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
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
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.
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.
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
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.
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
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 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
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
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...
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.
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!!
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.
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.
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.).
Greetings from Slovakia 🇸🇰
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.
Ahoj muj kamarad, jsem finsky
Ako sa mash from China
Naši priatel'ia! 🇸🇰🇷🇸
Greetings from 🇩🇪
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).
I also done in the same way...Thats great😃
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
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.
Cant you just say that a and b must be equal because x^2 + y^2 > 2xy if x and y are not equal?
that would have been much faster indeed, but you still would have needed plug a and b into the equation to see what is the result, I thinl
@@red0guy Yes but that's super quick, (a+c)(a^2+ac)=a(a+c)^2=4a => a=b=2-c, same result, much quicker.
Thanks for the solution. I suggest this slightly shorter way: 4a = a(b²+c²)+c(a²+b²) = a(b+c)²-2abc+c(a-b)²+2abc, thus (b+c)² = 4-((a-b)²/c)
I think that c=0,b=2,a=anything is also a working solution
All three must be strictly positive, i.e. 0 is not a valid value for any variable.
Kudos on the videos Michael, they are really fun to watch and thank you for educating us
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.
@@goodplacetostop2973 Very nice. How do you 'check' though that e.g. x^3+x^2-x+1 is not a possibility?
wait a min , you solve problems also (・o・)
{not being rude or anything}
@@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
@@jamescollis7650 With ax³+bx³+cx+d, the discriminant Δ is equal to b²c²−4ac³−4b³d−27a³d³+18abcd. If Δ
That's a job for my friend the method of Lagrange multipliers
Linear objective function subject to quadratic constraint: use the method of Lagrange .
i was looking for this! (:
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.
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.
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 solution for the triplet that solves for the original equation is: [ (p-k^2)/k , sqrt((p-k^2)(4-p)/p) , k] where, 0
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.
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.
Anyone else come here to get inspiration on doing their homework?
I can't understand any of this stuff. But his methodology is off the charts.
5:55 notice for the equality to hold, we can immediately deduce that a=b
Can u explain that
Notice he makes it into inequality by using a^2+b^2 >= 2ab, which comes from (a-b)^2 >= 0. And that holds when a=b
True, nice catch
WIth bc
b and c are Real, not Nautal, so for example you can have b = 1/2 and c = 2
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
We have bc
AM is greater not less
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. :)
You're forcing those who want to solve an olympiad to immediately recognize 1764 is the square of 42. That's not apparent to a normal people.
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
howd you get a^2 + b^2 = 2ab
@@angelmendez-rivera351 have we met before? In bprp comment section?
@@angelmendez-rivera351 oh right, thanks! i was trying to manipulate original statement
Thank you❤
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
a,b,c do not need to be integers, and isn't constrained to be so in the problem
Regarding the triples it also works for any a when b=2 and c=0.
c cannot be zero
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...
Very short and very well explained 😀
@@jofx4051
It's wrong...
I think
Wonderful factorizations here
Your videos are amazing.
(a,2,0) is also a triple that gives maximum value of b+c = 2
By hypothesis, c > 0, so that this is not OK.
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.
you can turn b^2 + c^2 into (b+c)^2 by adding and subtracting 2bc. Then group the minus term in with a^2 + b^2 to turn it into (a-b)^2. I loved that.
I got stuck because I tried to write the whole thing as a polynomial in b+c, with only a's and constants as coefficients.
and we will derive C? for the maxium value?
You only checked lhs on last equation. how about c = 0(rhs), so b = 2, and a will be (0, +infinity)
Thank you!
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!!
Lol aise thodi na suggest karte hai xD mail karo and also this channel mostly does algebraic and NT problems , not geometry xD (no offense tho)
@@chhabisarkar9057 Toke aami sob jayegaye dekhi
@@srijanbhowmick9570 lmao
@@chhabisarkar9057
Yeh Banda yahan bhi hai...
Wah Bhai...
Woww we all got united here.....but trying out a Geometry problem would be a new thing and Mike clearly said that we can give suggestions here
I tried the subst. x = a+b, y = b+c and z=a+c. x, y, z in R^3, x > 0, y > 0, z> 0. The constraining eq. looks worse in x, y, z. :(
Trump : nobody knows math better than me.
Biden : snooze.......
The Schweitzer Memorial math Competition problems are famously hard. Maybe you would give them a try.
Well I do not think it is a big mistake, but the triples (2-b,b,2-b) should also work but are different triples
Zdravím zo Slovenska
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.
Problem explicitly says a,b,c is in the open interval (0,inf) i.e. not containing 0
Don't you also have the triple (a,2,0) which is working for any positive a?
Yes but in the question all 3 variables must be greater than 0.
we cannot consider the case where c is 0 because all a,b,c are in the open interval 0 to infinity. so c cannot be 0.
@@TheQEDRoom thanks for this answer, with his notation I did notnotice the open interval
você é muito bom em lecionar
Yay, Slovakia!
Wait, what about c=0 a=0 b=2?
a,b or b cannot be 0 because they are in the open interval 0 to infinity.
@@TheQEDRoom I though it was a closed interval
@@gustavowadaslopes2479 [0, 2] would be a closed interval.
I just made everything 1 and it works... Am I doing this right?
It works but you should proof it with statement for stronger argument
Good
I like this question that it has some “dead ends”.
You forgot the c=0 case.
If c=0, then b=2 then any a works.
so the triple (a,2,0) a ∈ ℝ is solution too.
Problem explicitly says a,b,c is in the open interval (0,inf) i.e. not containing 0
Wonderful question as well as wonderful answer
From India
If a=0, b or c must be =0, too and there is no maximum for b+c.
Part 2: it seems like a=1, b=1 and c=1 works :p
such an underwhelming solution
Wishes from India 🇮🇳🇮🇳
2022 ayt matematik 8. soru
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.
From Cauchy-Buniakovski-Schwarz's inequality we have : (b+c)^2
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.
TUM SAALA HO GAYA MICHAEL PENN!!! 😡😡😡😡😡😡😡😡😡😡😡😡😡
Am I the only one that wanted to do this with a surface integral? Lol
Whats "Slovakian"? Adjective of "Slovakia" is "Slovak", right? "Slovakian" sounds odd as hell....
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.).
You can just put in a=1, b=1, c=1 and be done. Obviously what you have given is more and useful information, though.
Good