A problem from Denmark 2006 (Georg Mohr)
HTML-код
- Опубликовано: 2 окт 2024
- This problem is quite easy considering the algebra and level of reasoning involved in the solution I gave. After finding one set of solution, It was required to show that there were no other solutions.
![Find (x+y+z) [Harvard-MIT] Guts contest](http://i.ytimg.com/vi/Pe0s5N_jkLA/mqdefault.jpg)
I'm getting smarter by watching this channel.
True. Same here
When I have reached x^2 - 2x + (1 + z^2) = 0
I simplified the equation to get a sum of two squares, this became:
(x-1)^2 + z^2 = 0
And since squares can’t be negative therefore (1,1,0) is the only solution.
Yes. That's how I did it too
Sum of two squared terms is zero implies each term is zero. If term is equated to zero, we get 1,1,0 as solution
I did the same, trying to focus on circumference equation on x z plane. Center is x=1, z=0, radius=0 so x=1 z=0 is the only point in the circumference.
I solved it also this way.
In a general case this kind of system of equations (three variables, but only two equations) can lead to a volume in xyz space satisfying the equations, but here we are asked to find "all" triplets indicating there aren't infinite of them. So the way the problem has been set up indicates the size of the xyz space satisfying the given equations is zero (only one point/triplet).
From the upper equation we have y = 2-x. Substituting this into the lower equation we get x(2-x) - z² = 1. Lets write that as -x² +2x - 1 = z². Let f(x) = -x² +2x - 1 = z². This is a parabola opening down in x on the left side and a parabola opening up on the right side. Obviously the minimum of z² = 0 since x, y and z are real. Setting f'(x) = -2x + 2 = 0 we have the maximum of left parabola happen at x = 1 which gives f(1) = 0. Since this is the same as the minimum of z², there is only one solution. So, x =1, y = 2-x = 1 and z = 0. We have the triplet (1, 1, 0) as the only solution.
There were some issues with the sound around 4 minutes. Maybe the mic just rubbed against your shirt?
At 4:13 you can turn the equation into (x - 1)² + z² = 0, which implies x = 1 and z = 0. Then just solve (1) with x = 1 and you will get y = 1, hence, (x, y, z) = (1, 1, 0). That's a more convenient way.
I love that our video are easy enough that I can always work out the problem myself first, but not so simple that it feels trivial.
As for my solution to this problem, it started out the same (excet I didn't show (1,1,0) was a solution until the end) up to the point where you show that x^2-2x+1+z^2=0. Originally I was going to use the quadratic formula like you did, but I noticed that it's just a sum of squares: (x-1)^2+z^2=0. Because a real number can't square to give a negative, both (x-1) and z have to be 0. The solution is then obvious from there.
your way to solve this is pretty much the same as mine, but longer
i figured out that x+y=2 and xy=z^2+1 are in the vieta formula. Meaning x and y are the roots of the quadratic equation: t^2 - 2t + z^2 + 1 = 0 (t-1)^2 = -z^2
Because (t-1)^2 >= 0 >= -z^2 thus t=1 and z=0, in other words: x=y=1 and z=0
That equation can be rewritten as (x-1)^2 + z^2 =0, so both squares should be 0, thus x =1 and z=0, and y=1.
Here’s my quick solution:
In order for 2nd equation to have a solution we need xy>=1 since z^2 will always be non-negative. We see from first equation x+y=2 so then y=2-x and xy=x(2-x)=2x-x^2>=1 and subtracting 1 from both sides yields -x^2+2x-1>=0 and flipping the signs on both side (flip inequality when multiplying by negative 1) has x^2-2x+1
More challenging would, I think, be the same question for the system:
x + y = 2
xy = z^2 -1
for which I found, it seems to me, as solutions the points located on two lines symmetrical with respect to the plane z = 0
y = 2 - x
x(2 - x) - z² = 1
z² = x(2-x) - 1
z² = -x² + 2x - 1
-z² = (x-1)²
x - 1 must be z times ±i
we can only use real numbers, so, z has to be 0
which means x-1 = 0, so x = 1
and y = 1
alternatively
z² = -(x - 1)²
z = i(x-1), so x-1=0 and we get the same results.
0:19 sounds like one of this problems when you explain the solution i say yes make acutualy sence but i wouln´t finde a solution by my self
1:13 oh it is difficulter as i thought i thought we talking about N instead of R
2:33 ok that step would be posible for me too making and moving an equantion
3:39 some other idears here: xy = Z²+1 > 0 and xy-1 = Z² so xy and Z have to be positiv and if we look at xy it only can be positiv when both are postive ore negative bat both cant be negativ becauce ther sum is greater than 0 so we just have to look at positiv numbers haven´t we?
4:15 ok and a=1; b=2; c=1+Z²; ok that is even not so difficult
4:44 wait that is not posible becauce we need 4(1+Z²) 4+Z² 1 = 2y-y²|-1 ==> 0 = -y²+2y-1 ==> x = {1(+/-) W(4-4[-1][-1])}/-2 = 1/-2 no there have to be a mistake
7:17 yes make sense
9:45 ok now we knowe it is the only one! i like this tasks but wehn i saw sth like this i always don´t know what to do first xD
10:14 thanks for the advice!
LG
K.Furry
I found the value of XY from equation 1 and then substitute it in equation 2, that gives equation similar to hyperboloid: x^2 + y^2 +2Z^2 = 6 - cut by a plane x+ y = 2.
No Other point than (1 1 0).
Georg Mohr is the guy from compass only constructions
His works was rediscovered idependently by Lorenzo Mascheroni
Solution
x + y = 2 > -y
xy - z² = 1 > +z² -1
x = 2 - y
xy - 1 = z²
(2 - y)y - 1 = z²
2y - y² - 1 = z²
-1 * (y² - 2y + 1) = z²
-1 * (y - 1)² = z² |√
z = √(-1 * (y - 1)²)
z = (y - 1)i
The only way, that z is real, is if (y - 1) = 0, and therefore z = 0.
Therefore we have:
y - 1 = 0 > y = 1
x = 2 - y > x = 1
z = 0
The ONLY triplet is {1, 1, 0}
I really love the problems you solve in your videos, but why everytime you open brackets or multiple by a minus 1 it should be a whole ceremony of 3-4 lines and a whole minute?
10 min lesson and looking for solution give me think Denmark 2006 math is real good.
Puisque (1,1,0) est une solution, les autres solutions sont de la forme (1-t,1+t,z) si elles existent.
Or xy= (1-t)(1+t)= 1-t^2 et la seconde équation devient 1- (t^2 + z^2)=1
D’où t^2+z^2 =0 et t=z=0
Conclusion : (1,1,0) est la seule solution
Thanks Sir
*_x + y = 2_* ...①
*_xy - z² = 1_* ... ②
Multiply ① by _y:_
_xy + y² = 2y_
Subtract ②:
_y² + z² = 2y - 1_
⇒ _(y - 1)² + z² = 0_
∴ *_y = 1, x = 1, z = 0_*
I don't have any exam or use of maths in my field. I am still watching this :)
4:12 Here you have sum of two squares so
only way we get zero is when x - 1 = 0 and z = 0
AM>=GM :)
Since x and y are positive.
The only possible solution is AM=GM
But x or y could have been negative...
@@Matroskra
xy - z² = 1
xy = 1 + z²
1 + z² can never be negative
Therefore xy is positive, which means
Either x and y are both negative, or both positive.
If both are negative, x + y = 2 is not possible.
Hence both are positive.
@@Matroskra In the second equation, xy=1+z², and since z² will be positive(or 0), xy is positive meaning either both are positive or both negative. If both were negative the first equation would be false because two negatives cant make +2
@@Phantom_Reaper_ You are absolutely right, thanks!
@@Phantom_Reaper_lol did we both just say the same thing at the same time in different ways
I solved this problem in different way. After ispection finding the particular soluzion 1,1,0 because x + y =2 if exists a different solution for x and y, it will be in the form (1+ a) + (1 -a) = 2. But xy - z^2 =1 so (1+a)(1-a) - z^2 = 1 implies 1 - a^2 - z^2 = 1, then - a^2 - z^2 = 0. if and only if a = 0, z=0
I kind of want to challenge myself and put a twist to this question.
... What if instead x + y = 2
... It's x + y = 4
This might be harder than it looks, so i will write this down and save for later.
(x - y) ^2 = -4z ^2
Know that this can only hold if both side equals 0
Z must be 0
Then just solve for x and y
As x equals y, they must both be 1
for any of u that does not know how the firsy equation come from, just square the first given equation then deduct 4 times the second one
Sir I ❤ your videos and teaching
Sir can you solve this cubic polynomial equation in any video x^(3) - (1/4)x^(2) + 0x - 1 =0
Sir plss solve this !!!!
for xy - z^2 = 1 to be solved, wouldnt xy have to be greater than or equal to 1 since a squared real number isnt negative? and since xy cant exceed 1 if x + y = 2, you can derive that there is only 1 solution.
There are tons of ways to approach this question.
Good one!
(1,1,0)
Sir How can I send you a problem ? 😊
Which problem you wanna send?
He reads his comments. Put on there and if he thinks it’s cool enough for a video, he’ll do it.
@@Ron_DeForest ok bro 😊
Ler's try my favorite substitution: x=u+v ' y=u-v
so
x+y=2u=2 => u=1
x*y-z^2=(u+v)*(u-v)-z^2=u^2-v^2-z*2=1
u=1 so 1-v^2-z^2=1
v^2+z^2=0
v=0 , z=0
So:
x=1+0=1
y=1-0=1
z=0