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• 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
Message 1 of 6 , Jan 28, 2004
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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 ?

Thanks!
Sincerely,
Darij Grinberg
• Dear Darij ... 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.
Message 2 of 6 , Jan 28, 2004
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Dear Darij
> 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 ?

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.
Friendly. Jean-Pierre
• Dear Jean-Pierre, ... Thanks for the proof! 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
Message 3 of 6 , Jan 28, 2004
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Dear Jean-Pierre,

In Hyacinthos message #9148, you wrote:

>> [DG]
>> > 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.

Thanks for the proof!

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
• ... Dear Darij, Jean-Pierre and other colleagues! There is another interesting problem. Let given a quadrilateral ABCD, S is the common point of AC and BD, K,
Message 4 of 6 , Jan 28, 2004
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>
>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.
>
Dear Darij, Jean-Pierre and other colleagues!
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
• Dear 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
Message 5 of 6 , Jan 29, 2004
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Dear 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]
• Dear Eisso! ... You are right. This proof of Ponsele theorem for n=4 is typed in Sharygin s book Problems of planymetry . Sincerely
Message 6 of 6 , Jan 30, 2004
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Dear Eisso!
>
>* 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.
>
You are right. This proof of Ponsele theorem for n=4 is typed in Sharygin's
book "Problems of planymetry".

Sincerely Alexey
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