Problem Of The Week #434 September 14th, 2020

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anemone

MHB POTW Director
Staff member
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!

anemone

MHB POTW Director
Staff member
No one answered last week's POTW. 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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