For all those, who object, that you can't compare area units with length units: This is done on both sides of the ""! That means, you'll have to chose units first, then take the dimensionless values, then do the calculation, and it works! If you are smart, you chose the units, such that the numbers are pretty nice! To get it formally right, just multiply each RHS times 1 unit (length). Area = Perimeter * 1 unit (a + b - c) = 4 (csc(theta) + cot(theta)) * 1 unit or Area = Perimeter * 1 unit (a + b - c) / (1 unit) = 4 cot(theta/2) = 4 (1 + cos(theta)) / sin(theta) = 4 (csc(theta) + cot(theta)) (BTW: Works with any length, not just 1 unit!)
This is overspecified, since theta already is determined by (a, b, c). We could get an equation using only (a, b, c) by using Heron's formula, for example one nice way to write it is: 16(a+b+c)=(−a+b+c)(a−b+c)(a+b−c).
This isn't overspecified. If your "already is determined" argument was valid, then your formula would also be overspecified, since (−a+b+c)(a−b+c)(a+b−c) already is determined by (a,b,c). Of course, if (a,b,c) was given, then theta would also be determined. But we don't have (a,b,c) yet. Instead, we're looking for triangles, i.e. (a, b, c) triples, such that A = P. We then try to derive a criterion for such triples, and we come up with an equation. I like your equation though. Much simpler than the one derived in the video. Unfortunately, neither of the two equations actually help me calculate anything, do they? For example, given a and b and I want to find c such that A = P, how do I calculate c?
Not only is this quicker it's also better as your argument is reversible and shows that the given condition implies area = perimeter, which Michael claimed but didn't show (as he also used the area = perimeter condition in the middle of the argument).
I did something similar, starting by expanding P^2 and substituting the law of cosines and [1] to arrive at an equation like P(P - 2c -4(csc(theta) + cot(theta)) = 0. Since P is not zero, we can divide by it and rearrange to quickly arrive at the result. Your approach looks cleaner, however! I appreciate the insight.
Since area = rs for inradius r and semiperimeter s, we just want triangles with r = 2. But r = tan(α/2)(s-a) where α is the angle at A and a is the length of the side opposite A & the result follows directly (cot(α/2) = csc(α)+cot(α) and the tangent from A to the incircle has length s-a).
Here is a simple geometrical proof. A, B, C, vertex of triangle, a, b, c the sides opposed to A, B, C, I the center of the inscribed circle and r the radius, R, S, T, the tangent points of inscribed circle to a, b, c, sides, E the area of triangle and p the semiperimeter. Now area = semiperimeter × radius, so if area = perimeter then radius = 2 (2units used for perimeter). Examine the triangle CIR (or CIS that is congruent), it is rectangle in R, and its cathetus are IR = r (IS = r) and the tangent CR (CS), from elementary geometry CR = p-c (CS = p-c) The angle opposed to IR if C/2 because the center of inscribed circle is on the bisector of C angle, so we can write cot(C/2) = (p-c)/r = (p-c)/2. Observing that (p-c)/2 is (a+b-c)/4 and cot(C/2) is cos(C/2)/sen (C/2)=radq((1+cos(C))/(1-cos(C)))=radq((1+cos(C))^2/(1-cos(C)^2))=(1+cos(C))/sen(C)=csc(C)+cot(C), we achive the result.
Let L be the unit length (e.g. L = 1 foot if the units in question are feet). Then Area = L * Perimeter iff a + b - c = 4 * L * (csc theta + cot theta). Just in case anyone's actually wondering how to convert this theorem into one which respects units.
Hmm, quite a few souls from the "real" world of physics and engineering are upset with Prof. Penn's "pure math" problem that ignores dimensions and units! Let me reword it for you chaps, so that you don't miss out on the intellectual enjoyment of solving the problem: _Given a triangle with sides of length a, b, and c _*_in your preferred unit of length_*_ (cm, m, inch, foot, yard, mile, light year, Planck Length, whatever), with the angle opposite to c being θ, IF _*_the numerical value of_*_ its perimeter happens to be equal to _*_the numerical value of_*_ its area - _*_a unit area being defined as the area of a square of unit sides_*_ - THEN show that _*_the numerical value of_*_ a+b-c is equal to 4(csc θ + cot θ)._ You're welcome.
You can write a+b-c=2(p-c) with p semiperimeter and cot(x)+csec(x)=1/tan(x/2) so your formula is p-c=2/tan(C/2). you can achive it by Erone formula A=radq(p(p-a)(p-b)(p-c))=2p and Briggs Formulas or tangents theorem tan(C/2)=radq((p-a)(p-b)/(p-c)/p).
Hey dude, what about this: P.(a + b - c) = (a + b + c)(a + b - c) = (a + b)^2 - c^2 = a^2 + b^2 - c^2 + 2ab = 2ab.cos θ + 2ab = 2absin θ .(cot θ + cosec θ) = [(ab.sin θ)/2].4.(cot θ + cosec θ) = A.4(cot θ + cosec θ), whence, P = A => a + b - c = 4(cot θ + cosec θ) And a + b - c = 4(cot θ + cosec θ) => P = A
While we are on the subject of triangle areas and perimeters, why not this problem Prof? Find all Heronian triangles; i.e., all integer solutions, A, a, b, c, of the Diaphantine equation: 16.A^2 = (a + b + c)(a + b - c)(b + c - a)(c + a - b)
Guys, it's math. Length and area are represented by real numbers. That's all. There are no units. If you do insist on choosing units, the statement is still true. The area will equal the perimeter if and only if that other equality holds. It doesn't matter if you change the units, because no matter what units you choose, BOTH equalities will be true, or BOTH will be false. That's what "if and only if" means.
The area and the perimeter can only coincide in numerical value. The area can never equal the perimeter because they are measured in very different dimensions and units. Physics and mathematics developed together
How do we get the other direction? It’s presented as an if and only if but this video only shows one direction. Why does the formula holding imply area=perimeter?
Nice! Wolfram Alpha finds another solution for a=6, b=8: 2+4*sqrt(2) ≈ 7.65685... Input: 16(a+b+c)=(−a+b+c)(a−b+c)(a+b−c), a=6, b=8 (There's also the negative solution 2-4*sqrt(2) ≈ -3.65685..., but of course that doesn't make sense for a triangle.)
since area is square units and perimeter is length, i have to assume you mean that you end up with the same number even though the units are different )
is there any use for this? like simplifying an integral, for example if you had to integrate the RHS wrt theta, you'd come up with expression for c in theta and some a and b value to compute integral of a+b-c wrt theta?
I think that you should pause for a few seconds at the end where you give us your engaging valediction: "And that is a good place to stop." rather than cutting off so abruptly.
length can never equal area. To show the folly of this...just rephrase the question to " Draw a triange where the area in equal to the perimenter".....The answer to that is ANY TIRANGLE shape will work. Jus just have to pick the right units of length.
And the unit you end up with will determine the values of a, b, and c, and force the proven identity. It's math, not physics. The perimeter and area are just numbers. There's no reason we can't ask when those numbers are equal. Imagine I have three apples and you have three oranges. If I said we have the same number, would you really say "that's impossible, the units are different"?
If I measure my triangle first in metres then in centimetres, the area will change disproportionately. So (csc(theta) + cot(theta)) must include some unit-dependent factor which is being hidden here. I'd love to see a follow-up on what this means for angles and trig functions in terms of dimensional analysis!.
For a triangle of a given shape, only a single scaling factor will work to make area/perimeter equal. So the 4(csc(theta) + cot(theta)) term gives you that factor (loosely speaking).
Given that Perimeter scales in direct proportion to the scaling of the sides, while Area scales as the square of the scaling of the sides, each possible triangular shape (set of angles adding to Pi) can be made to satisfy Area = Perimeter if given the appropriate scaling which is equivalent to choosing an appropriate size of unit. The RHS of the given equation [i.e. 4(cscθ+cotθ) ] then tells you the scale at which a triangle with angle θ between sides of some fixed ratio (=a/b) satisfies Area = Perimeter.
I agree about the dimensionality of area vs perimeter. Csc and Cot are dimensionless as ratios of sides of triangle. Change your length units and area no longer equals perimeter. Or equivalently a+b-c changes but rhs doesn't.
Given that Perimeter scales in direct proportion to the scaling of the sides, while Area scales as the square of the scaling of the sides, each possible triangular shape (set of angles adding to Pi) can be made to satisfy Area = Perimeter if given the appropriate scaling which is equivalent to choosing an appropriate size of unit. The RHS of the given equation [i.e. 4(cscθ+cotθ) ] then tells you the scale at which a triangle with angle θ between sides of some fixed ratio (=a/b) satisfies Area = Perimeter.
The proof shows that the area equals the perimeter *if and only if* that other equation is true. If you change the units, you are changing the values of a, b, and c. So yes, the perimeter no longer equals the area, but the other equation will no longer be true either. Which is exactly what "if and only if" means. If one is true, so is the other. If one is false, so is the other.
@@RunstarHomer Yes, what you say is true. My comment was a half-joke regarding the dimensionality problem mentioned by many of the commenters. Consider the statement: a+b-c = 4(csc theta + cot theta). The dimension of the lhs is units of length while the rhs is dimensionless. The only way for this not to be nonsense is for the rhs to have a "hidden" length of one unit multiplied to it. The free variable here is the measure of one unit of length. It can be adjusted such that the statement is true regardless of the details of the triangle. You wouldn't normally work a problem this way, but you can. It also works if you multiply the perimeter by a hidden unit of length so it matches the dimensionality of the area.
@@TomFarrell-p9z It isn't nonsense. It is math, not physics. There are no units. a, b, c, and theta are just real numbers. The perimeter and area of the triangle are just real numbers.
@@RunstarHomer Of course there are units (actually dimensions)! What do you think those real numbers stand for? There is no such thing as a dimensionless triangle.
I have three apples and you have three oranges. If I say we have the same number of fruits, would you object that the units are different? This is math, not physics. There are no meters or kilometers or inches. A square can have a side length of 2. Not 2cm or 2mm, just 2. And that square has an area of 4. Just 4.
I know mathematicians ignore this often but the units don't fit ... an area can't be a length! Change units from meters to centimeters and the numbers won't fit anymore. Greetings from the real world ;-) Otherwise, I love your videos!
Notice that we are trying to find when is area = perimeter. Since area and perimeter have different units, they are never equal in usual sense. But we can consider their magnitudes and equate only magnitude. Therefore the equation still holds in the magnitude sense and can be true for the specific case.
I am not saying to do it so seriously. Just to find when do we get them equal. It's same as the fact that argument of a logarithm is always dimensionless but we still write log of volume in Boltzmann entropy in Statistical mechanics.
as long as the length is in m and the area in m^2, you’re fine. it’s a basic geometric fact, you just watched him prove it. how do you think the greeks did this stuff?
"Equal" in this sense just means the same number. If I have three apples and you have three oranges, then we each have an equal number of fruits. Objecting that the units are different doesn't change the fact that the numbers are the same.
It is impossible for the area of a triangle to equal its perimeter because the units are different. Areas are measured in square units (eg. square centimetres or square inches), whilst the perimeter is a length (eg. centimetres or inches). It gets worse: at 0:54, he drops a perpendicular from the apex to the base line, "which bisects the side a". NO! bisect means divide something into 2 EQUAL parts, which doesn't work here. I don't think I've ever seen a more amateurish presentation on RUclips. I you don't believe me, try changing the units: Assume that in the original problem, feet and square feet are used. Then convert the answer to inches and square inches - you'll find everything collapses.
I agree with the problem of the perpendicular bisecting the other side, but not the units. You don’t have to specify the unit to get the result, it works regardless of what units were used. It could be just a “unit” or inch, or meter, or mile. You created the problem by converting units AFTER finding the solution which was not the stated problem.
@@marcosmaldonado7890 and @bobh6728. There are at least fifteen other posts below screaming "units!" I understand that he's just talking about numerical values but he doesn't say that - it's sloppy science / teaching. If you study any scientific subject (electronics in my case), one of the first things that you learn is that in ANY equation, the units on either side must match. In fact, you can use this to help derive equations or to prove that an equation must be incorrect. So as soon as I saw this video, alarm bells starting ringing and it appears that I'm not the only one who had the same reaction.
Hm... Your videos are great, but here I don't understand the motivation. I guess I fail to see the beauty in a+b-c = 4(csc(theta)+cot(theta)). Doesn't look any more beautiful or useful to me than e.g. 2(a+b+c)/(ab) = sin(theta), which you (basically) derived in the first step...
I certainly think there's some aesthetic beauty in answering the question of when a triangle's perimeter is equal to its area. The end result isn't necessarily super nice, but it's quite an elegant problem.
@@RunstarHomer I agree, but the video is called "a beautiful result", and I don't see the beauty in the resulting formula. 😞As others have pointed out, the area is equal to the perimeter iff the radius of the inscribed circle is 2. That would have been a much more beautiful result. 🙂
For all those, who object, that you can't compare area units with length units: This is done on both sides of the ""!
That means, you'll have to chose units first, then take the dimensionless values, then do the calculation, and it works!
If you are smart, you chose the units, such that the numbers are pretty nice!
To get it formally right, just multiply each RHS times 1 unit (length).
Area = Perimeter * 1 unit (a + b - c) = 4 (csc(theta) + cot(theta)) * 1 unit
or
Area = Perimeter * 1 unit (a + b - c) / (1 unit) = 4 cot(theta/2) = 4 (1 + cos(theta)) / sin(theta) = 4 (csc(theta) + cot(theta))
(BTW: Works with any length, not just 1 unit!)
Interesting comment. Absolutely chock full of German commas!🤣
@@robert-skibelo Waiting for your elaborated long comment written in German... (see? No comma as required in German!)
This is overspecified, since theta already is determined by (a, b, c). We could get an equation using only (a, b, c) by using Heron's formula, for example one nice way to write it is: 16(a+b+c)=(−a+b+c)(a−b+c)(a+b−c).
Now try to nicely isolate any of the sides
This isn't overspecified. If your "already is determined" argument was valid, then your formula would also be overspecified, since (−a+b+c)(a−b+c)(a+b−c) already is determined by (a,b,c).
Of course, if (a,b,c) was given, then theta would also be determined. But we don't have (a,b,c) yet. Instead, we're looking for triangles, i.e. (a, b, c) triples, such that A = P. We then try to derive a criterion for such triples, and we come up with an equation.
I like your equation though. Much simpler than the one derived in the video. Unfortunately, neither of the two equations actually help me calculate anything, do they? For example, given a and b and I want to find c such that A = P, how do I calculate c?
I asked Wolfram Alpha to solve this for c, and the result was a mess...
To be honest I find your version more pretty.
@jcsahnwaldt You couldn't find c with just a and b, you would need another piece of information like an angle.
7:12 That should be -4*cot(theta), not 4*cot(theta)
Surely this is quicker. No sin squareds required
Perimeter = area =>
a + b + c = (1/2) ab sin(theta) [1]
Cosine rule =>
c^2 = a^2 + b^2 - 2ab cos(theta)
c^2 = (a+b)^2 - 2ab - 2ab cos(theta)
c^2 = (a+b)^2 - 2ab (1 + cos(theta))
(a+b)^2 - c^2 = 2ab (1 + cos(theta)) [2]
Divide [2] by [1]
(a+b)^2 - c^2 2ab (1 + cos(theta))
-------------------- = ---------------------------
(a+b) + c (1/2) ab sin(theta)
(a+b) - c = 4 (1 + cos(theta)) / sin(theta)
a + b - c = 4(csc(theta) + cot(theta))
Not only is this quicker it's also better as your argument is reversible and shows that the given condition implies area = perimeter, which Michael claimed but didn't show (as he also used the area = perimeter condition in the middle of the argument).
I did something similar, starting by expanding P^2 and substituting the law of cosines and [1] to arrive at an equation like P(P - 2c -4(csc(theta) + cot(theta)) = 0. Since P is not zero, we can divide by it and rearrange to quickly arrive at the result. Your approach looks cleaner, however! I appreciate the insight.
Also we could use the fact, that S = r * P/2, where r is a radius of inscribed circle. So S = P if r is 2
Is that inscribed circle surface ratio to the triangle’s limited to some triangles?
Since area = rs for inradius r and semiperimeter s, we just want triangles with r = 2. But r = tan(α/2)(s-a) where α is the angle at A and a is the length of the side opposite A & the result follows directly (cot(α/2) = csc(α)+cot(α) and the tangent from A to the incircle has length s-a).
Here is a simple geometrical proof.
A, B, C, vertex of triangle, a, b, c the sides opposed to A, B, C, I the center of the inscribed circle and r the radius, R, S, T, the tangent points of inscribed circle to a, b, c, sides, E the area of triangle and p the semiperimeter.
Now area = semiperimeter × radius, so if area = perimeter then radius = 2 (2units used for perimeter).
Examine the triangle CIR (or CIS that is congruent), it is rectangle in R, and its cathetus are IR = r (IS = r) and the tangent CR (CS), from elementary geometry CR = p-c (CS = p-c) The angle opposed to IR if C/2 because the center of inscribed circle is on the bisector of C angle, so we can write
cot(C/2) = (p-c)/r = (p-c)/2.
Observing that (p-c)/2 is (a+b-c)/4 and cot(C/2) is cos(C/2)/sen (C/2)=radq((1+cos(C))/(1-cos(C)))=radq((1+cos(C))^2/(1-cos(C)^2))=(1+cos(C))/sen(C)=csc(C)+cot(C), we achive the result.
Units: "am I a joke to you?"
Sincerely, an engineer.
That was my question too, I‘m a physicist
Let L be the unit length (e.g. L = 1 foot if the units in question are feet). Then Area = L * Perimeter iff a + b - c = 4 * L * (csc theta + cot theta).
Just in case anyone's actually wondering how to convert this theorem into one which respects units.
@@dustanlevenstein9477 yes, but nevertheless, an area is a length squared and a perimeter is a length. Whatever unit this is.
You have to meter half way.
@@wolliwolfsen291You do remember in measure theory class that the unit “unit(s)” can be used and you can exponentiate it as well 🧐
Hmm, quite a few souls from the "real" world of physics and engineering are upset with Prof. Penn's "pure math" problem that ignores dimensions and units! Let me reword it for you chaps, so that you don't miss out on the intellectual enjoyment of solving the problem:
_Given a triangle with sides of length a, b, and c _*_in your preferred unit of length_*_ (cm, m, inch, foot, yard, mile, light year, Planck Length, whatever), with the angle opposite to c being θ, IF _*_the numerical value of_*_ its perimeter happens to be equal to _*_the numerical value of_*_ its area - _*_a unit area being defined as the area of a square of unit sides_*_ - THEN show that _*_the numerical value of_*_ a+b-c is equal to 4(csc θ + cot θ)._
You're welcome.
But so what?
@@Vega1447 Man, if only everyone asked that question. Virtually every mathematician and scientist would be out of a job overnight!
Lmao
The interesting corollary is that for some unit of length, this is true for any triangle!
You can write a+b-c=2(p-c) with p semiperimeter and cot(x)+csec(x)=1/tan(x/2) so your formula is p-c=2/tan(C/2). you can achive it by Erone formula A=radq(p(p-a)(p-b)(p-c))=2p and Briggs Formulas or tangents theorem tan(C/2)=radq((p-a)(p-b)/(p-c)/p).
1/tan(x/2) is called cot(x/2)...
@@rainerzufall42 🤣
What's the motivation? If you wanted a,b,c on one side and theta on the other, you had it at the start 2(a+b+c)/ab = sin theta
You are missing grouping symbols: 2(a + b + c)/(ab) = sin(theta)
you're right, but that video is simply about other result, hes trying to prove other result
Hey dude, what about this:
P.(a + b - c) = (a + b + c)(a + b - c) = (a + b)^2 - c^2 = a^2 + b^2 - c^2 + 2ab = 2ab.cos θ + 2ab = 2absin θ .(cot θ + cosec θ) = [(ab.sin θ)/2].4.(cot θ + cosec θ) = A.4(cot θ + cosec θ), whence,
P = A => a + b - c = 4(cot θ + cosec θ)
And
a + b - c = 4(cot θ + cosec θ) => P = A
Yes.
2nd to the last line should be -4cot (theta)
While we are on the subject of triangle areas and perimeters, why not this problem Prof?
Find all Heronian triangles; i.e., all integer solutions, A, a, b, c, of the Diaphantine equation:
16.A^2 = (a + b + c)(a + b - c)(b + c - a)(c + a - b)
It’s not the area of a triangle equals the perimeter of the triangle!
It’s the numerical value of the area!
It's math, not physics or engineering. Area and perimeter *are* just numbers.
Guys, it's math. Length and area are represented by real numbers. That's all. There are no units. If you do insist on choosing units, the statement is still true. The area will equal the perimeter if and only if that other equality holds. It doesn't matter if you change the units, because no matter what units you choose, BOTH equalities will be true, or BOTH will be false. That's what "if and only if" means.
The area and the perimeter can only coincide in numerical value. The area can never equal the perimeter because they are measured in very different dimensions and units. Physics and mathematics developed together
Looks pretty weird until you realize csc(x) + cot(x) is equivalent to cot(x/2)... okay still interesting
How do we get the other direction?
It’s presented as an if and only if but this video only shows one direction. Why does the formula holding imply area=perimeter?
Desmos + chatgpt => the right angle triangle (6, 8, 10) satisfies this condition☺
Nice! Wolfram Alpha finds another solution for a=6, b=8: 2+4*sqrt(2) ≈ 7.65685...
Input: 16(a+b+c)=(−a+b+c)(a−b+c)(a+b−c), a=6, b=8
(There's also the negative solution 2-4*sqrt(2) ≈ -3.65685..., but of course that doesn't make sense for a triangle.)
Any triangle with inscribed circle radius of 2 will do it
since area is square units and perimeter is length, i have to assume you mean that you end up with the same number even though the units are different )
Yes, in mathematics, length, area, etc., are just numbers. There are no units.
@@RunstarHomer LOL right on
Please draw a triangle with this property.
is there any use for this? like simplifying an integral, for example if you had to integrate the RHS wrt theta, you'd come up with expression for c in theta and some a and b value to compute integral of a+b-c wrt theta?
Hi Michael! Love your videos
Whatchoo know about droppin some lines
Bisecting them sides
I think that you should pause for a few seconds at the end where you give us your engaging valediction: "And that is a good place to stop." rather than cutting off so abruptly.
Parabéns
length can never equal area. To show the folly of this...just rephrase the question to " Draw a triange where the area in equal to the perimenter".....The answer to that is ANY TIRANGLE shape will work. Jus just have to pick the right units of length.
And the unit you end up with will determine the values of a, b, and c, and force the proven identity.
It's math, not physics. The perimeter and area are just numbers. There's no reason we can't ask when those numbers are equal. Imagine I have three apples and you have three oranges. If I said we have the same number, would you really say "that's impossible, the units are different"?
you never did the left arrow, only the right arrow
If I measure my triangle first in metres then in centimetres, the area will change disproportionately. So (csc(theta) + cot(theta)) must include some unit-dependent factor which is being hidden here. I'd love to see a follow-up on what this means for angles and trig functions in terms of dimensional analysis!.
For a triangle of a given shape, only a single scaling factor will work to make area/perimeter equal. So the 4(csc(theta) + cot(theta)) term gives you that factor (loosely speaking).
Given that Perimeter scales in direct proportion to the scaling of the sides, while Area scales as the square of the scaling of the sides, each possible triangular shape (set of angles adding to Pi) can be made to satisfy Area = Perimeter if given the appropriate scaling which is equivalent to choosing an appropriate size of unit.
The RHS of the given equation [i.e. 4(cscθ+cotθ) ] then tells you the scale at which a triangle with angle θ between sides of some fixed ratio (=a/b) satisfies Area = Perimeter.
What about the minus at the end?
I agree about the dimensionality of area vs perimeter. Csc and Cot are dimensionless as ratios of sides of triangle. Change your length units and area no longer equals perimeter. Or equivalently a+b-c changes but rhs doesn't.
Given that Perimeter scales in direct proportion to the scaling of the sides, while Area scales as the square of the scaling of the sides, each possible triangular shape (set of angles adding to Pi) can be made to satisfy Area = Perimeter if given the appropriate scaling which is equivalent to choosing an appropriate size of unit.
The RHS of the given equation [i.e. 4(cscθ+cotθ) ] then tells you the scale at which a triangle with angle θ between sides of some fixed ratio (=a/b) satisfies Area = Perimeter.
The proof shows that the area equals the perimeter *if and only if* that other equation is true.
If you change the units, you are changing the values of a, b, and c. So yes, the perimeter no longer equals the area, but the other equation will no longer be true either. Which is exactly what "if and only if" means. If one is true, so is the other. If one is false, so is the other.
Works for any triangle! Simply adjust the length of your unit measure.
Haha!
No matter what units you choose, the proven statement is true. Either both equalities are true, or both are false, just like the statement says.
@@RunstarHomer Yes, what you say is true. My comment was a half-joke regarding the dimensionality problem mentioned by many of the commenters. Consider the statement: a+b-c = 4(csc theta + cot theta). The dimension of the lhs is units of length while the rhs is dimensionless. The only way for this not to be nonsense is for the rhs to have a "hidden" length of one unit multiplied to it. The free variable here is the measure of one unit of length. It can be adjusted such that the statement is true regardless of the details of the triangle. You wouldn't normally work a problem this way, but you can. It also works if you multiply the perimeter by a hidden unit of length so it matches the dimensionality of the area.
@@TomFarrell-p9z It isn't nonsense. It is math, not physics. There are no units. a, b, c, and theta are just real numbers. The perimeter and area of the triangle are just real numbers.
@@RunstarHomer Of course there are units (actually dimensions)! What do you think those real numbers stand for? There is no such thing as a dimensionless triangle.
I though the condition a +b +c = absintheta/2 was just fine. Stop there.
does not make sense that the lhs is a length while the rhs is unitless ...
I have three apples and you have three oranges. If I say we have the same number of fruits, would you object that the units are different? This is math, not physics. There are no meters or kilometers or inches. A square can have a side length of 2. Not 2cm or 2mm, just 2. And that square has an area of 4. Just 4.
I know mathematicians ignore this often but the units don't fit ... an area can't be a length! Change units from meters to centimeters and the numbers won't fit anymore. Greetings from the real world ;-)
Otherwise, I love your videos!
O certo é dizer: a área é numericamente igual ao perímetro.
@@professorrogeriocesar A 1m x 1m square has 1m^2 area, but a 100cm x 100xm square has a 10000cm^2 area
@@bernieg5874He meant that the right thing was to say: "The numerical value of the perimeter and the area coincide".
@@bernieg5874 No primeiro caso, bastava considerar, no início, o perímetro em metros. No segundo caso, bastava considerar o perímetro em centímetros.
that'd make the result less satisfying and harder to phrase. Of course it's absurd to equate area and length
You love breaking dimensional analysis don't you? Square meters equals meters ...
It's math, not physics. There are no units. I'm honestly dumbfounded at the number of people in these comments who don't understand this.
They never be equal because of different units
Notice that we are trying to find when is area = perimeter. Since area and perimeter have different units, they are never equal in usual sense. But we can consider their magnitudes and equate only magnitude. Therefore the equation still holds in the magnitude sense and can be true for the specific case.
@@PRIYANSH_SUTHAR You can't just strip a quantity of its units.
I am not saying to do it so seriously. Just to find when do we get them equal. It's same as the fact that argument of a logarithm is always dimensionless but we still write log of volume in Boltzmann entropy in Statistical mechanics.
as long as the length is in m and the area in m^2, you’re fine. it’s a basic geometric fact, you just watched him prove it. how do you think the greeks did this stuff?
It's math, not physics. There *are* no units.
I'd argue that the area of a shape never equals the perimeter of a shape, because the units are different.
"Equal" in this sense just means the same number. If I have three apples and you have three oranges, then we each have an equal number of fruits. Objecting that the units are different doesn't change the fact that the numbers are the same.
It is impossible for the area of a triangle to equal its perimeter because the units are different. Areas are measured in square units (eg. square centimetres or square inches), whilst the perimeter is a length (eg. centimetres or inches).
It gets worse: at 0:54, he drops a perpendicular from the apex to the base line, "which bisects the side a". NO! bisect means divide something into 2 EQUAL parts, which doesn't work here.
I don't think I've ever seen a more amateurish presentation on RUclips. I you don't believe me, try changing the units: Assume that in the original problem, feet and square feet are used. Then convert the answer to inches and square inches - you'll find everything collapses.
not physics bud. obviously he's talking about the numerical value
I agree with the problem of the perpendicular bisecting the other side, but not the units.
You don’t have to specify the unit to get the result, it works regardless of what units were used. It could be just a “unit” or inch, or meter, or mile. You created the problem by converting units AFTER finding the solution which was not the stated problem.
@@marcosmaldonado7890 and @bobh6728. There are at least fifteen other posts below screaming "units!" I understand that he's just talking about numerical values but he doesn't say that - it's sloppy science / teaching. If you study any scientific subject (electronics in my case), one of the first things that you learn is that in ANY equation, the units on either side must match. In fact, you can use this to help derive equations or to prove that an equation must be incorrect. So as soon as I saw this video, alarm bells starting ringing and it appears that I'm not the only one who had the same reaction.
Hm... Your videos are great, but here I don't understand the motivation. I guess I fail to see the beauty in a+b-c = 4(csc(theta)+cot(theta)). Doesn't look any more beautiful or useful to me than e.g. 2(a+b+c)/(ab) = sin(theta), which you (basically) derived in the first step...
I certainly think there's some aesthetic beauty in answering the question of when a triangle's perimeter is equal to its area. The end result isn't necessarily super nice, but it's quite an elegant problem.
@@RunstarHomer I agree, but the video is called "a beautiful result", and I don't see the beauty in the resulting formula. 😞As others have pointed out, the area is equal to the perimeter iff the radius of the inscribed circle is 2. That would have been a much more beautiful result. 🙂
I admire Michael's work but this production is not up to his usual very high standards..