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NMTC 2015 final-Junior level

These are the questions of the NMTC final stage.

Hope you people will help me and others by showing how to solve these questions. Different and innovative ways of solving the questions are appreciated:)

\(Q1\) a) 28 integers are chosen from the interval [104, 208]. Show that there exists 2 of them having a common prime divisor.

b) AB is a line segment. C is a point on AB. ACPQ and CBRS are squares drawn on the same side of AB. Prove the S is the orthocentre of the triangle APB.

\(Q2\) a) a,b,c are distinct real numbers such that \({ a }^{ 3 }=3({ b }^{ 2 }+{ c }^{ 2 })-25,\quad { b }^{ 3 }=3(c^{ 2 }+a^{ 2 })-25,\quad { c }^{ 3 }=3({ a }^{ 2 }+{ b }^{ 2 })-25\). Find the numerical value of abc.

b) \(a=1+\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ 3^{ 2 } } +\frac { 1 }{ { 4 }^{ 2 } } +..........+\frac { 1 }{ { 2015 }^{ 2 } }\)

find [a], where [a] denotes the integer part of a.

\(Q3\) The arithmetic mean of a number of pair wise distinct prime numbers is 27. Determine the biggest prime among them.

\(Q4\) 65 bugs are placed at different squares of a 9*9 square board. Abug in each moves to a horizontal or vertical adjacent square. No bug makes 2 horizontal or vertical moves in succession. Show that after some moves, there will be at least 2 bugs in the same square.

\(Q5\) f(x) is a fifth degree polynomial. It is given that f(x)+1 is divisible by \((x-1)^{ 3 }\) and f(x)-1 is divisible by \((x+1)^{ 3 }\) . Find f(x).

\(Q6\) ABC and DBC are 2 equilateral triangles on the same base BC. A point P is taken on the circle with centre D, radius BD. Show that PA, PB, PC are the sides of a right triangle.

\(Q7\) a,b,c are real numbers such that a+b+c=0 and \({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }=1\). Prove that \({ a }^{ 2 }{ b }^{ 2 }c^{ 2 }\le \frac { 1 }{ 54 } \). When does the inequality hold?

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Note by Satyajit Ghosh
10 months, 4 weeks ago

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Q7) \[a+b+c=0 \\ a+b = -c \\ a^2 + b^2 + (-a-b)^2 = 1 \\ a^2 + ab +b^2 = \dfrac{1}{2} \\ a^2 + ab + \left(b^2 - \dfrac{1}{2} \right) = 0\]

Similarly we get that \(c^2 + cb + \left(b^2 - \dfrac{1}{2} \right) = 0\) which implies that \(a\) and \(c\) are the roots of the equation \(x^2 +bx + \left(b^2 - \dfrac{1}{2} \right) = 0\). But \(a\), \(c\) are reals. So, its discriminant is greater than or equal to \(0\), which infers that \(b^2 \leq \dfrac{2}{3}\)

So, product of the roots i.e. \(ac = b^2 - \dfrac{1}{2}\).

So, \(a^2 b^2 c^2 = b^2 \times \left( b^2 - \dfrac{1}{2} \right)^2 \). But \(b^2 \leq \dfrac{2}{3}\). It implies that

\[\begin{align*} a^2 b^2 c^2 &= b^2 \times \left( b^2 - \dfrac{1}{2} \right)^2 \\ &\leq \dfrac{2}{3} \times \left(\dfrac{2}{3} - \dfrac{1}{2} \right)^2 \\ &= \dfrac{2}{3} \times \dfrac{1}{36} \\ &= \dfrac{1}{54} \end{align*} \]


It is observed that equality holds when \(b^2 = \dfrac{2}{3} \) i.e. \(b= \pm \dfrac{2}{\sqrt{6}}\). At the same time when \(b^2 = \dfrac{2}{3} \) which implies that discriminant \(= 0\) which infers that roots are equal i.e. \(a= c\). So the equality holds when \((a, b, c) = \left(\mp \dfrac{1}{\sqrt{6}} , \pm \dfrac{2}{\sqrt{6}}, \mp \dfrac{1}{\sqrt{6}} \right)\). Surya Prakash · 10 months, 3 weeks ago

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@Surya Prakash [This is a type of inequality question that I like, where a different perspective can help you approach the problem in a simpler way. I do feel that this approach is under appreciated, esp for cases where the equality case consists of 2 variables (out of 3) being equal.]

If you use the Cubic Discriminant, you can simplify the proof into a "one-liner".

Essentially, we have \( a + b + c = 0 , ab + bc + ca = - \frac{1}{2}, abc = S \). Then, \( a, b, c \) are the real roots to the cubic \( x^3 - 0 x^2 - \frac{1}{2} x - S = 0 \), which tells us that the cubic discriminant is \( \geq 0 \). This gives us:

\[ \Delta = 0.5 - 27 S^2 \]

And hence \( (abc)^2 \leq \frac{1}{54} \).

Equality holds when 2 of the roots are equal. Inequality holds otherwise (and depending on how much you think they want, you should hunt down the equality case, but I'm lazy.)


Looking at the graph of \( y = x^3 - \frac{1}{2}x\), this idea is made even clearer. We are told that there are 3 real roots, and want to determine the largest value of \( (abc)^2\), which is the square of the constant term, which determines how high or low the graph is shifted. Obviously, this occurs at the local max/min, and we can use calculus to determine this value (if we didn't know about the cubic discriminant!

Calvin Lin Staff · 10 months, 3 weeks ago

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@Calvin Lin Thank you for help! :) Nihar Mahajan · 10 months, 3 weeks ago

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@Calvin Lin Sir can you please look into the fourth question since nobody is answering me Satyajit Ghosh · 10 months, 3 weeks ago

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@Surya Prakash Thanks for the equality part too! Well can you tell me about the fourth question? Satyajit Ghosh · 10 months, 3 weeks ago

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Q-2 b) It's a well known series,\( \displaystyle \sum_{x=1}^{\infty} \dfrac{1}{x^2} = \dfrac{\pi^2}{6} \sim 1.64\).
Therefore, a = \(\displaystyle \sum_{x=1}^{2015} \dfrac{1}{x^2}\) will be definately larger than 1 and less than 1.64.
Hence, \([ a ] = 1\) Akhil Bansal · 10 months, 4 weeks ago

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@Akhil Bansal Thanks! Have a look at Q4 Satyajit Ghosh · 10 months, 4 weeks ago

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@Akhil Bansal How will you prove it? Kushagra Sahni · 10 months, 4 weeks ago

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@Kushagra Sahni Note that the expression is equal to zeta(2). Zeta(2)=π^2/6. See here Aditya Kumar · 10 months, 4 weeks ago

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@Aditya Kumar Thanks for the link. BTW did you solve other questions? Satyajit Ghosh · 10 months, 4 weeks ago

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@Satyajit Ghosh Q7 is direct use of am>=gm Aditya Kumar · 10 months, 4 weeks ago

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@Aditya Kumar Can you please show your solution because I tried and could not find the answer. Satyajit Ghosh · 10 months, 4 weeks ago

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@Aditya Kumar Please could u just post the solution ,coz i tried and couldn't find the answer Bala Vidyadharan · 10 months, 2 weeks ago

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@Bala Vidyadharan Multiple answers have been posted for this question. Please read the comments completely. Satyajit Ghosh · 10 months, 1 week ago

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@Kushagra Sahni There's no need to find the exact value of given summation.. Akhil Bansal · 10 months, 4 weeks ago

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@Akhil Bansal I mean how will you prove that it is equal to π^2/6 Kushagra Sahni · 10 months, 4 weeks ago

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Here's my approach .If u find any flaws in my proof please let me know.

Given:

a+b+c=0,\({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }\)=1

apply A.M G.M inequality for a,b,c

\(\frac { a+b+c }{ 3 } \ge \sqrt [ 3 ]{ abc }\)

\(\frac { 0 }{ 3 } \ge \sqrt [ 3 ]{ abc }\)

\(0\ge abc\Longrightarrow eq.1\\\)

\(apply\quad A.M\quad G.M\quad inequality\quad for\quad { a }^{ 2 },{ b }^{ 2 },{ c }^{ 2 }\)

\(\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 3 } \ge \sqrt [ 3 ]{ { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 } }\)

\(\frac { 1 }{ 3 } \ge \sqrt [ 3 ]{ { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 } }\)

\(\frac { 1 }{ 27 } \ge { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }\)

\(\frac { 1 }{ 27abc } \ge abc\Longrightarrow eq.2\)

\(eq.1+eq.2\Longrightarrow \frac { 1 }{ 27abc } \ge 2abc\)

\(\frac { 1 }{ 54 } \ge { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }\) Bala Vidyadharan · 10 months, 1 week ago

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@Bala Vidyadharan AM-GM is applicable for positive reals only. Although the answer may be correct but you have to first check that they are positive. Even dev sharma had said the same thing but it was not applicable. Satyajit Ghosh · 10 months, 1 week ago

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@Satyajit Ghosh Thanks for replying Bala Vidyadharan · 10 months, 1 week ago

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@Bala Vidyadharan read the question properly they also asked that when does the equality exists ....so we must answer that this inequality exists if a.b.c are positive integers .... Saran .P.S · 10 months, 1 week ago

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Q4 as currently phrased is wrong. We can have \( 8 \times 9 = 72 \) bugs placed in the top 8 rows, and then they move down and up and down and up.

It might require "no bug makes 2 vertical moves in succession", or that "bugs alternate between horizontal and vertical moves". Calvin Lin Staff · 10 months, 3 weeks ago

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@Calvin Lin Sir @Calvin Lin can you help me? Satyajit Ghosh · 10 months, 3 weeks ago

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@Calvin Lin Sir I am really sorry. The question is exactly what you said. No bug makes 2 horizontal or 2 vertical moves in succession. My apologies. Since I am using my phone, I will edit it tomorrow. Can you tell us about your answer Satyajit Ghosh · 10 months, 3 weeks ago

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i hav sol for Q1)b)... Saran .P.S · 10 months, 3 weeks ago

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@Saran .P.S Then please post it Bala Vidyadharan · 10 months, 2 weeks ago

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@Bala Vidyadharan i don't know how to post sry Saran .P.S · 10 months, 2 weeks ago

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7) \(ab + bc + ca = -1/2\)

let a,b,c be roots of

\(x^3 -(1/2)x - abc = 0\)

now we know to find discriminant of this

\(27(abc)^{2} < 4.(1/2)^{3} = 1/2\)

we're done! Dev Sharma · 10 months, 3 weeks ago

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Did anyone solved Q1?? Naitik Sanghavi · 10 months, 4 weeks ago

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@Naitik Sanghavi I have a solution. We need to prove that in our set of 28 integers at least 2 will exist having a commom prime divisor. So we will create a set of integers from 104 to 208 and try to insert as many co prime numbers as we can. We have 19 primes from 104 to 208 so we have 19 elements already in the set. Next we need other elements which are co- prime to each other and not necessarily prime. We can include product of two primes. Like 2×103( or any other prime), 3×67, 5×37, 7×29, 11×11( as no other prime can come, we can't take 13 or 17 as it will not allow us to add other elements in our set which is a prime multiples of 13 and 17, and all primes below 11 have been used), then we can add 13×13. Now no multiple of 17 can be added as any of its multiple must be less than 208 so we can product it with 7,8,9,...,12 only but all of these numbers will have a common divisor with pre existing numbers. Similarly no other multiples of primes greater than 17 can be added as they will not be co prime to other elements in the set. So we get 19+6=25 elements in our set all of which are co prime to each other. 26th element will have a common prime divisor with atleast one of them. So when we choose 28 obviously there will be at least 2 having a common prime divisor. I think that proves. Kushagra Sahni · 6 months, 2 weeks ago

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@Kushagra Sahni You have the right ideas. However, they are not clearly presented and you're making a lot of assumptions. For example, you seem to be saying that we must take "the numbers \( 2 \times 103\), \( 3 \times 67 \), etc ". Instead, you should find a way to paraphrase it such that we easily allow for the "any other prime" aspect of it. There is a much cleaner way to present this argument. Think about how to use the pigeonhole principle. Calvin Lin Staff · 6 months, 2 weeks ago

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@Calvin Lin I agree with you Sir, but is it right that 26 is the minimum number of elements we need to choose in order to find at least 2 numbers having a common prime divisor? Kushagra Sahni · 6 months, 2 weeks ago

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@Naitik Sanghavi the ans is 104,208 only .for proving i don't know how to explain my thinking ..im sorry.....

but i'll try.. let the 24 integers be the four multiples from 104 to 208. by prime factorizing 104 u will get 2^3/times13.. and by prime factorizing 208 u will get2^4/times 13.so in these 2 nos ony we have a common factor 2^3/times13.so it is the solution.

but if u ask why don't any other no will have a common factor means.... the other nos wont why because the starting no is 104 i.e2^#/times13 so the common factor must be this one..so if u multiply 2^3/times13 with any other integer other than one is the minimum value is 208 only i.e 2^3/times13/times2................that is my solution. if can understan it is well .but if u don't it is upto you..... Saran .P.S · 10 months, 3 weeks ago

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ok i may write the answers : Ques 2(a): 2 ;

2(b): 1:

Ques 3: 47 . Rakshit Joshi · 10 months, 4 weeks ago

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@Rakshit Joshi How did you solve 2(a)? Kushagra Sahni · 10 months, 4 weeks ago

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@Kushagra Sahni Let \( a^2 + b^2 + c^2 = k \) where \( k \) is constant.

Now, write the first equation as \( a^3 + 3a^2 = 3(a^2 + b^2 + c^2) - 25 = 3k - 25 \)

\( \implies a^3 + 3a^2 + 25 - 3k= 0 \)

Likewise, we can write the same in "b" and "c".

Therefore, we see that a,b and c are the solutions to the equation \( x^3 + 3x^2 +25 -3k = 0 \)

Now, by Vieta's formulas, we can see \( \sum a^2 = ( \sum a)^2 - 2 \sum ab = (-3)^2 - 2*0 = 9 \)

Therefore \( abc = -(25 - 3k) = -(-2) = 2 \) Siddhartha Srivastava · 10 months, 4 weeks ago

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@Abhijeet Verma You can ask your education institution or if you go to some coaching classes. If they do not conduct NMTC then you can check their site amtionline.com . Satyajit Ghosh · 10 months, 4 weeks ago

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can someone please tell me how to insert image? Rakshit Joshi · 10 months, 4 weeks ago

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@Rakshit Joshi Can you write it in LaTeX? Satyajit Ghosh · 10 months, 4 weeks ago

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@Dev Sharma @Kushagra Sahni you may tag others too but please refrain from mass tagging:p Satyajit Ghosh · 10 months, 4 weeks ago

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@Naitik Sanghavi Ohh I'm really sorry! I will do it tomorrow as I am currently not on my pc. Btw did you also give the final? How many did you solve? Satyajit Ghosh · 10 months, 4 weeks ago

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(Q3)There are only four pairs of primes whose mean is 27.

(47,7)(11,43)(41,13)(19,37)

So,the biggest prime among them is 47!! Naitik Sanghavi · 10 months, 3 weeks ago

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For Q7) I am getting \(a^2b^2c^2 \leq \dfrac{1}{36}\) :( Nihar Mahajan · 10 months, 3 weeks ago

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@Nihar Mahajan That's what many people are getting I guess. But I have been told by others that there is no problem with the question and you have to use jensons inequality. Satyajit Ghosh · 10 months, 3 weeks ago

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@Satyajit Ghosh (a-b)²≥0

a²+b²+ab≥3ab...(1)

a+b+c=0

c=-a-b

Also,a²+b²+c²=1=a²+b²+(-a-b)²

2a²+2b²+2ab=1

a²+b²+ab=1/2...(2)

From (1) and (2) we have,

ab≤1/6.....(3)

a²b²c²=a²b²(a²+b²+2ab)

=(ab)²(1/2+ab)

From(3)

a²b²c²≤1/36(1/2+1/6)≤1/36×2/3

a²b²c²≤1/54

Hence proved.

Sorry,I couldn't use latex as I'm preparing for NTSE!! Naitik Sanghavi · 10 months, 3 weeks ago

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@Naitik Sanghavi That's a nice solution which was based around what Dev Sharma had earlier told us taking ab<_ 1/6. But the problem was using AM-GM inequality. Thanks! Satyajit Ghosh · 10 months, 3 weeks ago

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@Satyajit Ghosh There is a proof without using that too.

Hint.Form a cubic polynomial. Dev Sharma · 10 months, 3 weeks ago

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@Dev Sharma Do you mean Qm-Gm? Satyajit Ghosh · 10 months, 3 weeks ago

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@Satyajit Ghosh see my proof Dev Sharma · 10 months, 3 weeks ago

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@Satyajit Ghosh I used Cauchy Schwarz ... Nihar Mahajan · 10 months, 3 weeks ago

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@Nihar Mahajan But isn't that applicable only for positive numbers? Satyajit Ghosh · 10 months, 3 weeks ago

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@Satyajit Ghosh It is for real numbers. Nihar Mahajan · 10 months, 3 weeks ago

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@Nihar Mahajan So you got 1/54 using Cauchy? Please show your answer Satyajit Ghosh · 10 months, 3 weeks ago

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@Satyajit Ghosh I got 1/36 using Cauchy ... I am not getting 1/54 Nihar Mahajan · 10 months, 3 weeks ago

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@Nihar Mahajan See my solution!!One thing is ,I couldn't figure out when does this inequality hold? Naitik Sanghavi · 10 months, 3 weeks ago

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I did Q1) b) using Pythagoras and similarity :P Nihar Mahajan · 10 months, 3 weeks ago

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@Nihar Mahajan Can you tell me about the fourth question and please tell your approach Satyajit Ghosh · 10 months, 3 weeks ago

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Problem. 7. Is easy.

a^2 + b^2 + ab = 1/2

by am gm ab<1/6

now its easy Dev Sharma · 10 months, 4 weeks ago

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@Dev Sharma Am gm cannot be applied as a,b,c are real numbers not necessarily positive. Mietantei Conan · 10 months, 4 weeks ago

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@Mietantei Conan there is one other idea too

form a cubic polynomial and find its discriminant Dev Sharma · 10 months, 3 weeks ago

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@Dev Sharma Thats why I was saying to show your AM-GM answer!They all are not +ve Satyajit Ghosh · 10 months, 3 weeks ago

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@Mietantei Conan Correct, AM-GM concept is applicable only for positive numbers Akhil Bansal · 10 months, 3 weeks ago

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@Mietantei Conan Exactly Kushagra Sahni · 10 months, 3 weeks ago

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@Dev Sharma From where did u get a^2 + b^2 + ab = 1/2? Aditya Kumar · 10 months, 4 weeks ago

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@Aditya Kumar use the fact a + b = -c Dev Sharma · 10 months, 4 weeks ago

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@Dev Sharma Hmm I c. I had ignored the fact that a+b+c=0 Aditya Kumar · 10 months, 4 weeks ago

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There is some problem in question 7.
It is given that \(a^2 + b^2 + c^2 = 1\) . We also have prove it..?
Question asks when does this inequality holds..but there is no inequality in question 7 Akhil Bansal · 10 months, 4 weeks ago

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@Akhil Bansal As pointed out by @naitik sanghavi , the question is what he has written. I will edit it tomorrow. I am currently using my phone . BTW how many did you solve? And please share your answers. Satyajit Ghosh · 10 months, 4 weeks ago

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@Satyajit Ghosh Tomorrow is my exam and my coaching class too... I will solve them on day after tomorrow..Thanks for sharing Akhil Bansal · 10 months, 4 weeks ago

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@Akhil Bansal Thanks a lot too! Satyajit Ghosh · 10 months, 4 weeks ago

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2)a) i am getting a = b = c = 5 Dev Sharma · 10 months, 3 weeks ago

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@Dev Sharma a,b,c are distinct Kushagra Sahni · 10 months, 3 weeks ago

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@Kushagra Sahni Oh, sorry!!

Let me know your approach. Dev Sharma · 10 months, 3 weeks ago

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Q7) Please find mistake in this:

Using Cauchy Schwarz inequality ,

\[(abc+abc+abc)^2 \leq (a^2b^2+b^2c^2+a^2c^2)(c^2+a^2+b^2) \\ \Rightarrow (3abc)^2 \leq (a^2b^2+b^2c^2+a^2c^2)\]

Setting \(x=ab \ , \ y=bc \ , z=ac\) , \(a^2b^2c^2 \leq \dfrac{x^2+y^2+z^2}{9} \dots (1)\)

Note that \(x+y+z=\dfrac{(a+b+c)^2-(a^2+b^2+c^2)}{2} = \dfrac{-1}{2}\)

\(xy+yz+xz=ab^2c+abc^2+a^2bc=abc(a+b+c)=0\)

Thus, \(x^2+y^2+z^2=(x+y+z)^2- 2 \times (xy+yz+xz) = \left(\dfrac{-1}{2}\right)^2 - 0 = \dfrac{1}{4}\)

Putting \(x^2+y^2+z^2=a^2b^2+b^2c^2+a^2c^2=\dfrac{1}{4}\) in \((1)\) , we get

\[a^2b^2c^2 \leq \boxed{\dfrac{1}{36}}\] Nihar Mahajan · 10 months, 3 weeks ago

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@Nihar Mahajan The inequality works for positive numbers. But due to the given condition the inequality must be edited accordingly. Aditya Kumar · 10 months, 3 weeks ago

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@Aditya Kumar It works for all real numbers. I didn't get your second sentence. Nihar Mahajan · 10 months, 3 weeks ago

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@Nihar Mahajan try using c=-(a+b) in the main equation Aditya Kumar · 10 months, 3 weeks ago

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@Nihar Mahajan Even my answer was like yours and I'm unable to find flaws it. Satyajit Ghosh · 10 months, 3 weeks ago

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@Satyajit Ghosh Once again, with inequalities, all that you have shown is an upper bound. You need to show that it is the least upper bound, in order to be a maximium. What’s your equality case?

Otherwise, you have not proven that there is a contradiction, since \( \frac{1}{54} \leq \frac{1}{36} \), so it is still possible for \( (abc)^2 \leq \frac{1}{54} \), and you just haven't shown it yet. Calvin Lin Staff · 10 months, 3 weeks ago

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@Calvin Lin Ok now I get you. Since my answer came different, I thought I was wrong and didn't check when the equality holds true Satyajit Ghosh · 10 months, 3 weeks ago

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@Calvin Lin Sir, I feel this arguement is incorrect. What if it was asked to find the maximum value of \({(abc)}^{2}\)? 1/36 won't work. Aditya Kumar · 10 months, 3 weeks ago

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