Maximum/minimum geometry, not easy

• Sep 2nd 2007, 01:22 PM
Ununuquantium
Maximum/minimum geometry, not easy
Given a point P in the interior of an angle with vertex O, find points A and B on the sides of the angle so that OA=OB and the sum PA+PB is minimal
• Sep 4th 2007, 09:14 AM
Ununuquantium
Could somebody help me with this problem, please?
I cannot get through it...
• Sep 4th 2007, 12:31 PM
topsquark
Quote:

Originally Posted by Ununuquantium
Given a point P in the interior of an angle with vertex O, find points A and B on the sides of the angle so that OA=OB and the sum PA+PB is minimal

Let's change the perspective here a bit and let O be the origin and OA be along the positive x-axis. (We may do this without loss of generality.) Let $\theta$ be the angle AOB. Then the coordinates of A will be $(a, 0)$ and for point B will be $(x_B, tan(\theta)x_B )$.

We want OA = OB, so
$\sqrt{a^2} = \sqrt{x_B^2 + tan^2(\theta)x_B^2}$
$x_B = \frac{a}{\sqrt{1 + tan^2(\theta)}}$

So the coordinates of A and B are:
$A \to (a, 0)$
and
$B \to \left ( \frac{a}{\sqrt{1 + tan^2(\theta)}}, \frac{a \cdot tan(\theta)}{\sqrt{1 + tan^2(\theta)}} \right )$

Let the coordinates of point P be simply $x, y$.

Now you want to set up an expression for the the distance
$d = PA + PB$

$d = \sqrt{(x - a)^2 + (y - 0)^2} + \sqrt{\left ( x - \frac{a}{\sqrt{1 + tan^2(\theta)}} \right )^2 + \left ( y -
\frac{a \cdot tan(\theta)}{\sqrt{1 + tan^2(\theta)}} \right )^2}$

Now you take it from here.

-Dan
• Sep 4th 2007, 01:17 PM
Ununuquantium
Than you very much Dan.

Could you please prove that angles BPO and APO are equal?

I cannot do it using coordinate system (I like classical geometry solutions :o, and I am not so good in coordinate system - I just do not like counting:o) So I thaught about normal solution, maybe using symmetry ? I think this problem is not so hard to do it using coordinate system - but if You want to use it Dan, please show me whole solution with proving that angles BPO and APO are equal. Could you please help me with whole solution :o??
• Sep 4th 2007, 01:57 PM
topsquark
Quote:

Originally Posted by Ununuquantium
Than you very much Dan.

Could you please prove that angles BPO and APO are equal?

I cannot do it using coordinate system (I like classical geometry solutions :o, and I am not so good in coordinate system - I just do not like counting:o) So I thaught about normal solution, maybe using symmetry ? I think this problem is not so hard to do it using coordinate system - but if You want to use it Dan, please show me whole solution with proving that angles BPO and APO are equal. Could you please help me with whole solution :o??

Too bad about not being able to use coordinates. Once you get an expression for x any y from the equation in my first post you should be able to prove that easily. There might be a short-cut around that (which would be nice, that distance equation is going to be a brute to minimize) but if there is I haven't spotted it yet.

Here's a hint for what to do with your distance equation in case it comes in handy:
Solve $\nabla d = 0$ to get the critical points.

-Dan
• Sep 5th 2007, 03:11 AM
Opalg
Here's a more geometric approach to the problem, which answers the question about angles.

Rotate the diagram through an angle $\theta$ about the point O (where $\theta$ is the angle AOB), so that A goes to A', B goes to B' and P goes to P'. Superimpose the rotated diagram on the original diagram. Then A' coincides with B (because OA=OB). We want to minimise AP+PB = BP'+PB. But this is easy: the minimum obviously occurs when PBP' is a straight line.

So the construction goes as follows: rotate P through the angle AOB about the point O, to get P'. Then B is the point where PP' crosses the line OB.

The triangle OPP' is isosceles (because OP=OP'). Therefore the angles OP'B and BPO are equal. But OP'B=OPA. So the angles BPO and OPA are equal.
. . . . . . . . . http://www.thelances.co.uk/imagesC/angles.png
• Sep 5th 2007, 08:02 AM
Ununuquantium
Ok, I understand - but tell me - You take hypothetical pionts A and B, and then you rotare everything and you say that this sum AP+PB is the smalles when PBP' is a straight line?
I mean the construction is like that:
We have an angle with vertex O and point P inside. We put hypothetical pionts A and B such that OA=OB and THEN we rotate "everything" - and then we say that this sum AP+PB is the smalles when PBP' is a straight line?

And thank You very much

Ununuquantium
• Sep 5th 2007, 08:45 AM
Opalg
Quote:

Originally Posted by Ununuquantium
Ok, I understand - but tell me - You take hypothetical pionts A and B, and then you rotare everything and you say that this sum AP+PB is the smalles when PBP' is a straight line?
I mean the construction is like that:
We have an angle with vertex O and point P inside. We put hypothetical pionts A and B such that OA=OB and THEN we rotate "everything" - and then we say that this sum AP+PB is the smalles when PBP' is a straight line?

Yes, I was being a bit casual about notation. We start with two lines making an acute angle $\theta$ at the point O where they meet, and a point P somewhere within the sector formed by this angle. I call the two lines OA and OB, but the positions of the points A and B on these lines are not initially determined. If you like, they are variable points on their respective lines, but subject to the condition OA=OB. We then rotate the picture to find the point P'. Then PA+PB=P'B+PB. This represents the distance from P to P', going via B. Since "a straight line is the shortest distance between two points", we can minimise this distance by choosing B to lie on the line PP'.
• Sep 5th 2007, 09:08 AM
Ununuquantium
OK, thank you very much