This problem has a crazy solution
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- Опубліковано 31 гру 2024
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I tried a purely algebraic approach for this problem.
(a²-2b²)+(2b²-3c²)=2(ab+bc+ca)
=> a²-2ca-3c²=2b(a+c)
=> (a+c)(a-3c-2b)=0
Put a=3c+2b in 3rd equation,
bc+(b+c)(3c+2b)=1=2b²-3c²
=> 6c(b+c)=0
If c=0, we get the solutions as professor Penn showed.
If b+c=0, then by putting it in the 3rd equation we get, b²=-1, and putting this in the 1st equation, we get, a²=-1. Thus we have 4 complex solutions,
(a,b,c)€{(i,i-i),(i,-i,i),(-i,i,-i),(-i,-i,i)}
If a+c=0, then also we get 2 new complex solutions, (-i,i,i) and (i,-i,-i)
I like this approach. An intersection of 3 degree 2 surfaces should have 2*2*2=8 solutions (by the generalized Bezout's Theorem), which as you have shown, is exactly what we get in the complex plane.
The symmetry of the complex solutions is interesting, you get complex conjugation as a symmetry (which you always will), but you can also permute a, b, and c, which is unexpected, but fits with the symmetry of the third equation.
The equation 2 was changed from a²-3c²=1 to 2b²-3c²=1 at 9:43
@@albertogarcia4177 Sure...i think that is wrong there...he approves that : csc^2(a)=2*csc^(b) , that doesnt mean that a^2=2*b^2 ....right ?
2:03 I like how the first attempt at drawing the graph is still visible. Never change your barely-hidden second takes. They make your videos so much more enjoyable knowing you're also not right the first time every time.
Out of curiosity I did it the "head on" algebraic way. If you set b=ax, c=ay you will get that 4x² = 3y²+1 and x+xy+y = 1-2x², so for example if 2x=√(3y²+1) we get (y+1)√(3y²+1)=1-2y-3y² which is scary looking but on close inspection we notice that it will be at most a quartic and on top of that the constants on both sides must be equal to 1. So if we square we land at a cozy y(y³+y²-y-1) = 0 from where it's not difficult to find the solutions. We can attend to another case similarly.
I promise I am borrowing the following language from a reputable source: Evan Chen, an IMO gold medalist, and Thomas J. Mildorf not infrequently mention dumbassing as an important approach to problem solving. I think this is the same thing as the “head on” algebraic way you referenced. I like it much more than “the bag of tricks” approach that is so prevalent nowadays. Chen even lamented that he only won the gold medal because he was able to regurgitate a few more tricks than his fellow competitors. He is being modest because it does take some ability to know when the tricks are applicable.
Nice solution. I want to note that from the point on where we have (y+1)√(3y²+1)=1-2y-3y², instead of squaring, we can simply divide both sides by (y+1) since y=-1 leads to no solution, which simplifies the rhs to -3y+1. Then squaring only results in a quadratic polynomial. A symmetric argument also works for the other case where 2x=-√(3y²+1).
Amazing how trigonometry helps solve such a complex algebra problem, great video as usual, thank you for being one of the best teachers out there, there's no other feeling like a watching a Michael Penn video😍
I hate it when the thumbnail isn't the same problem...
I know. It's fixed now thankfully but I spent 10 minutes trying to solve a completely different problem ☹
Waste of papier for me as well 😂
Me too.
Happened to me twice before. Now I never trust what I see in the thumbnail. I just watch the first 5 seconds of the video, then I go ahead.
yep me too' but now you have the cot hint come on Micheal get it right
1)To come up with this idea, you can divide the third equation by abc, and then we've got that the sum of 3 numbers is equal to their prodcut => these number are tangent of angles that sum up to 180 degrees.
2) You can solve this problem without trygonometry!! By adding a^2, b^2, c^2 to the last equation we get (a+b)(a+c) = a^2 +1, (b+a)(b+c) = b^2 + 1, (c+a)(c+b) = c^2 + 1 respectively. Now with using (1) we get that (a+b)(a-2b-c) = 0 and with usage of (2) we get that (b+c)(-a+2b-3c) = 0. If a+b = 0, then from (3) we get -a^2=1 , which is impossible. The same goes with b+c=0 case. So we have a-2b-c=-a+2b-3c=0, so a-c=2b=a+3c => c=0. The rest is easy.
Nice approach
That’s beautiful man. Tough knowing to divide third equation by abc. I would not have thought of it, but after that your approach is very clear.
you can't divide by abc until you know that a, b and c are all different from 0, but you can't prove this without solving the system, and in fact one of the three unknowns is just 0.
@@VideoFusco you make a good point I didn’t think of that. You could however assume that none are zero and check the cases later?
@@VideoFusco its pretty easy to check that abc is not equall to 0. Just chceck a=0, b=0, c=0
Thanks!
4:45 it's worth mentioning that we can divide by (cot alpha + cot beta), because it's not 0 - if it was, then a=-b, so the right side is 1+a^2>0, while the left is 0. But it's not so obvious, it took me a while to get to that.
yes thx
Knowing to start with equation three and relate it to angles of a triangle and onto trig functions is sort of a “skyhook” in my opinion.
Only very “well trained” (meaning be ready for a problem like this) OR someone who has already viewed the solution would know to start with equation three.
Still a wonderful solution albeit unmotivated.
I mean these are meant for students trying to compete in an olympiad. Yes the solution might seem unmotivated, but unless you know the tricks that the students are being taught, that's kind of the point for many of these questions and why they're difficult.
It's not completely unmotivated. In fact, the symmetric polynomial in equation 3 can be thought of as the inner product of a vector with itself after the "circular shift" operation has been applied to it, i.e., . Now, we restrict this inner product to 1. So in R3, you can think of it as transforming the space by rotating all the axiis about a line going through the middle by 90 degrees. You can then ask, when is the boundary of the polar body of these sets of vectors just a 90 degree rotation? Granted, that is not very computationally viable, but this problem is often extended to n-bit vectors, with different restrictions put on them after considering the 3rd equation.
Second equation was written wrong until he changed it at 9:49
Well spotted
If you can't answer the question as given, then just answer a completely different question. Michael must be a politician! 🤣
There was a switcheroo in equation 2 at 9:46. The earlier shown problem was quite a bit simplified since the relation between b and c popped right out of 1st two equations. Anyway, it was a very nice trick that you have shown here! I had honestly never used this trick before.
I have a question regarding the step:
cot(180 - gamma) = cot(alpha + beta) implies 180 - gamma = alpha + beta
I think extra justification is needed because 180 - gamma is in (0, 180), while alpha + beta is in (0, 360), since alpha, beta, gamma are all in the interval (0, 180 degrees). But cotangent isn’t one-to-one in (0, 360), we can’t get the arguments are equal. That is, there is a possibility that 180-gamma is in (0, 180), while alpha + beta is in [180, 360) and their cotangent values are equal
That would just lead to alpha+beta+gamma = 360, which is still a triangle in the sense that we can say a triangle with angles 120, 120, 120 is equivalent to a triangle with angles 60, 60, 60, or with angles 90, 130, 140 is equivalent to 40, 50, 90. Most importantly for this system of equations, these "equivalent" triangles obey the law of sines exactly how we need them to.
Great, I enjoyed the explanation! Congratulations from Brazil!
Fantastic problem.
Thank you, professor.
I think this video is below the usual quality level.
1. As others already pointed out, at 9:42 the second equation of the question is changed without mentioning or even puting a note there during editing. That's not cool, even if it was an honest mistake.
2. At 6:20 even if 180-gamma is staying in the (0, 180) domain, alpha+beta is in (0,360) so the condition of the invertibility of cot is not met. E.g if alpha=beta=gamma=120 then 180-gamma=60 and alpha+beta=240 and given that 240=60+180 cot(60)=cot(240). But really, they could be any value provided their sum is 360 and this branch of the solution is discarded without investigating it. (I don't say that it would've lead to valid triplet of values.)
3. Somewhat similarly, at 11:03 taking squareroots of an equation without taking care of the signs is a big no-no, and could lead to loosing solutions. [Edit: it was pointed out, that on the specified domain sin (and therefore csc) is always positive, so I officially revoke this complain - though this fact still would've worthed mentioning.]
I am a little sleep deprivated and hungover so I might be a little grumpy and too harsh. But the fact that I was capable noticing all of the above (despite my aforementioned condition) means that these are pretty obvious and big mistakes.
The first spot: the equation isn't exchanged this was the sum of the first and second, so there is no mistake.
The second spot you have said is right.
The third spot: at the domain of sin α, α there is in [0,π], is ever positive so the sign must be the same
Observation: Sorry for the writing if there is some mistakes
@@eduardomalacarne9024 no, the system was definitely changed. After he changed it, he took the sum of the 1st and the new 2nd equation, which gave something very close to the old 2nd equation but with 2 on the RHS, not 1.
@@synaestheziac please, stop at the moment 10:22, sum first with the second, it's right
@@eduardomalacarne9024 yes, but none of the equations on the board at that time are equivalent to the 2nd equation in the system at the beginning (or before 9:42)
@@synaestheziac oh, I've take it now, sorry if I've been or seen rute, that's because I read thumb and the blackboard at this moment and think it's all right, thank you
Nice problem and wonderful solution!
This trig trick with tangents and cotangents is a regular part of MOSP.
Note that if you divide the 3rd equation by abc, you get 1/a + 1/b + 1/c = 1/(abc), which tells us that there exists angles A, B, and C such that A + B + C is a multiple of pi and 1/a = tan A, 1/b = tan B, and 1/c = tan C.
Thanks for this example, Professor Penn!
I never would have come up with the idea with a trigonometric substitution.
What a beautiful solution!
wait, why did the second equation change at 9:49?
Same question
the trick is that he changed the question midway.
It helps that the thumbnail is fixed!
@@MichaelGrantPhD yep that clarifies it
Did equation 2 change around 9:40 ?
At the point that C=0 I would have questioned working the problem with a right triangle that in fact is not a right triangle since the hypotenuse is 0, and would not have trusted any results after that.
No, the solution is that c=0 (the value in the original equations) not C=0, the hypotenuse of the triangle
6 cyclic complex solutions exist with a pair of i's and one -i; and a pair of "-i"s and an i.
i = sqrt(-1)
Can I suggest Michael that sometimes you do a blind solve? Perhaps one of your peers could chose a problem and you see it for the first time on camera.
Super nice trip into trig, but if you just multiply (1) by 3c^2 and (2) by 2b^2 and subtract, you almost immediately get to c = 0 and then finish,
But since c ends up being zero, you are multiplying equation 1 by 0 on both sides.
@@bobh6728 ... which only _risks_ spoiling the solutions in the general case, but does not in this case.
You get: 3a^2c^2-4b^2 on the LHS
And: 3c^2-2b^2=-1 on the RHS
How does that lead to the conclusion that c=0?
@@alonskii You get 3a^2c^2 - 6b^2c2 = 3c^2 from (1) and 2a2c^2 - 6b^2c^2 = 2b^2 from (2). You subtract and get a^2c2 = 3c^2 - 2b^2 which is 0 by subtracting (1) from (2) immediately. This gives a*c = 0, but a cannot be 0 because of sign constraints in both (1) and (2).
@@IlTrojo You do not get that from (2) - you said multiply (2) by 2b^2 ...
Thia method takes a long time to solve. Id have started by assuming one of the variables is a "simple " value and tried 0 for each, then progressing to +-1, and seeing what solutions pop out or if I get a contradiction. Also worth trying all three pairs to see if any are negative, reciprocal, or negative reciprocal of each other.
The trouble with this approach is that you were asked to find all real solutions. You would have missed any solutions that didn't have a "simple" value or simple relationship.
Great problem! I love it when trig is involved!!!
Thank you to explain why eq2 was changed at 9.43 from a²-3c²=1 to 2b²-3c²=1... This is completely different, and the solutions are not the same...
Mike please help!
Very elegant solution
Michael's trigonometry approach is inefficient and would spend much more time on just one problem during a math contest. For this problem, looking at the third and the first equations: (a+b)c=ac+bc=1-ab=a^2-2b^2-ab=(a+b)(a-2b) which leads to c=a-2b since a+b=0 is ruled out by the first equation. Then looking at the second and the first equations (the second equation should always be 2b^2-3c^2=1): a(b+c)=ab+ac=1-bc=2b^2-3c^2-bc=(2b-3c)(b+c) which leads to a=2b-3c since b+c=0 is ruled out by the second equation. As long as we reach the two linear equations c=a-2b and a=2b-3c, we can easily arrive at c=0 and a=2b to get the two solutions of the original equations.
And what to do to those. who is used to trust what is written on the board?
I saw that the second equation a^2 - 3 *c^2 =1, paused the video, and began to solve it myself.
Then I decided to look at the author's solution, and I find that at 9:42 this equation instantly turned into 2b^2-3*c^2=1.
.............
However. We introduce the variable t= b/a, and for it, as a result of simple algebraic transformations, we obtain the equation
x± (x+1)*sqrt((4x^2-1)/3)=1-2x^2 => ±(x+1)*sqrt((4x^2-1)/3)=(x+1)(-2x-1)=>
Roots: x=-1 and 1(does not fit the original system of equations),
and x=1/2 => b/a=1/2..........
Never seen anything like this
Straightforward algebra very quikly (inbabout 2 minutes) gets to only real solution should be with c=0.
Changing of problem statement after 2/3 video duration makes all the thing wrong. I believe that Michael could add caption to show that initial problem statement is wrong
Dear Michael, please, in future videos, can you spend less time explaining that if three angles add up to 180⁰ they are the angles of a triangle, and then drawing a triangle to show us how that works and more time with the "well known trigonometric expressions" which may not be quite so well known to some of your audience.
How can the length of the hypotenuse be equal to zero?
It must be correct, but if in triangle C=0, should then it mean A=B ? But it would not affect anything here...
Is there no way to solve it without the geometry, using only algebra?
I can't believe the 2nd equation changed half way through the video, all my work was wrong (I still got c = 0 somehow though)
Anybody tried to solve this problem with the second equation a^2-3c^2=1? I didn't noticed the second equation was changed during his explanation...
When the title said crazy solution, I was hoping to see Michael solving a 5th polinomial.
Sqrt 2 is like, been there done that
6:00 sum angle formula? Do u have a video explaining that? I ve never heared ofcit
Seems that we do not have leared trig identities for tan or cot
Naughty edit at 9:40, substituting the more interesting 2.b^2 - 3.c^2 = 1 for the original a^2 - 3.c^2 = 1
I'm sure that the original title for this video was: "You'll never guess the trick for this problem". The Prof was right, I didn't. I hit upon another trick altogether.
My trick was this: substitute for "1" in the third eqn using the 1st eqn, to get, after a little regrouping and rearranging:
a^2 - a(b + c) - b(2b + c) = 0
Treat this as a quadratic in a and solve, to get
a = {(b + c) +/- sqrt [(b + c)^2 + 4b(2b + c)]}/2
We don't appear to be getting anywhere because the thing under the sqrt looks like junk, but when we remove the brackets we find (b + c)^2 + 4b(2b + c) = b^2 + 2bc + c^2 + 8b^2 + 4bc =
9b^2 + 6bc + c^2 = (3b + c)^2 - a perfect square!
leading to:
a = 2b + c or -b
Suppose first that a = 2b + c, then the first of the given eqns simplifies to:
2b^2 + 4bc + c^2 = 1
Substituting 2b^2 = 1 + 3c^2 from the second gives:
4c^2 + 4bc = 4c(b + c) = 0
Thus c = -b or c = 0. But if c = -b, the 2nd eqn becomes -b^2 = 1, which has no real solns. Thus c=0.
The 2nd of the given eqns is then 2b^2 = 1 => b = +/- (sqrt 2 / 2) and a = 2b + c = 2b = +/- (sqrt 2), where the sign of a is the sign of b
If, second, a = -b, then the first of the given eqns becomes -b^2 = 1, which again has no real solutions
Thus the real solutions are a = sqrt 2, b = (sqrt 2) / 2, c = 0 and a = -sqrt 2, b = -(sqrt 2) / 2, c = 0
As usual, the Prof wasn't aiming for the _simplest_ solution, but the solution that demonstrates a Mathematical technique - in this case a trig substitution suggested by the form of the 3rd eqn -, which is of more general application
9:41 The bait-and-switch of the question could have been called out - I was trying to solve it as originally stated before watching the video!
Beautiful!
Cool solution❤
How would you know to use the COT method when presented with the initial question?
I gasped at c needing to be zero. I wonder if there is another (even shorter) kind of shortcut to that result (realizing that the presented trigonometric route could be considered a shortcut already)
My solution (of course tremendously faster).
From (1) and (2) we get that 2b²=3c² then b=k.c where k can be equal to sqrt(3/2) or -sqrt(3/2).
We report this into (3): a.k.c+k.c²+a.c=1 a.c.(k+1)+k.c²=1.
If c=0 our original system is a²=1 and ab=1 then the solutions are (1,1,0) and (-1,-1,0).
If c is not equal to 0, our solutions for a given c are: ((1-k.c²)/[(k+1).c],k.c,c) with two possible values of k.
A nice solution but this problem can be solved with simple algebra. This method also finds all complex solutions.
Prof. Penn corrects eq(2) to read: 2b² - 3c² = 1. Hence,
(A) b² = (3c² + 1)/2 .
It follows from eq(1) and (A) that
(B) a² = 2b² + 1 = 3c² + 2 .
Suppose c² +1 = 0 (we're also looking for all complex solutions) .
Then b² = -1 and a² = -1, by eq(1) and (B). So, a = +-i, b = +-i, and c =+-i,
where i = sqrt(-1). Of these 8 possibilities, the 6 that are solutions are
(a, b, c) = (i, i, -i), (i, -i, i), (-i, i, i), (i, -i, -i), (-i, i, i), (-i,-i, i).
Therefore we can assume that c² +1 0.
(C) (a + b)c = 1 - ab ,
by eq. (3). Squaring both sides of (C), substituting from (A) and (B), and
simplifying yields:
(9c^4 + 5c²)/2 + 2abc² = (9c^4 + 9c² + 4)/2 - 2ab .
2(c²+1)ab=2abc²+2ab=(9c^4+9c²+4)/2 - (9c^4+5c²)/2=2(c²+1),
by transposing. Dividing this last equation by 2(c² + 1) gives
(D) ab = 1.
By (C),
(E) b = -a or c = 0.
Suppose c = 0. Hence, a = +-sqrt(2), by (B), and
b = +-sqrt(1/2) = +- sqrt(2)/2, by (A). Of these 4 possibilities, the 2 that are solutions are
(a, b, c) = (sqrt(2), sqrt(2)/2, 0), (-sqrt(2), -sqrt(2)/2, 0) .
Therefore, we can assume c 0. Thus, b = -a, by (D). Hence, a^2 = -1,
by (D). So, c² = -1, by (B). We've already dealt with this case. Therefore,
the eight complex solutions to the system are
(a, b, c) = (i, i, -i), (i, -i, i), (-i, i, i), (i, -i, -i), (-i, i, i), (-i,-i, i),
(sqrt(2), sqrt(2)/2, 0), (-sqrt(2), -sqrt(2)/2, 0) .
This is reminiscent of the fact that the sum of the tangents of a triangle is equal to their product.
"And that's a good place to stop."
Do we not need to consider the case where cot(alpha) + cot(beta) = 0
rationalize your denominators!
9:34 why tf did the equation change ?!?!??
Can we solve this problem by assuming a,b,c to be the roots of a cubic equation??? As we have ab+bc +ca =1 we might be able to use the theory of equations in some way. IDK this is just a guess, the way Micheal solved the problem was fun as always
Lovely problem
i totally bruteforced it, a=(1-bc)/(b+c) from 3rd equation (b+c=0 impossible) and from first two equations b= c*sqrt(6)/2 or b= (-c*sqrt(6)/2). Now combining these together and putting to first or second equation gives us equation with only one variable, and we can easily solve it since its quartic that reduces to quadratic (c^2=u or b^2=u) Then some boring calculations, two cases, and we are done.
EDIT: i solved it in version before this MAGIC change of second equation in 9:49. Now im gonna do it in second version
I'm curios how you pulled a, b, and c to equal cot functions. It's like pulling a hat from a rabbit!
Or you could do it without trig functions in half of the time.
I'm not a math wiz but i think he messed up somewhere. If i put the solutions from the end into the equations he starts with equation 2 reduces to 2 = 1 ....
Many students cannot deal with this unless they are trained. Amazing problem.
Such videos deserve way more views/likes than Mr. Beasts videos!!!
I agree 👍💯
The general population finds math unpalatable.
@@joeyvirrosai3574 that is true
If I may suggest... before starting the camera rooling, check that you have copied the problem correctly to the blackboard. Otherwise beautiful work.
Wow.
clean
Changing the given equation #2 almost 10 minutes into the video should have merited an acknowledgement that it was written incorrectly initially. You ignored that completely.
A really wonderful solution, but what bothers me here is that we actually end up with a degenerate triangle, because one of it's sides equal to zero....
And also, as mentioned above, the solution is not fully motivated
Wait But the thumbnail if different
I followed this, but wow!
SEE HIS SECRET TRICK! (Math Students hate him.) Lol, always pleasing to see a man do what he was born to do.
Very nice
The maddening thing about problems like this that every step is easily understandable, but it seems you need divine inspiration to come up with at least one of them.
Problem is purely algebraic.
LET'S CONSIDER THESE NUMBERS AS COTANGENTS OF ANGLES!
Come. The. F***. ON!
WOW!
If c=0 then in fact you do not have a triangle at all, much less a right angled one. Havent you in fact arrived at a contradiction? Usually when you arrive at a statement that contradicts previous reasoning, you halt, and change your approach to the problem. Not this guy though.
جيد.إنه عمل المحترفين
*Dislike. Thumbs-down for changing the problem as you were working on it.*