Can rectangles have the same area and perimeter?

[@nicholaslau3194]

Area of 0, perimeter of 0 Done

[@navarajpanday68]

Lol

[@debopamsil6965]

That's trivial

[@alan2here]

Width of 0, Length of -2? What are you doing trusty formula. Oh your telling me about the nature of rectangles with negative areas, thats ok then :)

[@alan2here]

"2 by (tends towards infinity in the positive direction)" as well.

[@nicholaslau3194]

Alan Tennant The formula still holds if w or l =0. This is solving for w given l or l given w, i.e. l(w) when w=0 or w(l) when l=0. All you need is elementary school math skills and the fact that anything multiplied by 0 is 0. No limits or infinities involved. For your example of a rectangle of 2 by x (when x->+inf), it is not correct. The perimeter will be lim(x->+inf)2x+4 and the the area will be lim(x->+inf)2x. 2x+4 is larger than 2x, while both tends towards infinity, they are not equal. Imagine a rectangle of infinite area, it must also have an infinite perimeter, but clearly the area is going to be much larger than the perimeter.

[@lucaslugao]

Te only integer solutions are (3,6) (6,3) and (4,4). This is true because the function f(x) = 2x/(x-2) has negative derivative for all x not equal to 2 which means the function always decreases. We also know that the limit of the function is 2 so the x> 3 part of the function is lower bounded by 2. Finally, since f(7)=2.8 and theres no integer between 2.8 and 2, the only integer solutions have 3

[@t_kon]

Or because 2x/(x-2) can be rewritten as 2 + 4/(x-2). And for the solution to be integer hence (x-2) | 4 which gives only 3 possibilities : x-2 = 1, x-2 = 2, x-2 = 4. You dont really need any kind of advanced math here Edit: Also x-2 = -1, x-2 = -2, x-2 = -4

[@pengin6035]

@@t_kon Exactly, just apply euclidean algorithm

[@admink8662]

How about 0x0?

[@t_kon]

@@admink8662 happened when x-2 = -2, hence x = 0.

[@MarcoMate87]

Not integer, but natural solutions you should say. An integer solution is for example w=1, l=-2.

[@RandomDays906]

You can rewrite 2w/(w-2) as 2 + 4/(w-2) and there are only 3 solutions where the term 4/(w-2) is a whole number. w = 3, 4, and 6. Therefore, there are only 3 whole number solutions for L = 2w/(w-2).

[@karthikrambhatla7465]

Speaking strictly, we have only one rectangle. Because w=4 implies l=4 that is a square. And w=6 gives l=3 that is again the same when w=3 l=6 just interchanged.

[@cptn_n3m012]

@@karthikrambhatla7465 I think a square IS a rectangle

[@karthikrambhatla7465]

@@cptn_n3m012 yeah.. so perhaps we have only two solutions

[@karthikrambhatla7465]

@densch123 2w can be written as 2w+4-4 and then (w-2)2+4 thereby it can be written so.. is that your question?

[@RandomDays906]

@densch123it can be written like 2 + 4/(w-2) because of this. 2w = 2w-4+4 = 2(w-2) + 4 After dividing through by w-2, we get 2w/(w-2) = 2 + 4/(w-2).

[@JefiKnight]

I know this is for fun and you specify the same *numerical value*, but I can hear my teachers yelling at me, "remember your units!" 18m ≠ 18m²

[@Selicre]

You can specify the task as the area of the rectangle being equal to the area of a rectangle with one side being the perimeter and the other side being 1.

[@DavyCDiamondback]

Selicre 1 what? You measure in meters, then convert to centimeters, the math stops lining up

[@NStripleseven]

True, true, value != amount.

[@darkdelphin834]

Absolutely but this was for fun Ig

[@GRBtutorials]

David Casto 1 unit, whichever you want, it works the same for any unit. This is maths, not physics.

[@mjones207]

One of my favorite Algebra 1 problems... the unique solutions are 3×6 and 4×4. Now step it up a bit... can you find the ten unique solutions to the same problem in three dimensions? That is, using only positive integer lengths of sides, find the dimensions of a right rectangular prism whose volumes and surface areas have equal values.

[@t_kon]

If you notice it can be done as many dimension as you want. In the end the finalized equation is 1/p + 1/l + 1/t + .... = 1/2. For 2 dimension it ends as in the video which is 1/p + 1/l = 1/2.

[@mjones207]

Good call, TkON [ ], and very nicely thought. I've given the 3-D problem to the brightest of my Algebra 2 students in the past, but had never considered the possibilities for higher dimensions until you mentioned it. Well done!

[@raykyledecastro6083]

Area : 16m² , perimeter : 16m. The rectangle is a square.

[@not_vinkami]

Square is a kind of rectangle. Problem solved

[@SinclairLocke]

If I say the square has a side of 4m, then the area is 16, and the perimeter is also 16. But if I say it's 400cm, which is the same as 4m, then the area is 160000, and the perimeter is 1600. So it both has and has not the same area and perimeter.

[@Seb135-e1i]

@@SinclairLocke Which is the whole point of units. Sure, 4m and 400cm are equivalent, but they have a different numerical value. Plug 4 units as the width into the formula and you get 4. Plug in 400 units as the width, and you get 800/398, or pretty much 2. What you presented is the same argument that if you have 0 degrees celcius and you're told it'll be "twice as cold" the next day, people convert to kelvin, double that amount, and back to celcius. Sure it gives an answer, but you can convert to any arbitrary unit and get any answer you want.

[@SinclairLocke]

@@Seb135-e1i no, the point is that an area can never be equal to a distance

[@Seb135-e1i]

@@SinclairLocke I never said that. I simply said that their numerical values can be equal.

[@ffggddss]

OK, let's see . . First of all, any solution will be purely numerical, i.e., unitless, as a physical length (perimeter) cannot equal an area. This also means that the answer will not scale by a constant factor. A = P ab = 2(a + b) ab - 2a - 2b = 0 . . . try to complete the product, (a-2)(b-2) (a-2)(b-2) = ab - 2a - 2b + 4 = 4 So take any two numbers whose product is 4, and add 2 to each, and that will be a solution. If you graph this, b vs a, it will be a rectangular hyperbola, above & right of its asymptotes, a=2 and b=2. Vertex is at (4, 4), representing a square of side 4; A = P = 16. The only other integer solution is (a-2, b-2) = (1,4); (a, b) = (3, 6); A = P = 18 - - - and, of course, the flip of that. There are infinitely many rational solutions. E.g., (3⅓, 5); A = P = 16⅔ (3.6, 4.5); A = P = 16.2 (2⅔, 8); A = P = 21⅓ (2½, 10); A = P = 25 . . . Now I watch... Very good! Yes, the only integer solutions are 4x4 (a square) and 3x6 (and obviously 6x3). This is because the only factorizations of 4 into a pair of positive integers, are 2·2 and 1·4. A related question: What are all the solutions for which A = P = an integer? I think there are infinitely many of those. The smallest value is 16. Choose an integer N ≥ 16. Use bprp's form of the relation: a = 2b/(b-2) N = ab = 2b²/(b-2) 2b² - Nb + 2N = 0 - - - a quadratic in b [remember: N has been chosen!] b² - ½Nb + N = 0 b = ¼(N ± √[N² - 16N] ) = ¼(N ± √[N(N - 16)] ) The ± is just giving us both sides of the rectangle (you can check this by multiplying both of those; quicker: use the fact that the product of roots of a QE is constant/leading!) Choosing b ≥ a, a = ¼(N - √[N(N - 16)] ) ; b = ¼(N + √[N(N - 16)] ) If you want to make it harder, ask how many of those solutions (integer N) have rational sides (a, b). Later, man! Fred

[@GreenMeansGOF]

I guess im late but yeah I was just about to say the same thing. (3,6) and (4,4) are the only integer solutions.

[@lucas29476]

Better than the original video and is what I'm looking for. The "number theory" tag is a bit misleading, since this is the actual NT approach. And you also see people in comments throwing calculus (hammer) at every tiny problem (nail)

[@martinzone8153]

TLDR? :)

[@MathNerd1729]

Well, for the final challenge, notice that N(N-16) has to be a perfect square. So: N(N-16)=k² (for some whole number k) N² - 16N = k² N² - 16N + 64 = k² + 64 (N - 8)² = k² + 64 (N - 8)² - k² = 64 (N - 8 + k) (N - 8 - k) = 64 The factors of 64 are: 64 × 1, N and k aren't integers 32 × 2, N = 25, k = 15, a = 2.5, b = 10 16 × 4, N = 18, k = 6, a = 3, b = 6 8 × 8, N = 16, k = 0, a = b = 4

[@badrunna-im]

It works with any rectangle that's not degenerate if you scale it's units right. e. g. a 10mm x10mm rectangle (square, but WLOG) has area 100 with perimeter 40 (A>P) while a 1cm x 1cm rectangle has area 1 but perimeter 4 (P>A). Intermediate value theorem says there's a certain unit between them that gets A=P.

[@chrisglosser7318]

Badrunnaim Al-Faraby yes - given a rectangle of arbitrary dimensions l and w, rescaling your units by a factor x = 2/l + 2/w will result in the numerical value for the area equaling that if the perimeter. So the correct answer is “this is true for any rectangle because you are comparing apples and oranges”

[@Quantris]

Actually they *always* do if you choose the right units.

[@unoriginalusernameno999]

My answer: It's a square cause square also rectangle. Perimeter = 4a Area = a^2 So a^2=4a and a = 4 Therefore Perimeter is 16 Area also is 16

[@ВасилийДрагунов-н8т]

Same

[@artix2468]

Let the rectangle have dimensions A and B. Area is AB and perimeter is 2(A+B). Suppose there exists a quadratic with roots A and B. Then it must have coefficients -(A+B) and AB. Therefore, we know the constant is double the coefficient of the x term. We have the quadratic x^2 -nx +2n=0. All we have to show is that this quadratic can have real roots for some values of n. Calculating the discriminant, we get n(n-4)>0 so infact there are infinitely many solutions.

[@joshmckay973]

Kinda related: For any regular polygon with N sides, S is the length of the side that will make the area equal to the perimeter. 4tan(180/N)=S

[@Altazor-fh9of]

This video: Problems in class Rest of the channel: Problems on the exam

[@Tapeshwarbisht]

NO I CAME HERE TO SOLVE MY CLASS X maths project hope no one read this comment

[@eleazaralmazan4089]

Awesome job BlackPenRedPen! Your videos are the best!

[@mig2698]

With algebraic manipulation you get 1/2 = 1/w + 1/l. Proving that you have found the solutions is hard to explain with just words but ordered pairs in the form of (l,w) are: (3,6);(4,4);(6,3). The real challenge is finding rectangular prisms with the same volume as they have surface area ;) It’s a really interesting problem with some excellent results!!

[@pepegasadge2977]

You can just substitute w-2 with some variable k. If we do that we get L=2*(k+2)/k=(2k+4)/k=2+4/k. For this to be a whole number k has to divide 4 which only happens for k=1, 2 and 4. Substituting back into k=w-2 we get that w= 3, 4 and 6. The respective lengths are 6, 4 and 3. This gives the three solutions (3,6) (4,4) and (6,3)

[@danielgarai-ebner1334]

Similarly, if we have a right angled triangle with legs x and y, we can use the formula 0.5xy=x+y+sqrt(x^2+y^2). We can simplify to (x^2)(0.25y^2-y) - x(y^2-2y)=0. This graph has four parts, however the two parts we need are the graph y=(x-2)/(0.25x-1). We only need to focus on the part entirely in the positive-positive quadrant if we are looking for integer solutions. Notice there is a vertical asymptote at x=4 and a horizontal asymptote at y=4. Also, notice that f(x)=f⁻¹(x). For integer solutions, we can simply test all values of 4

[@amitotc]

For 2w/(w-2) to be integer, we must have the following congruence true: 2w = 0 (mod (w-2)) 2(w-2) + 4 = 0 (mod (w-2)) 4 = 0 (mod (w - 2)) So (w - 2) | 4 , therefore w -2 = 1 or 2 or 4 and so w = 3 or 4 or 6. So only integer solutions are (3, 6), (4, 4) or (6, 3)

[@jagmarz]

So y=2x/(x-2) is a hyperbola, and has a horizontal asymptote at y=2, so there won't be any integer solutions below y=3. Similarly, there's a vertical asymptote at x=2, and there won't be any integer x

[@gncgenz5829]

There are only 3 unique values for the length and width, but (3, 6) and (6, 3) are interchangeable due to the commutative properties of area and perimeter. The last is “paired” with itself because l=w=4. (5, 10/3) is a solution but not an integer pair. The limit as the width approaches 2, so claiming another solution suggests the existence of another integer between 2 and 3. That is impossible thus, (3, 6), (4, 4), and (6, 3) are the only integer solutions.

[@danpost5651]

{ h, w } = { 3, 6 }, { 4, 4 }, { 6, 3 } are the only ones as w-2 will not divide evenly into any 2w > 6.

[@yaboylemon9578]

The limits of the x and y=2 asymptotes limit our options. L is y and w is x. The length for equal area and perimeter are 3,4,5,6.

[@casey206969]

I have seen this solution here but without full work. 2w/(w-2) = (w+w)/(w-2). Separating like terms. (w+w)/(w-2) = (((w-2)+(w-2))+4)/(w-2). Creating like terms via arithmetic. (((w-2)+(w-2))+4)/(w-2) = ((2(w-2))+4)/(w-2). Combining like terms. ((2(w-2))+4)/(w-2) = ((2(w-2)/(w-2)) + (4/(w-2)). Definition for addition of fractions. ((2(w-2)/(w-2)) + (4/(w-2)) = 2 + (4/(w-2)). 'Cancellation'. Since we want this quantity to be a whole number the fraction (4/(w-2)) must also be a whole number. (4/(w-2)) is a whole number iff w-2 is a factor of 4 or w-2 = 1. w-2 = 2. w-2 = 4. solving for w gives the values w=3, w=4, w=6. If a rectangle is listed as a coordinate pair of length (l) and width (w) denoted (l,w) we have the following rectangles. (3,6), (4,4), (6,3). In other words these are the only whole numbered length and width rectangles that have area equal to perimeter.

[@casey206969]

Regarding comments on nonpositive side length rectangles. When we talk about the length of a rectangle in a plane we are specifically referring to the image in the real numbers of the Euclidean metric (aka Euclidean distance) mapped from two points in the Euclidean plane (aka Euclidean 2-space) this is in fact the metric of a metric space and as such there are no negative lengths by definition. (see en.wikipedia.org/wiki/Metric_space#Definition) In fact the distance formula ((x2​−x1​)^2+(y2​−y1​)^2)^(1/2) that you learned in school is the metric (when taken as a function) that along with the plane makes a metric space. As for the 'rectangle' of side lengths 0 this is just a point a rectangle is a quadrilateral and as such must have 4 vertices and 4 edges. In general try not to forget the objects you are working with when you are analyzing some equation.

[@nepraos3151]

its not a factor of 4

[@casey206969]

If w-2=3 then (4/(w-2)) = (4/3). I was only looking for whole number solutions.

[@denikurnia9984]

Make a video about "math induction"

[@blackpenredpen]

here kzbin.info/www/bejne/bofcomeehqdoZqM

[@denikurnia9984]

Thanks

[@adamkadaban]

@@blackpenredpen when the video is private :'(

[@52.yusrilihsanadinatanegar79]

@@blackpenredpen private nooo

[@schizoseahorse]

If you take the formula l=2w/(w-2) and do polynomial long division, you get l=2+4/(w-2). If you want to set l to be an integer, we must require that 4/(w-2) is also an integer (because if it is not, adding 2 to the quantity will not change it to being an integer, which goes against our assumption that l is an integer). Because 4 only has three divisors (1, 2, and 4; technically it also has the negative versions of these as divisors as well, but each of them leads to a nonsensical result with non-positive dimensions), we can set w-2 to each of these independently to conclude that for both dimensions to be integers, w must equal either 3, 4, or 6. Plugging these back into the formula for l yields the following rectangles: 3x6, 4x4, and 6x3. Because the 6x3 rectangle is isomorphic to the 3x6 one over just a 90 degree rotation, the two examples of the 3x6 and 4x4 rectangles you gave in the video indeed are the only rectangles with integer dimensions such that their area and perimeter are equal.

[@firefist3684]

This is to answer the whole numbered length and width question: If you divide the original formula (2l+2w=lw) by 2lw you obtain 1/2=1/l+1/w. One of the two fractions on the right side of the equation must be greater than or equal to 1/4 and one of them must be less than or equal to 1/4. Without loss of generality assume that 1/l>=1/4 ----> l

[@h4c_18]

SInce l=2w/(w-2), we can divide that fraction to get 2 + 4/(w-2) and since 4 >= w-2 > 0 for that to be integer, you can find that w is in the interval (2,6]. The only values inside the interval where that is a positive integer are 3, 4 and 6. Therefore the only solutions are (l,w)=(6,3) and (3,6). (4,4) is out since it has to be a rectangle. I bet it'll be a lot interesting with a triangle having the same area and length ;) (The circle was boring, only r=2 lol)

[@Ismy64]

We can also show that the rectangle with the minimum area is the 4x4 square: just graph 2x^2/(x-2) which has minimum at x=4 for x>2.

[@i_am_anxious0247]

Well, I know this is old, but if it has length a and width b, then 2a+2b is the perimeter and ab is the area. So; ab=2a+2b Set a to 3 3b=2(3)+2b (3-2)b=6 b=6 So if length is 3 and width is 6, then the area and the perimeter are the same. So the answer is yes.

[@mike4ty4]

"Gaussian Integral" :) BPRP doesn't just love math, he's infatuated with it. :D

[@blackpenredpen]

: )

[@ffggddss]

Or you could say, he doesn't just love it, he wears it! Come to think of it, so does Burkard Polster (Mathologer). Fred

[@arghya5673]

We have (w-2)|2w And also (w-2)|2(w-2) So by simple arithmetic (w-2)|[2w-2(w-2)].....i.e. (w-2)|4 So w-2 can be ±1,±2,±4 As w,l>0 the only soln can be w=3,4,6 with corresponding l=6,4,3

[@hibye921]

I feel like a lot of the people are overcomplicating the problem. You just need to factor! We have l*w-2l-2w=0. l(w-2)-2(w-2)=4. (l-2)(w-2)=4. Notice that 4=1*4=2*2. (We only want l, w > 0). Thus, we have l-2=1, w-2=4 ==> (l, w)=(3, 6) l-2=2, w-2=2 ==> (l, w)=(4, 4) l-2=4, w-2=1 ==> (l, w)=(6, 3). Yay! We're done.

[@shadowatom]

lw = 2(l+w) => lw / (l+w) = 2 => lw = 2k, l+w = k for some real number k. Solving for l, we get l = k-w, and substituting this in, we get w(k-w) = 2k => w^2 - wk + 2k = 0. Using the quadratic formula we get w = (k +- sqrt(k^2 - 8k))/2, if you let w be the positive answer, and substitute into l = k-w, you'll find l is the other option, so l = (k -+sqrt(k^2 - 8k))/2. These give us every possible solution.... probably.

[@David-km2ie]

Whatever figure you take there is a coordinate system which will make them equal. This is due to the fact the perimeter grows linearly and the area quadratically. If you take one system with p perimeter and a area. Then pk=ak^2. The trivial solution k=0 and k=p/a will work

[@umka7536]

I am interested, where did you study maths? I graduated in Russia in the Novosibirsk State University. And what amazes me is that the notation you use, including even arrows to show continuity of your calculations, terms, names of variables are the same I used to study in my university. Why am I amazed by this? Because I saw maths lectures from other universities and the way material is presented there is slightly different. But yours is 100% the same from my past. Awesome. And easy for me to digest and follow. 👍

[@skenming]

Since l(w) is decresing w.r.t. w. and, l -> 2 when w->infinity, thus l(w=3) > l > 2 6 > l > 2 l = 2,3,6 only by try and error

[@vishnukadiri586]

There is a method for integer sides. After 2w/(w-2), simplify as 2+(4/(w-2)). Since this is an integer, w-2 should be a factor or 4. Hence just list the factors and equate to w-2.

[@johnnolen8338]

The only integer solutions are the 4 x 4 square and the 3 x 6 rectangle. Here's why: If you divide the perimeter by the area you get the equation, (2/W + 2/L) = 1. Dividing by two on both sides you obtain the equivalent expression: (1/W + 1/L) = 1/2. WLOG you may replace one of these dimensions by n and the other by n + k; doing so with a little algebraic manipulation yields the expression (2n + k)/(n^2 + kn) = 1/2. Since the product of the means equals the product of the extremes, 4n + 2k = n^2 + kn. This can be recast as a quadratic equation: n^2 +(k - 4)n - 2k = 0. Completing the square (and multiplying on both sides by 4) gives the expression: 4(n+(k-4)/2)^2 = k^2 + 16. Considering the right hand side, there's only one Pythagorean triple that has 4 as a side length, {3, 4, 5}. Therefore k = 3 is the only perfect square solution of k^2 + 16 (other than k = 0 of course); in other words, 3^2 + 4^2 = 5^2 is unique. When you substitute k = 3 into the original quadratic equation you obtain n^2 - n - 6 = 0, which can be factored as (n + 2)(n - 3) = 0. The solution n= - 2 is extraneous because n is the side length of a rectangle and cannot be negative for that reason. Thus n = 3 is the only satisfactory solution. When n = 3, n + 3 = 6 giving the dimensions of the 3 x 6 rectangle. Considering the case of k = 0, n = 4 and n + 0 = 4 as well, resulting in the dimensions of the 4 x 4 square.

[@BlokenArrow]

3,6 4,4 and -2,1 are the only integer solutions. At 4,4 you have a square. Increasing length decreases width and visa versa. When L goes to infinity, W goes to 0. The bounds on the size of the shape are (0,Infinity),[4,4], with minimum A and P =16. EDIT 1 All other nontrivial solutions are symmetrical. EDIT 2 This solution ignores dimensionality of units. EDIT 3 It would be interesting to see if a circle can hav the same area and circumference.

[@firefist3684]

the only time a circle can have the same area and circumference (ignoring dimensionality units of course) is when the radius equals 2 units

[@stevethecatcouch6532]

What does a -2 x 1 rectangle look like?

[@BlokenArrow]

Steve the Cat Couch very thin.

[@nancygoel1594]

What if we take the width=2? Let's check: L=2W ÷ W-2 L=2•2 ÷ 2-2 L=4÷0 L=? (Undefined)

[@kacper6149]

Ur a life saver i have homework and I need to find 6 shapes with the same area and perimeter and this helped me out a lot. Thank u😊 You will get a subscription for that

[@KW-12]

What about w

[@HerbertLandei]

I'm kinda disappointed you didn't use the example to go into Egyptian fractions. l*w = 2(l+w) gives 1/2 = (l+w)/lw or 1/2 = 1/l + 1/w. Solutions are 1/2 = 1/3 + 1/6 or 1/2 = 1/4 + 1/4.

[@weerman44]

That Gaussian shirt is so cool!

[@utsav8981]

Its easy AF, we know, l=2w/w-2 R.H.S. = 2(w-2+2)/w-2 = 2 + 4/w-2 Here, w-2 | 4, So, possible values are w-2=1,2,4 Hence, w=3,4,6 😁

[@crustyoldfart]

It's well known that Pure Mathematicians do not trust Applied mathematicians, but surely this piece is an example of how the reverse could become true. What is the meaning of statements in which dimensional units are ignored and/or mixed ? What can be the meaning of the quantity W-2 ?

[@diederickfloor4261]

How I solved it: (I use a and b instead of l and w) b = 2a / (a - 2) a - 2 = 2a / b therefore: b | 2a -> 2a = bk a = k + 2 b becomes: b = 2(k + 2) / k = 2 + 4/k integer must satisfy: 4/k to be integer. k = 1 or k = 2 or k = 4 -> [a = 3, b = 6] or [a = 4, b = 4] or [a = 6 and b = 3 this one doesn't really count] edit: I am ignoring negative values.

[@gartackpsdav4984]

3,6 and 4,4 are the only (and 6,3 of course) integer solutions. 2w/w-2 is an integer only if w-2 is a factor of 2w. This is impossible for any integers over 6.

[@davidecosciani9232]

Nel caso particolare del quadrato, si può risolvere così: x^2=4x x(x)=x(4) /÷x x=4

[@aayushdhungana360]

For the problem you left, (its really easy) 2w = 0 (mod w-2) 2(w-2) + 4 = 0 4 = 0 (mod w-2) So, w-2 is plus minus (1,2,4)....eliminate cases when w-2 is negative i.e. w=3,4,6....TADA!!!

[@yaboylemon9578]

The only cases of both l and w being real integers are l = 1,3,4,5,6 after that, when y approaches 2+, x approaches 2 with a vertical asymptote. Those are the only cases

[@profetorum7263]

Just consider a square (a rectangle) , so that a^2 = 4a ; a=4 works, so... :)

[@alexkidy]

Nice ! I love this channel !!!

[@nouration9685]

to get whole number, you can make some modifications to the equation L = 2W/(W-2), by adding and subtracting 4 in the nominator L = (2W - 4 + 4)/(W - 2), if we seperate 2W - 4 from + 4, we get L = (2W - 4)/(W - 2) + 4/(W-2) = 2 + 4/(W-2) so in order to get a whole number, 4/(W-2) must be equal to a whole number, so 4 must be divisible by W-2, since 4 is only divisible by 1, 2 and 4, then the only whole numbers that can be used for W to fulfill the equation are 3, 4 and 6 so that gives us two solutions, a 3x6 rectangle and a 4x4 square, there are no other solutions

[@Zonnymaka]

L=2W/(W-2) L and W cannot be both odd numbers, hence let's claim that W>2 is an even number, hence W=2N with N>1 Hence we got L/2=(N/N-1) Hence L is an integer only and only if N=2 or N-1=2, that is N=3

[@scorpion2.411]

I remember thinking about when I was 12 and what I found out was that for a rectangles perimeter to equal the area, )length -2) times (width -2) is always equal to 4. This can be proved algebraically to be the same as 2l +2w =lw but I knew no algebra at the time and what I thought is that if the rectangle gets separated into unit squares (and the perimeter into unit sides), there is 1 side per square on the edge, 2 sides per corner square and 0 sides per center square. Because a rectangle always has 4 corners each with a surplus of 1 side per square, the number of centers (with a surplus of -1 side per square) must also equal to 4 to balance it out and have the same number of unit sides and unit squares in overall.

[@shaktiprasannachand4319]

My question for you is can we do this with triangles with whole number

[@razor1869]

If w+l =C then 2(l+w)=lw So,x^2-cX+2c=0 so,∆=c^-8c....which is greater than 0 for C greater than 8

[@CountSacke]

I'm kinda wondering, have you done a video about finding the value of (-1)! I don't know I just thought it would be interesting to see how you would solve for it :)

[@bhavyaramakrishnan801]

Yes they can! It can be proven graphically that is why I know

[@ВасилийДрагунов-н8т]

At first i tried a square with a side equal to "d". Area = d*d, Perimeter = 4d. So, if Area = Perimeter, then d*d=4d => d=4. Solved.

[@leonardromano1491]

Next do this related problem: In what dimensions d is it possible that a d-hyperrectangle with d-hypervolume equal to the (d-1)-hypervolume of its edge and a d-hypersphere with the same property have the same area?

[@duggydo]

I would like to see you do a video (or a series) on gödel’s incompleteness theorem.

[@danielbenyair300]

A= (something)m^2 P= (something)m Therefore even if it has the same number it is not qualified just like hight and weight!!

[@deepanshkalra3860]

Please Make Challenging geometry problems

[@wilderuhl3450]

Okay. So l = 2w/(w-2) for w is odd, note that 2w is even while w-2 is odd therefor only w=3 produces an integer quotient. For w even, RHS reduces to (2n)/(n-1) where n = w/2. Again we have an even over odd situation which works for w=4 as n-1=1. Anyway. For l to be an integer 2w must be congruent to 0 mod w-2. For simplicity let m=w-2 so that 2m+4 ~ 0 mod m. This reduces to 4~ 0 mod m which only works for m= 1, 2, 4 this means that this works when w=3, 4, 6 for which the corresponding l values are 6, 4, 3.

[@darshandarshu5735]

Sir i was very much intrested in maths ur videos are increasing I love ur maths boss

[@blackpenredpen]

Darshan Darshu thanks

[@RichardChen]

1/L+1/W=1/2 is better for memory^^

[@qy9MC]

(4/x-2)+2

[@GourangaPL]

Here's my idea: for each odd w other than 3 we have even divided by odd so it's not whole number for each w = 4k we have 8k / (4k-2) = 4k / (2k-1) so again even divided by odd (except for k=1, w=4) for w = 4k + 2 we have (8k + 4) / 4k = (2k + 1) / k = 2k/k + 1/k = 2 + 1/k which is also whole number only for k=1, w = 6

[@abdullahsafarini4570]

Do it for a triangle please

[@Taterzz]

but can you find the a circle where the circumference and surface area are equal?

[@akshat9282]

I'd start by rewriting the equation as 4/(w-2) + 2 Now for whole number solutions, the one necessary and sufficient condition is that 4/(w-2) must be greater than 0 in our case. The least value it can take is when 4/(w-2) is equal to 1. this happens at w=6, l=3. This is the upper bound to the problem. It is also evident that the lower bound is 2. Hence, a finite number of solutions exist. w is an element of {3,4,5,6}. We can work out the values of l n each case. at w=5, l=3.33 (not a whole number. Since, the function is symmetric across l=w, unordered set (w,l): (3,6) (4,4)

[@aditidas9978]

I had reached the 5 the step then I can't proceed . now understood.A mind bobling question is this

[@pwootjuhs]

I was just thinking that if you have a circle with radius r, area A(r) and circumference C(r), then A'(r)=C(r)

[@gregorio8827]

I havent seen this solution so far. w-2 divides 2w an also divides 2w-4, then w-2 divides 4, therefore w can be 3, 4 and 6 (since w-2 must be positive unless w is 0) Edit: well is the same as the solution 2w/w-2 = 2 + 4/w-2 but whatever

[@ibrahimelashry1315]

Any value w that satisfies the following 2w=k(w-2) i.e. 2w mod (w-2)=0 will result in an integer l.

[@danielbenyair300]

How about the reletion? Build a function to their reletion

[@laurensiusfabianussteven6518]

Its only 3-6 and 4-4 that have this property Make l = 2 + 4/w-2 If want l as integer, w-2|4 Only integer statisfying this is 1,2 and 4 Then we get w is 3,4 or 6

[@prabalbaishya6179]

as per the eqn at 1:07 it implies that if the width of the rectangle tends to infinity then the length must tend to the value of 2 and vice versa in order to keep the area=perimeter. This is something that's worrying me.....if the length or width goes to infinity then the other must go to 2, but this doesnot make any sense as if either of the length or width becomes infinity then both area and perimeter become infinite. But why only the latter assumes the value of 2 as per our initial argument? whereas it can be any value as both the perimeter and area becomes infinity... but I think that taking infinite length or width is itself nonsensical :P

[@dudiobugtron]

The 2 comes from the fact that there are two long sides. If length is much bigger than width, then the perimeter is very close to twice the length. So you need the width to be 2 so the area is also twice the length.

[@harshavardhanharish4725]

Can you solve n^n = 2^4?

[@5vart5ol]

You are cool. Please show this with polar coordinate.

[@David-km2ie]

If one system has perimeter p and area a. Multiply every coordinare by p/a

[@kriswillems5661]

I've been thinking about a problem. Suppose you've 2 random parallel straight lines very close together (for instance at a distance of 1/1000 from each other). How do I find the point with integer coordinates between these 2 lines that is the closest to a given point on one of the lines. Any idea how to handle this problem? And I have the same question for 2 parallel planes.

[@tajpa100]

Hi. Unfortunatelly i have bought your t-shirt but i have never received it. How could you help me?

[@gergodenes6360]

Now what about an n-sided irregular shape instead of a rectangle?

[@chandankar5032]

See in the case of rectangles the numerical equality of area and peremeter though gives 3 whole number solution,but there is nothing to do this with space filling curves.If you take some other surfaces like right angled triangle and use pythagorus theorem then this area-perimeter numerical equality will lead to complex plane .After all it's not all about units and square units.😆

[@colep9247]

I want to see a “derivatives for you” tee shirt in addition to the sticker!

[@soliscrown1272]

I know this is entirely unrelated to the topic, but where did you get that math t-shirt? Gotta get one!

[@kobethebeefinmathworld953]

Just a bit surprise. I thought you would use the method of Putnam 2018 A1 to show the result

[@erikdiaz55]

Where can I buy that shirt??

[@kevinhermawan369]

Do with triangle please

[@HereWasDede]

James Maynard’s “small gaps between primes” series

[@yoavcarmel1245]

without watching, a square with side length 4

[@blackpenredpen]

Yoav Carmel yes

[@blackpenredpen]

And more?

[@yoavcarmel1245]

@@blackpenredpen now that i have watched the video i can give you infinite more solution ;)

[@danielw9542]

The inverse of this function is equal to the the function. (It's symetrical in y=x)

[@alexkidy]

I don't understand all he say, but I understand all he write ! Cause this Math is Amazing !!!!

[@metallicsilver1590]

Is the integration of x! not integrable?

[@pedrosantana8139]

W can't be 2 because with 2 (W-2) = 0 this can not happened so W have to be bigger than 2 (W > 2)

[@mathematicadeestremo6396]

Yeah...here appear a diophentine equation

[@sophiacristina]

Can you do the same with an ellipse?