BdMO National 2013: Higher Secondary 2

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
BdMO
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Joined: Tue Jan 18, 2011 1:31 pm

BdMO National 2013: Higher Secondary 2

Unread post by BdMO » Fri Jan 10, 2014 1:41 am

Let $g$ be a function from the set of ordered pairs of real numbers to the same set such that $g(x, y)=-g(y, x)$ for all real numbers $x$ and $y$. Find a real number $r$ such that $g(x, x)=r$ for all real numbers $x$.

ahmedulkavi
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Joined: Tue Feb 01, 2011 11:20 am

Re: BdMO National 2013: Higher Secondary 2

Unread post by ahmedulkavi » Mon Nov 28, 2016 10:03 am

How can I solve this problem?

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Thanic Nur Samin
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Joined: Sun Dec 01, 2013 11:02 am

Re: BdMO National 2013: Higher Secondary 2

Unread post by Thanic Nur Samin » Tue Dec 06, 2016 1:14 am

The statement essentially means that for any pair of real numbers $(x,y)$, we have $g(x,y)=-g(y,x)$. Note that, we can let $x,y$ be $\textbf{any}$ pair of real numbers. Substituting $(x,x)$ implies that,
\[g(x,x)=-g(x,x)\]
\[2g(x,x)=0\]
\[g(x,x)=0\]
Hence the problem is solved.

N.B. If you still have confusion in understanding, note that $(x+y)^2=x^2+2xy+y^2$. Plugging in $(x,x)$ implies $(x+x)^2=x^2+2xx+x^2$ or $(2x)^2=x^2+2x^2+x^2$ or $4x^2=4x^2$, which is true.
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