Quod Erat Demonstrandum

2017/04/05

答問

Filed under: Junior Form Mathematics — johnmayhk @ 11:11 下午
Tags:

網友問:

\displaystyle \frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}=0

\displaystyle \frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}

其實我沒有甚麼好方法,不過是死爆如下:

考慮

\displaystyle (\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b})

=\displaystyle \frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}+E

其中

E

=\displaystyle \frac{a}{b-c}(\frac{1}{c-a}+\frac{1}{a-b})+\frac{b}{c-a}(\frac{1}{b-c}+\frac{1}{a-b})+\frac{c}{a-b}(\frac{1}{b-c}+\frac{1}{c-a})

=\displaystyle \frac{a(c-b)}{(b-c)(c-a)(a-b)}+\frac{b(a-c)}{(c-a)(b-c)(a-b)}+\frac{c(b-a)}{(a-b)(b-c)(c-a)}

=\displaystyle \frac{a}{(a-c)(a-b)}+\frac{b}{(b-c)(b-a)}+\frac{c}{(c-a)(c-b)}

=\displaystyle \frac{a(b-c)+b(c-a)+c(a-b)}{(a-b)(a-c)(b-c)}

=0

又因

\displaystyle \frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}=0

故此

\displaystyle (\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b})=0

最後證明

\displaystyle \frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b} \ne 0

假設

\displaystyle \frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}=0

\Rightarrow \displaystyle \frac{1}{a-c}=\frac{1}{b-c}+\frac{1}{a-b}

\Rightarrow \displaystyle \frac{1}{a-c}=\frac{a-c}{(b-c)(a-b)}

\Rightarrow \displaystyle (a-c)^2=(b-c)(a-b)

\Rightarrow \displaystyle a^2+b^2+c^2-ab-bc-ca=0

\Rightarrow \displaystyle (a-b)^2+(b-c)^2+(c-a)^2=0

\Rightarrow \displaystyle a=b=c

這是不容許的,否則給定的

\displaystyle \frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}

是無意義。

於是

\displaystyle (\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b})=0

\Rightarrow \displaystyle \frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0

後記:開頭我搞甚麼 cyclic polynomials 之類的東西,很煩,最後只想到上法,估計應有更好方法,盼高手賜教。

廣告

2 則迴響 »

  1. According to the question, the answer must be a constant.
    So, just pre-assume a=0, b=1 and find the corresponding value of c which is (1/c^2)+c=0
    with the values of a, b and c, put them into the required equation show that the answer is indeed zero.
    Note that for LQ solution, one may show the required equation is a zero polynomial by considering combining the two equations.

    迴響 由 Simon YAU — 2017/04/06 @ 12:43 下午 | 回應

  2. Note that when a=0, b=1, we have c=-1, which gives the answer to be zero. For LQ sol, prove the polynomial is a constant and then prove it to be zero.

    迴響 由 Simon YAU — 2017/04/06 @ 7:04 下午 | 回應


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