four perpendiculars concurrent

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hollandman

112 posts

hollandman

#1 h Apr 11, 2009, 2:09 AM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a cyclic quadrilateral, and $M,N,O,P$ the midpoints of sides $AB,BC,CD,DA$ respectively. Prove that the perpendiculars from $M$ to $CD$ , from $N$ to $DA$ , from $O$ to $AB$ , and from $P$ to $BC$ are concurrent.

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erudito

61 posts

erudito

#2 h Apr 11, 2009, 2:38 AM • 2 Y

Y by Adventure10, Mango247

This is the fifth problem from OBM (Brazilian Mathematics Olympiad in 2001).

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mathVNpro

469 posts

mathVNpro

#3 h Apr 11, 2009, 11:13 AM • 2 Y

Y by Adventure10, Mango247

This problem is very classic. Here is my outline solution about this problem:

LEMMA: Let ABCD be the cyclic quadrilateral, AC intersects BD by the point E. If M is the midpoint of AB then ME is perpendicular to CD.

proof: Leave for the readers (This is very easy, just simple angle chasing then you will get the result

)
Apply this lemma 4 times in the problem, we shall get the result.

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hollandman

112 posts

hollandman

#4 h Apr 11, 2009, 1:12 PM • 2 Y

Y by Adventure10, Mango247

mathVNpro wrote:

This problem is very classic. Here is my outline solution about this problem:

LEMMA: Let ABCD be the cyclic quadrilateral, AC intersects BD by the point E. If M is the midpoint of AB then ME is perpendicular to CD.

proof: Leave for the readers (This is very easy, just simple angle chasing then you will get the result

)
Apply this lemma 4 times in the problem, we shall get the result.

I don't think your lemma is correct...

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mihai miculita

666 posts

mihai miculita

#5 h Apr 11, 2009, 1:42 PM • 2 Y

Y by Adventure10, Mango247

$\mbox{Let }\Omega \mbox{ the circumcenter of cyclic quadrilateral ABCD and } \{G\}=MO\cap NP,$
$\mbox{then the perpendiculars from to M, to CD, from N to DA, and from O to AB,}$
$\mbox{and to P to BC are concurrent in the symetric of point } \Omega \mbox{ respect to G.}$

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mathVNpro

469 posts

mathVNpro

#6 h Apr 11, 2009, 1:42 PM • 2 Y

Y by Adventure10, Mango247

Oops

. I am so sorry, I have thought that these diagnols of this ABCD are perpendicular to each other

. Sorry for my big mistake.

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livetolove212

859 posts

livetolove212

#7 h Apr 11, 2009, 2:20 PM • 1 Y

Y by Adventure10

hollandman wrote:

Let $ABCD$ be a cyclic quadrilateral, and $M,N,O,P$ the midpoints of sides $AB,BC,CD,DA$ respectively. Prove that the perpendiculars from $M$ to $CD$ , from $N$ to $DA$ , from $O$ to $AB$ , and from $P$ to $BC$ are concurrent.

Let I be the intersection of MO and PN. E be the circumcenter of ABCD, H be the symetric of E respect to I
Since $MNOP$ is a parallelogram, we get $MI=OI$ , moreover $HI=EI$ then $MHOE$ is a parallelogram
Hence $MH \perp DC$ . Similarly, $NH \perp AD, OH\perp AB, PH\perp BC$

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Luis González

4147 posts

Luis González

#8 h Apr 13, 2009, 1:24 PM • 2 Y

Y by Adventure10, Mango247

Let the perpendiculars from $M$ to $CD$ and $O$ to $AB$ meet at $U.$ If $K$ is the circumcenter of $ABCD,$ then $KM \perp AB$ and $KO \perp CD$ implies that $KMUO$ is a paralelogram $\Longrightarrow$ $U$ is the reflection of $K$ about the midpoint $G$ of $OM.$ But $G$ is also the midpoint of $NP.$ Hence, perpendiculars from $N$ to $DA$ and $P$ to $BC$ also pass through $U.$

Remark: It's easy to see that U coincides with the Euler point of ABCD.

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bloops

24 posts

bloops

#9 h Apr 16, 2009, 2:16 PM • 2 Y

Y by Adventure10, Mango247

hollandman wrote:

Let $ABCD$ be a cyclic quadrilateral, and $M,N,O,P$ the midpoints of sides $AB,BC,CD,DA$ respectively. Prove that the perpendiculars from $M$ to $CD$ , from $N$ to $DA$ , from $O$ to $AB$ , and from $P$ to $BC$ are concurrent.

1) The point of concurrency, the mid points of opposite sides and the circumcentre form a parallelogram.
Let the circumcentre be $G$ and the point of concurrency be $F$ . Since we can drop perpendiculars from $G$ to $AB$ , $CD$ at their midpoints $M$ and $O$ respectively, $MGOF$ is a parallelogram. Similarly, $NGPF$ is also a parallelogram.

2) Moving on to the Argand plane, with circumcentre as the origin, denote $z_1, z_2, z_3, z_4$ as the vertices of $ABCD$ . The point of concurrency is nothing but the vector sum of the midpoints of opposite sides.

So, $F = \frac {z_1 + z_3}{2} + \frac {z_2 + z_4}{2}$ which is the same point as $\frac {z_1 + z_4}{2} + \frac {z_2 + z_3}{2}$ . Thus, we arrive at the same point for each pair of opposite sides.

I think this proof can be generalized to a cyclic $n$ -gon, $n$ being even, if I'm not mistaken.

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mathVNpro

469 posts

mathVNpro

#10 h Apr 21, 2009, 1:19 AM • 2 Y

Y by Adventure10, Mango247

Here is my solution:
Denote $M_1$ , $N_1$ , $O_1$ , $P_1$ the projections of $M$ , $N$ , $O$ , $P$ onto $CD$ , $DA$ , $AB$ , $BC$ , respectively. Now, It is easy to see that $MNOP$ is parallelogram. Let $I$ be the intersection of its diagonals, $K$ be the circumcenter of $ABCD$ .
The symmetry through point $I$ maps $K$ into $G$ , maps $M$ into $O$ , hence, the line $KM$ is maped in to the line $OG$ . But KM is perpendicular to $AB$ , therefore, $OG$ is also perpendicular to $AB$ .
With the same argument, we also have $PG$ , $MG$ , $NG$ , respectively, is perpendicular to $BC$ , $CD$ , $AD$ .
Then we get the result of this problem.
Our proof is completed.

Attachments:

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jayme

9775 posts

jayme

#11 h Apr 21, 2009, 9:40 AM • 1 Y

Y by Adventure10

Dear Mathlinkers,
can some one explain why the meetpoint is called "anticenter".
Who gives this name?
Any reference after Honsberger?
Sincerely
Jean-Louis

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