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T. Popoviciu, Problem 5897, Gaz. Mat. 49 (1943), 322.

rmabry
Posts: 2
 Joined: 2009-05-06

Greetings from Louisiana!

First, I apologize if I am in the wrong forum. I cannot easily translate Romanian into English.

I'm hoping someone can provide me with a copy of the original printing of T. Popoviciu, Problem 5897, Gaz. Mat. 49 (1943), 322.

A translation into English would be most gratefully accepted, as well.

The problem is cited on pp 132-133 of "Geometric Inequalities" by O. Bottema, et al,
Wolters-Noordhoff Publishing, Groningen 1969.

Here is a snapshot that has a statement of the problem:

http://www.lsus.edu/sc/math/rmabry/corners/bottema.png

The trouble is, the statement by Bottema, "Equality only for a parallelogram", is false! I have an article soon to appear in Mathematics Magazine that discusses the problem (and related topics), and a referee alerted me to the citation by Bottema et al. So now I need to know, did Bottema get it wrong, or was Popoviciu in error?

Meanwhile, it is an easy exercise to show that  the ratio 1/5 holds if and only if the inner quadrilateral is a trapezoid, while the ratio 1/6 is approached for quadrilaterals tending toward triangles.

Many, many thanks for any help. I have searched for weeks for a copy of the problem, but no libraries on the WorldCat network have any issues in 1943 (a difficult year, no doubt). Perhaps someone here with access to the electronic version of Gazeta Matematica can post or send a copy of this one page. 

Also, is the problem cited elsewhere in GM? That is, do the statement of the problem and its solution appear separately? I would be grateful, of course, for both entries, if that is the case.

 

Cheers,

Rick Mabry

Department of Mathematics
Louisiana State University in Shreveport

 

 

 


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Publicat de rmabry la 2009, Mai 6 - 19:48.
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rmabry
Posts: 2
 Joined: 2009-05-06
Here is the statement of the

Here is the statement of the problem in Gazeta. Thanks to Marin Vlada (Bucharest) and Eugen Ionascu (Columbus, GA, USA) for independently sending me copies.

***************

5897. Fie ABCD un patrulater convex de arie S şi A', B', C', D' mijloacele laturilor BC, CD, DA, AB. Dreptele AA', BB', CC', DD', prin intersecţiile lor mutuale, formează un alt patrulater convex, interior patrulaterului ABCD şi de arie S'. Să se demonstreze, că avem

5S' ≤ S ≤ 6S'

şi să se determine cazurile în care una din aceste inegalităţi devine o egalitate.

***************

The article in which I discuss the problem (and variations of it) has not yet appeared. Meanwhile, I still do not know if a solution to the problem ever appeared in Gazeta.

With respect,

Rick Mabry

LSUS



2010, Aprilie 22 - 05:14
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