Macedonia National Olympiad 2016 Problem 4

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Macedonia National Olympiad 2016 Problem 4

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Source: Macedonia National Olympiad 2016

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#1 h Apr 9, 2016, 2:38 PM • 2 Y

Y by Adventure10, Mango247

A segment $AB$ is given and it's midpoint $K$ . On the perpendicular line to $AB$ , passing through $K$ a point $C$ , different from $K$ is chosen. Let $N$ be the intersection of $AC$ and the line passing through $B$ and the midpoint of $CK$ . Let $U$ be the intersection point of $AB$ and the line passing through $C$ and $L$ , the midpoint of $BN$ . Prove that the ratio of the areas of the triangles $CNL$ and $BUL$ , is independent of the choice of the point $C$ .

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#2 h Apr 9, 2016, 4:49 PM • 2 Y

Y by Adventure10, Mango247

Is it real national olympiad ?
Let $M$ is midpoint of $CK$
We apply a bunch of Menelaus get $AK:KU:UB=2:1:1$ , $UL:LC=1:2$ , $AN:NC=2:1$ and $NM:ML:LB=1:1:2$
So we can find $\frac{[ULB]}{[BMK]}$ and $\frac{[AUC]}{[CNL]}$ are fixed constant
And $\frac{[BMK]}{[AUC]} =\frac{1}{2} \cdot \frac{[BMK]}{[AMU]} =\frac{1}{2} \cdot \frac{2}{3}$ which is also a fixed constant

This post has been edited 1 time. Last edited by ThE-dArK-lOrD, Apr 9, 2016, 4:49 PM

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#3 h Apr 9, 2016, 6:37 PM • 2 Y

Y by Adventure10, Mango247

ThE-dArK-lOrD wrote:

Is it real national olympiad ?
Let $M$ is midpoint of $CK$
We apply a bunch of Menelaus get $AK:KU:UB=2:1:1$ , $UL:LC=1:2$ , $AN:NC=2:1$ and $NM:ML:LB=1:1:2$
So we can find $\frac{[ULB]}{[BMK]}$ and $\frac{[AUC]}{[CNL]}$ are fixed constant
And $\frac{[BMK]}{[AUC]} =\frac{1}{2} \cdot \frac{[BMK]}{[AMU]} =\frac{1}{2} \cdot \frac{2}{3}$ which is also a fixed constant

Unfortunately, yes.

Contestants are getting easily over 30 out of 40.

The problem is too trivial. $K$ can be any point on $AB$ , as long as it is fixed. Also $C$ can be any point in the plane. If we let $\frac{AK}{KB} = x$ , then $\frac{[BMK]}{[AUC]} = \frac{x+3}{x+1}$ .

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#4 h Jun 19, 2016, 5:15 AM • 2 Y

Y by Adventure10, Mango247

This problem falls easily to coordinate geometry as well. (For simpler calculations, let $A = (-1, 0)$ , $B=(1, 0)$ , $C=(0, 2c)$ .)

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#5 h Dec 25, 2017, 3:30 PM • 2 Y

Y by Adventure10, Mango247

ThE-dArK-lOrD wrote:

Is it real national olympiad ?
Let $M$ is midpoint of $CK$
We apply a bunch of Menelaus get $AK:KU:UB=2:1:1$ , $UL:LC=1:2$ , $AN:NC=2:1$ and $NM:ML:LB=1:1:2$
So we can find $\frac{[ULB]}{[BMK]}$ and $\frac{[AUC]}{[CNL]}$ are fixed constant
And $\frac{[BMK]}{[AUC]} =\frac{1}{2} \cdot \frac{[BMK]}{[AMU]} =\frac{1}{2} \cdot \frac{2}{3}$ which is also a fixed constant

$\frac{[BUL]}{[CNL]}=\frac{UL}{LC}=\frac{1}{2}$

This post has been edited 1 time. Last edited by StefanS, Dec 25, 2017, 3:31 PM

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#6 h Sep 28, 2019, 3:50 PM • 2 Y

Y by Adventure10, Mango247

After some Menelaus it's :
$NA=2CN$
$UA=3UB$
$LC=2LU$
$UK=UB=\frac12{AK}$

Note that by Thales $CU\parallel{KN}$

Hence $\frac{[LUK]}{[CNL]}=\frac{[BUL]}{[CNL]}=\frac12$ .

This post has been edited 1 time. Last edited by Kamran011, Sep 28, 2019, 3:50 PM

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