106Q6. Let a, b, c be positive real numbers such that a3 + b3 = c3. Prove that a2 + b2 - c2 > 6(c-a)(c-b)
Q1. Let ABC be a triangle and let P be an interior point such that <BPC=90°, <BAP=<BCP. Let M,N be the midpoints of AC,BC respectively. Suppose BP=2PM. Prove that A,P,N are collinear
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106
3412nd ones an absolute sitter. an = 0 when the number is a perfect square. So a1a2... is aperiodic making the number irrational.
62Q3. Find all real numbers x such that
[x2 +2x] = [x]2 + 2[x]
[x2 +2x]+1 = [x]2 + 2[x]+1
[x2 +2x+1] = [x]2 + 2[x] + 1
[(x+1)2] = ([x]+1)2
[(x+1)2] = ([x+1])2
x+1=t
[t2]=[t]2
now is it solvable? (by graphs?)
106bhaiyya i am very poor at graphs.
What is the solution??
I got x = all integers and x belongs to [n, [√(n+1)2+1] -1) for all n belongs to integer and [-1,∞)
when i had solved in the exam
62i havent seen the -ve side of it properly.. but that is what i feel..
I dont know how you got the expression that you did..
Could you justify ur expression...
106Q2. Define a sequence <an>n=1 to ∞ as follows:
an = 0, if the number of positive divisors of n is odd
= 1, if the number of positive divisors of n is even.
(The positive divisors of n include 1 as well as n). Let x=0.a1a2a3... be the real number whose decimal expansion contains an in the n-th place, n>=1. Determine, with proof, whether x is rational or irrational.
341From rahul's post I gather that we have to prove that if a3+b3 = c3, then
a2+b2-c2> 6(c-a)(c-b)
WLOG, we can assume c=1
So now we have to prove that if a3+b3 = 1, then a2+b2-1 > 6(1-a)(1-b)
Now I substitute as follows p = a+b and q = ab
The inequality now becomes, given
p(p2-3q) = 1, prove that
p2+6p-7>8q
Now since a,b<1, we have a3<a and b3<a. So a+b>a3+b3 = 1. So, we have p>1
From Power Mean Inequality (a3+b3/2)1/3> (a+b/2).
So p<22/3
Now 3q = p - 1/p2
So the inequality becomes to prove that
3p (p+7)(p-1)>8(p3-1)
Since p>1, we can cancel on both sides to reduce it to proving that
3p(p+7)>8(p2+p+1) or 5p2-13p+8<0
This is true if p lies between the roots that is between 1 and 8/5
We have already proved that p>1. Now since p<22/3, we have to prove that 22/3<8/5
or 5 . 22/3<23
or 5 < 27/3 or
125<128 which is obviously true which proves the inequality. This also gives us a very good idea of how tight the given inequality is.
At least one nice problem was there in the INMO!
106nishant bhaiyya for Q3.
i used x = [x] + {x} and then substituted in both sides.
[x2 + 2x] = [ [x]2 + 2[x]{x} + {x}2 + 2[x] + 2{x}]
= [x2] + 2[x] + [2[x]{x} + {x}2 + 2{x}]
[x2 + 2x] = [x2] + 2[x] is given
So, [2[x]{x} + {x}2 + 2{x}] = 0
0<=[2[x]{x} + {x}2 + 2{x}] <1
Considering both inequalities, i got the expression given.
62salute ur genious Prophet... :)
Yup it is such a very "tight" inequality
Great work...
btw i have to ask this to you.. (these can be solved at a class xii level.. but how do u manage pure mathematical proofs!!! I sometimes have a feeling that you have graduated from mathematics or in some pure maths 1st/2nd year course propeht...)
341One point that has to be cleared is how i assumed c = 1
We were given a3+b3 = c3 and asked to prove that
a2+b2-c2>6(c-a)(c-b)
I can write equivalently
(a/c)3+(b/c)3 = 1 and have to prove that (a/c)2+(b/c)22 - 1 > 6(1-a/c)((1-b/c)
So I can substitute x = a/c and y = b/c and the inequality then is
x3+y3 = 1 and x2+y2-1 > 6(1-x)(1-y)
Instead of all this rigmarole, you can simply state WLOG c = 1 and make life simpler
341got to admit that i am actually an ex-iitian. In fact i am hsbhatt from goiit. But i am not any maths graduate. But i just got rehooked on math recently. I like solving probs
341i hope nishant bhaiya doesnt have any probs with me posting here?
62okie... gr8 man...
I wanted to ask this to you a couple of times before when you got a couple of pure maths proofs......
now it makes more sense :) [1]
I hope this is not conflict of interest for you... (goiit and targetiit)
I have seen a few users here who have fought about this and a user who did that at goiit ... i sometimes feel ashamed but that is not the point of starting this site :)
62and you dont have to call me bhaiya i guess i should start doing that... :)
62ashish i see a mistake in ur method.. at one step you have used
[x^2+2x]=[x^2]+[2x]
Or did i miss/mess up something?
341As long as we can help students i dont see any problem. I dont gain commercially any way. This is something I do apart from my regular job. I just want to be useful to those attempting JEE.
I wanted to be incognito coz it gave me some fun moments like taking on eureka123 who is my friend in goiit. Hope u dont mind bro (!!!)