P is any point inside a triangle ABC. The perimeter of the triangle AB +BC+CA =2s. Prove that s< AP+BP+CP<2s.

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- Jun 14th 2006, 09:36 PMmalaygoelThe semi-perimeter 's'
P is any point inside a triangle ABC. The perimeter of the triangle AB +BC+CA =2s. Prove that s< AP+BP+CP<2s.

- Jun 15th 2006, 10:35 AMThePerfectHackerQuote:

Originally Posted by**malaygoel**

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That point is called the "Fermat Point".

Consider triangle ABC and let P be inside this triangle.

We want that AP+BP+CP to be minimized.

Rotate 60 degrees triangle APB around point B to outside.

You now have a new triangle A'P'B

Triangle P'BP is equilateral cuz, P'B=PB with <P'BP=60

Thus AP+BP+CP=A'P'+P'P+PC

But A'P'+P'P+PC is the path from A' to C

It is minimal when A'P'+P'P+PC is a sraight line.

Thus, BPC=180-P'PB=180-60=120

Thus, the "Fermat Point" form lines of 120 degrees. - Jun 17th 2006, 05:56 PMThePerfectHacker
I have completed the solution to this problem. As I suspected my first post on this problem was necessary to complete the proof. Thus, to understand my solution you will not to understand my first post denomstrating that the minimized point has the property that, <CPB= <BPA= <CPA= 120 degrees.

Below is a picture showing the geometric construction in case you were no able to draw one yourself due to my bad explantion. - Jun 17th 2006, 05:57 PMThePerfectHacker
I will demonstrate the first part of the inequality that,

That is simple to answer.

1)Triangle ABC

3)Point P is inside triangle ABC

2)Draw AP, BP, CP

3)Three triangle, APB, APC, BPC.

4)By the triangular inequality you have,

Add them,

Thus,

Thus,

And the first part is complete.

Note: I made no use of the fact that P was a "Fermat Point". - Jun 17th 2006, 05:58 PMThePerfectHacker
I appologize that I have not been able to find an elegant geometric method to demonstrate the second part of this inequality. You need to be familiar with Lagrange Multipliers.

Begin by P being a "Fermat Point", Meaning that the triangles,

APB, APC, BPC are have angle P all 120 degress (this is my first post).

Take triangle APC and let side and sides

Now, we will find the necessary conditions that is maximized.

Since angle P=120 we can use the law of cosines,

.

In analytic terms we need to,

Thus, maximize with constraint curve, where .

By Lagrange's Theorem,if is a extrenum point, then

But,

Also,

=

Thus,

Thus solve,

A unique solution exists,

Thus the extrerum value for is (an isoseles triangle with vertex 120).

Question is whether this is maximum or minimum.

Select any triangle with longest side and angle and see whether the sum of its sides exluding exceeds if it does then we found that the minimized value for otherwise we found the maximized value.

For example, select a triangle with sides,

Thus, by the law of cosines,

thus, and in sum you have,

Thus, is a maximum point!

Now, the rest is easy.

By this argument we have shown that the maximum value that AP+PC can have is

Similarly, the max value of the AP+PB and BP+CP is and where represent the other sides of triangle.

Thus,

= =

Thus,(the largest possible value of)

Because,

Since its max is less than then certainly an other lengths (because they are less than max which is less than and use transitive property.) - Jun 17th 2006, 11:12 PMmalaygoelTry, it is simpleQuote:

Originally Posted by**ThePerfectHacker**

Try to work it out, you will have fun and also it will strenghthen your belief in simplicity as it has done mine.

:) - Jun 18th 2006, 08:56 AMThePerfectHackerQuote:

Originally Posted by**malaygoel**

and then,

But solving these equations is probably messy and the solution is not as simple as through Lagrange Multipliers. - Jun 18th 2006, 06:23 PMmalaygoelQuote:

Originally Posted by**ThePerfectHacker**

Draw a triangle ABC

Take a point P in the interior of ABC

Join AP, BP and CP

Extend AP to meet BC at D(this can be done since P is an interior point) :)

Keep Smiling - Jun 18th 2006, 06:41 PMThePerfectHackerQuote:

Originally Posted by**malaygoel**

1)You need to demonstrate that AP intersects BC. Actually I made a thread about this, but cannot find it :(

2)You forgot to state the second postulate, how dare you! - Jun 18th 2006, 06:50 PMmalaygoelQuote:

Originally Posted by**ThePerfectHacker**

I didn't forget to state the second postulate, you can see my first post in this thread.

Nevertheless, if it not the condition you cannot prove it because it is not valid then.

Keep Smiling - Jun 18th 2006, 07:17 PMmalaygoelQuote:

Originally Posted by**malaygoel**

In triangle ABD

Adding these four equations we get

Let BP intersect AC at E, CP intersect AB at F

Similarly we can form three eqution. adding three gives

Proof complete

Do you like it?

Keep Smiling

Malay