Break

[Sujeet Kumar]

Following problem will give break from my boring problems-
Q.Find the triangle of the least area which can cover any triangle with sides not exceeding 1.

Above problem has just one connection with my previous problems.

[Avni Pllana]

Hi Sujeet,

the sought triangle is an equilateral triangle with side length a = 2*sqrt(3)/3 .

Best regards,
Avni

[Philippe 92]

>> Q.Find the triangle of the least area which can cover
>> any triangle with sides not exceeding 1.
>>

> the sought triangle is an equilateral triangle with
> side length a = 2*sqrt(3)/3 .

Dear Avni,

This triangle is not the smallest.

The smallest is a triangle with base 1 and altitude 1, and both base
angles >=60 deg, hence area = 0.5, instead of your equilateral
triangle with altitude 1, hence side 2/sqrt(3), which has an area of
1/sqrt(3) = 0.57735 > 0.5

The base of this triangle must be >= 1 to accept equilateral triangle
with side 1 inside.
The altitude is >= 1 to accept triangles with base ~0 and
altitude ~1.cos(0)
However the *smaller* side of the inside triangle must be on the
base of the enclosing triangle to guarantee that this triangle fits.
Your wrong solution is with some other side on the base !
The placing of the inside triangle is then not optimal, resulting
into a unnecessary larger solution.

Regards.

Attachment available from http://mathforum.org/kb/servlet/JiveServlet/download/129-2226526-7354369-663744/smallesttriangle.gif

[Sujeet Kumar]

To Anni and Philippee

I found this problem in an old russian geometry book.Acording to this book.More stronger result was obtained by M.D. Kovalev.It states- Among all the convex
figures covering any triangle with sides not exceeding
unity, the smallest area is posseessed by the triangle
ABC in which angle A=60 degree, AB=1, and the altitude
drawn to AB is equal to cos 10 deg.The area of this triangle equals 0.5*cos(10)=0.4924.
Solution is very long and I will write it later on.

Best Regards from Sujeet