Convex quadrilateral and midpoints [RMO2-2011, India]

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Convex quadrilateral and midpoints [RMO2-2011, India]

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Potla

#1 h Dec 31, 2011, 12:19 PM • 1 Y

Y by Adventure10

Let $ABCD$ be a convex quadrilateral. Let $E,F,G,H$ be the midpoints of $AB,BC,CD,DA$ respectively. If $AC,BD,EG,FH$ concur at a point $O,$ prove that $ABCD$ is a parallelogram.

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547 posts

jatin

#2 h Dec 31, 2011, 1:32 PM • 1 Y

Y by Adventure10

Let $l$ be the line joining the midpoints of the diagonals.
Note that $EFGH$ is a parallelogram. (Use midpoint theorem)
Hence $O$ is the midpoint of $EG$ and $FH$ .
It is well-known that $EG,FH$ and $l$ are concurrent. (Use analytic geometry)
And the result follows.

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661 posts

Sahil

#3 h Jan 1, 2012, 6:19 AM • 2 Y

Y by Adventure10, Mango247

outline of my proof:
Click to reveal hidden text

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#4 h Jan 1, 2012, 11:10 AM • 1 Y

Y by Adventure10

Let $EG\cap FH=O$ .Fix the origin of the argand plane at O.It's easy to see that the midpoints of $EG$ and $FH$ coincide and it is $0=\frac{a+b+c+d}{2}$ .so $a+b+c+d=0$
Now since $b,0,d$ are collinear we have $\frac{b}{\overline{b}}=\frac{d}{\overline{d}}$ .Similarly, $\frac{a}{\overline{a}}=\frac{c}{\overline{c}}$
Now, $a\overline{c}=c\overline{a}\Rightarrow a(-\overline{b}-\overline{a}-\overline{d})=\overline{a}(-a-b-d)$
If $b+d\neq 0$ , $\frac{a}{\overline{a}}=\frac{b+d}{\overline{b}+\overline{d}}=\frac{b}{\overline{b}}=\frac{d}{\overline{d}}$ .But this means $A$ , $B$ , $D$ are collinear.so $b+d=0$ .From this the conclusion follows.
$QED$

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346 posts

#5 h Jan 1, 2012, 12:47 PM • 1 Y

Y by Adventure10

its the direct application for $Varignon$ $Theorem$

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#6 h Jan 1, 2012, 2:06 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Potla wrote:

Let $ABCD$ be a convex quadrilateral. Let $E,F,G,H$ be the midpoints of $AB,BC,CD,DA$ respectively. If $AC,BD,EG,FH$ concur at a point $O,$ prove that $ABCD$ is a parallelogram.

Let the midpoint of OB be X, and midpoint of OD be Y. Now, its easy to see that X lies on EF and Y lies on GH.

Consider now, the half turn about O. This takes E to G and F to H. Therefore, it takes the segment EF to segment GH. Therefore, it takes X to Y, since X,O,Y are collinear. Thus, OX = OY, resulting in OB = OD. Similarly, OA = OC.

lol. I didn't even need Latex.

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#7 h Jan 10, 2012, 1:34 PM • 2 Y

Y by Adventure10, Mango247

It becomes trivial if we use projective geometry. Just consider perspectivities centered at the point of intersection of diagonals.

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#8 h Jan 10, 2012, 3:28 PM • 2 Y

Y by Adventure10, Mango247

[Euclide] wrote:

its the direct application for $Varignon$ $Theorem$

See the first solution.

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547 posts

jatin

#9 h Jan 11, 2012, 4:40 AM • 2 Y

Y by Adventure10, Mango247

Pascal96 wrote:

It becomes trivial if we use projective geometry. Just consider perspectivities centered at the point of intersection of diagonals.

Please elaborate.
I'm interested.

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547 posts

jatin

#10 h Mar 5, 2012, 10:00 AM • 3 Y

Y by aayush-srivastava, Adventure10, Mango247

Pascal96 sent me this:

I just noticed that you had asked me to elaborate on the projective geometry solution to the RMO 2011 question. Sorry I didn't see this earlier, or I would have replied. Well here it is:
Let ABCD be the quadrilateral, and let E,F,G,H be the midpoints of AB, BC, CD, DA. Let O be the intersection of the diagonals (and EG and FH). Let P be the point at infinity, which we temporarily add to the Euclidean plane.
Consider the perspectivity centered at O taking A,B,E to C,D,G respectively. Note that A,B,E,P form a harmonic division. Hence, since perspectivities preserve harmonic divisions, C,D,G,Q is harmonic where Q is the image of P in the perspectivity. But G is the midpoint of CD, hence Q is at infinity. But Q is the intersection of CD and the line through O parallel to AB, and since Q is at infinity, this line must be parallel to CD. Thus the line through O is parallel to both AB and CD, implying AB is parallel to CD. Similarly AD is parallel to BC and we are done.

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1979 posts

#11 h Oct 1, 2015, 11:02 AM • 2 Y

Y by Adventure10, Mango247

Just use the fact that the mid-points of $AC,BD,AB,CD$ form a parallelogram and that $EG,FH$ bisect each other.

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239 posts

#12 h Oct 3, 2015, 2:05 PM • 2 Y

Y by Adventure10, Mango247

Let $X$ lie on $AC$ such that $HX$ is $parallel$ to $CD$ .
Let $Y$ lie on $BD$ such that $FY$ is $parallel$ to $CD$ .
Notice that $HYFX$ is a $parallelogram$ .
Let $XY\cap HF$ = $P$ . See that $P$ is the $midpoint$ of $HF$ .
Note that as $EFGH$ is a $parallelogram$ and $P$ is $midpoint$ of $HF$ , $P$ lies on $EG$ .
So , $P$ = $EG\cap FH$ .
So , $P$ lies on $AC$ . , $X$ lies on $AC$ imply $Y$ lies on $AC$ .
By definition of $Y$ , $Y$ lies on $BD$ .
So , $Y\equiv O$ implying $O$ is $midpoint$ of $BD$ .
Analogously show that $O$ is $midpoint$ of $AC$ .
This proves that $ABCD$ is a $parallelogram$ .

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#13 h Oct 30, 2021, 10:30 AM

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My solution, complete and beginner-friendly

This post has been edited 1 time. Last edited by Geoclid, Oct 30, 2021, 10:31 AM

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#14 h Mar 29, 2024, 6:20 AM

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It is well known that $EFGH$ is a parallelogram. So $EG$ and $FH$ bisect each other.
Now, we have lines $AB,HF$ and $DC$ which are bisected by the lines $BC,EG$ and $AD$ .
By the converse of intercept theorem, we have $CD\parallel HF\parallel AB$ .
Analogously, we have $AD\parallel EG\parallel BC$ .
Thus $ABCD$ is a parallelogram.
Our proof is complete.

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#15 h Mar 27, 2025, 9:13 PM

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