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Math Help - Maximum/minimum geometry, not easy

  1. #1
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    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
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  2. #2
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    Could somebody help me with this problem, please?
    I cannot get through it...
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ununuquantium View Post
    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}
    leads to
    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 - <br />
\frac{a \cdot tan(\theta)}{\sqrt{1 + tan^2(\theta)}} \right )^2}

    Now you take it from here.

    -Dan
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  4. #4
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    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 , and I am not so good in coordinate system - I just do not like counting) 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 ??
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ununuquantium View Post
    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 , and I am not so good in coordinate system - I just do not like counting) 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 ??
    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
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  6. #6
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    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.
    . . . . . . . . .
    Last edited by Opalg; September 5th 2007 at 02:28 AM.
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    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
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  8. #8
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    Quote Originally Posted by Ununuquantium View Post
    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'.
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  9. #9
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    OK, thank you very much
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