[KLMN]/[ABCD]<8/27

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[KLMN]/[ABCD]<8/27

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Source: Turkey TST 2003 - P2

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#1 h Apr 6, 2013, 11:03 PM • 1 Y

Y by Adventure10

Let $K$ be the intersection of the diagonals of a convex quadrilateral $ABCD$ . Let $L\in [AD]$ , $M \in [AC]$ , $N \in [BC]$ such that $KL\parallel AB$ , $LM\parallel DC$ , $MN\parallel AB$ . Show that $\dfrac{Area(KLMN)}{Area(ABCD)} < \dfrac {8}{27}.$

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1479 posts

#2 h Apr 8, 2013, 6:06 PM • 1 Y

Y by Adventure10

Suppose $LD=y,AL=x$ now so $[LKD]=\frac {x^2}{(x+y)^2}[ABD],[ALM]=\frac {x^2}{(x+y)^2}[ACD]$ . So $[LKM]=\frac {x^2[ACD]-xy[ABD]}{(x+y)^2}$ . Now as $\frac {LD}{AL}=\frac {MC}{AM}$ so $LK=MN$ and that implies $MNKL$ is nothing but a parralelogram.Now take $[AKD]=py,[AKB]=px,[BKC]=xt\implies [CKD]=yt$ .Now so certainly $[ABCD]=(p+t)(x+y)$ . And also $[KLMN]=2(x^2y(p+t)-xy(x+y)p$ . So finally now we're required to $\frac {xy}{(x+y)^3}(\frac {xt-yp}{p+t})<\frac {4}{27}$ .Now we can take $t\geq p$ . So now keep $x,y,p$ constant, then $f(t)=(\frac {xt-yp}{p+t})$ is increasing. So $f(t)\leq \lim_{t\to\inf} f(t)= x$ .Thus it's now remain to show $\frac {x^2y}{(x+y)^3}<\frac{4}{27}$ . Which is obvious just by $AM-GM$ inequality,So done.

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#3 h Apr 9, 2013, 3:32 AM • 1 Y

Y by Adventure10

I dont see how the AM-GM gives you that result.

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#4 h Apr 9, 2013, 11:24 AM • 1 Y

Y by Adventure10

NgoNgang wrote:

I dont see how the AM-GM gives you that result.

Just use on $\frac {x}{2},\frac{x}{2},y$

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#5 h Apr 11, 2013, 5:21 AM • 2 Y

Y by Adventure10, Mango247

Another way is to use ${ \frac {x}{x + y}, \frac {x}{x + y}, \frac {2y}{x + y} }$

This post has been edited 1 time. Last edited by Vo Duc Dien, Apr 11, 2013, 11:28 PM

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#6 h Apr 11, 2013, 5:25 AM • 2 Y

Y by Adventure10, Mango247

$\frac{x^2y}{(x+y)^3}<= \frac{4}{27}$ and not < (smaller). There is a positive element that is greater than zero on the denominator to make it smaller than zero.

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