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Problem Of The Week #434 September 14th, 2020

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
Here is this week's POTW:

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Let $P_1(x)=ax^2-bx-c,\,P_2(x)=bx^2-cx-a,\,P_3(x)=cx^2-ax-b$ be three quadratic polynomials where $a,\,b$ and $c$ are non-zero real numbers. Suppose there exists a real number $k$ such that $P_1(k)=P_2(k)=P_3(k)$, prove that $a=b=c$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
No one answered last week's POTW.(Sadface) However, you can find the suggested solution below:

We have three relations:

$ak^2-bk-c=m\\ bk^2-ck-a=m\\ ck^2-ak-b=m$
where $m$ is the common value.

Eliminating $k^2$ from these, taking these equations pair-wise, we get:

$(ca-b^2)k-(bc-a^2)=m(b-a)\\(ab-c^2)k-(ca-b^2)=m(c-b)\\(bc-a^2)k-(ab-c^2)=m(a-c)$

Adding these three we get

$(ab+bc+ca-a^2-b^2-c^2)(k-1)=0$

Thus, either $ab+bc+ca-a^2-b^2-c^2=0$ or $k=1$.

In the first case

$ab+bc+ca-a^2-b^2-c^2=0\\ \dfrac{1}{2}((a-b)^2+(b-c)^2+(c-a)^2)=0$
shows that $a=b=c$.

If $k=1$, then we obtain $a-b-c=b-c-a=c-a-b$, once again we obtain $a=b=c$.
 
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