Triangle Problem: Max/Min PD+PE+PF

[ehrenfest]

Homework Statement

A point P is located in the interior of an equilateral triangle ABC. Perpendiculars drawn from P meet each of the sides in points D,E,F, respectively. Where should P be located to make PD + PE + PF a maximum? What about a minimum?

Homework Equations

The Attempt at a Solution

From symmetry considerations. I think that the maximum will be at the incenter. I am not sure about the minimum. Maybe at a vertex, but that is not really in the interior... I tried the standard reflections. I am not sure if I actually need to set up a coordinate system with P at the origin... That looks like it could get messy? I am also not sure how to use the fact that this is an equilateral triangle.

 

[ehrenfest]

anyone?

 

[marlon]

anyone?

Isn't this one of them Lagrange multiplier things ?

Having said that, i don't know how to set up :

1)the funtion that needs to be minimized/maximized

2) the function that is going to be the constraint.

I would reallly like to see the solution to this problem, though.

Marlon

 

[jacobrhcp]

I think it's my english that's bad here, but how can you draw 3 perpendicular lines in a plane?

 

[chickendude]

They are not perpendicular to each other.

They are perpendicular to each side of the triangle.

 

[chickendude]

How about this approach: (sorry for the double post)

Draw segments from P to each vertex of the triangle. You have split the triangle into three new triangles.

The area of each triangle is given by $$\frac{1}{2}sh_i$$ where s is the side length of the equilateral triangle and h_i is the height of the perpendicular on that side

Through this, we can set up the equality

$$\frac{1}{2}sh_a + \frac{1}{2}sh_b + \frac{1}{2}sh_c = \frac{s^2 \sqrt{3}}{4}$$

$$h_a + h_b + h_c = \frac{s \sqrt{3}}{2}$$

meaning that the sum of the three perpendiculars is always constant regardless of the location of P

 

[jacobrhcp]

well, take the situation that P is at point a, then if you move D a distance d, for instance, you can still hold E still, but you'll have to move F to keep P one point.

now P moves in the direction of E with a distance: 2d/root[3], this is the distance it shrinks.
the distance it grows is: 2d tan(30) = 2d/root[3], this is equal.

because any motion of P can be described as the sum of the motion in the direction of perpendiculars, the distance is always the same.

 

[jacobrhcp]

I must have done something wrong, because chickendude seems right.

EDIT: I've edited my post and saw what i did wrong.

 

[Rainbow Child]

Thinking about a coordinate system is a good idea, but the choise of te origin is bad

Let a vertex of the triangle be at the origin, say B and vertex C, lie on $x$ axe. The coordinates of the vertexes are

$$A(\frac{\alpha}{2},\frac{\sqrt{3}\,\alpha}{2}, \,B(0,0), \, C(\alpha,0)$$

where $\alpha$ is the side length of the equilateral triangle.
With this arragement the points $$E \in AB,\, F \in AC,\, D \in BC$$ have coordinates

$$E(\frac{\alpha}{4},\frac{\sqrt{3}\,\alpha}{4}),\,F(\frac{3\,\alpha}{4},\frac{\sqrt{3}\,\alpha}{4}),\,D(\frac{\alpha}{2},0)$$

Now let the coordinates of P be $$(x,y)$$. The function you want to extremize is

$$f(x,y)=|PD|+|PE|+|PF|\Rightarrow f(x,y)=\sqrt{(x-\frac{\alpha}{2})^2+y^2}+\sqrt{(x-\frac{\alpha}{4})^2+(y-\frac{\sqrt{3}\,\alpha}{4})^2}+\sqrt{(x-\frac{3\,\alpha}{4})^2+(y-\frac{\sqrt{3}\,\alpha}{4})^2}$$​

By finding the critical points of $$f$$, i.e. $$\vec{\nabla}\,f=0$$ you arrive to

• The minimum value of $$f$$ correspons to the point $$P(\frac{\alpha}{2},\frac{\sqrt{3}\,\alpha}{6})$$ which is nothing else but the circumcenter.
• The maximum value of $$f$$ corresponds to the vertex A, thus from symmerty every vertex maximizes $$f$$. There is no interior point P, which makes $$f$$ maximum.

Hope I helped!

 

[ehrenfest]

By finding the critical points of $$f$$, i.e. $$\vec{\nabla}\,f=0$$ you arrive to

• The minimum value of $$f$$ correspons to the point $$P(\frac{\alpha}{2},\frac{\sqrt{3}\,\alpha}{6})$$ which is nothing else but the circumcenter.
• The maximum value of $$f$$ corresponds to the vertex A, thus from symmerty every vertex maximizes $$f$$. There is no interior point P, which makes $$f$$ maximum.

Hope I helped!

Hmmm. I did not see anything wrong with chickendude's post, which means there is probably something wrong here.

 

[NateTG]

Hmmm. I did not see anything wrong with chickendude's post, which means there is probably something wrong here.

If you consider the point $p$ to be the tip of three triangles, each of which has one of the sides of the original equilateral triangles as it's base then the total area of those triangles will be:
$$\frac{1}{2} l_1 s + \frac{1}{2} l_2 s + \frac{1}{2} l_3 s = \frac{1}{2} (l_1 + l_2 +l_3) s$$
which is also the area of the original triangle, which is clearly constant, so
$$(l_1+l_2+l_3)$$
must be constant.

 

[ehrenfest]

If you consider the point $p$ to be the tip of three triangles, each of which has one of the sides of the original equilateral triangles as it's base then the total area of those triangles will be:
$$\frac{1}{2} l_1 s + \frac{1}{2} l_2 s + \frac{1}{2} l_3 s = \frac{1}{2} (l_1 + l_2 +l_3) s$$
which is also the area of the original triangle, which is clearly constant, so
$$(l_1+l_2+l_3)$$
must be constant.

Yes. You just repeated chickendude's proof. I am just saying if that is right, then something is wrong with Rainbow Child's proof.

 

[Rainbow Child]

My mistake!

In my proof I took the points

$$E(\frac{\alpha}{4},\frac{\sqrt{3}\,\alpha}{4}),\,F (\frac{3\,\alpha}{4},\frac{\sqrt{3}\,\alpha}{4}),\ ,D(\frac{\alpha}{2},0)$$

which are the midpoints of the sides AB, AC and AD! (nothing to do with the original problem )

So I gave the answer to the question:

"Which point extemizes the sum $$|PD|+|PE|+|PF|$$ when D, E and F are the midpoints of the sides!"

Sorry!