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?

Please reshare this note so that it can reach to the experienced and other brilliantians who will help us :)

## Comments

Sort by:

TopNewestQ7) \[a+b+c=0 \\ a+b = -c \\ a^2 + b^2 + (-a-b)^2 = 1 \\ a^2 + ab +b^2 = \dfrac{1}{2} \\ a^2 + ab + \left(b^2 - \dfrac{1}{2} \right) = 0\]

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 · 1 year, 4 months ago

Log in to reply

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 · 1 year, 4 months agoLog in to reply

– Nihar Mahajan · 1 year, 4 months ago

Thank you for help! :)Log in to reply

– Satyajit Ghosh · 1 year, 4 months ago

Sir can you please look into the fourth question since nobody is answering meLog in to reply

– Satyajit Ghosh · 1 year, 4 months ago

Thanks for the equality part too! Well can you tell me about the fourth question?Log in to reply

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 · 1 year, 4 months ago

Log in to reply

– Satyajit Ghosh · 1 year, 4 months ago

Thanks! Have a look at Q4Log in to reply

– Kushagra Sahni · 1 year, 4 months ago

How will you prove it?Log in to reply

here – Aditya Kumar · 1 year, 4 months ago

Note that the expression is equal to zeta(2). Zeta(2)=π^2/6. SeeLog in to reply

– Satyajit Ghosh · 1 year, 4 months ago

Thanks for the link. BTW did you solve other questions?Log in to reply

– Aditya Kumar · 1 year, 4 months ago

Q7 is direct use of am>=gmLog in to reply

– Satyajit Ghosh · 1 year, 4 months ago

Can you please show your solution because I tried and could not find the answer.Log in to reply

– Bala Vidyadharan · 1 year, 4 months ago

Please could u just post the solution ,coz i tried and couldn't find the answerLog in to reply

– Satyajit Ghosh · 1 year, 4 months ago

Multiple answers have been posted for this question. Please read the comments completely.Log in to reply

– Akhil Bansal · 1 year, 4 months ago

There's no need to find the exact value of given summation..Log in to reply

– Kushagra Sahni · 1 year, 4 months ago

I mean how will you prove that it is equal to π^2/6Log in to reply

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 · 1 year, 4 months ago

Log in to reply

– Satyajit Ghosh · 1 year, 4 months ago

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.Log in to reply

– Bala Vidyadharan · 1 year, 4 months ago

Thanks for replyingLog in to reply

– Saran .P.S · 1 year, 4 months ago

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 ....Log in to reply

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 · 1 year, 4 months ago

Log in to reply

@Calvin Lin can you help me? – Satyajit Ghosh · 1 year, 4 months ago

SirLog in to reply

– Satyajit Ghosh · 1 year, 4 months ago

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 answerLog in to reply

Just in a few hours ill appear 2016 nmtc finals!! – Ayush Pattnayak · 5 months, 1 week ago

Log in to reply

Did anyone get q6?? – Ayush Pattnayak · 5 months, 1 week ago

Log in to reply

i hav sol for Q1)b)... – Saran .P.S · 1 year, 4 months ago

Log in to reply

– Bala Vidyadharan · 1 year, 4 months ago

Then please post itLog in to reply

– Saran .P.S · 1 year, 4 months ago

i don't know how to post sryLog in to reply

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 · 1 year, 4 months ago

Log in to reply

Did anyone solved

Q1?? – Naitik Sanghavi · 1 year, 4 months agoLog in to reply

– Kushagra Sahni · 1 year ago

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.Log in to reply

pigeonhole principle. – Calvin Lin Staff · 1 year ago

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 theLog in to reply

– Kushagra Sahni · 1 year ago

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?Log in to reply

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 · 1 year, 4 months ago

Log in to reply

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

2(b): 1:

Ques 3: 47 . – Rakshit Joshi · 1 year, 4 months ago

Log in to reply

– Kushagra Sahni · 1 year, 4 months ago

How did you solve 2(a)?Log in to reply

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 · 1 year, 4 months ago

Log in to reply

Log in to reply

amtionline.com . – Satyajit Ghosh · 1 year, 4 months ago

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 siteLog in to reply

can someone please tell me how to insert image? – Rakshit Joshi · 1 year, 4 months ago

Log in to reply

– Satyajit Ghosh · 1 year, 4 months ago

Can you write it in LaTeX?Log in to reply

@Dev Sharma @Kushagra Sahni you may tag others too but please refrain from mass tagging:p – Satyajit Ghosh · 1 year, 4 months ago

Log in to reply

Log in to reply

– Satyajit Ghosh · 1 year, 4 months ago

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?Log in to reply

(

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 · 1 year, 4 months ago

Log in to reply

For Q7) I am getting \(a^2b^2c^2 \leq \dfrac{1}{36}\) :( – Nihar Mahajan · 1 year, 4 months ago

Log in to reply

– Satyajit Ghosh · 1 year, 4 months ago

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.Log in to reply

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 · 1 year, 4 months ago

Log in to reply

– Satyajit Ghosh · 1 year, 4 months ago

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!Log in to reply

Hint.Form a cubic polynomial. – Dev Sharma · 1 year, 4 months ago

Log in to reply

– Satyajit Ghosh · 1 year, 4 months ago

Do you mean Qm-Gm?Log in to reply

– Dev Sharma · 1 year, 4 months ago

see my proofLog in to reply

– Nihar Mahajan · 1 year, 4 months ago

I used Cauchy Schwarz ...Log in to reply

– Satyajit Ghosh · 1 year, 4 months ago

But isn't that applicable only for positive numbers?Log in to reply

– Nihar Mahajan · 1 year, 4 months ago

It is for real numbers.Log in to reply

– Satyajit Ghosh · 1 year, 4 months ago

So you got 1/54 using Cauchy? Please show your answerLog in to reply

– Nihar Mahajan · 1 year, 4 months ago

I got 1/36 using Cauchy ... I am not getting 1/54Log in to reply

– Naitik Sanghavi · 1 year, 4 months ago

See my solution!!One thing is ,I couldn't figure out when does this inequality hold?Log in to reply

I did Q1) b) using Pythagoras and similarity :P – Nihar Mahajan · 1 year, 4 months ago

Log in to reply

– Satyajit Ghosh · 1 year, 4 months ago

Can you tell me about the fourth question and please tell your approachLog in to reply

Problem. 7. Is easy.

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

by am gm ab<1/6

now its easy – Dev Sharma · 1 year, 4 months ago

Log in to reply

– Mietantei Conan · 1 year, 4 months ago

Am gm cannot be applied as a,b,c are real numbers not necessarily positive.Log in to reply

form a cubic polynomial and find its discriminant – Dev Sharma · 1 year, 4 months ago

Log in to reply

– Satyajit Ghosh · 1 year, 4 months ago

Thats why I was saying to show your AM-GM answer!They all are not +veLog in to reply

– Akhil Bansal · 1 year, 4 months ago

Correct, AM-GM concept is applicable only for positive numbersLog in to reply

– Kushagra Sahni · 1 year, 4 months ago

ExactlyLog in to reply

– Aditya Kumar · 1 year, 4 months ago

From where did u get a^2 + b^2 + ab = 1/2?Log in to reply

– Dev Sharma · 1 year, 4 months ago

use the fact a + b = -cLog in to reply

– Aditya Kumar · 1 year, 4 months ago

Hmm I c. I had ignored the fact that a+b+c=0Log in to reply

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 · 1 year, 4 months ago

Log in to reply

@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 · 1 year, 4 months ago

As pointed out byLog in to reply

– Akhil Bansal · 1 year, 4 months ago

Tomorrow is my exam and my coaching class too... I will solve them on day after tomorrow..Thanks for sharingLog in to reply

– Satyajit Ghosh · 1 year, 4 months ago

Thanks a lot too!Log in to reply

2)a) i am getting a = b = c = 5 – Dev Sharma · 1 year, 4 months ago

Log in to reply

– Kushagra Sahni · 1 year, 4 months ago

a,b,c are distinctLog in to reply

Let me know your approach. – Dev Sharma · 1 year, 4 months ago

Log in to reply

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 · 1 year, 4 months ago

Log in to reply

– Aditya Kumar · 1 year, 4 months ago

The inequality works for positive numbers. But due to the given condition the inequality must be edited accordingly.Log in to reply

– Nihar Mahajan · 1 year, 4 months ago

It works for all real numbers. I didn't get your second sentence.Log in to reply

– Aditya Kumar · 1 year, 4 months ago

try using c=-(a+b) in the main equationLog in to reply

– Satyajit Ghosh · 1 year, 4 months ago

Even my answer was like yours and I'm unable to find flaws it.Log in to reply

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 · 1 year, 4 months ago

Log in to reply

– Satyajit Ghosh · 1 year, 4 months ago

Ok now I get you. Since my answer came different, I thought I was wrong and didn't check when the equality holds trueLog in to reply

– Aditya Kumar · 1 year, 4 months ago

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.Log in to reply