Here is another nice problem from our olympiad k is nonzero integer. Show that equation: k=(x^2-xy+2y^2)/(x+y) has odd number of integer solutions (x,y) if and only if k is divisible by 7.
I think can be done even simpler as follows: We can frame the equation as a(b+c)^2+c(a-b)^2=4a We got above one by adding and subtracting 2abc in LHS and taking common ones out. This gives us,(b+c)^2=4-(c(a-b)^2)/a), so max value of b+c=2 which occurs when a=b. This the triplet is (2-c,2-c,c).
Question I have, how would you dream up that approach in the time constraints of a math contest? I would never think of that. I'll chalk this one up as "one I didn't get" , LOL My approach was to find a in terms of b and c but nothing obvious popped out.
@@Tiqerboy the idea is to get relation for b+c. Since we have ab^2+ ac^2 term in LHS. We can get (b+c)^2 in LHS by adding and subtracting 2abc. And thus get our relation by taking square root both side.
@Tiqerboy many times it would depend on the questions too..but yeah this method first comes in mind obviously when there are 3 different variables and nothing seems relatable to each other because somehow we have to relate the terms (b+c) in some other expressions to find out maximum value of it
The way I did this problem was to rewrite (a+c)(b^2+ac) as a(b+c)^2 + c(b-a)^2. so we have a(b+c)^2 + c(b-a)^2=4a (b+c)^2 =4 - (c/a)(b-a)^2 and we notice that (c/a)(b-a)^2 is always greater or equal than 0, so the maximum value of (b+c)^2 is 4 and the maximum value of b+c is 2. We get this maximum value only when (c/a)(b-a)^2=0. Since a and c cannot be 0, we know this only happens when a=b. b+c=2 so b=2-c. We know a=b so a=2-c. And so, all triples (a,b,c) are of the form (2-c,2-c,c) where 0
At 5:52 we not only have b+c=2 as a maximum, but also a^2+b^2=2ab implying a=b 'to ensure equality in the inequality'. Direct substitution of a=b, and c=2-b yields that the triple (b,b,2-b) for 0
For the achievability, one can alterantively argue that x^2+y^2=2xy if and only if x=y and so in order to have equality in the first part you need a=b. Once you have this you can rewrite the initial equation as (b+c)(b^2+bc)=4b, we can simplify it by b to obtain that (b+c)^2=4 hence b=2-c by positivity of b and c.
My solution (actually pretty much the same as the guy before me): as a quadratic equation of a, the given equation can be written as ca^2 + (b^2 + c^2 - 4)a + cb^2 = 0. Note that, since all of a, b, c are positive, (b^2 + c^2 - 4) must be negative. From this fact and the root formula for the quadratic equation, it can be seen that a positive root exists as long as the determinant is non-negative. The determinant for the quadratic equation is (b^2 + c^2 - 4)^2 - 4b^2c^2, and this being non-negative is equivalent to (b+c)^2
11:57 To je dobré miesto na zastavenie This is from Putnam 1968 so probably die hard fans of the competition may have already done this problem. Answer in the comments... Find all finite polynomials whose coefficients are all ±1 and whose roots are all real.
Answer= ±(x + 1), ±(x - 1), ±(x² + x - 1), ±(x² - x - 1), ±(x³ + x² - x - 1) and ±(x³ - x² - x + 1). The linear and quadratic polynomials are easy to find (and it is easy to show that they are the only ones). Suppose the polynomial has degree n ≥ 3. Let the roots be k_i. Then ∑ k_i² = (∑ k_i)² - 2 ∑(k_i × k_j) = a² ± 2b, where a is the coefficient of x^(n-1) and b is the coefficient of x^(n-2). Hence the arithmetic mean of the squares of the roots is (a² ± 2b)/n. But it is at least as big as the geometric mean which is 1 (because it is an even power of c, the constant term). So we cannot have n > 3. For n = 3, we must have b the opposite sign to the coefficient of x^n and it is then easy to check that the only possibilities are those given above.
@@hamiltonianpathondodecahed5236 Usually I give the answer few hours after I post the homework but today I’ve decided to change it. I post directly the answer so people don't have to wait
We can find the triples more easily:- When b+c = 2, Then going back tells us that a²+b² = 2ab Which means (a-b)² = 0 hence a-b = 0 or a = b and we are done
My solution: let x=b+c, then substitute in to cancel c, then treat a as the main variable and b&c as parameters, then we can get a quadratic equation of a, then we have the determinant >=0 since a is a real number, that will give us the upper bound of x is no greater than 2. The rest part is the same, that is construct the generic solutions with a free variable which in passing also proves that x=2 can really be reached which completes the first part logically.
Well, I think AM-GM inequality is well known and you can use it to solve the problem. (a+c)(b²+ac)=4a By AM-GM Inequality, we have: (ab²+a²c+b²c+ac²)/4≥(ab²•a²c•b²c•ac²)^1/4 ==> a≥abc ==> 1≥bc For the expression to achieve maximum value, bc=1. Now again use AM-GM inequality (intuitive as you see √(bc)) on b and c: (b+c)/2≥√(bc) b+c≥2 So, b+c=2. Edit: This is wrong... Huh...
You can get a=b from the first part more easily, because if not, a^2+b^2>2ab and thus you have strict inequality if b+c=2. So, if b+c=2, the only way 4a=... is if a=b.
I noticed that it possible to do the multiplication on the left side and then divide by abc to get a/b + b/a + b/c + c/b = 4/bc. This looks a lot like a number plus its inverse twice, which is always at least 2+2=4. So we have bc must be less or equal to 1. Now I forgot what was I thinking, sorry.
This problem is well posed and devious. It's a great test question. Brilliant solution. My only suggestion for the problem is that the equation be changed such that the maximum of b+c is 42. :)
We can easily deduce that b^2 = 4a/(a + c) - ac so we need to find the maximum of sqrt{4a/(a + c) - ac} + c. It is difficult to guess a maximum value so let us just suppose that sqrt{4a/(a + c) - ac} + c
My method was to rewrite the initial equation as a quadratic in a that looks like this ca²+(b²+c²-4)a+cb²=0 now the discriminant is b⁴+c⁴-2b²c² -8b²-8c²+16 which factors into (b-c-2)(b-c+2)(b+c-2)(b+c+2) now we set this ≥0 in order to have real solutions well in this inequality there are a bunch of cases but it's not so hard to go trough all so finally we see that the maximum achievable value of b+c is 2 and the rest is like in the video.
6:02 No! Please don't wipe important stuff away. Whenever one derives / uses inequalities, it is best to do bookkeeping about all "... with equality iff (some condition)" notes. -- Here: (x-y)^2 >= 0 with equality iff x=y. Hence x^2+y^2 >= 2xy with equality iff x=y, and later ... + (a^2+b^2)c >= ... + 2abc with equality iff a=b. And of course $b+c
A tiny loss of rigour there. Showing b+c can't exceed 2 gives us AN UPPER LIMIT for b+c not THE MAXIMUM. It's only proven to be the maximum once you come up with a b+c=2 example. I did maths at Uni and these little losses of rigour got stamped on by our pure maths supervisors like walnuts at Christmas when there were no nutcrackers in the house.
If you multiply it out and divide by a you get b^2 +cb^2/a +ac +c^2=4. We know cb^2/a must be an integer because all the other terms are integers. At least one term (a,b,c) must be equal to zero or otherwise all these products must be equal to 1. From there a solution is easy
Hi Michael.. I would like u to do one question from the 2015 Indian National Math Olympiad Problem 1 , I tried it and my solution and all the other solutions I have seen to the problem were too bashy ..Let ABC be a right-angled triangle with ∠B = 90◦ . Let BD be the altitude from B on to AC. Let P, Q and I be the incentres of triangles ABD, CBD and ABC respectively. Show that the circumcentre of of the triangle PIQ lies on the hypotenuse AC. I HOPE U DO THIS PROBLEM ... LOVE FROM INDIA!! KEEP THE GOOD WORK ON!!
Had to sleep on it, but got it this morning. Rewrote it as (b+c)^2 + (c/a)(a-b)^2 = 4. So the max for b+c is 2, and this occurs when a-b=0, therefore (a,a,2-a), a in (0,2), is the solution set.
The initial observation does not need the condition that x and y are positive, as (x-y)^2>=0 holds for all reals. You are simply reproving the arithmetic mean-geometric mean inequality. A more straightforward solution is as follows: The given equation is quadratic in "a": a^2*c+a(b^2+c^2-4)+b^2c=0, so use the quadratic formula to find its solutions for "a" in terms of b and c. It has a real solution for "a" as long as its discriminant is nonnegative. This inequality is (b^2+c^2-4)^2-4*b^2*c^2>=0, which simplifies to (b^2-c^2-4)^2 >= 16c^2. Using the positivity of a,b,c, we get that the coefficient of the linear term must be negative, thus b^2+c^2=4c, which in turn simplifies to (2-c)^2>=b^2, which simplifies to 2-c>=b (since c
11:16 If c (a+(c-2))^2 = 0, can't we just set c =0 for a to be any positive real number and b = 2 from b = 2-c? that seems to be a part of solution missing at the end.
at 3:45, the sum of squares is replaced by twice the product, so the equality must be less than not greater than. i.e. < 2 abc + 2 abc. not > than as mentioned. Because the product is the smaller number than the sum of Square.
To me, a native English speaker, it sounds normal. But upon further research, it seems to just be a common mistake for a number of Eastern Europeans (e.g. Serbians instead of Serbs, etc.).
