# The semi-perimeter 's'

• June 14th 2006, 09:36 PM
malaygoel
The 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.
• June 15th 2006, 10:35 AM
ThePerfectHacker
Quote:

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

I did not do this inequality, but since you seem knowledgable this should greatly help you finish it.
------
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.
• June 17th 2006, 05:56 PM
ThePerfectHacker
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.
• June 17th 2006, 05:57 PM
ThePerfectHacker
I will demonstrate the first part of the inequality that,
$s
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,
$\left\{ \begin{array}{c}
AP+BP> AB\\
BP+PC> BC\\
AP+PC> AC\\
$

$2AP+2BP+2CP> AB+BC+AC=2s$
Thus,
$2(AP+BP+CP)> 2s$
Thus,
$AP+BP+CP>s$
And the first part is complete.

Note: I made no use of the fact that P was a "Fermat Point".
• June 17th 2006, 05:58 PM
ThePerfectHacker
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 $AP=a$ and sides $AP=x,PC=y$
Now, we will find the necessary conditions that $AP+PC=x+y$ is maximized.
Since angle P=120 we can use the law of cosines,
$a^2=x^2+y^2-2xy\cos 120=x^2+xy+y^2$.

In analytic terms we need to,

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

Thus, maximize $f(x,y)=x+y$ with constraint curve, $g(x,y)=a^2$ where $g(x,y)=x^2+xy+y^2$.
By Lagrange's Theorem,if $(x,y)$ is a extrenum point, then
$\nabla f(x_0,y_0)=k\nabla g(x_0,y_0)$
But,
$\nabla f(x,y)=\frac{\partial (x+y)}{\partial x}\bold{i}+\frac{\partial (x+y)}{\partial y}\bold{j}=\bold{i}+\bold{j}$
Also,
$\nabla g(x,y)=\frac{\partial (x^2+xy+y^2)}{\partial x}\bold{i}+\frac{\partial (x^2+xy+y^2)}{\partial y}\bold{j}$= $(2x+y)\bold{i}+(2y+x)\bold{j}$
Thus,
$\bold{i}+\bold{j}=k(2x+y)\bold{i}+k(2y+x)\bold{j}$
Thus solve,
$\left\{ \begin{array}{c}
2x+y=1/k\\
2y+x=1/k\\
x^2+xy+y^2=a^2$

A unique solution exists,
$(x,y,k)=\left( \frac{a}{\sqrt{3}},\frac{a}{\sqrt{3}}, a\sqrt{3} \right)$

Thus the extrerum value for $x+y$ is $x=y=\frac{a}{\sqrt{3}}$ (an isoseles triangle with vertex 120).
Question is whether this is maximum or minimum.
Select any triangle with longest side $a$ and angle $120$ and see whether the sum of its sides exluding $a$ exceeds $x+y=\frac{2a}{\sqrt{3}}$ if it does then we found that the minimized value for $x+y$ otherwise we found the maximized value.
For example, select a triangle with sides, $d,2d,a$
Thus, by the law of cosines,
$d+2d^2+4d^2=a^2$ thus, $d=\frac{a}{\sqrt{7}}$ and $2d=\frac{2a}{\sqrt{7}}$ in sum you have,
$\frac{3a}{\sqrt{7}}<\frac{2a}{\sqrt{3}}$
Thus, $x+y$ 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 $\frac{2a}{\sqrt{3}}$
Similarly, the max value of the AP+PB and BP+CP is $\frac{2b}{\sqrt{3}}$ and $\frac{2c}{\sqrt{3}}$ where $a,b$ represent the other sides of triangle.

Thus,
$2(AP+BP+CP)$= $(AP+PB)+(BP+CP)+(AP+PC)=\frac{2a}{\sqrt{3}}+\frac{ 2b}{\sqrt{3}}+\frac{2c}{\sqrt{3}}$= $\frac{2(a+b+c)}{\sqrt{3}}$
Thus,(the largest possible value of)
$AP+BP+CP=\frac{a+b+c}{\sqrt{3}}=\frac{2s}{\sqrt{3} }<2s$
Because,
$\frac{1}{\sqrt{3}}<1$

Since its max is less than $2s$ then certainly an other lengths (because they are less than max which is less than $2s$ and use transitive property.)
• June 17th 2006, 11:12 PM
malaygoel
Try, it is simple
Quote:

Originally Posted by ThePerfectHacker
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.

I am not familiar with Langrange Multipliers, nevertheless I have been able able to work out a geometric method to demonstrate the secont part of the inequality, although I don't know whether it is elegant or not. The method makes use of the triangle inequality and the condition stated in the question(could you figure out the condition?).
Try to work it out, you will have fun and also it will strenghthen your belief in simplicity as it has done mine.
:)
• June 18th 2006, 08:56 AM
ThePerfectHacker
Quote:

Originally Posted by malaygoel
I am not familiar with Langrange Multipliers, nevertheless I have been able able to work out a geometric method to demonstrate the secont part of the inequality, although I don't know whether it is elegant or not. The method makes use of the triangle inequality and the condition stated in the question(could you figure out the condition?).
Try to work it out, you will have fun and also it will strenghthen your belief in simplicity as it has done mine.
:)

If, $x,y,z$ represents the lengths of AP,BP,CP
and $a,b,c$ then,

$\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.
• June 18th 2006, 06:23 PM
malaygoel
Quote:

Originally Posted by ThePerfectHacker
If, $x,y,z$ represents the lengths of AP,BP,CP
and $a,b,c$ then,

$\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.

I will post the second proof I have worked out, but to do it a diagram a needed.Please post the following diagram and I will follow it with the solution.
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
• June 18th 2006, 06:41 PM
ThePerfectHacker
Quote:

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

Two mistakes (since I am a proud formalist ;) )
2)You forgot to state the second postulate, how dare you!
• June 18th 2006, 06:50 PM
malaygoel
Quote:

Originally Posted by ThePerfectHacker
Two mistakes (since I am a proud formalist ;) )
2)You forgot to state the second postulate, how dare you!

Is there any condition in which AP doesn't intersect BC?
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
• June 18th 2006, 07:17 PM
malaygoel
Quote:

Originally Posted by malaygoel
I will post the second proof I have worked out, but to do it a diagram a needed.Please post the following diagram and I will follow it with the solution.

I will use only triangle inequality so I will not state the name of ineequality used in solution.
In triangle ABD
$AB + BD > AP + PD$
$AC + CD > AP + PD$
$PD + BD> PB$
$PD + CD> PC$
Adding these four equations we get
$AB + BC + CA > 2AP + PB + PC$
Let BP intersect AC at E, CP intersect AB at F
Similarly we can form three eqution. adding three gives
$4(AB + AC + BC) > 4(AP + BP + CP)$
$2s > AP + BP + CP$
Proof complete
Do you like it?
Keep Smiling
Malay