# 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?

Please reshare this note so that it can reach to the experienced and other brilliantians who will help us :) Note by Satyajit Ghosh
4 years, 1 month 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{aligned} 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{aligned}

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 $(a, b, c) = \left(\mp \dfrac{1}{\sqrt{6}} , \pm \dfrac{2}{\sqrt{6}}, \mp \dfrac{1}{\sqrt{6}} \right)$.

- 4 years, 1 month ago

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

- 4 years, 1 month ago

[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! Staff - 4 years, 1 month ago

Thank you for help! :)

- 4 years, 1 month ago

Sir can you please look into the fourth question since nobody is answering me

- 4 years, 1 month ago

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, $[ a ] = 1$

- 4 years, 1 month ago

Thanks! Have a look at Q4

- 4 years, 1 month ago

How will you prove it?

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Note that the expression is equal to zeta(2). Zeta(2)=π^2/6. See here

- 4 years, 1 month ago

Thanks for the link. BTW did you solve other questions?

- 4 years, 1 month ago

Q7 is direct use of am>=gm

- 4 years, 1 month ago

- 4 years, 1 month ago

Please could u just post the solution ,coz i tried and couldn't find the answer

- 4 years ago

- 4 years ago

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".

Staff - 4 years, 1 month 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 answer

- 4 years, 1 month ago

Sir @Calvin Lin can you help me?

- 4 years, 1 month ago

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

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

$\frac { 0 }{ 3 } \ge \sqrt [ 3 ]{ abc }$

$0\ge abc\Longrightarrow eq.1\\$

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

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

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

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

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

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

$\frac { 1 }{ 54 } \ge { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }$

- 4 years 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.

- 4 years ago

- 4 years 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 ....

- 4 years ago

@Dev Sharma @Kushagra Sahni you may tag others too but please refrain from mass tagging:p

- 4 years, 1 month ago

can someone please tell me how to insert image?

- 4 years, 1 month ago

Can you write it in LaTeX?

- 4 years, 1 month ago

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

2(b): 1:

Ques 3: 47 .

- 4 years, 1 month ago

The question was unclear. Whether we are looking for a sequence of primes or pairwise primes (coprimes). In both cases we can have larger values than 47. For example, the sequence 3, 5, 73. So keep trying!

- 2 years, 2 months ago

How did you solve 2(a)?

- 4 years, 1 month ago

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 $abc = -(25 - 3k) = -(-2) = 2$

- 4 years, 1 month ago

Did anyone solved Q1??

- 4 years, 1 month ago

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.....

- 4 years, 1 month 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.

- 3 years, 9 months 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 the pigeonhole principle.

Staff - 3 years, 9 months 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?

- 3 years, 9 months ago

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!

- 4 years, 1 month ago

i hav sol for Q1)b)...

- 4 years, 1 month ago

- 4 years ago

i don't know how to post sry

- 4 years ago

Did anyone get q6??

- 3 years, 1 month ago

Just in a few hours ill appear 2016 nmtc finals!!

- 3 years, 1 month ago

Q1.b) CR is parallel to AP extended. So altitude to AP is perpendicular to CR. Thus it passes through S (as it has to be the other diagonal). PC already goes through S. QED

Q2.a) Generate a symmetric function whose roots are a,b,c. Use Vieta's Formulae and proceed. abc=2

Q2.b) Use Basel Result of the sum of inverse squares (Zeta(2)). As this sum is smaller than it but bigger than 1, its integer part is equal to 1.

Q6. Use basic coordinate geometry and prove that (PA)^2 = (PB)^2 +(PC)^2.

Q7. Form a quadratic and make discriminant non negative to get the stronger version of the inequality.

Equality occurs when a= -+ 1/sqrt6 b= +- sqrt(2/3) c=a

Q1.a) , Q4 are based on the application of the Pigeon-Hole principle. Hope I helped.

- 2 years, 2 months ago

- 2 years, 2 months ago

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

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

Tomorrow is my exam and my coaching class too... I will solve them on day after tomorrow..Thanks for sharing

- 4 years, 1 month ago

Thanks a lot too!

- 4 years, 1 month ago

I did Q1) b) using Pythagoras and similarity :P

- 4 years, 1 month ago

- 4 years, 1 month ago

For Q7) I am getting $a^2b^2c^2 \leq \dfrac{1}{36}$ :(

- 4 years, 1 month 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.

- 4 years, 1 month ago

I used Cauchy Schwarz ...

- 4 years, 1 month ago

But isn't that applicable only for positive numbers?

- 4 years, 1 month ago

It is for real numbers.

- 4 years, 1 month ago

- 4 years, 1 month ago

I got 1/36 using Cauchy ... I am not getting 1/54

- 4 years, 1 month ago

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

- 4 years, 1 month ago

There is a proof without using that too.

Hint.Form a cubic polynomial.

- 4 years, 1 month ago

Do you mean Qm-Gm?

- 4 years, 1 month ago

see my proof

- 4 years, 1 month ago

(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!!

- 4 years, 1 month 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!

- 4 years, 1 month ago

Problem. 7. Is easy.

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

by am gm ab<1/6

now its easy

- 4 years, 1 month ago

From where did u get a^2 + b^2 + ab = 1/2?

- 4 years, 1 month ago

use the fact a + b = -c

- 4 years, 1 month ago

Hmm I c. I had ignored the fact that a+b+c=0

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Correct, AM-GM concept is applicable only for positive numbers

- 4 years, 1 month ago

there is one other idea too

form a cubic polynomial and find its discriminant

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Exactly

- 4 years, 1 month ago

(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!!

- 4 years, 1 month ago

2)a) i am getting a = b = c = 5

- 4 years, 1 month ago

a,b,c are distinct

- 4 years, 1 month ago

Oh, sorry!!

- 4 years, 1 month ago

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

$a^2b^2c^2 \leq \boxed{\dfrac{1}{36}}$

- 4 years, 1 month ago

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

- 4 years, 1 month ago

It works for all real numbers. I didn't get your second sentence.

- 4 years, 1 month ago

try using c=-(a+b) in the main equation

- 4 years, 1 month ago

There is no flaw here. Its just that you have come up with a weak inequality. Use Cauchy-Shwarz equality condition and you'll see that the equality doesn't occur. We need to find the stronger version ( as 1/54 < 1/36). For example, one may also conclude here that abc<1. It is true, but weak.

- 2 years, 2 months ago

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

- 4 years, 1 month ago

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.

Staff - 4 years, 1 month ago

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

- 4 years, 1 month 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.

- 4 years, 1 month ago