Third International Mathematics Olympiad #2
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- Опубліковано 4 чер 2024
- This problem is from the third IMO held in Hungary 1961. I found it relatively easy compared to other problems I see these days. It required a basic knowledge of triangle areas, binomial expansion, and inequalities. For the second part of the problem, a=b also means a=b=c since a,b were chosen arbitrarily. so it's an equilateral triangle.
We have T = (absin𝜃)/2 where 𝜃 is the angle between a and b. We also have a²+b²-2abcos𝜃 = c², so a²+b²+c² = 2a²+2b²-2abcos𝜃 = 2ab(a/b+b/a-cos𝜃) [1]. Now, 4√3T = 2ab(√3sin𝜃) [2]. Since ab ≠ 0, we can divide [1] and [2] by 2ab, so we need to compare a/b+b/a-cos𝜃 and √3sin𝜃, so a/b+b/a and √3sin𝜃+cos𝜃. But a/b+b/a ≥ 2 (AM-GM) and √3sin𝜃+cos𝜃 = 2(sin(𝜋/3)sin𝜃+cos(𝜋/3)cos𝜃) = 2cos(𝜃-𝜋/3) ≤ 2, thus: a/b+b/a ≥ 2 ≥ 2cos(𝜃-𝜋/3) which proves the inequality. Equality holds when a/b+b/a = 2, so a = b and 𝜃-𝜋/3 = 0, so 𝜃 = 𝜋/3. Since a, b were chosen arbitrarily, the same holds for c. In other words, the triangle must be equilateral.
Thanks @randomjin9392 for your proof! Really impressive!
nice solution. Thanks!
very nice alternative demonstration of the inequality and correction to the equality condition, just a one amendment: a/b+b/a >=2 is not due to the AM-GM inequality but because (a-b)^2>=0 ( a,b € lR+). 🤗👏👍
@@vladimirrodriguez6382 In this case one can .. choose. For n = 2 and positive numbers a, b the AM-GM and the non-negativity of the square of difference are equivalent. So my conclusion is based on the thing identical to the one you mentioned. One doesn't have to derive AM-GM from the square of difference (for instance, it is possible to do so geometrically).
I've just noticed that (a,b € lR+):
a/b+b/a>=2
a^2+b^2>=2ab
(a-b)^2>=0
a^2+b^2-2ab>=0
a^2+b^2+2ab>=4ab
(a+b)^2>=4ab
a+b>=2sqrt(ab)
(a+b)/2>=sqrt(ab)
AM-GM inequality
Thus I rectify the indicated amendment, and congratulations for your contribution to solving the problem

You can use Heron radical T=sqrt(p*(p-a)*(p-b)*(p-c )). Then we want to prove that a^2+b^2+c^2 >= 4*sqrt(3) * sqrt( (a+b+c)/2 * (b+c-a)/2 * (a+c-b)/2 * (a+b-c)/2 ) = sqrt(3) * sqrt( (b+c-a) * (a+c-b) * (a+b-c)) * sqrt(a+b+c). Squaring both sides, what we want to prove is (a^2+b^2+c^2)^2>= 3* (b+c-a) * (a+c-b) * (a+b-c) * (a+b+c). Here we use the well-known inequality that abc>=(b+c-a) * (a+c-b) * (a+b-c) . It is easy to prove by AM-GM . Set x+y>=2*sqrt(x*y), set x+z>=2*sqrt(x*z), set y+z>=2*sqrt(y*z), if you multiply them you will have 8xyz=(b+c-a) * (a+c-b) * (a+b-c). From this last inequality we want to show that (a^2+b^2+c^2)^2>=3abc(a+b+c). Here we use cauchy-schwarz inequality. We have that 3*(a^2+b^2+c^2)>=(a+b+c)^2. Squaring both sides we have 9*(a^2+b^2+c^2)^2>=(a+b+c)^4. Now in the previous expression we multiply everything by 9 to have 9*(a^2+b^2+c^2)^2>=27abc(a+b+c). Using the result from Cauchy-Schwarz we want to prove that (a+b+c)^4>=27abc*(a+b+c). This is equivalent to (a+b+c)^3>=27abc, since a,b,c are the sides of the triangle. This last expression is exactly the AM-GM inequality for a,b,c which is true. The equality holds for a=b=c, which is the equilateral triangle.
Well, I think it's pretty obvious that the equals situation is when the triangle is equilateral. Area of an equilateral triangle is √3s²/4 (where s is the side length of the triangle). Multiply that by 4√3 you get 3s². And in an equilateral triangle, a² = b² = c² = s², so a²+b²+c² = 3s².
Prime Newtons: uses T =
0.5absinθ
Me, an intellectual: HERON’S FORMULA
I was hoping he would use that
I am sorry I didnt understand the second part of the questions, the answer is probably correct but the explanation didnt make sense to me
Even though you sometimes fail in your reasonings I commend you for trying things your way. Good channel.
When the equality holds , it should be an equilateral triangle , a = b = c.
I had it at school about few weeks ago... I have flashbacks.... I hate geometry and trigonometry, but maybe I will change my mind :)
I could see since the beginning that the equality holds when the triangle is equilateral, because a = b = c, and the area of a equilateral triangle is (a²+a²+a²)/(4√3) = 3a²/(4√3) = √3a²/4. Now for the general case let us say 'a' is the longest side (or one of the longest sides). Then a² + a² + a² >= a² + b² + c², and we prove the general case.
I don't think your solution to part 2 is correct. Counterexample: 1-1-sqrt(2) right triangle, area is 1x1x1/2=1/2, pluging everything in you get 1^2+1^2+sqrt(2)^2 = 4sqrt(3)x1/2 => 4 = 2*sqrt(3) ->
That's interesting 🤔. I see. Equality holds when it's an equilateral triangle a=b=c. Thanks
@@PrimeNewtons I think the erroneous assumption was that the triangle was a right triangle, because you were looking at a different inequality (ab/2 >= T) and assuming that equality needed to hold in that case to hold in the original inequality. When you found that a^2+b^2+c^2>=8T, this was with respect to sin theta being maximized previously, so it doesn't hold for a general triangle. In fact, because 8 is strictly greater than 4 sqrt 3, equality can never hold in the original inequality under these conditions. I suspect that the Law of Cosines would be the useful link between the ab expression for area and the a^2+b^2+c^2 expression we're comparing it to.
@nidoking042 I got confused when it went from the general case to the right angled triangle and then back to the general case.
Excellent solution sir
I don’t understand.
If a^2 + b^2 + c^2 >= 8T, this expression never can be equal 4sqrt(3). It is not true?
Indeed, you accurately pointed out one of the logical fallacy of his « proof ». This one is the « least » and « most obvious » absurd one. Furthermore, even his hard core « result » pretending a^2+b^2+c^2 >=8T, is generally false. So he has not proven anything, totally missed all points, and made a mess of the problem. Such problem can be rigorously solved by different approach.
The fastest is by use of Euler-Lagrange equation with Lagrange multiplier to solve this optimization problem under « Heron’s formula » constraint. Minimizing Q=a^2+b^2+c^2 under Heron’s constraint : Q^2 = (4T)^2 + 2(a^4+b^4+c^4), leads indeed (in a straightforward few lines of easy computation) to a=b=c equilateral configuration, where such Heron’s identity furthermore gives the wanted expression with sqrt(3) : Q=3a^2=4sqrt(3)T.
One can also use the path of Cauchy-Schwartz inequality
Or take the path of a purely geometric powerful « elementary » proof based on Euclide theorem about parallelogram area invariance under sliding (which defines « determinants »). Which proves that a triangle of « base » a and « height » b, has maximum area when those are orthogonal (since any slide increases b, further from the circle of its fixed constant value). And by reason of symmetry, on the contrary, the minimum area is obtained for isosceles configuration with c=b. Finally, since the quantity Q (which is NOT the area! Only closely related to), that is under minimization, is fully symmetric on a, b and c, it hits an extreme value on the equilateral configuration. Such canot be a maximum since the right angle configuration plays this role. Thus it’s indeed a minimum. QED.
Let x = a+b-c, y = b+c-a, z = c+a-b. Notice that x+y+z = a+b+c. By Heron’s formula, 4T = sqrt(xyz(x+y+z)). By the AM-GM inequality, 4T
I assume the formula is known (easily demonstrable by setting a=kb)
a^2 + b^2 > 2ab (= for a=b).
I call p, q, r the angles respectively opposite to the sides a, b, c.
By Carnot's theorem we have:
c^2 = a^2 + b^2 -
2ab cos r
while T = 1/2 ab sin r.
I replace these two formulas in the formula given at the beginning and with simple steps I obtain:
a^2 + b^2 >= 2ab (1/2 cos r + rad3 /2 sin r)
i.e.
a^2 + b^2 >= 2ab sin(30° + r)
Since the sin is always less than one the inequality is verified and the equal is easily found for r = 60°, and a=b i.e. for an equilateral triangle.
You can prove something stronger. You can give ab+bc+ca instead a²+b²+c²
Very good. Thanks 🙏
Wasn't the pythagorean step only valid for right angled triangle. Where did we generalize the result to any triangle?
i enjoy your vidieos
Congratulations
can we use herons formula for area of triangle
What about ab+bc+ca?
We can use cosin theorem
this still feels wrong tho. substituting those things could cause it to create exceptions later right?
Somewhere the deduction that since 8>4*sqrt(3) , the ineuality also hold correct. If that is the case, then we can replace 4*sqrt(3) by 1 also, and the process will look correct but actually wrong. This solution do not convinces me.
Doesn't the fact you did it with a right triangle prove the area only for right triangles? I think you would have to do it with a l triangle but do a bunch of trig instead so it proves it for any triangle.
Was wondering about this...
Please sir, please can you explain proof by contradiction? I am having my final year exam tomorrow. 😭
His entire « proof » is totally false. And gives you perfect exemple of « proof by contradiction ». Since he pretends to have proven that Q=a^2+b^2+c^2 >= 8T and that 8>4sqrt(3) (which is the only true statement of his reasoning), and finally that the « equality is hit for a=b ». Which is obviously absurd and self contradictory, since 8 STRICTLY GREATER THAN 4sqrt(3) ! Thus he has proven by contradiction, without noticing it, that his « proof » is false !
Moreover his entire « proof » is false since Q=a^2+b^2+c^2 is NOT >= 8T , in general. That’s obviously false since in a general case, c^2 can be greater or smaller than a^2+b^2, depending on the context.
One correct way of reasoning with elementary means is to use any useful variant of Heron’s formula. The most straightforward one is to square it (at least) : T^2=area^2 = s(s-a)(s-b)(s-c) where s is the semi perimeter s=(a+b+c)/2.
Formula that can easily be rewritten in a more convenient form : (4T)^2 = Q^2 - 2(a^4+b^4+c^4)
Such formula gives easily the desired equality in the equilateral configuration a=b=c
To prove finally that such targeted configuration actually minimizes Q can be done by different ways. The fastest is to use Euler-Lagrange equations with Lagrange multiplier. Another is to derive the desired inequality using Cauchy-Schwartz. Or more elementary use Euclide invariant theorem on parallelogram area and symmetric arguments, since Q is fully symmetrical on a, b, c.
i wrote a long comment and vorget to sent it up wow
so short form here again:
it was interesting all is fine nice video!
LG K.Furry!
I tried this with Heron's formula, I simplified and ended up with 16T^2 = - (a^2 + b^2 + c^2)^2. Why is it negative? I checked on Wolfram Alpha, it also gives me a negative value or imaginary value for the area. Can someone explain?
You surely made computational mistakes since the secular irrational Heron’s formula can strait forwardly be expressed in a purely algebraic form (without irrational quantities) : (4T)^2 = (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) , or in other useful algebraic variants. But none is of « your (erroneous) form ». Check it out
The proof is quite easy. But what’s far more interesting IMHO is: Can you derive it? If you didn’t know this was true, would you get the RHS?
Well, yes and no. If I was doing a thing and I found myself with a^2 + b^2 + c^2 and wanted to compare it to T, I would expect that there would be something of the form mT
You had to use schwartz inequality
This proof doesn't seem justified. The inequality comes from the area formula for arbitrary theta, but later you use pythagoras, i.e. a right angle triangle, to find a connection between a, b, and c.
Using the cosine formula for the connection between a, b, and c should give a valid proof.
This cabbage needs to be boiled again.
ig The IMO was easy back then.
I think you proved a^2+b^2+c^2>=8T for a right triangle only, not for any triangle. Also, if a^2+b^2+c^2>=8T, it can never equal 4*sqrt(3)*T.
I do enjoy your videos.
A couple of thoughts:
-as mentioned elsewhere, you honestly can't use too many onions...with a long braise, they disappear
-the very first step is to make stock from the chicken carcass(es)...homemade stock (instead of store-bought broth) really elevates this dish.
-definitely use sweet Hungarian paprika.
-what I remember from the old folks (this was a classic wedding/funeral supper) is that they would dredge the chicken pieces in flour heavily seasoned with paprika and then brown off the chicken.
-our Slovak family served with knedlíky, a sort of bread dumpling that is steamed in the shape of a loaf and sliced.
Sorry I think you have the wrong number. UA-cam mobile does this sometimes when you switch videos. The comment sections don’t change properly.
@@cnjdevwrong number?
@@Careerhumresource I was referencing something that used to happen a lot (and still happens to a lesser extent)
This time you are wrong!!!!😮
😁🤪👍👋
Your entire « proof » is totally false! First, the equality DOESN’T necessarily holds for a=b, since a=b=1 and c=sqrt(2) gives a counter example : the area is indeed T=1/2, thus 4sqrt(3)T=2sqrt(3)=8T (which is false by the way!), and on the other hand you correctly remark that 8>4sqrt(3). But despite that, you still pretend that the « equality (with 4sqrt(3) », holds for a=b. This is absurd since you « proved » that Q>=8>4sqrt(3) !
So your « proof » is totally false everywhere ! Your « game playing » on a particular case to « prove » a general result contains the fraud. Indeed, for a non right angle triangle, one cannot say what you pretend about c^2 compared to a^2+b^2. It may be smaller or larger, depending on the situation. And thus Q=a^2+b^2+c^2 is NOT equal to 2(a^2+b^2). That’s why you have NOT established that Q>=8T. Such a « result » is false.
So what have you done ? Well quite a naive little mess! You didn’t listen to the sqrt(3) bell ! You couldn’t figure out where this precise boundary was coming from, and thus you forced a « proof » that is an entire forgery. You could have guessed that the symmetric equilateral triangle would pump out the desired sqrt(3), even if simply checking such special case is NOT a proof that other configurations doesn’t give the same answer.
So your approach to such problem is globally much to naive. First you are trained to use transcendental trig functions, which is a conceptual weak tool. It hides what is actually going on under transcendental stuff covering a geometric problem governed by ALGEBRAIC identities. You could have at least used the irrational Heron formula for the triangle area. And make use of its square. Or even a better formulation of Heron, purely algebraic : (a^2+b^2+c^2)^2 = (4T)^2 +2(a^4+b^4+c^4).
Which gives 3a^2 = 4sqrt(3)T in the EQUILATERAL case, and a easy rough lower bound : Q=a^2+b^2+c^2 >= 4T
But this last one is not enough and one needs the full « Heron » CONSTRAINT of this optimisation problem (who didn’t say his true name), to optimize Q=a^2+b^2+c^2. It’s thus an optimization problem under constraint. And to solve such problem, one needs to use the tool built by the 19th century great French mathematician Louis Lagrange : Euler-Lagrange equations with Lagrange multiplier.
The computation is simple and straightforward. It gives in a few lines (of partial derivatives) : a=b=c. Which correspond indeed to the EQUILATERAL case which minimizes the problem, and where the 4sqrt(3)T minimum is actually hit.