International Mathematical Olympiad 1962 Problem 2
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- Опубліковано 14 чер 2021
- #Math #IMO #Algebra
In this video we are going to solve Problem 2 in IMO 1962..
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I share Maths problems and Maths topics from well-known contests, exams and also from viewers around the world. Apart from sharing solutions to these problems, I also share my intuitions and first thoughts when I tried to solve these problems.
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A quicker solution:
The √ function takes in non-negative arguments, so -1≤x≤3
The function f(x)=√(3-x)-√(1+x) is strictly decreasing for x∈[-1,3], so by IVT there exists a unique a such that f(a)=1/2
It can be shown that f(1-√(31)/8)=1/2, therefore the range of solutions is [-1,1-√(31)/8)
same
You’re make my day ! This is a good videos for me .keep it up sir
Grammar 100
can you make video about explaining inequalities in arbitrary function in which can you explain tangent line trick, Jensen/karmata and n-1EV trick please... its seems difficult i m just mugged it up and using now but i want some feel of these so can you explain
thanks for the video though...
I drew y= (3-x)^0.5 and the other function, y=(x+1)^0.5 + 0.5 We expect one intersection negative-point. [ One is increasing and the other is decreasing ]. It should be basic to draw such functions.
Easier is to graph root(3-x) and root(1+x) by hand to observe [_1,3] is domain. Then set root(3-x) - root(1+x) =1/2 to get right endpoint of [_1,0.3)
That's the only problem on your channel that I was able to solve...
1°.- Sqrt(3-x)-sqrt(1+x)>1/2
2°.- Sqrt(3-x)>1/2+-sqrt(1+x)
3°.- (sqrt(3-x)^2>(1/2+sqrt(1+x))^2
4°.- 3-x>1/4+sqrt(1+x)+1+x
5°.- 12/4+-1/4>sqrt(1+x)+2x
6°.- 49/16>(sqrt(1+x)+2x)^2
7°.- 49/16>1+x+4x^2+4x*sqrt(1+x)
Si dices que sqrt(1+x) = t, y 5°=7° sale que t es igual a un cierto valor de grado distinto a 1/2, despejas y resuelves
Substituting v=1-x makes life much easier while solving this one.
U indian??
You should have shown that inequality at the very beginning when you squared both sides. Because, that is where it arises from.
When you squared both sides extra solutions from √(1+x)-√(3-x)>1/2
came into it.
This aspect comes more clearly in an Olympiad question which video I shall put in my channel in a day (how extraneous solutions creep in(). Best wishes
You are entitled to eat your own opinions. :?
Sqrt(3-x)-sqrt(1+x) > 1/2
Sqrt(3-x) > 1/2 +sqrt(1+x) same way as:
Sqrt(A) > p+ sqrt(B)
A > p^2 +B +2psqrt(B)
A-B-p^2 > 2psqrt(B)
{ (A-B-p^2) ^2 > 4Bp^2 ●
A-B-p^2 >= 0 and B>=0 ■ }
● gives x < x1= [8 - sqrt(31)] /8
or x > x2= [8+sqrt(31)] /8
■ gives -1
@@touhami3472
Sorry you are wrong at one point. Sqrt(A) > p + sqrt(B) does not
A> p^2 + B + 2*p*sqrt(B)
This is completely true only when modulus of LHS> modulus of RHS.
If p was very negative this double implication would falter.
For ex:-
√2> -10 + √2
But when you square both sides
2> 100+2-20√2
Which is clearly not true.
@@oskarjung6738
you're right, but here p is Positive reel ( stricly).
I thought to precise it but it seems evident because if p= -1/2, i would write :
Sqrt(3-x)- sqrt(x+1) > -1/2
sqrt(x+1) < 1/2 + sqrt(3-x) wich is in same general case:
Sqrt(A) < p + sqrt(B) where the both sides are POSITIVES : necessery condition before squaring.
Thanks.
I really enjoy your videos, and probably someone already pointed this out. Yet, here we are.. in my opinion the condition 3-x > x+1 should have been enforced the first time you squared both sides. Indeed, the function f(x)=x^2 is injective only on subintervals of either (-infty, 0] or [0, +infty). Thus, being the left hand side (1/4) positive, in order to square both sides and obtain a equivalent inequality you must have sqrt(3-x) - sqrt(1+x), which is equivalent to 3-x > 1+x (of course, subject to the existence conditions 3-x>=0 and 1+x >=0).
Still, loving your videos: keep up the good work! 😀
Nice..
I love it🎶🎵🌞
I used a different method: √(3-x) - √(1+x) > 1/2 can be written as √(3-x)> 1/2 + √(x+1), when adding (x+1) to both sides. There are 2 cases for 1/2 + √(x+1). When 1/2 + √(x+1)
There was a typo. I meant to say √(x+1).
try to SUB x =0, 1 , 100 etc, then find out 1 positive and 1 negative value and compare the trend to find out the answer.
Isn't this question different from the actual one that appeared in the exam?
This same kind of question was asked in indian high school level exam,
Known as NTSE STAGE 1, and i have qualified that exam.
This question does not seems to be of IMO Level, i am confused 🤔
It is also asked in Viet Nam high school exam too with even more difficult level.
@@buithihai817 yeah, this is a very basic question.
this is one of the first few IMOs, so thats why
I like
make it easier by multiplying by 64? no! easier by completing the square. xx-2x +1> 31/64, (x-1)^2>31/64
gives x> 1+ sqrt(31)/8 or x< 1-sqrt(31)/8, Then eliminate spurious soln. and apply original sqrt contraints.
What about letting u = sqrt(3-x), s.t. u^2=3-x, x=3-u^2, x+1=4-u^2, sqrt(x+1)=sqrt(4-u^2) and now let u=2cos(t) s.t. sqrt(4-u^2)=2sin(t). Now solve cos(t) - sin(t) > 1/4
Finally a cool solution!
You'd need arccos in this case though. So gotta have a calculator ready.
@@anonymous_4276 not necessarily though, if you solve for cos(t). Let sin(t)=sqrt(1-cos^2(t)). The quadratic will give one valid answer for cos(t) and since u=2cos(t) no inverse function is needed.
@@ytsimontng but then there's no point in making the trigonometric substitution.
@@anonymous_4276 True
So ez for me with graphic
Easy peasy for jee aspirants
Разве это сложно?
02:40 Really???
All you had to do is move the fraction to the RHS and complete the square!!! 🤦♀️🤦♂️
I did the same lol
Can you explain Putnam 1996 problem A6 ?
No thanks?
@@particleonazock2246 What do you mean ?
It was so easy for owr years…
F(x)-1/2>0
Can anyone kindly explain why x is smaller than the smaller root or larger than the larger root
Because √3-x mist be larger than √x+1 if that doesn't happen will have something negative that's great than 1/2 which is not true. We need a set of solutions for which √3-x is going to outake √x+1 so the left side of the equation is positive and greater than 1/2.
This answer seems to be incomplete. I think you should also consider the domain of the x, which is -1
-1
We can use delta 🔼
You are from morroco 😅😅😅🤣
@@youssef9795 yees 😂
😂 😂 😂 😂
@@youssef9795 hit glte delta 😂😂😂 z3ma hdrt b fr 😂😂😂
@@youssefch4712 anglais
Its gets more easier if you find allowed values for x which is = [-1,3]
And then solve by inequality.
x must be in [-1; 7/8]
If x in }-1;3] the intervalle (1+sqrt(31)/8; 3] would be solution too.
Sqrt(3-x)-sqrt(1+x) > 1/2
Sqrt(3-x) > 1/2 +sqrt(1+x) same way as:
Sqrt(A) > p+ sqrt(B)
A > p^2 +B +2psqrt(B)
A-B-p^2 > 2psqrt(B)
{ (A-B-p^2) ^2 > 4Bp^2 ●
A-B-p^2 >= 0 and B>=0 ■ }
● gives x < x1= [8 - sqrt(31)] /8
or x > x2= [8+sqrt(31)] /8
■ gives -1
@@touhami3472According to rule of square root we cannot put the values of x which gives negative result.
Hence we can put all values of x where it gives either positive result or zero thats why
x belongs to [-1,3]
if still you have any doubt you can check by putting values of x in between [-1,3] 🙃
@@anoopmishra2695
The inequality gives two sets of solutions s1=[-1, 1-sqrt(31)/8) and s2=(1+sqrt(31)/8, 3]
Notice that both sets are include in [-1,3].
So why s1 works but not s2?
@@touhami3472 because if you try any number in s2,the inequation does not exict,because all s2 numbers will be 1/2
@@user-gg1ds1bq5n
The solutions must.be.in [-1, 7/8] , no in [-1, 3]. ( see proof above).
So all x in [-1, x1=1-sqrt(31)/8 ] work because [-1, x1] is INCLUDED IN [-1,7/8].
But why [x2=1+sqrt(31)/8 , 3] who is included in [-1,3] does not work ?
Answer: because it is not included in [-1,7/8] , the good intervalle .
Here's link to a video of fractional calculus:
ua-cam.com/video/yI6GAWcrKfY/v-deo.html
Rahim
How are you.
@@rahimfeni6024 Binod
0?
I don't understand as to why was this an IMO problem: it has nothing tricky or elegant about it, just any 7th grader should be able to solve it ...
I agree with you!
Dude, IMO has got harder for the past decades, in 1960 things were easier than today. Do you think IMO level questions are this easy nowdays? The level has risen
x=2 is not proof .
Thanks for calculus.
Its too easy
sir i think there is a mistake .. u cannot suare the lhs of the inequality coz we dont have adequate info about its sign on lhs .so what i did was to take it to rhs and then square the entire expression
We know it's > 1/2, and thus it has to be positive.
The mistake is in first step:
Before squaring, sqrt(3-x)- sqrt(x+1)>=0 ===> 3-x>= x+1>=0 .
Then x must be in [-1; 1]
Under this restriction, thd set of solutions is only [-1, 1-sqrt(31)/8).
Another way:
Sqrt(3-x)-sqrt(1+x) > 1/2
Sqrt(3-x) > 1/2 +sqrt(1+x) same way as: p>0,
Sqrt(A) > p+ sqrt(B)
A > p^2 +B +2psqrt(B)
A-B-p^2 > 2psqrt(B)
{ (A-B-p^2) ^2 > 4Bp^2 ●
A-B-p^2 >= 0 and B>=0 ■ }
● gives x < x1= [8 - sqrt(31)] /8
or x > x2= [8+sqrt(31)] /8
■ gives -1
@@oenrn we know it is >1/2 but it DEPENDS of x.
So, sqrt(3-x)-sqrt(x+1)>=0 ==> x-3>=x+1>=0 ==> -1
@@touhami3472 LHS = Left Hand Side ✋👍😊
@@aprilalbert4589 Thank you very much.
Bruh
You could have shortened substantially. Please re consider the solution
Here's my method: √(3-x) - √(1+x) > 1/2 can be written as √(3-x)> 1/2 + √(x+1), when adding (x+1) to both sides. There are 2 cases for 1/2 + √(x+1). When 1/2 + √(x+1)
Shorter solutions are not necessarily better. Longer solutions can be easier to apply to other problems so more instructive, and you’re often less likely to go wrong under competition conditions (rather than trying to be overly elegant).
@@theevilmathematician ""which means x< 1-. (√ 31 )/8" !! How did you straightaway conclude in one step after adding ""x+1 "" as you said ? Even if you meant adding ""√(x+1)'" how is conclusion so obvious ? Please explain.
@@AlephThree Another reasoning could mean shorter solution !! Or simpler !! Why inequalities, quadratic etc etc ? 2 solutions will be put in a video by my name in few hours.
@@Dharmarajan-ct5ld I did mean √(x+1). It is because when you arrive at √(3-x)> 1/2 + √(x+1), there is one identity I used. If √f(x)> g(x), then g(x)
This one wasn’t hard
Asian statement
It was 60 years ago dude.
@@failsmichael2542
I know
@@dfdxdfdydfdz Still, international math olympiad should not have been that easy, even 60 years ago, the difficlulty is supposed to be around the same since it's not about high level math but about difficult math (with arounf high school level question).
This is an early IMO test in the 1960s, just a few years after the contest was formed in 1959. Earlier questions might be easy now, but back then, they were hard. IMO Questions now might be very hard, but probably within years, it would be very easy.
A SIMPLE SOLUTION shall be given in a day in my video quoting logic wherein all these complexity can be omitted. Kindly consider. We need to be like JAPANESE"to shorten, miniaturise".
This is a quick solution: √(3-x) - √(1+x) > 1/2 can be written as √(3-x)> 1/2 + √(x+1), when adding (x+1) to both sides. There are 2 cases for 1/2 + √(x+1). When 1/2 + √(x+1)
@@theevilmathematician 1-sqrt(31)/8 excluded !
@@touhami3472 i said that answer. -√(31)/8+1 = 1- √(31)/8. the case was not excluded.
@@theevilmathematician yes i see. You're right .
Sorry
@@theevilmathematician
Here a complete solution using same your begining statement:
Sqrt(3-x)-sqrt(1+x) > 1/2
Sqrt(3-x) > 1/2 +sqrt(1+x) same way as:
Sqrt(A) > p+ sqrt(B)
A > p^2 +B +2psqrt(B)
A-B-p^2 > 2psqrt(B)
{ (A-B-p^2) ^2 > 4Bp^2 ●
A-B-p^2 >= 0 and B>=0 ■ }
● gives x < x1= [8 - sqrt(31)] /8
or x > x2= [8+sqrt(31)] /8
■ gives -1
Today one (or few) more solutions, by different tools will be put by video in my channel (displayed name). Kindly consider. Best wishes
This INEQUATION has form of:
Sqrt(A)-sqrt(B) > 1/2 (1)
From which one, you get:
Sqrt(A)+ sqrt(B) < 4(1-x) (2)
After squaring (1) and (2), you get:
A+B -2sqrt(A)*sqrt(B) > 1/4 ●
A+B +2sqrt(A)*sqrt(B) < 16(1-x)^2 ■
ADDING BOTH sides, you get:
2(A+B) = 1/4+16(1-x)^2 !!!
So, why did you use '=' instead '>' or '' used in ●
'
Má sao dễ thế
Alternate solution video uploaded just now. Reference my comments yesterday. Kindly consider. Best wishes. Different people think differently and this is what makes world beautiful.
Better solutions exist
.
Can you explain why some comments on your new video has been removed ?
Thank you very much pro fess or .
You say " Different peoole think differently and this what makes world beautiful"
Very nice, but why did you remove some comments in your video ?
A few days ago, they are 26 and today 9 !!!!!!!
it not good
Poor explanation