Math Is Fun Forum

juantheron

Member

Registered: 2011-10-19

Posts: 312

Triangle with minimum area

Among all the triangles containing a given circle, find the one whose area is the smallest

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: Triangle with minimum area

hi juantheron,

Made the diagram below in Sketchpad and it looks like it has to be equilateral. 

Can I prove it?

Not yet, but I'm working on it.

The sides ( AD, DF and FA) can be split into 6 segments with AB = AC, DB = DE and FE = FC.

The total area = ½AB.BO +  ½AC.CO +   ½DB.BO +  ½DE.EO + ½FE.EO + ½FC.CO

                     = ½radius(AB + BD + AC + CF + DE + EF) = ½radius . perimeter.

So the problem reduces to finding the smallest perimeter.

Bob

Last edited by Bob (2012-01-09 06:30:20)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

juantheron

Member

Registered: 2011-10-19

Posts: 312

Re: Triangle with minimum area

Thanks bibbundy (This will surely help me)

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: Triangle with minimum area

hi bob

i just found a new nickname for you.it was inspired by juantheron's last post.i shall call you bib!

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“Here lies the reader who will never open this book. He is forever dead.
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: Triangle with minimum area

hi juantheron,

Here's a proof that the triangle has the smallest area when it is equilateral.

It is a bit 'messy' but it seems to work.

I'll repeat the diagram below.

The circle BCE is fixed.  Let radius = R.

For the moment, fix A as well.

That means angle AOB + angle AOC is fixed.

So reflex BOC is fixed.

As OD bisects angle BOE and FO bisects COE, this means that angle DOF is fixed. 

Let angle DOF = P (fixed) and angle EOF = Q (variable as E moves).

Lemma.

DE + EF is minimum when Q = ½P

Proof.

DE depends on R and DO.  EF depends on R and FO.

So DE + EF will be minimum when DO + OF is minimum.   ***

A diagram will easily show that this total  length gets smaller as E is moved into a position so that EO bisects P.  ***

Let V = DO + OF

So turning points when this is zero:

This will happen when Q = ½P.  ***

.........................................

So with A fixed, the minimum is when EO bisects angle DOF.

Now repeat this argument with these angles, but now fixing D

The new minimum will be when  CO bisects angle AOF.  It will be smaller than the previous minimum.

By continuing this process a sequence of diminishing minimums will be generated as the angles at the centre tend towards equal values.

When all are 60 degrees the triangle ADF will be equilateral and no further diminishing is possible.

This proof is a bit 'dodgy' at the points marked ***.  I've had to take a few liberties in order to reach a conclusion.  I've checked with a Sketchpad model and all these steps are true.  The reader is invited to improve on what I've written!

Bob

Last edited by Bob (2012-01-10 07:33:25)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob