A geometry quickie!
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3 жыл бұрынWe look at a nice quick geometry problem.
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@AlexandreRibeiroXRV7 3 жыл бұрын
You could have made this a bit easier by noticing that the center of the top right circle is r*sqrt(2) away from the top right corner... Add that to 3r, which accounts for the remaining length of the diagonal that is contained within the circles, and then to 2r*sqrt(2), which is the length of the diagonal smaller square. Our setup looks like this: (3+3sqrt(2))r = sqrt(2), thus r equals (2 - sqrt(2))/3.
@Ya666Wa 3 жыл бұрын
Exactly my thoughts
@Chriib 3 жыл бұрын
I did it this way also
@JesusP7 3 жыл бұрын
Same, no need to know cosines. Only pythagoras
@ananintesarbinfaiz863 3 жыл бұрын
yep did it in that way
@LarsHHoog 3 жыл бұрын
That's the way I solved it too
@johannesh7610 3 жыл бұрын
The diagonal with length √2 goes through the rightmost circle's midpoint, and from there to the corner it is √2 r. Adding the other three radii and the square's diagonal we get √2 = 2 √2 r + 3 r + √2 r = 3(√2+1)r. Therefore r = 1/3 √2/(1+√2) = 1/3 (2-√2).
@anonymoususer9837 3 жыл бұрын
That's how I figured it out too.
@pa28cfi 3 жыл бұрын
Yeah, much quicker way to do it, and the way I did it as well.
@calunsagrenejr 3 жыл бұрын
Same, I used the diagonals too.
@PlutoTheSecond 3 жыл бұрын
Same.
@mathieubonnet8991 3 жыл бұрын
This is also the way I came up with the answer, and I think it is a shorter way than the one in the vidéo.
@phasm42 3 жыл бұрын
I solved this by building an equality for the length of the diagonal of the outer square, since it passes through the centers of the circles.
@paulchapman8023 3 жыл бұрын
You do have to account for the small segment of the diagonal of the large square that is not inside the small square or the two circles (in the upper right corner), but once you realize that that is equal to r(sqrt(2)-1), it is pretty simple.
@Aaron-hv7pr 3 жыл бұрын
@@paulchapman8023 yeah or you could just use the distance from the center of that circle to the right corner with is half of the diagonal of the 2r length square. This would mean r*sqrt(2) since the bottom square's diagonal length is 2r*sqrt(2). So you could just set an equality of 2r*sqrt(2) + 2r + r + r*sqrt(2) = sqrt(2). Combine terms and you have 3r*sqrt(2) + 3r = sqrt(2). You can see I basically equaled the smaller parts to the big one.
@abdulkabeer7313 3 жыл бұрын
a quick question does the diagonal of square of side length 2r make the bottom left angle 45° or not if yes cant we just sum the diagonal of small square (2r✓2) with the diameter of two circles?
@MarcoLiedekerken 3 жыл бұрын
The only tricky thing was the orange shape which somehow was defined to be a square :p
@ethanJ496 3 жыл бұрын
It's alright. It can identify itself as a square. We must appreciate shape equality.
@frandelajungla3620 3 жыл бұрын
H570hn9o 7b
@christopherellis2663 3 жыл бұрын
@@ethanJ496 not all squares are created equal, yet each has four sides. I love it when they rationalise the roots!
@meestyouyouestme3753 3 жыл бұрын
All definitions are created equal... by definition they are definitive.
@timothyelicada2630 3 жыл бұрын
He should have put line marks on each side and put a square on the corner to properly indicate it as a square. It's a proper math thing.
@santiagoaguilera7289 3 жыл бұрын
you'll always have our respect, thanks for taking your time to help us!!
@isaactrapala 3 жыл бұрын
You can also use pythagoras to define the big diagonal (which is equal to √2) in terms of r In the top right corner there is a square with sides r defined by the circles radius, then a 3r diagonal, then the large square with sides 2r √2 = r•(√2 + 2√2 + 3) r = √2 / (3√2 + 3) Rationalize to r = (6 - 3√2)/9 Then r = (2 - √2)/3 No trig.
@arhankamdar7923 3 жыл бұрын
I didnt know Niel Patrick Harris was so good at math.
@kalin83 3 жыл бұрын
He's nowhere near looking like NPH
@saxbend 3 жыл бұрын
I did it another way. The total diagonal is root 2. The distance between the top right corner and the centre of nearest circle is root 2 r (triangle with two radii parallel to the top and right hand sides). That plus three more circle radii and the small square diagonal which is 2root2(r) gives you an equation with the total length of root 2 that you can solve for r.
@marcosziadi9059 3 жыл бұрын
it gave me 1/6 doing it that way, i did something wrong ?
@alessandropereira4976 3 жыл бұрын
Marcos Ziadi probably, I did this way and found the same that the video
@karim1029 3 жыл бұрын
@@marcosziadi9059 evidently
@RexxSchneider 2 жыл бұрын
The equation is (√2 + 3 + 2√2)r = √2. (3√2 + 3)r = √2 r = √2 / (3√2 + 3) We multiply top and bottom by the conjugate (3√2 - 3) to clear surds from the denominator: r = √2*(3√2 - 3) / (3√2 + 3)*(3√2 - 3) r = (3*2 - 3√2) / ((3√2)² - (3)²) r = 3(2 - √2) / (18 - 9) r = 3(2 - √2) / 9 That gives r = (2 - √2)/3.
@josue8429 3 жыл бұрын
Discovered your channel yesterday and just wanted more! Great explanations!
@russelltownsend6105 3 жыл бұрын
Fantastic content as usual. Old school chalk on board is a delight. Thanks for reviving my maths passion some 30 years after highschool.
@moonlightcocktail 3 жыл бұрын
I'm overjoyed that I could solve it while viewing the thumbnail. Well, it was a quickie. It's simple: 2sqrt2 r + 2r + sqrt2 r + r = sqrt2 r = sqrt2 / 3 (1 + sqrt2) = (-2 + sqrt2)/ -3 = ( 2 - sqrt2) / 3
@karolpalion2883 3 жыл бұрын
You can't tell from the thumbnail that these are squares. It could be two rectangles, each with only one side known.
@moonlightcocktail 3 жыл бұрын
@@karolpalion2883 I guess, but how would you solve it in that case? Hmm.. actually that is a pretty interesting question, thanks for suggesting it. I'm thinking about something with tan...
@user_2793 3 жыл бұрын
Congratulations, you are not alone.
@RexxSchneider 2 жыл бұрын
@@karolpalion2883 If it were, there would be no unique solution, so we know that they are squares. Although it is explicitly stated in the problem, that reasoning also applies to the fact that the centres of the circles must lie on the diagonal of the square. If you know that a solution exists, you can find it from the thumbnail.
@bobthesir1467 3 жыл бұрын
You should do more geometry videos! Keep up the awesome content!
@xwtek3505 3 жыл бұрын
I solved this quite differently. First I noticed that the diagonal of the inner square, outer square, and the center of two circles are coincident because the picture is a mirror image along that diagonal. Then I found the diagonal of the inner triangle and the line from the corner of triangle to the center of the upper circle. After that, I made a square so that the corner is the outer square's upper right corner and the upper circle's center. I found the diagonal, and added all the lines. The result is the formula for the outer square's diagonal, then we can use that to find r easily.
@Aaron-hv7pr 3 жыл бұрын
I basically just used the diagonal instead to solve for r. Since the circles have a diameter of 2r, they basically are able to fit the square below perfectly. This also means that using the isosceles right triangle rule, the entire diagonal is sqrt(2) and the small square diagonal is 2r*sqrt(2). Then the circle in between is 2r and the extra 1r from the top circle makes 3r. Then since the square and circle are same in length/diameter of 2r, the distance from the center to the corner is also the same. This is half the diagonal of the small square with length of 2r*sqrt(2). That means from the center of that circle to top right corner is r*sqrt(2). You combine like terms and set it equal to sqrt(2) which is the big square's diagonal length. So basically, 3r*sqrt(2) + 3r = sqrt(2). If solved correctly, r is the same as yours.
@zuhail339 3 жыл бұрын
I love how convincing those thumbnails are
@calebferguson2767 3 жыл бұрын
Right? They are offset in the thumbnail, instead of diagonal - likewise, there was no sidelength of 1 on the big square.
@TimothyReeves 3 жыл бұрын
Of course there’s “more than one way to skin this cat”.
@lxathu 3 жыл бұрын
As my good ol' and beloved math teacher stated: we know roads that don't lead to Rome but we still know that not only one road leads there.
@rahulronaldo6813 3 жыл бұрын
Very very interesting.Thank u Michael...
@tetrisman11 3 жыл бұрын
Love your geometry proofs!
@NirousPlayers 3 жыл бұрын
You can do this easily by noticing that the diagonal of the big square goes through the center of all the other polygons. Notice too that the center of the top circle to the top right vertice of the big square makes another square of side r, i'll call it square r So you have Bottom square diagonal : r * 2 * sqrt(2) Middle circle diameter: r * 2 Top circle radius : r * 1 Square r diameter: r * sqrt(2) Just sum it all and it'll be the big square diameter : sqrt(2) You'll have then, after manipulating a little r = (2 - sqrt(2) )/3
@mariobrito427 3 жыл бұрын
Love these videos, very engaging! Also, i can almost smell the chalk as you write :) Cheers!
@goodplacetostop2973 3 жыл бұрын
5:16 Road to 100K subs, lets’s keep the momentum, folks
@harshranjan5655 3 жыл бұрын
@Adam Romanov give me 1-2 week until I anylse his subscriber growth rate. Then I will plot graph and predict an approx date for his 100k sub.
@isaacanwarwatts8844 3 жыл бұрын
I subscribed only about a few weeks ago and he was at 50k then so I definitely think November December
@malawigw 3 жыл бұрын
y' > 0
@goodplacetostop2973 3 жыл бұрын
Adam Romanov According SocialBlade future projections, Michael will reach 100K subs on December 12. That Silver Creator Award would be an early Christmas gift.
@jhevarnsurajpal6423 3 жыл бұрын
Nice once ! Really enjoying these
@abdallahmohamedelhady6668 3 жыл бұрын
Iam greatful to you for helping me. Thank you very much
@luna9200 3 жыл бұрын
Paused the video at the start and found this solution. A bit of a simpler way without having to draw the triangle in the middle would be to notice that the square's diagonal, plus 4 radii, plus a little bit extra at the top right, equals the diagonal of the whole square. In an equation, 2r√2 + 4r + x = √2, where x is the little line segment that goes from the top circle to the upper-right corner of the square. Fortunately, just doing another little triangle at the top right leads to a value of "x" of r(√2 -1). Solving for r becomes fairly simple to give the result you got at the end. A bit of a similar method, but just a different way of looking at it. Figured I'd give my solution just to get it out there. Btw, loving your abstract algebra playlist; as someone taking a second-semester abstract algebra course, they're becoming very useful to explain what my textbook/professor cannot. Cheers.
@Quantiad 3 жыл бұрын
This is easier solved by directly working out the large diagonal in terms of r (pythagoras), equate it to √2 and then solve.
@11thNite 3 жыл бұрын
The same answer can be reached by finding the length of the diagonal of the outermost square. The missing bit in the top right can be put in terms of r by taking the difference between the radius of the circle there and half the diagonal of an imagined square in which the circle is inscribed. This gets us (starting from bottom left) -> 2root(2)r + 2r + 2r + (root(2)r - r) = root(2)
@henryremy649 3 жыл бұрын
You can also prove that there's a π/4 triangle then using parallel lines find that r=1/(4+✓2), is around the same in decimal
@mackk123 3 жыл бұрын
root2 ~ 1.41 *The hypotenuse of 1 smaller square plus 2 diameters having radii equal to half that smaller square's side added with the following circle's inscribed corner of a surrounding outer square frame containing all 3 of these shapes with a perimeter of 4,* _is equal to 7.23 times the size of the radius of a single of those circles._ 2(1.41)R+4R+.41R=7.23R 1.41"=7.23"R .195
@yal2983 3 жыл бұрын
I solved it that way too
@Grizzly01 3 жыл бұрын
@@yal2983 _Approximately_ solved
@djsnowpdx 3 жыл бұрын
Totally get it! Thank you!
@jaimecortes1185 3 жыл бұрын
I instead used that the diagonal of the square was root 2, then equaled it to the diagonal of the square of length 2r plus 4r plus x. X=r sqrt(2) -r. You end up with the same result. Fun problem!!!
@dgarrido3381 3 жыл бұрын
Your channel is pretty cool , I guess you research a lot of nifty math cool problems. Your exposition is nice and humble. Great presentation too!
@Monkeyman249 3 жыл бұрын
Interesting, I tried it before the video and got it a different way. Instead of finding a side length in terms of r, I found the diagonal length in terms of r and knew that it was equal to sqrt(2). Fun quickie!
@evanmagill9114 3 жыл бұрын
I love geometry puzzles that are just a simple image. Solved this in the thumbnail before watching. (Simplifying the fraction was the hardest part)
@tyzonemusic 3 жыл бұрын
The construction of that first triangle assumes that its hypothenuse passes through the center of the leftmost circle. I think It would be possible, using an arbitrary value of r, to construct a similar shape where the left circle is not aligned with the other shapes; so do we assume that its center lies on the segment between the corner of the square and the center of the other circle, or is there really no way for it to be misaligned?
@tyzonemusic 3 жыл бұрын
Nevermind, that's specified in the intro ("the diagonal of the unit square coincides with the diameters of the two circles"), I just skimmed over it.
@Scrogan 3 жыл бұрын
Unlike the guys using the square’s diagonal to solve this, I saw the vertical side length 1 = 3r + 3sqrt(2)r. The answer is the same.
@akshitkumar9402 3 жыл бұрын
maybe thats wrong because the square sides dont perfectly align with the cirlee?
@GammaFZ 3 жыл бұрын
you forgot to divide your answer by sqrt(2) because 3r + 3sqrt(2)r = sqrt(2) since it’s a diagonal of a square with sidelength 1. The answer is (3r + 3sqrt(2)r)/sqrt(2) = (6r + 3sqrt(2)r)/2
@christinafacts444 3 жыл бұрын
I did it from thumbnail by adding up to get the diagonal of the whole box. Diagonal of that 2r box + all 4 radius + [half of (diagonal of 2r box - 2 radius) to account for that little corner] = diagonal of the overall box 1. 2r√2 + 4r + [1/2 * ( 2r√2 - 2r)] = √2 (2-√2)/3
@coseacaso3629 3 жыл бұрын
I considered that the circle have the center that is the center of the big square too, so I assumed that r+2r*sqrt(2)cos45 was equal 1/2, that gives to me the same result, more or less
@kevinbihari 3 жыл бұрын
Measure the diagonal Take the diagonal of the left square. Add 2r for the middle circle. Add 1r to get to half the circle. Make a square in the top right with side length r. Divide by root 2
@Austin101123 3 жыл бұрын
Diagonal of triangle is 2r*sqrt2, then the rest of the diagonal is 3r+rsqrt2, from making a traingle with sidelengths r and r for the last bit. So the total diagonal is also equal to sqrt2. so sqrt2=3r+3rsqrt2 --> sqrt2=r(3+3sqrt2) --> r= sqrt2/(3+3sqrt2) You can then simplify similarly to what you did.
@superkaj1001 3 жыл бұрын
I used an assumption that the centers of the circles are on the diagonal of the square. No idea why but It made sense in my head and got the same anwser:
@definitelynotobama6851 3 жыл бұрын
You're absolutely right. That's true specifically because all the internal shapes are tangent to each other.
@suranjanlk 3 жыл бұрын
Well 👍good explanation
@vaibhavborale204 3 жыл бұрын
There's a simple way. The length of square with side 1 is sqrt(2) Which can be equated to the addition of diagonal of smaller square, diameter of middle circle, radius of 2nd circle and diagonal length of square of side r. Eventually we get r = sqrt(2) ÷ {3(1+sqrt(2)}
@geonalugala 3 жыл бұрын
A square of side length 1 has a diagonal of √2 (by Pythagorean squaring). A square of side length 2r has a diagonal of √8r, or 2√2r The small square at the top right tip (based on radius of the top small circle) has a diagonal of √2r. Total diagonal, therefore: √2 = 2√2r + 3r + √2r √2 = 3√2r + 3r = 3r + 3r√2 √2 = 3r(1 + √2) Inverting, to factor r. 3r = √2/(1 + √2); rationalize. r = ⅓(2-√2). r = 0.19526214587
@zuthalsoraniz6764 3 жыл бұрын
I went at it a little differently. The two circles are the diagonal of a smaller square, of side length 1-2r. Thus, from Pythagoras, 4r=sqrt(2*(1-2r)^2). Expand, solve for r, and bingo boingo you got your solution
@patrickjane5796 3 жыл бұрын
Put a circle of radius r inside the smaller square on the bottom left, it would be very simple from there.
@acr1327 3 жыл бұрын
assuming you know (or show) that the diagonal of a square is rt2 of its side length, you have 2rt2*r on the diagonal of the square, 2r on the middle circle, and r+rt2*r connecting the circles' intersection to the opposite corner, and obviously the square diagonal is r2, so (3+3sqrt2)r = sqrt2 so r = rt2/(3+3rt2), or (2-rt2)/3
@JLvatron 3 жыл бұрын
Solved it. Great Q, thanks!
@udic01 3 жыл бұрын
The square diagonal.is 2sqrt(2)r+3r+sqrt(2)r And is also equal sqrt(2).
@yal2983 3 жыл бұрын
yeah I solved it that way too
@SlidellRobotics 3 жыл бұрын
I just calculated the diagonal in terms of r and set it equal to √2; a bit easier.
@rogerkearns8094 3 жыл бұрын
Blimey, thank you, one I can do.
@mrbear8462 3 жыл бұрын
No need for Trig or the intermediate triangle. Using pythag on a square in the upper right corner gives the answer as Sqt(2)/(3+Sqt(2)+Sqt(8)) - an equivalent formulation.
@ratnachoudhary888 3 жыл бұрын
Just find the equation of diagonal of square in terms of R and equate it to root2. And their comes your answer . Love from India .
@Thomas0so0kool 3 жыл бұрын
That's cool
@RobertShane 3 жыл бұрын
The diagonal of the smaller square is 2sqrt(2)*r and from the center of the top circle to the top right corner is sqrt(2)*r. This means the entire diagonal of the larger square is (after simplifying) 3r + 3r*sqrt(2). But that's also equal to sqrt(2) but it's the diagonal of a square with side length 1. So set them equal to each other and solve for r. 3r*sqrt(2) + 3r = sqrt(2) 3r(sqrt(2) + 1) = sqrt(2) r=sqrt(2) / 3(sqrt(2) + 1) (This simplifies to the final answer)
@math-4-science32 3 жыл бұрын
Can you please find a close form of r_n defined as: (r_1)^2 = 1/12 and [2(r_1 + r_2 + ... + r_(n-1)) + 3r_n]^2 = 3/4 ? r_n is in fact the radius of the nth circle stacked on the height of an equilateral triangle of side 1 and that is tangent to two sides of the triangle and the (n-1)th circle
@emmeeemm 3 жыл бұрын
Like other commenters, I did this by finding the big diagonal in terms of big side length (sqrt(2)) and in terms of r (4r+2r*sqrt(2)+r*(sqrt(2)-1)), and then solve for r. Finding that piece of the big diagonal not accounted for by small square or circles wasn't terrible. If you build a square around that top circle, it has side length 2r. Then, you find its diagonal (2r*sqrt(2)), subtract the part of that diagonal inside the circle (2r), and cut the difference in half. That expression simplified a bit, to r*(sqrt(2)-1). Now that we know every segment in the big diagonal, add them together and set the sum equal to sqrt(2). It's about this easy to solve for r, but also, I feel like I avoided invoking trigonometry.
@RexxSchneider 2 жыл бұрын
Or you could build a square with corners at the centre of the top circle and the top corner of the big square. That clearly has side r and so its diagonal is r √2. You now know that the centre of the top circle is r √2 from the top corner of the big square. Then you just have 3 radii and the diagonal of the square in the bottom left corner to add in order to make the main diagonal. I'm pretty certain summing along the main diagonal is more straightforward that summing along the side of the big square.
@newmanthurairatnam1471 3 жыл бұрын
Can the height/altitude of the triangle that you draw be 1-3r ?
@mr.nkundu5729 3 жыл бұрын
Simply using Pythagoras.....first find the distance from the centre of the top circle to the top right corner of the big square & is equal to root(2) times the radius of the circle, then just simple calculation you can get the value of r & which is equal to (2 - √2)/3 = 0.19526 (approx).
@spoilofthelamb 3 жыл бұрын
Looks like others did the same, but: If diagonal of the large square is sqrt(2) by pythagorean theorem, then: box in the upper right corner using the radius of the upper right circle, creating another pythagorean diagonal of from the center of the upper right circle to the upper right corner of the larger rectangle of sqrt(2)r . The diagonal of the lower left rectangle is 2sqrt(2)r, also by pythagorean's theorem. this gives: diagonal of large square = diagonal of lower left square + diameter of center circle + radius of upper-right circle + distance from center of upper-right circle to upper right corner sqrt(2) = 2sqrt(2)r + 2r + r + sqrt(2)r sqrt(2) = 3sqrt(2)r + 3r sqrt(2) = 3r ( sqrt[2] + 1) (1/3) (sqrt[2])/(sqrt[2] + 1) = r Which is the same result.
@watcher8582 3 жыл бұрын
How come you tend to not resolve the algebraic expressions to decimals to get a feel for how big the numbers are?
@CedderNo.1 7 ай бұрын
so very similar to a multiple choice question in Oxford MAT 2023
@josem138 3 жыл бұрын
I thought in using the principal diagonal which is sqrt(2) = 2r*sqrt(2)+3r+r*sqrt(2) sqrt(2) = 3r*sqrt(2) + 3r sqrt(2) = 3r*(sqrt(2)+1) r = sqrt(2)/(3*(sqrt(2)+1)) r = 2/(3*(2+sqrt(2))) which obviously gives the same answer, but no trigonometry needed
@edgoogle9297 3 жыл бұрын
Or just calculate the square root of (rr plus rr)+2r + square root of (2r sq+ 2rsq) = 1 only r in the equation! Coz u know both side of a right angle and r is the length of the small square and 2r is the larger one
@sergiolucas38 3 жыл бұрын
in decimal mine give 0,196 and yours -0,196 v2/2v2+3+v2, i made different, will it be wrong?
@fokstephen3642 3 жыл бұрын
why the acute angle is pi over 4?
@billy.7113 3 жыл бұрын
This is much easier than the usual ones you did.
@GiampieroSalvi 3 жыл бұрын
Perhaps a simpler solution would be to note that the whole diagonal is √2 but is also the sum of the diagonal of the little square (2√2 r), the diameter of the first circle (2r), the radius of the second circle (r) and the diagonal of a square with side r (√2 r).
@parameshwarhazra2725 3 жыл бұрын
I really like u among all the math KZfaqr. With love from India.❤️❤️❤️❤️❤️❤️❤️
@bloxio3001 3 жыл бұрын
it can be resolved only using pythagorean theorem... the square inside has a diagonal of 2r*sqrt(2)... the middle circunference, we take de diameter to be 2r... the upper circunference we only take one radius r and consider the rest as the distant from the center to the diagonal and as we can make a square out of its tangent points the diagonal is just r*sqrt(2)... then we just add up all of that and we got that the diagonal is (3sqrt(2)+3)r... and as the large square has a diagonal of sqrt(2)... then (3sqrt(2)+3)r=sqrt(2) and by simply dividing both parts we end up with r=(2-sqrt(2))/3
@sreenivaslv4271 3 жыл бұрын
Why can't we calculate it by moving the circles on the same level as the square and then make It (2r +2r+2r = 1) ? and then r = 1/6 which is almost equivalent to the answer you have found? @michaelPenn ?
@denverstrong473 3 жыл бұрын
I used the diagonal but was off a bit because apparently 1^2+1^2=1^2 for me.
@Grizzly01 3 жыл бұрын
🤣
@wydadiyoun 3 жыл бұрын
Is the small square a square or rectangle?
@redskunkify 3 жыл бұрын
Using the diagonals you get. Sqrt(2r^2+2r^2) + 3r + sqrt(r^2+r^2) = 4r + 3r + 2r = sqrt(1^2+1^2) = 9r = sqrt(2) so r = sqrt(2)/9
@code_pi2289 3 жыл бұрын
Why not deduce the length of the diagonal in terms of r and then set it equal to root 2 Seems simpler and faster, no trig needed (except Pythagorean's theorem)
@erickherrerapena8981 3 жыл бұрын
Otra solución era: (2squart(2)+4)r, que es la diagonal principal, sobre squart(2). Igualado a 1.
@CM63_France 3 жыл бұрын
De acuerdo con los otros arriba.
@ethanJ496 3 жыл бұрын
Fird r? But I found 4 of them. Which one do you want me to find? :)
@erickherrerapena8981 3 жыл бұрын
Buen video.
@ghaithghazi6748 3 жыл бұрын
I used a very different and easier method. I noticed that we have a 90 degree triangle starting from the center of the upper circle moving right to the tangent between the circle and the square then move up to the corner of the square then back to the circle center. so r^2+r^2=(r+ the distance between the circle and the square corner)^2 that's the first relation the other one is simply the diagonal of the square. the rest is just Algebra.
@celesteschott6030 3 жыл бұрын
Wait, how is that angle measure pi/4? In order to have pi/4 or a 45 degree angle in a right triangle, the two non-hypotenuse sides must be equal. The side he labels as x is at least 2r long, while the side opposite the angle must be less than 2r. The x side is longer, therefore the angle must be less than pi/4. Maybe I'm missing something or being mislead by how he drew it. Can anyone explain where I might be going wrong?
@johnnychin9598 3 жыл бұрын
how did he get the 45 degrees?
@Aadvey 3 жыл бұрын
Luv u keep it up
@shawondutta7881 3 жыл бұрын
The easiest way is to apply Pythagoras theorem twice First, on the small square to find it's diagonal Second, on the large square ( diameter of the two circles + diagonal of the small square =the diagonal of the large square)
@PlutoTheSecond 3 жыл бұрын
Can you find the general formula where s = side length of big square (s = 1 in example), kr = side length of small square (k = 2 in example) and c = number of circles (c = 2 in example) with rationalized denominator? What if k < 0, i.e. the bounds of the small square extend outside the bounds of the big square, with the circles extending far enough to coincide with the far corner of the small square? Then, suppose we take some combination of (s, k, c) (assume k > 0) and rearrange the shapes along the diagonal (adjusting r to fit) and/or add more small squares of varying side lengths. What other general formulas describe these situations? Hint: only two more need to be found to cover all scenarios - explain why!
@kowsarmohamed2042 2 жыл бұрын
1:51 why is the angle measure 45°?
@michaelkaye4998 3 жыл бұрын
Why didn’t you make the hypotenuse extend to the upper right corner?
@marvinsaintnasisr.siarot9756 3 жыл бұрын
My solution is just using pythagorean theorem, First is getting the diagonal of the small square by using Pythagorean theorem which is r(squaroot)(8) then the diameter of the center cirlce which is 2r , then add r, and lastly add the diagonal of both radius which is equal to r(squareroot)(2) all equal to the diagonal of the large square which is squareroot(2) Then shift solve; voila 0.19526
@haydenreynolds9617 3 жыл бұрын
Why would you cancel when you have 3-2 on the bottom, that just leaves you with a 1 on the denominator
@HDitzzDH 3 жыл бұрын
I got r = 0.18469 approximately, not sure what I got wrong, perhaps I assumed something that wasn't true in the problem. The length of the diagonal "d" in the smaller square at the bottom left is r*sqrt(8) using simple pythagoras, so d+r is the length of half the diagonal for the entire big square, but we also know that half the length of the big diagonal "D" is sqrt(2)/2 since each side is 1, aka 1^2 + 1^2 = D^2 so half of D is sqrt(2)/2, we can set these two equal to each other and solve for r: d + r = sqrt(2)/2 ( sqrt(8)*r ) + r = sqrt(2)/2 r * ( sqrt(8)+1 ) = sqrt(2)/2 r = sqrt(2) / ( 2*( sqrt(8)+1 ) ) ≈ 0.18469 but the real answer from this video should get us 0.19526
@RexxSchneider 2 жыл бұрын
Why must (d+r) be half the length of the diagonal of the entire square? It isn't, and that's why your answer isn't correct.
@tovekauppi1616 3 жыл бұрын
So 2/(3sqrt(2)+6)?
@Volka0071 3 жыл бұрын
Nice, but it's seems pretty easier to consider the diagonal of square and notice the equality: sqrt(2)(1 - r) = 3r + 2*sqrt(2)*r
@YK-ef5rs 3 жыл бұрын
Are they(2 centers , 2 tangent pts) Collinear?
@RexxSchneider 2 жыл бұрын
Yup. Michael Penn states that the centres are collinear with the main diagonal in his introduction. The collinearity of the tangent points follows through symmetry about the main diagonal. Of course, you could also reason that if the centre of the lower circle did not lie on the main diagonal, then r would not have an unique value.
@mohamedbabe7324 3 жыл бұрын
Good
@Vasunchik 3 жыл бұрын
if a triangle has one angle of 90 degrees, the second angle is 45 degrees, then the third angle is also 45 degrees, it follows that the triangle is isosceles, but in the diagram it is not. what is the problem?
@bhavinmoriya9216 3 жыл бұрын
Hi How do you get an angle Pi/4
@Grizzly01 3 жыл бұрын
The diagonal of a square is at 45° (π/4 radians) to its sides.
@bhavinmoriya9216 3 жыл бұрын
@@Grizzly01 thanks 😊
@jack-jt2lm 3 жыл бұрын
nice
@pedroferreiradasilva5450 3 жыл бұрын
I haven’t done any math in 3 months but this is kinda easy
@saymashampa784 3 жыл бұрын
why the angle is π/4???
@user-wg7iv1dk2p 3 жыл бұрын
Another answer should be 1/3*(2+(2^0.5)) no?
@sanesanyo 3 жыл бұрын
Well just start summing the distance along the x axis. You have 2r, followed by 2*rcos(π/4) and followed by rcos(π/4)+r. This should sum to 1. Rest you can calculate. :)
@zohar99100 3 жыл бұрын
I solved in two minutes and only four rows. The diagonal of the big square (=sqrt2) can be expressed by r ! I got sqrt2/(3sqrt2+3) Same result as here
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