# Olympiad Practice Problems Part 1

Some problems from NMTC level 2 junior contest, conducted in India by the AMTI. Try to solve 'em all! And please post solutions, like, and re-share!

1 [NMTC 2011]

If $${a=2011}^{2010}$$, $$b={2010}^{2011}$$, $$c={4021}^{4021}$$, and $$d=2011$$, find the value of $\frac{bc(a+d)}{(a-b)(a-c)} + \frac{ac(b+d)}{(b-a)(b-c)} + \frac{ab(a+d)}{(c-a)(c-b)}$

2 [NMTC 2011]

$x$, $y$, $z$ and $a$ are real numbers such that $x+y+z=3$ and $xy+yz+zx=a$. The difference in maximum and minimum possible value of $x$ is $8$. Find $a$.

3 [NMTC 2012]

Find all positive integral solutions $(x,y)$ to the equation $4x^3-3x-1=2y^2$

4 [NMTC 2012]

Consider the set of numbers $A={1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ........, \frac{1}{2012}}$. We delete any two numbers $a$ and $b$ from the set $A$, and replace them with a single number $ab+a+b$. What number will be left at the end, after $2011$ such operations. Will the number always be the same?

5 [NMTC 2013]

Find positive real solutions $(x,y,z)$ which satisfy both these equations: $x^3+y^3+z^3=x+y+z$ and $x^2+y^2+z^2=xyz$.

6 [NMTC 2013]

Do there exis $10$ distinct integers, such that the sum of any nine of them is a perfect square?

Note by Satvik Golechha
6 years, 4 months ago

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For the second one, I got $\boxed{a=-9}$

Take the second equation and differentiate it with respect to $x$, to get $f'(x)=y+z$. Extrema is reached when this is equal to zero, and as there is no constraint on interval, $y+z=0$. Plugging into the first equation, we get $x=3$. This must be the minimum, since the other extreme of $x$ is $5$, which is the maximum. For the maximum, $y+z=-2$ or $z=-y-2$

Now, manipulating the second equation,

$xy+z(x+y)=a$

$xy-(y+2)(x+y)=a$

$-(y^2+2y+10)=a$

Completing the square, we get $(y+1)^2+9=-a$

Plugging in $z$, we get $(z+1)^2+9=-a$

Here my approach was a bit unfounded.

I found the minimum value of $a$ in both expressions, which is achieved when $(y+1)^2,(z+1)^2=0$

Thus, $x=5, y=-1, z=-1, \boxed{a=-9}$, is a solution set where $x$ is maximized.

- 6 years, 4 months ago

This HAS to be solved without Calculus-So this method would be better. $y+z=3-x$ and $yz=a-x(3-x)$. Manipulate these , form quadratics, use the discriminant and you're done. :D

- 6 years, 4 months ago

What is the answer you arrived at?

- 6 years, 4 months ago

$-9$ only

- 6 years, 4 months ago

I have done it like:

${x}^{2} + {y}^{2} + {z}^{2} = 9 - 2a$

${y}^{2} + {z}^{2} = {(3-x)}^{2} - 2(a-x(3-x))$

$9-2a - {x}^{2} = {(3-x)}^{2} - 2(a-x(3-x))$

This is the approach you are talking about?

- 6 years, 4 months ago

Hey @Nanayaranaraknas Vahdam how do you know calculus at 15 i.e in 10th class, in rajasthan the study pattern's diffrerent. But what abt. there please tell me?????Please!(But i know only Minima and Maxima)

- 6 years, 4 months ago

I didn't learn it through school, I did some independent studying, and learned some basics in Calculus.

P.S. I am in 11th

- 6 years, 4 months ago

Oh u are in 11th ,,,actually i am in 10th

- 6 years, 4 months ago

There exist an infinite number of solutions for equation in problem nr.3,given by:

$x=2a^{2}+1$

$y=a(4a^{2}+3)$ where $a$ can be any positive integer.

- 6 years, 4 months ago

Here's how i came up with that:

$3x^{3}-3x+x^{3}-1=2y^{2}$.

$=3x(x^{2}-1)+(x-1)(x^{2}+x+1)=2y^2$

$=(4x^{2}+4x+1)(x-1)=2y^2$.

$=\frac{(2x+1)^{2}(x-1)}{2}=y^{2}$. Since $(2x+1)^{2}$ is always odd,then $x-1$ must be divisible by $2$ or $x$ must be an odd number.Also we have the product of two numbers where one of them is a square and that the whole product is a square,thus $\frac{x-1}{2}$ must be a perfect square therefore $x=2a^{2}+1$ where $a^2=\frac{x-1}{2}$.Substituting $x$ we have $y=a(4a^2+3)$.

- 6 years, 4 months ago

Brilliant! *That is the only thing I have to say.

- 6 years, 4 months ago

Looks great!

Staff - 6 years, 4 months ago

Yes, that is a family of solutions.

Is that the only family? Can there be other solutions which are not of that form?

Staff - 6 years, 4 months ago

No, I have a proof of that

- 6 years, 4 months ago

No,because the way this is derived shows that there is no room for other family of solutions.

- 6 years, 4 months ago

for 4 take product of each number increased by 1 as invariant initial value is 2013 so answer is at final step 2013-1=2012

- 6 years, 4 months ago

- 6 years, 4 months ago

Consider product (a+1)*(b+1) when a and b are deleted and ab+a+b is added but it is equal to product.so this will remain constant at all steps?hope u got it

- 6 years, 4 months ago

^6) Is already discussed on Brilliant.

- 6 years, 4 months ago

Where?

- 6 years, 4 months ago

For #4)-The second part of the question was not tested in the paper. $2012$ as the answer is easily guess-able but quite hard to prove.

- 6 years, 4 months ago

Will you please prove it @Krishna Ar ? Thanks.

- 6 years, 4 months ago

After bashing a lot, I have simplified the 1st to $d - \frac{ab}{c-b}$. Is it correct? I am really not sure about it.

- 6 years, 4 months ago

That expression is algebraically correct.

How would you proceed to do the calculation of the final value?

Staff - 6 years, 4 months ago

I think this much simplification would do. It's value is 2010.99999999999999....., so they will accept 2011.

- 6 years, 4 months ago

Yeah! Absolutely! I was about to write that $\frac{ab}{c-b}$ is almost equal to 1.

BTW, @Satvik Golechha are you planning to make more such sets like for NT or geometry?

- 6 years, 4 months ago

@Kartik Sharma Did you like it? These were some of my doubts. And yes, if people like it, I would definitely continue. Geometry is my Achilles heel, I need to study.

- 6 years, 4 months ago

In 5 note x y z all can't be greater than 1. Due to first equation if all are 1 it is one tulle is pf any of them is less than 1 say x then x^2+y^2+z^2>y^2+z^2>yz so. Second fails so only 1,1,1 is solution

- 6 years, 4 months ago

Another approach maybe ${x}^{3} + {y}^{3} + {z}^{3} \geq 3xyz$

But, this - $x + y + z \geq 3({x}^{2} + {y}^{2} + {z}^{2})$ for positive integer (x,y,z) as $3({x}^{2} + {y}^{2} + {z}^{2})$ will increase more than $x + y + z$ for positive integers(x,y,z)

- 6 years, 4 months ago

yes by cauchy scwarz inequality (x+y+z)^2<=3(x^2+y^2+z^2).so your method is also elegant

- 6 years, 4 months ago

But ${1}^{2} +{1}^{2} + {1}^{2} \neq 1*1*1$

- 6 years, 4 months ago

Exactly.

- 6 years, 4 months ago

5actually doesn't possess a solution

- 6 years, 4 months ago

sorry for the mistake. so in case 1,1,1 too no solution exist hence there is no solution to 5.

- 6 years, 4 months ago

the third one , (x-1)(2x+1)^2=2y^2 ,y=(2x+1)sqrt{(x-1)}/2 for y to be an integer (x-1)/2=m^2 ,x=2*m^2+1 then for m=0,1,2,3,4,5........ we have infinite number of solutions

- 6 years, 3 months ago

thank you very much!!!

- 6 years, 3 months ago

Try the second part too!

- 6 years, 3 months ago

try a angular kinematics problem from me . rotating ring.

- 6 years, 3 months ago