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 \[\dfrac{Area(KLMN)}{Area(ABCD)} < \dfrac {8}{27}.\]
Problem
Source: Turkey TST 2003 - P2
Tags: limit, inequalities, geometry proposed, geometry
08.04.2013 21:06
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.
09.04.2013 06:32
I dont see how the AM-GM gives you that result.
09.04.2013 14:24
NgoNgang wrote: I dont see how the AM-GM gives you that result. Just use on $\frac {x}{2},\frac{x}{2},y$
11.04.2013 08:21
Another way is to use $ { \frac {x}{x + y}, \frac {x}{x + y}, \frac {2y}{x + y} }$
11.04.2013 08:25
$ \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.