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,

$\displaystyle s<AP+BP+CP$

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,

$\displaystyle \left\{ \begin{array}{c}

AP+BP> AB\\

BP+PC> BC\\

AP+PC> AC\\

$

Add them,

$\displaystyle 2AP+2BP+2CP> AB+BC+AC=2s$

Thus,

$\displaystyle 2(AP+BP+CP)> 2s$

Thus,

$\displaystyle AP+BP+CP>s$

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 $\displaystyle AP=a$ and sides $\displaystyle AP=x,PC=y$

Now, we will find the necessary conditions that $\displaystyle AP+PC=x+y$ is maximized.

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

$\displaystyle a^2=x^2+y^2-2xy\cos 120=x^2+xy+y^2$.

In analytic terms we need to,

$\displaystyle \mbox{Maximize }x+y$

$\displaystyle \mbox{Subject to }x^2+xy+y^2=a^2, x,y>0$

Thus, maximize $\displaystyle f(x,y)=x+y$ with constraint curve, $\displaystyle g(x,y)=a^2$ where $\displaystyle g(x,y)=x^2+xy+y^2$.

By Lagrange's Theorem,if $\displaystyle (x,y)$ is a extrenum point, then

$\displaystyle \nabla f(x_0,y_0)=k\nabla g(x_0,y_0)$

But,

$\displaystyle \nabla f(x,y)=\frac{\partial (x+y)}{\partial x}\bold{i}+\frac{\partial (x+y)}{\partial y}\bold{j}=\bold{i}+\bold{j}$

Also,

$\displaystyle \nabla g(x,y)=\frac{\partial (x^2+xy+y^2)}{\partial x}\bold{i}+\frac{\partial (x^2+xy+y^2)}{\partial y}\bold{j}$=$\displaystyle (2x+y)\bold{i}+(2y+x)\bold{j}$

Thus,

$\displaystyle \bold{i}+\bold{j}=k(2x+y)\bold{i}+k(2y+x)\bold{j}$

Thus solve,

$\displaystyle \left\{ \begin{array}{c}

2x+y=1/k\\

2y+x=1/k\\

x^2+xy+y^2=a^2$

A unique solution exists,

$\displaystyle (x,y,k)=\left( \frac{a}{\sqrt{3}},\frac{a}{\sqrt{3}}, a\sqrt{3} \right)$

Thus the extrerum value for $\displaystyle x+y$ is $\displaystyle x=y=\frac{a}{\sqrt{3}}$ (an isoseles triangle with vertex 120).

Question is whether this is maximum or minimum.

Select any triangle with longest side $\displaystyle a$ and angle $\displaystyle 120$ and see whether the sum of its sides exluding $\displaystyle a$ exceeds $\displaystyle x+y=\frac{2a}{\sqrt{3}}$ if it does then we found that the minimized value for $\displaystyle x+y$ otherwise we found the maximized value.

For example, select a triangle with sides, $\displaystyle d,2d,a$

Thus, by the law of cosines,

$\displaystyle d+2d^2+4d^2=a^2$ thus, $\displaystyle d=\frac{a}{\sqrt{7}}$ and $\displaystyle 2d=\frac{2a}{\sqrt{7}}$ in sum you have,

$\displaystyle \frac{3a}{\sqrt{7}}<\frac{2a}{\sqrt{3}}$

Thus, $\displaystyle x+y$ 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 $\displaystyle \frac{2a}{\sqrt{3}}$

Similarly, the max value of the AP+PB and BP+CP is $\displaystyle \frac{2b}{\sqrt{3}}$ and $\displaystyle \frac{2c}{\sqrt{3}}$ where $\displaystyle a,b$ represent the other sides of triangle.

Thus,

$\displaystyle 2(AP+BP+CP)$=$\displaystyle (AP+PB)+(BP+CP)+(AP+PC)=\frac{2a}{\sqrt{3}}+\frac{ 2b}{\sqrt{3}}+\frac{2c}{\sqrt{3}}$=$\displaystyle \frac{2(a+b+c)}{\sqrt{3}}$

Thus,(the largest possible value of)

$\displaystyle AP+BP+CP=\frac{a+b+c}{\sqrt{3}}=\frac{2s}{\sqrt{3} }<2s$

Because,

$\displaystyle \frac{1}{\sqrt{3}}<1$

Since its max is less than $\displaystyle 2s$ then certainly an other lengths (because they are less than max which is less than $\displaystyle 2s$ 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 $\displaystyle a,b,c$ then,

$\displaystyle \left\{ \begin{array}{c}

x^2+xy+y^2=a^2\\

x^2+xz+z^2=b^2\\

y^2+yz+z^2=c^2\\

$

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

$\displaystyle AB + BD > AP + PD$

$\displaystyle AC + CD > AP + PD$

$\displaystyle PD + BD> PB$

$\displaystyle PD + CD> PC$

Adding these four equations we get

$\displaystyle AB + BC + CA > 2AP + PB + PC$

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

Similarly we can form three eqution. adding three gives

$\displaystyle 4(AB + AC + BC) > 4(AP + BP + CP)$

$\displaystyle 2s > AP + BP + CP $

Proof complete

Do you like it?

Keep Smiling

Malay