
get a life dude
@@elie.makdissi it is Penn himself
@@chessematics you kidding right?
@@chessematics No, he has already cleared he's not Michael. Go to some of the older videos and read his replies
@@elie.makdissi Why are you obsessed about what the OP is doing with their life? One of us _should_ get a life and I'm not talking about Good Place To Stop or myself
I found it out using power of a point
Draw EF such that EF=AB with AB parallel to EF
Extend EF to make diameter EG
Let 2s be the side of the square with EF bisecting BC.
EF=2s, BF=FC=s
Let EG=2r
By power of a point
(BF)(FC)=(EF)(FG)
(s)(s)=(2s)(2r-2s)
s=2(2r-2s)
5s=4r
Same ratio that led to the same finish as him
Edit: thanks for the videos
I ended up using the fact that the radius of a circumscribed circle for an isosceles triangle is a^2 / 2h where a is the length of the equal sides of the triangle and h is the triangle's height, easy variables to calculate here.
x, y and r are a Pythagorean Triple (3, 4, 5)
@das- Neves Because the ratio between y and r (of the right triangle) was 4 to 5 (as explained in the video).
and, r is the hypotenuse, so it has to be a 3, 4, 5 triangle
IOW, the only option left is for x is to be the "3"
Of, course, if you are asking why the universe works that way, well.....
Most excellent lesson! But we must not forget this important theorem - pie are not squared, pie are round.
Nice little problem!
Another approach (more geometric) is to use the power of a point theorem from a corner of the square outside the circle. With the Pythagorean Theorem, you get that the side of the square is 8/5 the radius of the circle. (Then you know that 2*pi < 6.4.)
Thanks for sharing!
So I just watched a short geometry lecture
Easier is:
Assume the square's side length = 1. This gives the base of the triangle drawn at 1:09 would be 1-r. So we have (1 - r)² + (½)² = r². Expanding, cancelling the r² and simplifying we get r = 0.625. The circle's circumference is then 2πr ≈ 3.927. Since the square's perimeter is 4, the square has a larger perimeter. But it's close!
In problems like this, when it's obvious that the answer will not change when scaling up or down, it makes life easy to assume one of the dimensions is 1.
If he was my teacher I'd never miss a class!
Nice problem! I might have to throw this one at my geometry kids later in the year.
Nice now can you show us how to construct a circle and square with the same area would appreciate it 🤙
A neat solution to finding the radius of the circle is using the chord chord power theorem.
Let the sidelength of the square be x. Draw a horizontal line through the middle of the square. This line is exactly the diameter of the circle. This has diameter has a length of y+x. By the chord chord power theorem; y/2*y/2=x*y. It follows from this that the diameter of the circle is 1.25 times bigger than the sidelength of the square.
The rest follows the same argument as in the video.
I solved this using the chord law in a circle.
I drew a horizontal diameter from E, which intersected side BC. So if the square side is length 's', then the 4 lengths are: s & x (horizontal), and s/2 & s/2 vertical.
The chord law states intersecting chords in a circle have equal products of their 2 section lengths. So xs= (s/2)squared.
This gives x=s/4. So the diameter is 5s/4.
This gives us the perimeters for the square & circle, in terms of s, and shows the square Perim is slightly bigger.
damn it took all of that
I did it by placing point E at the origin and point B at (a, a/2).
Then the circle has equation (x-r)^2 + y^2 = r^2 because it goes through the origin and is symmetric about the x-axis.
Plug in the point (a, a/2) for (x,y) and solve for r in terms of a.
You get r = 5a/8.
The perimeter of the square is 4a.
The circumference of the circle is 2πr = 2π(5a/8) = (5π/4)a.
Comparing 4 to 5π/4, the square is bigger.
Very nice problem :) definitely recommend viewers try it before watching
Geometry is fun! When you’re first learning math you need to have so many values given to you beforehand to work out the remaining variables, but as you learn more and more theorems and algebra techniques you can find a solution with far less given!
how can you assume PE where p is the centre is a bisector?
Square ABCD with AB = x, and
circumscribed circle around a triangle KBD (K - center of AC) with radius r. r = abc/4S, where a=b=KB and c = x. KB = sqrt(a^2 + 1/4 a^2) = a*sqrt(1.25). S(KBD) = 1/2 * S(ABCD) = a^2/2. r = 5a/8. P1 = 4a. P2 = Pi * r^2 = Pi * 25a^2/64. P1>P2.
Finally a question I know how to solve!
How to answer this question if we do not know the formula of a perimeter of a circle?
With simple math: triangle EAB has cos = 2/sqrt(5). Triangle EOB has same angle and relates to r. O is center of circle. Length EB is known y*sqrt(5). Then very quickly arrive at r=5y/4 and the final ratio 5*pi/16
If you call the point that IDCI and the circle intersect point F you can see that EDF CDE and BEF are all similar, 1-2-5^(1/2) ratio, right-angle triangles with a common side when pairing them. To see this easily notice along with the 90 degrees angle, that they each have the angle of the EF chord. From this, the answer becomes solvable without even doing quadratics. you can essentially eyeball it. if IDFI =1; IEDI = 2, IDCI= 4 ,IEFI = root5 , IFBI=2r=5.
Dang this is some satisfying math
The angle CEB is half of COB (O being center of the circle). Same is true for halves, so if F is on the opposite side of the square from E, angle FEB twice FOB and also a triangle inscribed into half of the square, (r+x)/y = 0.5 (because square!) , FEB=atan(0.5). Now y/r = sin FOB, r = y/sin(FOB) = y/sin(2 FEB) = y/sin(2 atan(0.5)) ; perimeter of square is 8y, of circle 2pi y/ sin(2 atan(0.5))=7.853y < 8y
Nice sol!
find r depending from a leads to 3 equations with 3 unknown dimensions, xm, ym,r
Let z is the distance from |BC| to BC arc. So (triangles are similar) z/y = y/(x+r) = 1/2 => z = |AB|/4 => D = r+x+z = 5|AB|/4.
good job
No need for pythagoras. Give the corners of the square coordinates (+/-2,+/-2) hence the origin is the centre of the square.
Draw EB (gradient 0.5, centre coordinates (0,1)) and its perpendicular bisector (which is a diameter with gradient -2.)
The perpendicular bisector hits a horizontal line from E (equation y=0) at (0.5, 0). Hence the distance from centre of the circle to E is 5/8 the sidelength of the square.
finally, one that i could solve by myself in under 10 mins.
excellent !
Probability/Combinatorics would be awesome
Elegant!
SOLVED IT BEFORE WATCHING THE VIDEO!!! Pretty nice problem tbh
Does that make xyr a 3 4 5 right triangle?
Nice question
nice to work the math out....circle/sphere is more efficient than square/rectangle....same problem working volume will show circle has more...cheers
Great..Keep Working hard..Love from India
Finally, something i understand
Ok i definitely need a chalkboard
Using DCA as a base
I defined the circle as :
(x - x_0)^2 + (y - y_0)^2 = r^2
I have three points which I know are on this circle : E,B and C. Three equations with three unknowns, solved it and got a circle or perimeter 5pi/4 compared to the 4 of the square
Micheal, I tried to construct this diagram geometrically (ruler, compass) and could not figure out a way to do it. Any hints?
Drawing the square ABCD, then finding the midpoint (E) of side AD is all stuff you know how to do, I presume.
You can then use the method here: www.mathsisfun.com/geometry/construct-circle3pts.html
to find the centre of the circle from points E, B & C.
Solution without pythagoras or algebra.
Let the side length of the square=2, Ps=8
From the tangent point, bisect the square and continue to the opposite edge of the circle.
This line bisects the opposite side of the square.
Now use power of a point
a,b are two half sides of the square. c is the width of the square. d is the bit between the square and the circumference.
ab=1×1=1=cd=2×d
So d =1/2
So diameter =c+d=2+1/2
So perimeter of circle =pi×5/2 ~= (22/7)×(5/2)=55/7
Pc= 55/7 but Ps=56/7
So the circle has the shorter perimeter.