Here's my approach: First notice that tan is a continuous, strictly monotonous function on the interval [0, π/2). Since for a triangle we have 0 < α, β, γ < π and α+β+γ = π, wlog we can set α
@@kamilrichert8446 I don't agree that it goes without saying. Michael found 3 angles but he didn't prove that the 3 angles sum to 180degrees. Without proving that, we don't know that the 3 angles actually make up a triangle. This bit added by Mr. Housman, that arctan2 + arctan3 = 135 degrees, is exactly what we need. (but Maybe the fact that Tan(a+b+c) = 0 is the proof we need that angles a + b + c = 180 and therefore that it's a triangle)
@@armacham Michael only got to the main equation by using the fact that the sum of the angles is 180º, so, by finding a solution, the sum holds without need to check.
@@m4rielCorrect me if I'm wrong, but I'm not sure it holds without need to check, because he just showed that if these three angles a,b,c satisfy tan(a+b+c) = 0. But that doesn't mean a+b+c = pi (because tan is not injective). It could mean a+b+c = n*pi where n is any integer. But you can simply check that atan(1)+atan(2)+atan(3) = pi by using a calculator.
Case 1 not yielding a solution also makes sense because the triangle with two tangents equal to 1 is the isosceles right triangle. The tangent of a right angle is undefined.
arctan(2)>60°, so we can't have all arctans 2 or bigger. So one angle has to be 45°. For the other two angles there are 135° remaining. You can't have another 45° angle as for the remaining angle we would have 90°, and tan(90°) is not an integer. Arctan(3)>67.5°, so we can't have both remaining arctans 3 or bigger. Thus one of the arctans must be 2. With 2 of the angles fixed we can calculate the remaining angle as arctan(3).
This is a really nice solution, I had a slightly alternate idea which comes up with the same solution but does not use case analysis. First, alpha, beta, gamma are all positive because they are angles in a triangle. Thus, since their sum is pi, alpha, beta, and gamma are all less than pi. Moreover, the tangent function is positive over (0, pi/2) and negative over (pi/2, pi). Since we need the tangent of all angles to be positive integers, this means alpha, beta, gamma are all in the interval (0, pi/2). Next, let a=tan(alpha), b=tan(beta), c=tan(gamma) and wlog, suppose a
Of course, we should check the possibility of zero. It is quickly dismissed to be fair (having an angle of zero in a triangle is a degenerate case) but worth mentioning as a footnote.
This would actually be an easier problem... since we know the formule for tan(x+y) we can get the formula: arctan(x)+arctan(y)=arctan((x+y)/(1-xy))±pi, thus, by plugging x=2,y=3 we get arctan(2)+arctan(3)=arctan(-1)+pi, and thus arctan(1)+arctan(2)+arctan(3)=arctan(1)+(arctan(-1)+pi)=pi Q.E.D P.S: It's a nice idea but you should notice that you removed a bit of generallity of the question, because I didn't need to show that those are the *only* integer values with that property, so I could immidiatly know this is an easier problem.
I think that's not actually hard, and it hides the geometrical interpretation behind interpreting atan() as angle. Applying tan() yields tan(atan 1 + (atan 2 + atan 3)) = 0, and thus, by the tan(x+y) expression, tan(atan 2 + atan 3) = -1. The left side is, by the tan expression again, (2+3)/(1 - 2×3) = -1.
@@sternmg numberphile did a video ( ruclips.net/video/m5evLoL0xwg/видео.html ) years ago about a geometry problem that amounts to "arctan(1) + arctan(1/2) + arctan(1/3) = pi/2"... but the diagram can be used to prove "arctan(1) + arctan(2) + arctan(3) = pi" by using angles AED, DEH and FEH @ 10:00
Not all too hard: Draw the following polygon: PABC with vertices P=(1,0), A=(1,1), B=(-1,3), C=(-10,0) We now have three right angled triangles (check using pythagoras!): OPA with sides 1, 1, and sqrt(2). If angle α = POA then tan(α) = 1/1=1 OAB with sides sqrt(2), sqrt(8) and sqrt(10). If angle β = AOB then tan β = sqrt(8) / sqrt(2) = 2 OBC with sides sqrt(10), sqrt(90) and 10. If angle γ = BOC then tan γ = sqrt(90) / sqrt(10) = 3. Notice that α + β + γ = π because POC are on a straight line.
The angle addition rule for tan might be simpler if you multiplied (1 + i tan A)(1 + i tan B), and similarly to add 3 angles, (1 + i a)(1 + i b)(1 + i c), for which we then need the imaginary part to be equal to zero.
I LOVE the catch that a=1, c=2, b=3 is not a solution. It seems like such a minor point, but that kind of precision is immensely important in mathematics. There's a difference between "no solutions" and "no additional solutions."
Moreover, there is also only one solution over integers. Suppose there are solutions in which a, or b, or c takes a non-positive integer value. First, a, b, and c cannot be equal to 0, since a triangle cannot have an angle of 0 or pi. Secondly, a triangle cannot have more than one obtuse angle, so there is only one negative value. And by the condition a = 1, which is a contradiction. Sorry for my english, translated by google.
I looked at it this way. The three angles total pi so they average to pi/3. Therefore if the angles are not all pi/3 then the smallest angle must be less than pi/3. It's well known that tan(pi/4) = 1. But the atan of 2 is greater than pi/3 so the smallest angle must be pi/4. This leaves only one possibility for two integer-tan angles totalling 3pi/4 which are atan2 and atan3.
This is a very good approach, it's much faster, I like it. All angles are positive and sum to pi, so each angle is at most pi. Their tangents are positive real, so the angles are strictly less than pi/2. tan(pi/3)=sqrt(3)
A nice follow-up question, then, is what are the ratios of the sides of triangles where the arctangents of its angles are 1, 2, and 3? The law of sines means that the ratios of the sides are proportional to the ratios of the sines of the angles, but here we have ratios of arctangents so it seems like with some fiddling you could maybe work out a set of ratios for the sides here (maybe, I haven’t tried 🤷♂️)
We know that sin(arctan(x))=x/(sqrt(x²+1)) for all x, and thus by using the law of sines we get: a/sin(alpha)=b/sin(beta), so a/b=sin(alpha)/sin(beta) but sin(alpha)=sin(arctan(1))=1/sqrt(1²+1)=1/sqrt(2)=sqrt(2)/2 and sin(beta)=sin(arctan(2))=2/(sqrt(2²+1))=2/sqrt(5) Hence we achive a/b=(sqrt(2)/2)/(2/sqrt(5))=sqrt(10)/4 and in a similar approch you can derive a/c or b/c.
I enjoy these Trig videos so much. Thank you. I have been slowly getting back into doing some Recreational Trig, as, I have been putting off learning Navigational Trig due to *LIFE*... Who knows, maybe I'll make a video series on Navigational Maths, and put it on RUclips...
I've got to say, tan(A) + tan(B) + tan(C) = tan(A)tan(B)tan(C) for the angles in a triangle ABC is a neat little result that I don't think I've seen before. e: though it does fall apart for right angled triangles. Still, I guess in a way a+b+infinity *does* equal a*b*infinity...
It starts with a nice review of the complex numbers multiplication. Very well ! If not mathematician, you can always can say - I found this formula in the book.
Hi. To prove that arctan (1) + arctan (2) + arctan (3) = pi, we can multiply the complex numbers 1 + i, 1 + 2i and 1 + 3i and observe that the result is a negative real. Obviously, you do more than that in your video (you show the uniqueness of such a triangle), but it would be interesting if you could make another video, but this time using complex numbers. Congratulations on your RUclips channel.
i think “ a good place to stop” is sometimes a good place to start (investigating). In this case, how would you draw a triangle with that shape. Using squared paper, join the originO to point A (0,3), also join O to point B(-1,2) draw a horizontal line from B to the line OA, label the intercept as C. OBC is the required triangle with angles at O,B,C being α,β,γ
That's quite a convoluted way to draw a A(0; 0) B(3; 0) C(2; 2) triangle. Indeed you need only AH = CH = 2·BH, where CH is an altitude from C to AB. Interestingly, cos A = √2/2, cos B = √5/5 and cos C = √10/10.
I have an interesting calculus question from IIT-JEE advanced 2018 the famous exam, as follows Let f be a function R -> R so that f(0) is equal to 1 and f(x+y)=f’(x)f(y) + f(x)f’(y) Find the ln(f(4)) Thank you! P.S: Btw keep up with the great work, you are one of my favorite math channels on youtube!
@@ravirajshelar250 yea thats the point, i solved it before commenting, and initially plugged values x=y=0 finding f’(0)=1/2. Then plugging y=0 you would find f’(x)= 1/2 f(x). This is a separable differential equation dy/dx= y/2. dy/y= dx/2. Integrate both sides ln|y| = x/2 + c for x=0 y=1; so you would find c=0 plug y= f(4), hence x=4 which gives the answer 2
For those unsatisfied with the final answers: It's approximately 63.435° and 71.565° (and PI/4 is 45° exactly). EDIT: Found a way to draw it it using only a compass and straightedge. By the definition of tangent ` γ=90° → tan(α)=a/b ` we can draw two triangles {(-2,0); (0,2); (0,0)} and {(0,0); (0,2); (1,0)} and combine them into {(-2,0); (0,2); (1,0)}.
The identity for tangent can be proved with sin(a+b)/cos(a+b). Using the addition formulas with stuff canceling out just like your proof. You are mighty fond of using complex numbers for about everything!
@Michael Penn, the final answer cannot exist. It is impossible to have a triangle with sides 1,2,3. Otherwise the triangle inequality does not hold (recall that is must be an strict inequality). That means that there are no solutions
since sin^2=tan^2/(1+tan^2) the sin^2 values are then 1/2, 4/5, 9/10 i.e. 5/10, 8/10, 9/10 with 10 as common denominator. Because of the sine rule the side length of such a triangle should then have the relation: sqrt(5):sqrt(8):sqrt(9)
The identity tan(a) + tan(b) + tan(c) = tan(a)*tan(b)*tan(c) when a+b+c=π is a (lesser known) standard identity known as the triple tangent identity. It would have been nice to have mentioned that in the video. There's a similar identity for the cotangent when the angles sum to an odd multiple of π/2.
My brain: So basically when you pulkerwhast the wostabroobaton and poorlaglan the Wolla wolla wumptingclan that gives you the cramonched vestercyulinga.
8:10 as soon as you wrote this eqn I knew that the product abc has to be a perfect number. 1,2,3 in any permutation gives the smallest valued solution. I didn't immediately see that larger perfect numbers are ruled out
That's because you're looking for a natural number which has a product of three proper factors equal to their sum. That can only be a perfect number if it has an unique factorisation (apart from itself and 1), and only the first perfect number fits that. Another way of looking at it is that is only numbers that are the product of two primes have just three proper factors, so that eliminates all of the perfect numbers except 6.
One thing...You said that that [tan( a + b ) + tan( g )] / [ 1 - tan( a + b ) tan( g ) ) is zero only if the numerator tan( a + b ) + tan( g ) = 0. Doesn't it also go to zero when a + b = g = +/- pi / 2? Then you have a zero of "type" (infty + infty) / ( infty * infty). Not that it matters for anything.
I actually proved that even with tan(A), tan(B), tan(C) are integers (allowing obtuse triangles), that triangle of tan(A) = 1, tan(B) = 2, tan(C) = 3 Here goes: ASSUME WLOG a pi/2, arctan(a) + arctan(b) > pi which breaks the triangle. So a must be the only negative. now we have: a < 0 < b
To be more rigorous we should make sure the denominator is also not 0 for our solution. Even if the numerator is 0, if the denominator is 0 the ratio is undefined and we have no solution so that should be checked.
01:17 *_"... we're in flat space ..."_* Aww. I _really_ wanted Michael to solve it for curved spaces 😢. You see, I've only been taught Euclidean geometry, and that's all I've used all my life. Now it's true that a major chunk of our modern lives is engineered using Euclidean geometry. But you need more if you want to grok General Relativity. So Michael, a follow-up video perhaps?
There's no general solution for curved spaces because the solution relies on Euclid's fifth postulate holding. If parallel lines no longer stay the same distance apart, then the angle sum of a triangle ceases to be 180 degrees.
In every triangle Tan(a) + Tan(b) +Tan(c) = Tan(a)*Tan(b)*Tan(c) Simplify to a+b+c = a* b *c The only integers that respect this condition are 1 2 3 And 1 3 2 And 2 1 3 And 2 3 1 I forgot 3 1 2 And dammn 3 2 1 😛
Engineering students have to do one additional step: They're expected to look up the values of arctan(2) and arctan(3) and write the values of all 3 angles *in degrees* up to the specified precision. Otherwise I grade their answers as incomplete and deduct points. Shouldn't slack off like math students! 🤣 Whaddyamean why? Because the bloody machinists aren't going to look up arctans. And their tools aren't calibrated in radians. That's why!
it's goddamn weird that the mathematicians don't ever bother with the actual results, like that's for the bottom feeders. so anticlimactic every time. I imagine sex between mathematicians is like that: a good teaser, moving up the ladder, final crescendo, and then at the last minute he's already smoking and she's in her pajamas. "and that's a good place to stop" wait wut?! where's my goddamn triangle?!
Hey I have stuck with problem that might involve inequalities maybe anyone can help, x,y,z are non negetive integers satisfying 2(x³+y³+z³)=3(x+y+z)² then find the maximum value of x+y+z
You can put any angle into a right-angled triangle to calculate its sine, cosine or tangent. But the values of those functions exist independently of where you found your angle: it doesn't have to be in a triangle at all, much less a right-angled triangle.
In fact, you don't need any triangles to calculate sin(x) or cos(x): there are infinite series for sine and cosine: cos x = 1 - x²/2! + x⁴/4! ... with alternating signs and even powers over the corresponding factorial. sin(x) = x - x³/3! + x⁵/5! ... and you don't see a triangle anywhere (unless you look really hard, that is).
Hey Michael, great video. I have a question tho, if the triangle had an obtuse angle, wouldn't the tangent of that angle be negative? And then maybe there would other cases besides those 3?
If the triangle has an obtuse angle, the other two acute angles must sum to less than 90°. So at least one of them must be less than 45°. But arctan(x) < 1 for all 0 < x < 45°. So we can't have both acute angles with integral tangents. Therefore, a solution involving an obtuse angle is not possible.
Although I understand each step individually I could never solve a problem like this. In the end mathematical rigor is worthless without the human intuition on how to string the operations together to get the desired result.
Excellent, but unsatisfying. Alpha has a "simple" form (a factor of pi), but what about beta and gamma? Are they just weird (presumably irrational and maybe transcendental) numbers?
Well they sum to 3π/4 so they can't both be rational, but that's pretty obvious. I just read online that arctan(2) is not a rational multiple of pi either (which implies arctan(3) also isn't) but I was unable to find a satisfactory proof
I spent most of the day doing it, but I managed to prove that there is no rational number, r, such that tan(r pi) = 2. Proof: By way of contradiction, assume that there is such an r. This means that there is a pair of positive integers, m and n, such that tan(m/n pi) = tan(r pi) = 2. From e^(i th) = cos(th) + i sin(th), we get e^(i r pi) = e^(i m/n pi) = cos(m/n pi) + i sin(m/n pi). From e^(i th)^n = e^(i th*n) = cos(th*n) + i sin(th*n), e^(i m/n pi)^n = cos(m pi) + i sin(m pi), and, since m is an integer, e^(i m/n pi)^n = 1. This means that (cos(m/n pi) + i sin(m/n pi))^n = 1. From cos(th) = 1/sqrt(1 + tan(th)^2), sin(th) = tan(th)/sqrt(1 + tan(th)^2) (for angles th in the first quadrant), we get cos(m/n pi) = 1/sqrt(5), sin(m/n pi) = 2/sqrt(5), so v = e^(i m/n pi) = (1 + 2 i)/sqrt(5). Some algebra gives the recurrence relations: Re(v^j) = - 6/5 Re(v^(j-2)) - Re(v^(j-4)) Im(v^j) = - 6/5 Im(v^(j-2)) - Im(v^(j-4)) with Re(v^2) = - 3/5, Re(v^4) = - 7/25, Im(v^2) = - 4/5, Im(v^4) = -24/25. Now consider just Re(v^(2 k)), and assume that for some k >= 2, we have Re(v^(2 k-2)) = s/5^(p+1) and Re(v^(2 k-4)) = t/5^p, where p is a nonnegative integer, and neither s nor t is divisible by 5. From the recurrence, we have: Re(v^(2 k)) = - 6/5*s/5^(p+1) - t/5^p = - (6 s + 25 t)/5^(p+2) Considering just the expression in parenthesis in the numerator, reduce it modulo 5: u = 6 s + 25 t == s (mod 5) Since s is coprime with 5, so is u. And since this holds for k = 2, it holds for all k > 2, as well, which means that the numerator of Re(v^(2 k)) can never be divisible by 5, let alone equal to 5^q for any positive integer q. And this means that Re(v^(2 k)) can never equal 1. But this means there is no k such that v^(2 k) = 1 is ever true (note that a proof by contradiction can show that there is no odd j such that v^j = 1 is true, either - if there were, then (v^j)^2 = (1)^2 = 1 = v^(2 j), a contradiction). But this means that there is no rational number, r = m/n, such that tan(m/n pi) = 2 This is probably not the simplest proof. If anybody has a pointer to a less verbose proof, I'd appreciate it.
you should stream on twitch if possible and it will give you a chance to interact with your audience too. If you do it on this channel, it will be cool too.
I understand that there is a proof of why every triangle has a total internal angle measurement of 180° or π radians, but isn't that something we can just inherit from geometry? Who doesn't know that the internal angles of all triangles sum to π radians?
@@TJStellmach in addition to your comment. It's useful for a fundamental theorem of arithmetic which states that any positive integer (except one) can be represented by a product of prime numbers ONLY BY ONE WAY. If "one" was a prime number, that the theorem would not hold, cause e.g. , 4 could be represented as 1*2*2 and also 2*2 and 1*1*1*2*2 etc... So in this case we would have infinitely many cases which is not useful at all
@@IoT_ This makes a definition of "unit" as being a number that has an inverse. In the integers 1 and -1 are the only units, in the rationals every nonzero number is a unit. Units are not primes. 2 and -2 are both primes, as -2 = 2 * (-1), that is, 2 times a unit. Multiplication by a unit gives something known as a "conjugate".
I really appreciate that you're using "arctan" notation instead of "tan^-1" :D
We all love the continental notation. :P
Yes, he always does that.
Same, I don't like the (...)^-1 notation for inverse trig functions
Nat notation is better check black pen red pen video
@@advaykumar9726 link?
He has made quite a few videos...
Here's my approach:
First notice that tan is a continuous, strictly monotonous function on the interval [0, π/2).
Since for a triangle we have 0 < α, β, γ < π and α+β+γ = π, wlog we can set α
Also a+b+c=abc suggests trace = determinant for certain similar 3x3 matrices with off-diagonal zeros.
Also 1+2+3=6 is first perfect number.
... and adding angles is equivalent to multiplying the complex numbers .... _spoooooooooooky_ !
@@simonmultiverse6349 invoking complex number was redundant.
@@janami-dharmam 2x2 matrices and complex numbers have a large overlap. You can use either of them to analyse this problem.
We don't have a closed form for arctan(2) and arctan(3) but we do have a closed form of arctan(2) + arctan(3). It is 3*pi/4
those arctans are just angles in a triangle and one angle is pi/4, so it kinda goes without saying
@@kamilrichert8446 I don't agree that it goes without saying. Michael found 3 angles but he didn't prove that the 3 angles sum to 180degrees. Without proving that, we don't know that the 3 angles actually make up a triangle. This bit added by Mr. Housman, that arctan2 + arctan3 = 135 degrees, is exactly what we need.
(but Maybe the fact that Tan(a+b+c) = 0 is the proof we need that angles a + b + c = 180 and therefore that it's a triangle)
@@armacham micheal technically stated arctan(1)+arctan(2)+arctan(3)=pi at the beginning of the solution I guess
@@armacham Michael only got to the main equation by using the fact that the sum of the angles is 180º, so, by finding a solution, the sum holds without need to check.
@@m4rielCorrect me if I'm wrong, but I'm not sure it holds without need to check, because he just showed that if these three angles a,b,c satisfy tan(a+b+c) = 0. But that doesn't mean a+b+c = pi (because tan is not injective). It could mean a+b+c = n*pi where n is any integer. But you can simply check that atan(1)+atan(2)+atan(3) = pi by using a calculator.
That's why inequalities are important not just for algebra, but number theory and geometry as well!
What a nice coincidence. arctan(1) + arctan (2) + arctan(3) = pi is a really flashy result
I loved that, too!
log1 + log2 + log3 = log(1+2+3) for similar reasons
Regarding 12:15, any odd permutation represents a reflection, while even permutations are congruent triangles up to rotation.
Case 1 not yielding a solution also makes sense because the triangle with two tangents equal to 1 is the isosceles right triangle. The tangent of a right angle is undefined.
This.
arctan(2)>60°, so we can't have all arctans 2 or bigger. So one angle has to be 45°. For the other two angles there are 135° remaining. You can't have another 45° angle as for the remaining angle we would have 90°, and tan(90°) is not an integer. Arctan(3)>67.5°, so we can't have both remaining arctans 3 or bigger. Thus one of the arctans must be 2. With 2 of the angles fixed we can calculate the remaining angle as arctan(3).
Nice solution
Yes, this was my first thought. I looked into the comments to see if someone else had the same approach. Happy to see that I'm not alone.
That assumes that an answer actually exists.
@@timanderson5717 The last sentence takes care of proving the assumption.
This is a really nice solution, I had a slightly alternate idea which comes up with the same solution but does not use case analysis.
First, alpha, beta, gamma are all positive because they are angles in a triangle. Thus, since their sum is pi, alpha, beta, and gamma are all less than pi. Moreover, the tangent function is positive over (0, pi/2) and negative over (pi/2, pi). Since we need the tangent of all angles to be positive integers, this means alpha, beta, gamma are all in the interval (0, pi/2).
Next, let a=tan(alpha), b=tan(beta), c=tan(gamma) and wlog, suppose a
Of course, we should check the possibility of zero. It is quickly dismissed to be fair (having an angle of zero in a triangle is a degenerate case) but worth mentioning as a footnote.
This really shows how different branches of mathematics can be used together, wow! Videos like these inspire me to share my maths content!
13:12
So a harder way to pose the problem would have been to ask: "prove that atan1+atan2+atan3=pi"
I really thought about it in the end of the video.
This would actually be an easier problem...
since we know the formule for tan(x+y) we can get the formula: arctan(x)+arctan(y)=arctan((x+y)/(1-xy))±pi, thus, by plugging x=2,y=3 we get arctan(2)+arctan(3)=arctan(-1)+pi, and thus
arctan(1)+arctan(2)+arctan(3)=arctan(1)+(arctan(-1)+pi)=pi
Q.E.D
P.S: It's a nice idea but you should notice that you removed a bit of generallity of the question, because I didn't need to show that those are the *only* integer values with that property, so I could immidiatly know this is an easier problem.
I think that's not actually hard, and it hides the geometrical interpretation behind interpreting atan() as angle.
Applying tan() yields tan(atan 1 + (atan 2 + atan 3)) = 0, and thus, by the tan(x+y) expression, tan(atan 2 + atan 3) = -1. The left side is, by the tan expression again, (2+3)/(1 - 2×3) = -1.
@@sternmg numberphile did a video ( ruclips.net/video/m5evLoL0xwg/видео.html ) years ago about a geometry problem that amounts to "arctan(1) + arctan(1/2) + arctan(1/3) = pi/2"... but the diagram can be used to prove "arctan(1) + arctan(2) + arctan(3) = pi" by using angles AED, DEH and FEH @ 10:00
Not all too hard:
Draw the following polygon: PABC with vertices P=(1,0), A=(1,1), B=(-1,3), C=(-10,0)
We now have three right angled triangles (check using pythagoras!):
OPA with sides 1, 1, and sqrt(2). If angle α = POA then tan(α) = 1/1=1
OAB with sides sqrt(2), sqrt(8) and sqrt(10). If angle β = AOB then tan β = sqrt(8) / sqrt(2) = 2
OBC with sides sqrt(10), sqrt(90) and 10. If angle γ = BOC then tan γ = sqrt(90) / sqrt(10) = 3.
Notice that α + β + γ = π because POC are on a straight line.
The angle addition rule for tan might be simpler if you multiplied (1 + i tan A)(1 + i tan B), and similarly to add 3 angles, (1 + i a)(1 + i b)(1 + i c), for which we then need the imaginary part to be equal to zero.
I LOVE the catch that a=1, c=2, b=3 is not a solution. It seems like such a minor point, but that kind of precision is immensely important in mathematics. There's a difference between "no solutions" and "no additional solutions."
Moreover, there is also only one solution over integers.
Suppose there are solutions in which a, or b, or c takes a non-positive integer value.
First, a, b, and c cannot be equal to 0, since a triangle cannot have an angle of 0 or pi.
Secondly, a triangle cannot have more than one obtuse angle, so there is only one negative value.
And by the condition a = 1, which is a contradiction.
Sorry for my english, translated by google.
Thank you, professor.
This was a great, if not overly complicated, problem.
I looked at it this way. The three angles total pi so they average to pi/3. Therefore if the angles are not all pi/3 then the smallest angle must be less than pi/3. It's well known that tan(pi/4) = 1. But the atan of 2 is greater than pi/3 so the smallest angle must be pi/4. This leaves only one possibility for two integer-tan angles totalling 3pi/4 which are atan2 and atan3.
This is a very good approach, it's much faster, I like it.
All angles are positive and sum to pi, so each angle is at most pi.
Their tangents are positive real, so the angles are strictly less than pi/2.
tan(pi/3)=sqrt(3)
A nice follow-up question, then, is what are the ratios of the sides of triangles where the arctangents of its angles are 1, 2, and 3? The law of sines means that the ratios of the sides are proportional to the ratios of the sines of the angles, but here we have ratios of arctangents so it seems like with some fiddling you could maybe work out a set of ratios for the sides here (maybe, I haven’t tried 🤷♂️)
We know that sin(arctan(x))=x/(sqrt(x²+1)) for all x, and thus by using the law of sines we get:
a/sin(alpha)=b/sin(beta), so
a/b=sin(alpha)/sin(beta)
but sin(alpha)=sin(arctan(1))=1/sqrt(1²+1)=1/sqrt(2)=sqrt(2)/2
and sin(beta)=sin(arctan(2))=2/(sqrt(2²+1))=2/sqrt(5)
Hence we achive
a/b=(sqrt(2)/2)/(2/sqrt(5))=sqrt(10)/4
and in a similar approch you can derive a/c or b/c.
@@luckycandy4823 Thanks! 👍 Simplifying it further then they form a triangle similar to one with edge lengths a= 5√2 , b= 4√5, and c= 3√10 .
Fiddled with it a bit more and got a slightly nicer looking triplet:
a = √5 , b = 2√2, c = 3
I think that's about as simple as it'll get. 🙂
I enjoy these Trig videos so much.
Thank you.
I have been slowly getting back into doing some Recreational Trig, as, I have been putting off learning Navigational Trig due to *LIFE*...
Who knows, maybe I'll make a video series on Navigational Maths, and put it on RUclips...
I've got to say, tan(A) + tan(B) + tan(C) = tan(A)tan(B)tan(C) for the angles in a triangle ABC is a neat little result that I don't think I've seen before.
e: though it does fall apart for right angled triangles. Still, I guess in a way a+b+infinity *does* equal a*b*infinity...
It starts with a nice review of the complex numbers multiplication.
Very well !
If not mathematician, you can always can say - I found this formula in the book.
Hi. To prove that arctan (1) + arctan (2) + arctan (3) = pi, we can multiply the complex numbers 1 + i, 1 + 2i and 1 + 3i and observe that the result is a negative real. Obviously, you do more than that in your video (you show the uniqueness of such a triangle), but it would be interesting if you could make another video, but this time using complex numbers. Congratulations on your RUclips channel.
i think “ a good place to stop” is sometimes a good place to start (investigating). In this case, how would you draw a triangle with that shape. Using squared paper, join the originO to point A (0,3), also join O to point B(-1,2) draw a horizontal line from B to the line OA, label the intercept as C. OBC is the required triangle with angles at O,B,C being α,β,γ
That's quite a convoluted way to draw a A(0; 0) B(3; 0) C(2; 2) triangle. Indeed you need only AH = CH = 2·BH, where CH is an altitude from C to AB.
Interestingly, cos A = √2/2, cos B = √5/5 and cos C = √10/10.
I have an interesting calculus question from IIT-JEE advanced 2018 the famous exam, as follows
Let f be a function R -> R so that f(0) is equal to 1 and f(x+y)=f’(x)f(y) + f(x)f’(y)
Find the ln(f(4))
Thank you!
P.S: Btw keep up with the great work, you are one of my favorite math channels on youtube!
Hey answer is 2
2 easy
Tried by partial first but was stuck at a step.So just put y=0 in initial rule and integrated.f'(0) can also be easily found.
@@ravirajshelar250 up
@@ravirajshelar250 yea thats the point, i solved it before commenting, and initially plugged values x=y=0 finding f’(0)=1/2. Then plugging y=0 you would find f’(x)= 1/2 f(x). This is a separable differential equation dy/dx= y/2. dy/y= dx/2. Integrate both sides ln|y| = x/2 + c for x=0 y=1; so you would find c=0 plug y= f(4), hence x=4 which gives the answer 2
For those unsatisfied with the final answers: It's approximately 63.435° and 71.565° (and PI/4 is 45° exactly).
EDIT: Found a way to draw it it using only a compass and straightedge.
By the definition of tangent ` γ=90° → tan(α)=a/b ` we can draw two triangles {(-2,0); (0,2); (0,0)} and {(0,0); (0,2); (1,0)} and combine them into {(-2,0); (0,2); (1,0)}.
The identity for tangent can be proved with sin(a+b)/cos(a+b). Using the addition formulas with stuff canceling out just like your proof. You are mighty fond of using complex numbers for about everything!
All of your videos are completely inspiring. Thank you very much.
@Michael Penn, the final answer cannot exist. It is impossible to have a triangle with sides 1,2,3. Otherwise the triangle inequality does not hold (recall that is must be an strict inequality).
That means that there are no solutions
The sides aren't 1, 2, and 3, those are the tangent of the angles
since sin^2=tan^2/(1+tan^2) the sin^2 values are then 1/2, 4/5, 9/10 i.e. 5/10, 8/10, 9/10 with 10 as common denominator.
Because of the sine rule the side length of such a triangle should then have the relation:
sqrt(5):sqrt(8):sqrt(9)
So glad I’ve discovered this channel it’s so good
The identity tan(a) + tan(b) + tan(c) = tan(a)*tan(b)*tan(c) when a+b+c=π is a (lesser known) standard identity known as the triple tangent identity. It would have been nice to have mentioned that in the video.
There's a similar identity for the cotangent when the angles sum to an odd multiple of π/2.
My brain: So basically when you pulkerwhast the wostabroobaton and poorlaglan the Wolla wolla wumptingclan that gives you the cramonched vestercyulinga.
8:10 as soon as you wrote this eqn I knew that the product abc has to be a perfect number. 1,2,3 in any permutation gives the smallest valued solution.
I didn't immediately see that larger perfect numbers are ruled out
That's because you're looking for a natural number which has a product of three proper factors equal to their sum. That can only be a perfect number if it has an unique factorisation (apart from itself and 1), and only the first perfect number fits that.
Another way of looking at it is that is only numbers that are the product of two primes have just three proper factors, so that eliminates all of the perfect numbers except 6.
One thing...You said that that [tan( a + b ) + tan( g )] / [ 1 - tan( a + b ) tan( g ) ) is zero only if the numerator tan( a + b ) + tan( g ) = 0. Doesn't it also go to zero when a + b = g = +/- pi / 2? Then you have a zero of "type" (infty + infty) / ( infty * infty).
Not that it matters for anything.
I knew about the identity before but the fact that this is the only solution is interesting
"If ab=1, then... 2=0? That feels wrong!"
I actually proved that even with tan(A), tan(B), tan(C) are integers (allowing obtuse triangles), that triangle of tan(A) = 1, tan(B) = 2, tan(C) = 3
Here goes:
ASSUME WLOG a pi/2, arctan(a) + arctan(b) > pi which breaks the triangle.
So a must be the only negative.
now we have:
a < 0 < b
Wouldn't the WLG equation be a
Video suggestion: prove by induction that for all natural numbers n, pi is jot equal to n
I think you should also test if these angles make a triangle
I don't subscribe to channels that add random animations in their video to try to remind me. That has no place in a video like this.
I thought the solution will be some random numbers, this (1,2,3) solution is so satisfying
such a triangle would have respective vertices (0,0), (2,2), (3,0) or be similar
To be more rigorous we should make sure the denominator is also not 0 for our solution.
Even if the numerator is 0, if the denominator is 0 the ratio is undefined and we have no solution so that should be checked.
Is it proven that there are no closed expressions for arctan(2) and arctan(3)?
01:17 *_"... we're in flat space ..."_*
Aww. I _really_ wanted Michael to solve it for curved spaces 😢. You see, I've only been taught Euclidean geometry, and that's all I've used all my life. Now it's true that a major chunk of our modern lives is engineered using Euclidean geometry. But you need more if you want to grok General Relativity. So Michael, a follow-up video perhaps?
There's no general solution for curved spaces because the solution relies on Euclid's fifth postulate holding. If parallel lines no longer stay the same distance apart, then the angle sum of a triangle ceases to be 180 degrees.
perfect number is such that sum of their devisers equal product of their desisers.First such number is 6.
If three natural numbers have sum *and* product 6, can't we say that the numbers can only be 1, 2, and 3?
Very good problem .You are a genius , dear professor
In every triangle
Tan(a) + Tan(b) +Tan(c) = Tan(a)*Tan(b)*Tan(c)
Simplify to a+b+c = a* b *c
The only integers that respect this condition are 1 2 3
And 1 3 2
And 2 1 3
And 2 3 1
I forgot 3 1 2
And dammn 3 2 1 😛
3! combinations
Nice question and nice solution. Thanks
it’s the case of the perfect numbers!
To be complete, tans can be negative too.
Engineering students have to do one additional step: They're expected to look up the values of arctan(2) and arctan(3) and write the values of all 3 angles *in degrees* up to the specified precision. Otherwise I grade their answers as incomplete and deduct points. Shouldn't slack off like math students! 🤣
Whaddyamean why? Because the bloody machinists aren't going to look up arctans. And their tools aren't calibrated in radians. That's why!
Yes, that's why
it's goddamn weird that the mathematicians don't ever bother with the actual results, like that's for the bottom feeders.
so anticlimactic every time. I imagine sex between mathematicians is like that: a good teaser, moving up the ladder, final crescendo, and then at the last minute he's already smoking and she's in her pajamas.
"and that's a good place to stop"
wait wut?! where's my goddamn triangle?!
@@milanstevic8424 "he's already smoking" - My mind went to pipe, and he's Hugh Hefner.
Real content starts at 5:05
Hey I have stuck with problem that might involve inequalities maybe anyone can help, x,y,z are non negetive integers satisfying
2(x³+y³+z³)=3(x+y+z)² then find the maximum value of x+y+z
I never expected to see condom advertisements while watching a maths video!
It asks for "all possible triangles". What about non-planar triangles? Their sum of angles isn't usually pi.
If not specified, you can assume “triangle” means “triangle in a Euclidean space”, unless you’re angling to be pointedly obtuse.
Doesn't the triangle have to be a right-triangle to use trig functions like tangent?
You can put any angle into a right-angled triangle to calculate its sine, cosine or tangent. But the values of those functions exist independently of where you found your angle: it doesn't have to be in a triangle at all, much less a right-angled triangle.
In fact, you don't need any triangles to calculate sin(x) or cos(x): there are infinite series for sine and cosine: cos x = 1 - x²/2! + x⁴/4! ... with alternating signs and even powers over the corresponding factorial. sin(x) = x - x³/3! + x⁵/5! ... and you don't see a triangle anywhere (unless you look really hard, that is).
Can we count an isoceles right triangle as a solution? Angles are 45, 45 & 90.. so tan become 1, 1 and infinity.
looks reasonable
infinity is not a number
Hey Michael, great video. I have a question tho, if the triangle had an obtuse angle, wouldn't the tangent of that angle be negative? And then maybe there would other cases besides those 3?
If the triangle has an obtuse angle, the other two acute angles must sum to less than 90°. So at least one of them must be less than 45°. But arctan(x) < 1 for all 0 < x < 45°. So we can't have both acute angles with integral tangents. Therefore, a solution involving an obtuse angle is not possible.
nice work, thanks
So, the taylor series for arctan(1) + taylor series for arctan(2) + taylor series for arctan(3) = taylor series for pi. Interesting!
Although I understand each step individually I could never solve a problem like this. In the end mathematical rigor is worthless without the human intuition on how to string the operations together to get the desired result.
Is this video in red/blue 3d?
Excellent, but unsatisfying. Alpha has a "simple" form (a factor of pi), but what about beta and gamma? Are they just weird (presumably irrational and maybe transcendental) numbers?
Well they sum to 3π/4 so they can't both be rational, but that's pretty obvious. I just read online that arctan(2) is not a rational multiple of pi either (which implies arctan(3) also isn't) but I was unable to find a satisfactory proof
I spent most of the day doing it, but I managed to prove that there is no rational number, r, such that tan(r pi) = 2.
Proof: By way of contradiction, assume that there is such an r. This means that there is a pair of positive integers, m and n, such that tan(m/n pi) = tan(r pi) = 2. From e^(i th) = cos(th) + i sin(th), we get e^(i r pi) = e^(i m/n pi) = cos(m/n pi) + i sin(m/n pi). From e^(i th)^n = e^(i th*n) = cos(th*n) + i sin(th*n), e^(i m/n pi)^n = cos(m pi) + i sin(m pi), and, since m is an integer, e^(i m/n pi)^n = 1. This means that (cos(m/n pi) + i sin(m/n pi))^n = 1. From cos(th) = 1/sqrt(1 + tan(th)^2), sin(th) = tan(th)/sqrt(1 + tan(th)^2) (for angles th in the first quadrant), we get cos(m/n pi) = 1/sqrt(5), sin(m/n pi) = 2/sqrt(5), so v = e^(i m/n pi) = (1 + 2 i)/sqrt(5). Some algebra gives the recurrence relations:
Re(v^j) = - 6/5 Re(v^(j-2)) - Re(v^(j-4))
Im(v^j) = - 6/5 Im(v^(j-2)) - Im(v^(j-4))
with Re(v^2) = - 3/5, Re(v^4) = - 7/25, Im(v^2) = - 4/5, Im(v^4) = -24/25. Now consider just Re(v^(2 k)), and assume that for some k >= 2, we have Re(v^(2 k-2)) = s/5^(p+1) and Re(v^(2 k-4)) = t/5^p, where p is a nonnegative integer, and neither s nor t is divisible by 5. From the recurrence, we have:
Re(v^(2 k)) = - 6/5*s/5^(p+1) - t/5^p = - (6 s + 25 t)/5^(p+2)
Considering just the expression in parenthesis in the numerator, reduce it modulo 5:
u = 6 s + 25 t == s (mod 5)
Since s is coprime with 5, so is u. And since this holds for k = 2, it holds for all k > 2, as well, which means that the numerator of Re(v^(2 k)) can never be divisible by 5, let alone equal to 5^q for any positive integer q. And this means that Re(v^(2 k)) can never equal 1. But this means there is no k such that v^(2 k) = 1 is ever true (note that a proof by contradiction can show that there is no odd j such that v^j = 1 is true, either - if there were, then (v^j)^2 = (1)^2 = 1 = v^(2 j), a contradiction). But this means that there is no rational number, r = m/n, such that tan(m/n pi) = 2
This is probably not the simplest proof. If anybody has a pointer to a less verbose proof, I'd appreciate it.
so the sum of the tans is equal the product of the tans for any triangle ...wow
You could've proven that the tangent of the phase angle is the imaginary part over the real part.
ah, by using the three rectangles problem (arctan1 + arctan2 +arctan3 =pi ) gives a way to construct that triangle ig
you should stream on twitch if possible and it will give you a chance to interact with your audience too. If you do it on this channel, it will be cool too.
I understand that there is a proof of why every triangle has a total internal angle measurement of 180° or π radians, but isn't that something we can just inherit from geometry? Who doesn't know that the internal angles of all triangles sum to π radians?
Everybody knows that the internal angles of all triangles sum to π radians, but only a very pure mathematician would feel the need to prove it.
İlköğretim matematik öğretmenligi okumak istiyorum meraklıyım bakalım video da neler olucak seviliyorsun Michael penn
I see you don't have the bandage on your arm any more. I hope it's better now. Did you do one of your back flips and land badly?
excuse me, but I think you failed to consider the possibility that 2=0 for sufficiently large values of 0 :-)
You are right: I never reflected on that possibility...
great one
Amazing.
This is nice 👍
chiamate le tangenti m,n,p il vincolo e' m+n+p=mnp.....un esempio e' 1,2,3
I need You mitre Michel pleas
Michael , you do gym , right ? I can see the pumps
he has a gym at home. His hobby, besides math, is mountain climbing
This is as easy atan(1)+atan(2)+atan(3).
Come on 99.999% know and don't need a proof that the sum of the angles of a triangle is π!!!
We can get that tan(γ)=tan(a+b) by changing the γ to the (π-(a+b)), so it is faster and easier)
That's what I was thinking too:
a + b + c = pi => a + b = pi - c => tan(a+b) = tan(pi - c) => tab(a+b) = -tan(c)
Poeple who disliked this are really sick.
cool
you really look like Adam Groff from sex education but older
If you want to deal with triangles without all the trig functions take a look at wildbergers rational trigonometry.
nth
Hang on, since when is 1 not a prime? I mean it's not a composite for sure.
It's part of the definition of prime numbers, simply because that definition is more useful.
Prime numbers only have two factors, 1 and the number itself. 1 has less than that.
@@TJStellmach Interesting. I was always taught that a prime is "whatever is not composite". I was like 8 at the time tho I may not remember much.
@@TJStellmach in addition to your comment. It's useful for a fundamental theorem of arithmetic which states that any positive integer (except one) can be represented by a product of prime numbers ONLY BY ONE WAY. If "one" was a prime number, that the theorem would not hold, cause e.g. , 4 could be represented as 1*2*2 and also 2*2 and 1*1*1*2*2 etc... So in this case we would have infinitely many cases which is not useful at all
@@IoT_ This makes a definition of "unit" as being a number that has an inverse. In the integers 1 and -1 are the only units, in the rationals every nonzero number is a unit. Units are not primes. 2 and -2 are both primes, as -2 = 2 * (-1), that is, 2 times a unit. Multiplication by a unit gives something known as a "conjugate".
Sorry, hit the dislike by accident. Great video.
3rd
4th
6th
2nd