N is inside the quadrilateral abcd

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#1 h Oct 22, 2005, 9:56 PM • 3 Y

Y by harshu926, NathalieShwarz, Adventure10

$ABCD$ is a quadrilateral with $BC$ parallel to $AD$ . $M$ is the midpoint of $CD$ , $P$ is the midpoint of $MA$ and $Q$ is the midpoint of $MB$ . The lines $DP$ and $CQ$ meet at $N$ . Prove that $N$ is inside the quadrilateral $ABCD$ .

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#2 h Oct 11, 2012, 7:16 PM • 2 Y

Y by Adventure10, Mango247

I'm not sure why no one has posted a solution here yet. The problem is not very hard. Let the lines through $P,Q$ parallel to $\overline {BC}$ and $\overline{AD}$ intersect $\overline {CD}$ in $P',Q'$ and let the parallel line through $N$ intersect $\overline{AB},\overline{CD}$ in $Q,L$ . Assume WLOG that $CD=4$ , and let $r=BC$ , $s=AD$ , $y=CL$ .

We need to show that $LN \le LQ$ . Because of the similarity of triangles $CQQ'$ and $CNM$ , we have that $LN=ry/2$ . Similarly, $LN=(4-y)s/2$ . Solving yields $y=4s/(r+s)$ and $LN=2rs/(r+s)$ . Now $LQ=(sy+r(4-y))/4 =(r^2+s^2)/(r+s)$ , so the desired result follows from the AM-GM inequality.

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#3 h Feb 12, 2014, 6:27 PM • 1 Y

Y by Adventure10

Let's denote $S_{F}$ the area of figure $F$ .
It is obvious one of the angles $BCD$ or $CDA$ is acute.
Therefore we easily obtain that point $N$ is between lines $CD$ , $BC$ , $AD$ .
Suppose point $N$ is outside or the quadrilateral $ABCD$ $\Rightarrow$ $S_{TBC}+S_{TCD}+S_{TDA}>S_{ABCD}$ .
$BQ=QM\Rightarrow S_{BNQ}=S_{NQM}$ , $S_{BQC}=S_{QCM}$ $\Rightarrow S_{NBC}=S_{NCM}$
Similarly $S_{NDM}=S_{NDA}$ .
$CM=MD \Rightarrow S_{NBC}=S_{NCM}=S_{NMC}=S_{NDA}=X$
$S_{TBC}+S_{TCD}+S_{TDA}=4X$
Let the distance between $N$ and $AD$ be $y$ and between lines $BC$ and $AD$ be $h$ .
$S_{AND}=X=\frac{1}{2}yAD=\frac{1}{2}(h-y)BC \Rightarrow y=\frac{h BC}{BC+AD}$
$S_{ABCD}<S_{TBC}+S_{TCD}+S_{TDA}=4X=4\frac{1}{2}\frac{hBC}{BC+AD}AD=\frac{2h AD BC}{AD+BC}<h\frac{AD+BC}{2}=S_{ABCD}$ we used here AM-GM. Contradiction.

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#4 h Apr 8, 2017, 10:41 AM • 2 Y

Y by Adventure10, Mango247

polya78 wrote:

I'm not sure why no one has posted a solution here yet. The problem is not very hard. Let the lines through $P,Q$ parallel to $\overline {BC}$ and $\overline{AD}$ intersect $\overline {CD}$ in $P',Q'$ and let the parallel line through $N$ intersect $\overline{AB},\overline{CD}$ in $Q,L$ . Assume WLOG that $CD=4$ , and let $r=BC$ , $s=AD$ , $y=CL$ .

We need to show that $LN \le LQ$ . Because of the similarity of triangles $CQQ'$ and $CNM$ , we have that $LN=ry/2$ . Similarly, $LN=(4-y)s/2$ . Solving yields $y=4s/(r+s)$ and $LN=2rs/(r+s)$ . Now $LQ=(sy+r(4-y))/4 =(r^2+s^2)/(r+s)$ , so the desired result follows from the AM-GM inequality.

you said that $\triangle PQQ'\sim \triangle PMN$ means $QQ'||MN\implies MN||AD$ how do you proved that?

This post has been edited 1 time. Last edited by TRYTOSOLVE, Apr 8, 2017, 10:41 AM

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#8 h May 24, 2019, 2:09 PM • 2 Y

Y by Myriam2003, Adventure10

we will use the coordinates $(D;\vec{DA},\vec{DC})$
we put
D $(0,0)$
A $(1,0)$
C $(0,c)$
B $(b,c)$
and thus
M $(0,c/2)$
P $(1/2,c/4)$
Q $(\frac{b}{2},3c/4)$
solve for N
N $(\frac{2b}{b+1},\frac{bc}{b+1})$
now it's clear the the coordinates of $N$ are positive
so it suffices to show that $N$ is to the left to $AB$ and down of $CB$
the equation of $AB$ is $y=\frac{cx-c}{b-1}$
so the if a point is to the left of that line it must satisfy the inequality:
$y<\frac{cx-c}{b-1}$ if $b<1$
or
$y>\frac{cx-c}{b-1}$ if $b>1$
Click to reveal hidden text
which is clear if we put $N$ into the inequality
and because $c>\frac{bc}{b+1}$
$N$ is down of $CB$
and we win

Click to reveal hidden text

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#9 h May 19, 2022, 4:28 PM • 1 Y

Y by Mango247

It's obviously sufficient to show that $N$ lies on the same side of $AB$ as $D$ . We use barycentric coordinates with respect to $\triangle ABD$ . Let $C=(-c,1,c)$ . Take $c>0$ , then we have to show that the $z$ -coordinate of $N$ is non-negative. Indeed, $M=(-c:1:1+c),P=(2-c:1:1+c),Q=(-c:3:1+c)$ , so $\overline{DP}: x=(2-c)y$ and $\overline{CQ}: 0=(1-2c)x+cy-2cz$ . Intersecting these two lines yields $N=\left(\frac{c(2-c)}{c+1},\frac{c}{c+1},\frac{(c-1)^2}{c+1}\right)$ and the last coordinate is $\geq 0$ , as desired.

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#10 h Sep 11, 2023, 3:10 PM

Y by

A nice problem for coordinates!
anyway D=(0,0) A=(1,0) and C=(0,m) and B=(n,m). now M=(0, m/2) and P=(1/2, m/4) and Q=(n/2, 3m/4) so DP is y=mx/2, and CQ slope is m/2n, now if we solve for N, we get N=(2n/(n+1), mn/(n+1)), now we need to show that N is below BC, so we need to prove m>(mn)/(n+1) which follows because 1>n/(n+1)

yay

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