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=20112010{a=2011}^{2010}, b=20102011b={2010}^{2011}, c=40214021c={4021}^{4021}, and d=2011d=2011, find the value of bc(a+d)(ab)(ac)+ac(b+d)(ba)(bc)+ab(a+d)(ca)(cb)\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]

xx, yy, zz and aa are real numbers such that x+y+z=3x+y+z=3 and xy+yz+zx=axy+yz+zx=a. The difference in maximum and minimum possible value of xx is 88. Find aa.

3 [NMTC 2012]

Find all positive integral solutions (x,y)(x,y) to the equation 4x33x1=2y24x^3-3x-1=2y^2

4 [NMTC 2012]

Consider the set of numbers A=1,12,13,14,........,12012A={1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ........, \frac{1}{2012}}. We delete any two numbers aa and bb from the set AA, and replace them with a single number ab+a+bab+a+b. What number will be left at the end, after 20112011 such operations. Will the number always be the same?

5 [NMTC 2013]

Find positive real solutions (x,y,z)(x,y,z) which satisfy both these equations: x3+y3+z3=x+y+zx^3+y^3+z^3=x+y+z and x2+y2+z2=xyzx^2+y^2+z^2=xyz.

6 [NMTC 2013]

Do there exis 1010 distinct integers, such that the sum of any nine of them is a perfect square?

Note by Satvik Golechha
5 years ago

No vote yet
1 vote

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

Take the second equation and differentiate it with respect to xx, to get f(x)=y+zf'(x)=y+z. Extrema is reached when this is equal to zero, and as there is no constraint on interval, y+z=0y+z=0. Plugging into the first equation, we get x=3x=3. This must be the minimum, since the other extreme of xx is 55, which is the maximum. For the maximum, y+z=2y+z=-2 or z=y2z=-y-2

Now, manipulating the second equation,

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

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

(y2+2y+10)=a-(y^2+2y+10)=a

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

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

Here my approach was a bit unfounded.

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

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

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

Krishna Ar - 5 years ago

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What is the answer you arrived at?

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@Nanayaranaraknas Vahdam 9-9 only

Krishna Ar - 5 years ago

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I have done it like:

x2+y2+z2=92a{x}^{2} + {y}^{2} + {z}^{2} = 9 - 2a

y2+z2=(3x)22(ax(3x)){y}^{2} + {z}^{2} = {(3-x)}^{2} - 2(a-x(3-x))

92ax2=(3x)22(ax(3x))9-2a - {x}^{2} = {(3-x)}^{2} - 2(a-x(3-x))

This is the approach you are talking about?

Kartik Sharma - 5 years ago

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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)

Mehul Chaturvedi - 5 years ago

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

P.S. I am in 11th

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@Nanayaranaraknas Vahdam Oh u are in 11th ,,,actually i am in 10th

Mehul Chaturvedi - 5 years ago

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

x=2a2+1x=2a^{2}+1

y=a(4a2+3)y=a(4a^{2}+3) where aa can be any positive integer.

lawrence Bush - 5 years ago

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Here's how i came up with that:

3x33x+x31=2y23x^{3}-3x+x^{3}-1=2y^{2}.

=3x(x21)+(x1)(x2+x+1)=2y2=3x(x^{2}-1)+(x-1)(x^{2}+x+1)=2y^2

=(4x2+4x+1)(x1)=2y2=(4x^{2}+4x+1)(x-1)=2y^2.

=(2x+1)2(x1)2=y2=\frac{(2x+1)^{2}(x-1)}{2}=y^{2}. Since (2x+1)2(2x+1)^{2} is always odd,then x1x-1 must be divisible by 22 or xx 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 x12\frac{x-1}{2} must be a perfect square therefore x=2a2+1x=2a^{2}+1 where a2=x12a^2=\frac{x-1}{2}.Substituting xx we have y=a(4a2+3)y=a(4a^2+3).

lawrence Bush - 5 years ago

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

Kartik Sharma - 5 years ago

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Looks great!

Calvin Lin Staff - 5 years ago

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Yes, that is a family of solutions.

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

Calvin Lin Staff - 5 years ago

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No, I have a proof of that

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

lawrence Bush - 5 years ago

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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

Gourav Bhattad - 5 years ago

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Please explain. Thanks... :D

Satvik Golechha - 5 years ago

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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

Gourav Bhattad - 5 years ago

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^6) Is already discussed on Brilliant.

Krishna Ar - 5 years ago

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Where?

Satvik Golechha - 5 years ago

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For #4)-The second part of the question was not tested in the paper. 20122012 as the answer is easily guess-able but quite hard to prove.

Krishna Ar - 5 years ago

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Will you please prove it @Krishna Ar ? Thanks.

Satvik Golechha - 5 years ago

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

Kartik Sharma - 5 years ago

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That expression is algebraically correct.

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

Calvin Lin Staff - 5 years ago

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

Satvik Golechha - 5 years ago

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Yeah! Absolutely! I was about to write that abcb\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?

Kartik Sharma - 5 years ago

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@Kartik Sharma @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.

Satvik Golechha - 5 years ago

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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

Gourav Bhattad - 5 years ago

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Another approach maybe x3+y3+z33xyz{x}^{3} + {y}^{3} + {z}^{3} \geq 3xyz

But, this - x+y+z3(x2+y2+z2)x + y + z \geq 3({x}^{2} + {y}^{2} + {z}^{2}) for positive integer (x,y,z) as 3(x2+y2+z2)3({x}^{2} + {y}^{2} + {z}^{2}) will increase more than x+y+zx + y + z for positive integers(x,y,z)

Kartik Sharma - 5 years ago

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

Gourav Bhattad - 5 years ago

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But 12+12+12111{1}^{2} +{1}^{2} + {1}^{2} \neq 1*1*1

Kartik Sharma - 5 years ago

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Exactly.

Satvik Golechha - 5 years ago

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5actually doesn't possess a solution

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

Gourav Bhattad - 5 years ago

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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

ahmed abdo - 5 years ago

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thank you very much!!!

Gaurav Jain - 5 years ago

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Try the second part too!

Satvik Golechha - 5 years ago

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try a angular kinematics problem from me . rotating ring.

Gaurav Jain - 4 years, 12 months ago

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