RMO practice problems from past year papers (polynomials)

{P1(x)=ax2bxcP2(x)=bx2cxaP3(x)=cx2axb \begin{cases} P_{1}(x) = ax^{2}-bx-c \\ P_{2}(x)=bx^{2}-cx-a \\ P_{3}(x)=cx^2-ax-b\ \end{cases} Let the above be three quadratic polynomials where a,b,c are non-zero real numbers. Suppose there exists a real number α \alpha such that P1(α)=P2(α)=P3(α)P_{1}(\alpha)=P_{2}(\alpha)=P_{3}(\alpha), then

  • Prove that a=b=ca=b=c.

Please post your answers after at least one week (although this one is quite simple).

Note by A Former Brilliant Member
4 years, 1 month ago

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just put alpha at x. I am taking alpha as y.

ay2byc=kay^2 - by - c = k

by2cya=kby^2 - cy - a = k

cy2ayb=kcy^2 - ay - b = k

removing y2y^2 from these equations, we get three other equations and add them all the we get

ab+bc+caa2b2c2=0ab+bc+ca-a^2-b^2-c^2 = 0

so a = b = c

Dev Sharma - 4 years, 1 month ago

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Nice!

A Former Brilliant Member - 4 years, 1 month ago

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Thanks. Are you preparing for olympiad?

Dev Sharma - 4 years, 1 month ago

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@Dev Sharma Yes

A Former Brilliant Member - 4 years, 1 month ago

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How do you simply "remove" y2y^2 from these equations?

Mehul Arora - 4 years, 1 month ago

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Make the coefficient of y2y^2 same and subtract to eliminate it. For eg: multiply first equation by bb and second by aa and observe the coefficient of y2y^2 and thereby subtract them.

Nihar Mahajan - 4 years, 1 month ago

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@Nihar Mahajan Yeah, I got it :)

Mehul Arora - 4 years, 1 month ago

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Simple standard solution.

Nihar Mahajan - 4 years, 1 month ago

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Change your Profile pic :3 xD

A good troll anyway :P

I got trolled xD

Mehul Arora - 4 years, 1 month ago

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@Mehul Arora i got confused when i was reading comment. Lol.

Dev Sharma - 4 years, 1 month ago

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Hey Nihar and Mehul, try this problem.

If x3=x+1x^3 = x + 1 then determine integer a, b and c such that x7=ax2+bx+cx^7 = ax^2 + bx + c

Dev Sharma - 4 years, 1 month ago

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@Dev Sharma x3=x+1x^3=x+1

Squaring both sides yields

x6=(x+1)2=x2+2x+1x^6=(x+1)^2=x^2 + 2x+1

Since xx is non zero we can multiply both sides by xx and get

x7=x3+2x2+xx^7=x^3+2x^2+x

x7=x+1+2x2+xx^7=x+1+2x^2+x

x7=2x2+2x+1x^7=2x^2+2x+1

So we get a=2,b=2a=2,b=2

Ravi Dwivedi - 4 years, 1 month ago

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@Ravi Dwivedi nice

Dev Sharma - 4 years, 1 month ago

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Since, the three polynomials have same root. So, one of them can be expressed as linear combination of other two. So, P1(x)=P2(x)+λP3(x)P_{1} (x) = P_{2} (x) + \lambda P_{3} (x), where λ\lambda is a real number. Comparing the coefficients of x2,xx^{2},x and constant, we get different values of λ\lambda. They are

λ=cab=abc=bca=ab+bc+caa+b+c=0\lambda = \dfrac{c-a}{b} = \dfrac{a-b}{c} = \dfrac{b-c}{a} = \dfrac{a-b+b-c+c-a}{a+b+c} = 0 which implies that a=b=c\large a=b=c

Surya Prakash - 4 years, 1 month ago

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i did same.....

Sourav Kumar Surya - 4 years, 1 month ago

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How do u know that the roots are equal?

Sneha Iyer - 4 years, 1 month ago

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Thanks for mentioning😛

Swapnil Das - 4 years, 1 month ago

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Thanks for mentioning me! @Svatejas Shivakumar

@Silas Hundt There is some problem. I am not receiving notifications when someone @mentions me

Mehul Arora - 4 years, 1 month ago

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Which kind of notifications? The ones in the corner on Brilliant or emails or...?

Silas Hundt Staff - 4 years, 1 month ago

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@Silas Hundt The ones in the corner, sir :)

Mehul Arora - 4 years, 1 month ago

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@Mehul Arora Is this still happening? I'm still seeing notifications...

Silas Hundt Staff - 4 years, 1 month ago

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@Silas Hundt It happened with me twice. Hasn't happened very recently. I'll be sure to inform you in any such event :)

Mehul Arora - 4 years, 1 month ago

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try this

P(x)=4x32x215x+9P(x) = 4x^3 - 2x^2 - 15x + 9 and Q(x)=12x3+6x27x+1Q(x) = 12x^3 + 6x^2 - 7x + 1 then prove that each polynomial has 3 distinct real roots and let a and b be greatest roots of p(x) and q(x), respectively then prove that a2+3b2=4a^2 + 3b^2 = 4

Dev Sharma - 4 years, 1 month ago

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@Dev Sharma Can you post the solution?

A Former Brilliant Member - 4 years, 1 month ago

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I still did not understand how to solve the question

Arnav Goel - 2 years, 1 month ago

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Bt why

ARGHYA SENGUPTA - 2 years, 1 month ago

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Thnx

ARGHYA SENGUPTA - 2 years, 1 month ago

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