- View SourceDear Darij
> Please help me with the following problem (I need it as

Let A' and D' be the reflections of A and D about O.

> a lemma):

>

> Given a cyclic quadrilateral ABCD with the circumcenter

> O. The perpendicular to BD through B meets the

> perpendicular to AC through C at E. The perpendicular

> to BD through D meets the perpendicular to AC through A

> at F. Finally, let X be the intersection of the lines

> AB and CD. Then, the points O, E, F, X are collinear.

>

> Well, it is easy to show that O is the midpoint of EF;

> hence, O, E, F are collinear, but how about X ?

O = AA' inter DD', E = CA' inter BD' and X = AB inter DC are

collinear by Pascal's theorem.

Friendly. Jean-Pierre - View SourceDear Jean-Pierre,

In Hyacinthos message #9148, you wrote:

>> [DG]

Thanks for the proof!

>> > Please help me with the following problem (I need it as

>> > a lemma):

>> >

>> > Given a cyclic quadrilateral ABCD with the circumcenter

>> > O. The perpendicular to BD through B meets the

>> > perpendicular to AC through C at E. The perpendicular

>> > to BD through D meets the perpendicular to AC through A

>> > at F. Finally, let X be the intersection of the lines

>> > AB and CD. Then, the points O, E, F, X are collinear.

>> >

>> > Well, it is easy to show that O is the midpoint of EF;

>> > hence, O, E, F are collinear, but how about X ?

>>

>> [JPE]

>> Let A' and D' be the reflections of A and D about O.

>> O = AA' inter DD', E = CA' inter BD' and X = AB inter DC

>> are collinear by Pascal's theorem.

And here is how I came up with the problem above:

Some time ago I solved the problem 3 in the 2nd Round

of the Bundeswettbewerb Mathematik (German National

Mathematics Competition) 2003. The problem was:

Given a cyclic quadrilateral ABCD, let S be the meet

of the diagonals AC and BD, and K and L the

orthogonal projections of S on the sides AB and CD.

Show that the perpendicular bisector of the segment

KL bisects the sides BC and DA.

Well, the proof is not very easy, but not too

complicated, too. However, I found an interesting

additional result: The circumcircles of triangles

SBC, SDA and SKL and the circle with diameter SO are

coaxal, i. e. they have a common point different from

S. Hereby, O means the circumcenter of our

quadrilateral ABCD.

Now, it is easy to show that the circumcircle of

triangle SBC has the segment SE as diameter, the

circumcircle of triangle SDA has the segment SF as

diameter, and the circumcircle of triangle SKL has

the segment SX as diameter. Finally, it remains to

show that the circles with diameters SE, SF, SX and

SO are coaxal.

This is very easy using the collinearity of E, F, X,

O you have proven.

Thanks again, and I will send a paper on the

Bundeswettbewerb problem, together with my solution,

the above result with its proof and some other

extensions to the little German mathematics

periodical "Die Wurzel" with the corresponding

credit to you in the proof of the above result. I

can send you a PDF file of the paper if you are

interested (it is in German, however).

Sincerely,

Darij Grinberg - View Source
>

Dear Darij, Jean-Pierre and other colleagues!

>Given a cyclic quadrilateral ABCD, let S be the meet

>of the diagonals AC and BD, and K and L the

>orthogonal projections of S on the sides AB and CD.

>Show that the perpendicular bisector of the segment

>KL bisects the sides BC and DA.

>

There is another interesting problem. Let given a quadrilateral ABCD, S is

the common point of AC and BD, K, L, M, N - the projections of S on AB, BC,

CD, DA. Then if we have only the points K, L, M, N we can restore the

quadrilateral ABCD. So we have a bijection of quadrilaterals ABCD - KLMN.

The problem is: what properties of KLMN corresponds to given properties of

ABCD. Some results are known.

1. KLMN is circumscribed if and only if ABCD is inscribed.

2. KLMN is inscribed if and only if AC and BD are perpendicular.

3. KL*MN=ML*NK if and only if PS and SQ are perpendicular, where P is common

point of AB and CD, Q - common point of AD and BC.

4. KL and MN are parallel if and only if the angles A and C are equal.

5. The angles K and M are equal if and only if AD and BC are parallel.

Does anybody know another facts?

My paper concerning this problem was typed in "Kvant" (1998, N 4).

Sincerely Alexey - View SourceDear Alexey,

Just a few quick observations (without proof):

* Your properties 1 & 2 give rise to an easy and/or elegant

construction of a bi-centric quadrilateral, something I have been

looking for for a while. All you need to do is to take a circle and a

point P inside the circle. The four points of intersections of any pair

of perpendicular lines through with the circle will form a cyclic

quadrilateral and the KLMN that you construct will be bi-centric. When

you construct a bi-centric KLMN in this way, it turns out that both

``centers'' of KLMN and the center on the circumscribed circle of ABCD

are collinear. In fact, the circumcenter to KLMN lies exactly halfway

between the incenter of KLMN and the circumcenter of ABCD. Perhaps this

property can be used to give an alternative proof of Krafft's (?)

formula for the relation between the radii of incenter and excenter of

a bicentric quadrilateral and the distance between the centers of the

circles.

* For ABCD cyclic, A,B,C,D, are the centers of the ex-circles of KLMN.

This also means that the point of intersection of KL and MN and that of

KN and LM each lie on a diagonal of ABCD.

Eisso

On Thursday, January 29, 2004, at 01:33 AM, Alexey.A.Zaslavsky wrote:

> There is another interesting problem. Let given a quadrilateral ABCD,

> S is

> the common point of AC and BD, K, L, M, N - the projections of S on

> AB, BC,

> CD, DA. Then if we have only the points K, L, M, N we can restore the

> quadrilateral ABCD. So we have a bijection of quadrilaterals ABCD -

> KLMN.

> The problem is: what properties of KLMN corresponds to given

> properties of

> ABCD. Some results are known.

> 1. KLMN is circumscribed if and only if ABCD is inscribed.

> 2. KLMN is inscribed if and only if AC and BD are perpendicular.

> 3. KL*MN=ML*NK if and only if PS and SQ are perpendicular, where P is

> common

> point of AB and CD, Q - common point of AD and BC.

> 4. KL and MN are parallel if and only if the angles A and C are equal.

> 5. The angles K and M are equal if and only if AD and BC are parallel.

> Does anybody know another facts?

> My paper concerning this problem was typed in "Kvant" (1998, N 4).

--

Eisso J. Atzema, Ph.D.

Department of Mathematics & Statistics

University of Maine

Orono, ME 04469

(207) 581-3928 (office)

(207) 866-3871 (home)

atzema@... or FirstClass

[Non-text portions of this message have been removed] - View SourceDear Eisso!
>

You are right. This proof of Ponsele theorem for n=4 is typed in Sharygin's

>* Your properties 1 & 2 give rise to an easy and/or elegant

>construction of a bi-centric quadrilateral, something I have been

>looking for for a while. All you need to do is to take a circle and a

>point P inside the circle. The four points of intersections of any pair

>of perpendicular lines through with the circle will form a cyclic

>quadrilateral and the KLMN that you construct will be bi-centric. When

>you construct a bi-centric KLMN in this way, it turns out that both

>``centers'' of KLMN and the center on the circumscribed circle of ABCD

>are collinear. In fact, the circumcenter to KLMN lies exactly halfway

>between the incenter of KLMN and the circumcenter of ABCD. Perhaps this

>property can be used to give an alternative proof of Krafft's (?)

>formula for the relation between the radii of incenter and excenter of

>a bicentric quadrilateral and the distance between the centers of the

>circles.

>

book "Problems of planymetry".

Sincerely Alexey