APMO 1996 - Problem 1

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Labib
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APMO 1996 - Problem 1

Unread post by Labib » Mon Jun 13, 2011 12:35 am

Let $ABCD$ be a quadrilateral $AB = BC = CD = DA$. Let $MN$ and $PQ$ be two segments
perpendicular to the diagonal $BD$ and such that the distance between them is $d > BD/2$,
with $M \in AD$, $N \in DC$, $P \in AB$ and $Q \in BC$. Show that the perimeter of hexagon
$AMNCQP$ does not depend on the position of $MN$ and $PQ$ so long as the distance between
them remains constant.
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User avatar
Labib
Posts:411
Joined:Thu Dec 09, 2010 10:58 pm
Location:Dhaka, Bangladesh.

Re: APMO 1996 - Problem 1

Unread post by Labib » Tue Jun 14, 2011 12:34 am

Here's how I did it ::
The quadrilateral $ABCD$ is a rhombus.

I denote $M_1$ and $P_1$ the intersection point of $MN$ and $BD$ and the intersection point of $PQ$ and $BD$ respectively. I also denote O the intersection point of $AC$ and $BD$.

It's easy enough to notice that the pentagons- $AMM_1P_1P$ and $CNM_1P_1Q$ are congruent. (Use symmetry or reflection)

So I fold the figure $ABCD$ with respect to the segment $BD$ and get $\triangle ABD$.

Now if we can prove that the perimeter of the pentagon $AMM_1P_1P$ remains constant when $d$ remains constant, we're done!

There must be a point $D_1$ on $P_1M_1$ for which $P_1D_1=d/2$

I place $D_1$ on $O$.
Looking carefully, it's noticeable that now, $PP_1=MM_1$.
Again we've congruent figures $APP_1O$ and $AMM_1O$.

So we again fold the figure with respect to the segment $AO$.

Now we've $\triangle ABO$ with a parallel line $XX_1$ to its base $AO$.

Because of folding, The line $MM_1$ and $PP_1$ also lies on $XX_1$.

Now if we move $PP_1$, $k$ distance in any direction, $MM_1$ moves $k$ distance in other direction.

Now if we can prove that for any new position of $PP_1$ and $MM_1$ their length's sum would equal $2XX_1$ , we are done.

We can achieve this target using simple similarity... try yourselves.
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.


"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes

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