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a very nice Pythagorean triangle puzzle
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Fun problem! I didn't use the parameterization so I just solved a+b+sqrt(a^2+b^2)=ab/2 and ended up getting the condition that a=(4b-8)/(b-4). Since a must be a natural number you can analyse the cases mod 4 and get b=8 & b=12 for 0 mod 4, b=5 for 1 mod 4, and b=6 for 2 mod 4 (no solutions for 3 mod 4). There are no extraneous solutions (this is because the original equation and a,b being natural numbers guarantees sqrt(a^2+b^2) is also a natural number, but it's possible to also just check all the cases).
Alternatively, if you dont want to check the cases mod 4, restricting b
I did the same and ended up with (a-4)(b-4)=8 or b = 8/(a-4) + 4
@@Skandalos Nice! That's equivalent to mine, but the decomposition into (a-4) and (b-4) makes all the possible cases super easy! I'm just usually bad at spotting polynomial decompositions
@@alucs6362 I find those factorizations always neat problems by themselves.
"Since a must be a natural number you can analyse the cases mod 4 ..."???
What? You got
a = (4b-8)/(b-4)
You can
a = (4b-16+8)/(b-4)
= 4+8/(b-4)
The only possible values for b are given by
b-4 b
1 5
2 6
4 8
8 12
The parametrisation (2pq, p^2-q^2, p^2+q^2) doesn't yield all the Pythagorean triples by itself, only up to ratio (though it does yield the primitive triples). E.g. (9,12,15) can't be made this way.
That said, you might as well work with a and b as parameters.
ab/2 = a + b + c
c = ab/2 - (a + b)
c^2 = a^2b^2/4 - ab(a + b) + a^2 + b^2 + 2ab
= a^2 + b^2
a^2b^2/4 - ab(a + b) + 2ab = 0
ab /4 - a - b + 2= 0
ab - 4a - 4b = - 8
(a - 4)(b - 4) = 8
The only solutions with positive a and b are
a - 4 b - 4 a b c
1 8 5 12 13
2 4 6 8 10
Was looking for this, you need a multiple k also in N and a=2pqk; b= k(p^2-q^2); c= k(p^2+q^2)
P and q need not be integers here
I just looked at the title and solved XD
a+b+\sqrt{a^{2}+b^{2}} = ab/2
2a+2b+2\sqrt{a^{2}+b{2}} = ab
2a +2b - ab = -2\sqrt{a^{2}+b{2}}
(2a +2b - ab)^{2} = 4a^{2} + 4b^{2}
4a^{2} + 4b^{2} + a^{2}b^{2} + 8ab - 4a^{2}b - 4b^{2}a = 4a^{2} + 4b^{2}
simplifying you have
4a+4b-ab=8
using SFFT
(a-4)(b-4)=8
So the integer solutions are when 8 is factored which is 2*4 or 1*8
so a=6, b=8
or a=5, b=12
and vice versa
**And that's a good place to stop**
Dimensional analysis says... No.
Heron of Alexandria begs to differ ... but also consider the relationship between the area of a circle and its perimeter ... it's an integral/derivative thing, more generally an n D boundary sweeps out an n+1 D volume.
but yeah area = perimeter is click-bait lol
Indeed a meter will never be a square meter
this means the magnitude of area and perimeter are same. he did not ue dimesions anywhere in the equations
Just consider the area of the rectangles extruded 1 unit outwards from the sides and the dimensions work out
Just asume that the inradius is r=2 using the units you prefer.
And since the area A = r * P/2 we have A = P × 1 without messing up the units
As a physicists "area = perimeter" bothers me a little because of units so I tried my hand at "area = perimeter squared" and there seems to be infinite solutions to that one
Just asume that the inradius is r=2 using the units you prefer.
And since the area A = r * P/2 we have A = P × 1 without messing up the units
You should find some "better" things to bother you.
There is a nice visualization of this. The area of the triangle equals the 1/2(perimeter)(radius of the incircle). The problem then becomes finding all triangles with incircle 2, which generalizes to finding triangles with incircle n.
Exactly. And the problem remains valid even for physicists who are unnecessarily concerned about preserving the units.
A fun solution is to square the perimeter P = a+b+c and use the Pythagorean theorem and the hypothesis to get a quadratic in the perimeter: P^2 = (2c+4)P. Since the perimeter is nonzero, the only solution is a+b = c+4. Eliminating c gives (a-4)(b-4) = 8, from which the solutions (5,12,13) and (6,8,10) fall out as the only integral ones.
a+b+sqrt(a^2+b^2)=ab/2
ab/2-(a+b)=sqrt(a^2+b^2)
(ab/2)^2-ab(a+b)+a^2+2ab+b^2=a^2+b^2
(ab)^2/4-ab(a+b-2)=0
(b^2/4-b)a^2+(2b-b^2)a=0
ab=0 provides two degenerate triangle cases. Otherwise,
(b/4-1)a-b+2=0
(b-4)a=4b-8
a=(4b-8)/(b-4)
a=4+8/(b-4)
(a-4)(b-4)=8
example: the triangle has equal area and perimeter
area: 6•8/2=24
perimeter: 6+8+10=24✅
Naughty mathematicians! Cheekily ignoring that area and length have different units, so they can never be equal 🙂
Just asume that the inradius is r=2 using the units you prefer.
And since the area A = r * P/2 we have A = P × 1 without messing up the units
But q and p can be rational and even irational too,not only natural, bc you don't need only natural numbers so 2pq,p²-q²,p²+q² are natural
If p or q is rational but not integral, the side lengths of the triangle might not be integral, although of course you could multiply by a factor to fix that. If one of them is irrational, it might not be normalizable.
I enjoyed doing this. Using standard algebra tricks, I boiled the prob down to (a - 4).(b - 4) = 8, where a and b are the legs of the RAT
If we allow a and b to take real values, what is the locus of, say, the vertex opposite the base as the base varies in length? Would it be something like the hyperbola y = 4 + 8/(x - 4)? [right-angle here located at (a, 0), legs parallel to the axes]
This problem also has a geometric solution, b/c we have A=pr where A is the area, p is the semiperimeter & r is the inradius. From the condition and this property we have that r=2 and that allows us to restict a & b and finish the soln.
Yes, and by thinking this way, the equation becomes valid even for those physicists who complained about preserving the units.
To all who argue about dimensional analysis:
The equation AREA = PERIMETER looks silly, and it certainly would be if one were looking for arbitrary right-angled triangles. But here, one is looking for Pythagorean triangles, which introduces a length scale into the problem, since all lengths are restricted to discete values.
Put differently: Choose a length scale u and look for right-angled triangles whose sides are integer multiples of u and whose area equals that of a rectangle formed by u and by the perimeter of the triangle.
Just asume that the inradius is r=2 using the units you prefer.
And since the area A = r * P/2 we have A = P × 1 without messing up the units
@@UltimaGaina So, all triangles with an incircle of radius 2 have the property that their area equals the perimeter. I didn't think of it that way, but of course you're right. Still, this is just another way of introducing a length scale into the problem.
3:04 I think it's not true that choosing p and q to be relatively prime generates primitive triples. If p=7 and q=5, then 2pq=70, p^2-q^2=24, and p^2+q^2=74, which is a triple, of course, but not primitive. All three of those expressions are divisible by 2 if p and q have the same parity. So maybe you need them to be relatively prime *and* of opposite parity.
It is a necessary-but-not-sufficient. I..e. primitive triples require p and q to be coprime, but many coprime p and q do not yield a primitive triple. It is however the case that any primitive triple is generated by some coprime p and q. There is a proof of this on the Wiki page for Pythagorean triples.
I was waiting for the insane integral.
Let x,y and c be the sides’ lengths with c being the hypothenuse. Solving for integer c^2 in the area=perimeter equation we get that one of the sides must have even length. Suppose then x=2k for some positive integer k. Substituting in the previously obtained equation and then substituting in the Pythagorean equality we may rearrange to get the quadratic diophantine equation ky-4k-2y+4=0. A formula for solving this kind of quadratic diophantine equation is known, and yields only three admissible solutions, that in turn yield only two essentially unique triangles.
There are only two distinct triangles with integer sizes that hold this property.
(5,12,13)
(6,8,10)
If you had asked a question where some g(x) was equal to the integral of some f(x) nobody would blink.
As an applied mathematician, I looked at the title, and immediately thought, "That can't be right, the dimensions don't match"
Fantastic video! I got stuck without the parameterization, but once I saw that, then I got it. I'm seeing people saying that they solved it without the parameterization, so I'm going to go back and try it again!
I know another way to parameterize pythagorean triples in two degrees of freedom. A is of the form A. B is of the form B. And C is of the form sqrt(A^2 + B^2)
a=3k, b=4k, c=5k
ab/2=a+b+c
3k.4k/2=3k+4k+5k
(12k^2)/2=12k
12k=24
k=2
a=3k=6
b=4k=8
c=5k=10
I would never think that I can start this problem by parametrizing the length of the sides this way
I wrote a python program to brute force this problem upon seeing the thumbnail and it was very satisfying to see it spit out only 4 possibilities despite running about 100 million iterations
Interesting. I experimented from the thumbnail and saw that any right triangle can be scaled by k such that ak + bk + ck = (ab/2)k^2. But to get a Pythagorean triple out of this, k must be an integer, so the base case area must be less than or equal to the base case perimeter. I see that cannot happen with larger-valued PTs as the products grow faster than the sums. So it seems that 5 - 12 - 13 is the only primitive PT that has equal perimeter and area (30 units, 30 sq units) and the only other example of a non-primitive PT would be 6 - 8 - 10 (24 units, 24 sq units).
Extension for the physicists: Find all Pythagorean triangles whose area is proportional to its perimeter squared, with a fixed proportionality constant C>0. That is, Area=C(perimeter)^2.
Very nice!
An amazing way to derive the p q parameterization is to square the complex number p+qi
And that's a good place to stop❤❤
Nice, but I missed how you got the parameterization. Could have been nice having it as an appendix.
Yes, I agree. How the devil did he obtain that parameterisation. Where is the discourse?
Call the leg lengths x and y so we want it that x + y + sqrt(x^2 + y^2) = (xy) / 2. We can solve for y to get y = 4(x - 2) / (x - 4) to get a vertical asymptote of x=4 and solve for x to get x = 4(y - 2) / (y - 4) to get a horizontal asymptote of y = 4 so legs of length 6 and 8 or 12 and 5 are the only solutions for natural number side lengths.
Don't the leg lengths of 5 and 12 also fit those equations?
@RexxSchneider Yes I forgot to include those 2.
I agree with the main result, but of course not absolutely all Pythagorean triples can be expressed in that form, so the argument does need a little tidying. (A well-known counterexample is (9,12,15); there are many others, all of them non-primitive.)
In fact, your investigation of n q (p - q) at the end takes care of all multiples of primitive triples, showing there are no additional examples to worry about.
My result was that, aside from permutations, there are only those two triples a, b, c that satisfy the requirements: 6, 8, 10 and 5, 12, 13. Is that wrong?
My solution path: a = m^2 - n^2, b = 2mn, c = m^2 + n^2 (Euclid's formula).
We want: ab/2 = a + b + c => mn (m^2 - n^2) = 2m (m + n) => n (m - n) = 2
And this can only work if either n = 1 and m = 3 or n = 2 and m = 3.
ps: nevermind, I found the bug, after adding an additional factor k to Euclid's formula - so it also generates all non-primitive triples - the final formula becomes n (m - n) = 2k, which obviously has more solutions.
A Michael Penn that I could follow. I think that this is the second time.
Physicists and engineers be like, what about the units?
Can only hope that if I watch your videos long enough some of your intellect will rub off on me.🤣
In geometry, it would be unpossible for some exams to contain this problem since the units of periment are not the same to the area's ones.
In geometry? IN ... GEOMETRY? Hahahaha.
@@samueldeandrade8535 What?
Has anyone else noticed that every Pythagorean triple has one number that is divisible by 5?
The possible values for
n² (mod 10)
are
0, 1, 4, 5, 6, 9
When n²=0 or n²=5 (mod 10), n is divisible by 5. So let's see when x²+y² can be a square z² (mod 10), with x and y NOT divisible by 5:
1 4 6 9
1 2×
4 5✓ 8×
6 7× 0✓ 2×
9 0✓ 3× 5✓ 8×
In all possible cases, z is divisible by 5.
p=3 and q=1, which yield 6, 8 and 10 as the side lengths, don't produce a primitive Pythagorean triangle obviously, though gcd(p, q) =1. Is the relative primeness of p and q a necessary but insufficient condition? Is there any sufficient condition?
Congrats on 300K!
There are two conditions for primitive Pythagorean triples, you stated one, but missed the other. You said they have to have to be relatively prime, but the generators also need opposite parity (one even, one odd)
@ingiford175 - You have garbled something. A Pythagorean triple is still a Pythagorean triple if both of the smaller numbers are odd. It's the INPUTS going into Euclid's formula for FINDING Pythagorean triples that must be one even natural number and one odd natural number. When both of them are odd, all three of the natural numbers that are outputted by the formula/method will be even, and thus NOT a "primitive" Pythagorean triple, even though they will still satisfy the condition that the sum of the two smaller ones squared is equal to the largest one squared. But that's a condition for the inputs to the formula to work, i.e. to result in a primitive Pythagorean triple, not a condition that a triple must meet to be considered a primitive Pythagorean triple.
You are correct: to guarantee a unique primitive, 0 < q < p, gcd(p, q) = 1, and p - q is not cong to 0 mod 2. However, you can plug in any positive real values for p and q and still get a right triangle, such as p = sqrt(3) and q = 1 which yields a 2, 2sqrt(3), 4 right triangle. So long as p > q (otherwise you might get a leg with length 0) this always works.
And you can plug in any natural numbers and still get a Pyth triple that may or may not be primitive. Here Michael is looking for primitives and non-primitives so no need to worry about the parity and divisibility. Again, the 0 < q < p remains critical as we want to avoid degenerate triangles and impossible lengths.
If p and q are both odd, you get the same triangle as if you chose (p + q)/2 and (p - q)/2 except it's doubled in length. Ex. (3, 1) is the same as ((3 + 1)/2, (3 - 1)/2) or (2, 1) except sides are twice the length.
If p and q are both even OR have any prime factors in common (say the gcd = s --- if they're both even then 2 divides into s), then it's a little more hairy to describe but if you divide p and q by s, you'll either have your preferred even and an odd OR you'll be in the above case.
@@topherthe11th23 I said primitive. if both generators are odd, then both a and b are even.
@@ingiford175 I am talking about Penn's definition of a primitive Pythagorean triple, given at 1:13 to 1:26. What you are talking about is where Penn said much later, at 2:57, that the numbers p and q that he could use to generate Pythagorean triples must be have no common factors, and then he did indeed forget the part where p and q must be an odd and an even natural number. In my error I was aided by what YOU SAID in your original comment. It didn't say that he omitted a 2nd condition required for the numbers p and q. Your original comment said that's a condition on Pythagorean triples to qualify them as primitive, and that's not true. Your later comment said it's an additional condition on the "generators", not the resulting Pythagorean triple itself, and that IS true. When I read your first comment it seemed to me that you must be talking about the two short sides of the right triangle having a requirement that one be odd and one be even. Maybe you could edit your original comment.
And that's a good place to stop ❤
Heron’s formula
Why can't you just do a+b+c=ab/2=a^2+b^2+c^2 and then solve the system of equations.
Area = Perimeter*(1 unit of length). That solves the dimensionality problem. Frustrating that mathemicians don't seem to take dimensionality as seriously as they ought to. It is an easy check to catch many errors.
Just asume that the inradius is r=2 using the units you prefer.
And since the area A = r * P/2 we have A = P × 1 without messing up the units
Haha. Just shut up.
That's my tshirt 😊
Nice video!
should be clear up front that an area can't equal a perimeter because they are measured in different units. It would be more correct to say that the numerical value of the area equals the numerical value of the perimeter, sans uniits
There are no units this is math
Even if there were no units, the original commenter is right because without mentioning numerical value, it wouldn't make sense to have an area (space) equal to a perimeter (length). Not bashing on the video though
@@MushookieMan Your comment is ridiculous. There is no way to conceive of a triangle as having a length instead of an area, and no way to conceive of a perimeter as having an area instead of a length. The result that is proven in this video is that the number of units of length of the perimeter is equal to the number of units of area of the triangle, where one unit of area of the triangle is the SQUARE of the unit of length.
Just asume that the inradius is r=2 using the units you prefer.
And since the area A = r * P/2 we have A = P × 1 without messing up the units
@@UltimaGaina A is something in square units. R is in linear units. P is in linear units. When you multiply the linear units of r and the linear units of P on the right, you will get something in SQUARE units. So far so good. Square units on both sides. But then how do we get from there to A = P x 1? If r equals 2 units (meaning, r doesn't equal 2, but, rather, r equals 2 units) then when we divide r by 2 we get 1 UNIT. So what you SHOULD have said is A square units is equal to P linear units times 1 linear unit, NOT the same as P linear units times 1.
Perimieter and area can NEVER be equal. DIfferent units.
Just asume that the inradius is r=2 using the units you prefer.
And since the area A = r * P/2 we have A = P × 1 without messing up the units
@@UltimaGaina no, a radius can't be"2". It has to be 2 Units, and an area has to be 2 units squared. THey can not be equal, ever.
@@lynnrathbun that's what I said: set r to 2 units of your liking, and you'll have to solve the same equation as proposed in the clip.
Otherwise, rest assured that the teacher and the vast majority of his followers (probably 100%) are fully aware of this very basic rule everybody learned in the first elementary school physics lessons.
You bring nothing new on the table.
@@UltimaGaina An Inch can never be equal to an inch squared. A CM can never be equal to a CM squared. Never.
@@lynnrathbun I never said that.
Read more carefully and you will understand than when you set the inradius to 2 inches (for example), the left side of the equation A is expressed in square inches, while the right side of the equation r * P/2 is expressed in inches * inches = square inches.
Once you do that, you get to solve the very same equation as proposed by the OP, while also satisfying your elementary school requirement about preserving the units on both sides.
6,8,10
vous ne pouvez pas écrire périmètre = surface...par contre vous pouvez écrire périmètre x 1m = surface...
The area can never equal the perimeter because they have different units. 😜
Sorry I studied physics. ;)
Just asume that the inradius is r=2 using the units you prefer.
And since the area A = r * P/2 we have A = P × 1 without messing up the units
@@UltimaGaina Area and perimeter fundamentally can't have the same units. One is the square of the other. For example, if you picked meters, then the perimeter would be in meters and the area would be in meters^2. Different units.
@@stevenschmidt
You didn't get my point.
One more time: asume that the inradius is r=2m.
And since the area A = r * P/2 we have A (m^2)= 1/2 * P (m) × 2m
Both sides are now expressed in m^2 and you must solve exactly the same equation without having any reason to complain about units.
@@UltimaGaina The whole point was to find a triangle where the Perimeter (which has units of "meters") is equal to the Area (which has units of meters^2). I understand that ignoring the units you can do that, but with units they are fundamentally different. Your equation doesn't show that. It does relate the perimeter to the area, but by multiplying by the inradius -- which is how the meters becomes meters^2. Yes, your equation has meters^2 on both sides, because A = r*P/2, so on the left A has units of m^2, and on the right we have r (with units of 'm') multiplied by P/2 (which has units of 'm'), making m^2 on the right. But like I said, this still doesn't show that P = A, because P is in units of meters while A is in units of m^2. I feel like I'm just repeating myself over and over now.
@@stevenschmidt
Obviously, I didn't manage to make my point clear. One more try:
Basically all triangles with an inradius equal to 2 units have the area NUMERICALLY equal to the perimeter.
He could have simply stated the problem as follows:
"Find the length (in meters) of the sides of a right triangle, if its inradius is equal to 2 (in meters)", and all sides are natural numbers (in meters)."
You would be forced to solve exactly the same equation without having any reason to complain about the units integrity.
I realize I am not the only commenter to have stated this, but the premise of the question, namely that an area and perimeter could ever be equal, is flawed. And the whole exercise is just artificial and (sorry Pauli) "not even math".
Just asume that the inradius is r=2 using the units you prefer.
And since the area A = r * P/2 we have A = P × 1 without messing up the units
Not having what p and q are in this same video is pointless. For me that is.
p and q is the standard parametrisation for Pythagorean triples.
Take p and q arbitrary integers, p>q. Then a=2pq, b=p²-q², c=p²+q² is a Pythagorean triple. (Check that.)
One can obtain all irreducible Pythagorean triples ("irreducible" means that a, b, c do not have a common factor) in this way, and for that one does not even need all possible pairs (p, q), but only those which do not have a common factor and where either p or q is even. (Actually, if these conditions on p and q aren't fulfilled, the Pythagorean triple obtained is not irreducible.)
@@user-gd9vc3wq2h Ahhhh! Right! Thank yo for being so kind. Now I can follow like I wished. Thanks man :) A 3,4,5 is a triple :).....I see!
The area of a triangle is NEVER equal to its perimeter. The AREA of anything can't be equal to its perimeter. Area is measured in square units of length, perimeter is measured in merely units of length. A triangle whose perimeter is x units and whose area is x square units may well be found, but its length is in no way shape or form equal to its area simply because x is equal to x. Let's grant that we find a triangle whose perimeter is x units and whose area is x square units. Can we say those are equal? No, we can't. We can start with just x equals x, but then we multiply the LEFT side of the equation by a length (say, 1 unit) and multiply the RIGHT side of the equation by the SQUARE of the thing by which me multiplied the left, i.e we multiply the right by 1 SQUARE unit, to allege the falsehood that 1 unit is the same thing as 1 square unit. Since when does arithmetic allow us to multiply one side of an equation by apples and the other side by oranges? Not in my world. Such an idea can have disastrous results if it's taken out of geometrical playing around and moved into real-world applications.
Technical right. I assumed the "magnitude of the area" and the "magnitude of the perimeter" was implied.
Calm down dude. It's just a slip of the tongue. He OBVIOUSLY meant *the numerical value of the area* and the *numerical value* of the perimeter.
You don't need to write an angry treatise for such a trivial thing.
@@RafaelSCalsaverini My "treatise" wasn't even close to angry. I'd say it was just a statement of the facts. Given that I'm responding to a PhD and faculty-member (who should therefore know better), I'd say it's a DEFIANT statement of the facts, because a person must advocate REALLY HARD for their opinion when it's against someone with actual credentials (because I'm anticipating a lot of push-back and pre-responding to it so I don't have to come back). But "angry"? No. "Firm"? Yes.
Just asume that the inradius is r=2 using the units you prefer.
And since the area A = r * P/2 we have A = P × 1 without messing up the units
@@UltimaGaina You end up with A square units = r units times P/2 units. So far so good, you have square units on both sides. But you can't get from there to A square units = P units. It's P units times 1 unit, so that on the right there would also still be square units. Professor Penn should have said "Find a right triangle that has an area "A". "A" divided by the unit of area is no longer an area, but is just a natural number, "x". The triangle has a perimeter "P". "P" divided by the unit of length is no longer a length, but is just a natural number "y". Find the triangles where the natural numbers "x" and "y" are the same.
What practical application can this have when perimeter has the units of length and the units of area are length squared?
So asking when is the perimeter of a right angle triangle equal to the area of the same triangle is nonsensical.
You can compare numbers as a purely theoretical exercise but the quantities cannot be equated or made to be equivalent.
You may as well ask “at what radius of a sphere will the temperature of that sphere be equal to surface area of the sphere in square millimetres if the temperature is 25 deg C?
You can find the numerical answer to this question but it doesn’t mean anything in the real world.
That’s why Mathematics is not a science
Whats your point? Nobody is allowed to do math because they like it? That everything MUST be applied math?
Without these, according to you useless, exercises we would not have most of the math that gets applied. Like the entire branch of Number Theory was in the realm of pure math right up until the late 20th century when computers became common and things like digital cryptography came about and use a lot number theory.
Yeah the numerical values of length and area were the things equated in the video. I get your point though 👍🏼
Respectfully, while maths does have practical applications, maths isn't centered around practicality. I mean, those Olympiad number theory and geometry problems serve more as brain ticklers and food for thought than a practical application. So I don't think "What practical application can this have?" is the right question to ask.
That said, I once again recognise the importance of mentioning that the numerical values are the things to be compared, which I believe was the point you were trying to make.
Application: having fun. Tax dollars go to mathematicians so we can have fun discovering, and every once in a while something useful pops up. Not this time though, thanks for the funding.
I can't believe that 3^2+4^2=5^2 and 3,4,5 are consecutive. Every day we are graced by the coincidences that small numbers grant us. We should be thankful for universal truths like this.
Just asume that the inradius is r=2 using the units you prefer.
And since the area A = r * P/2 we have A = P × 1 without messing up the units