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Trigonometry, right triangles

  1. Dec 13, 2007 #1
    Two points, A and B, are given in the plane. Describe the set of points X such that:

    [tex]AX^2 + BX^2 = AB^2[/tex]

    I'm not really sure how to start this off. I've drawn the circle and a line going from A to B with a midpoint AB at the origin (AB).

    I also, solved for X which is:

    [tex]X=\frac{AB}{\sqrt{A+B}}[/tex]

    Also, I could probably even use the Distance and Mid-point formula but, I really don't know what to do from here.
     
    Last edited: Dec 13, 2007
  2. jcsd
  3. Dec 13, 2007 #2

    Dick

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    Your notation could really use some work here. What does sqrt(A+B) mean if A and B are points in the plane?
     
  4. Dec 13, 2007 #3

    Gib Z

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    Let [tex]A= (a_x , a_y), B = (b_x , b_y), X= (X_x, X_y)[/tex] and relate them using the distance formula.
     
  5. Dec 14, 2007 #4
    Since AB is the midpoint on the origin, it's: (a_x + b_x)/2 , (a_y + b_y)/2

    [tex]A=\sqrt{(a_{x}-\frac{a_{x}+b_{x}}{2})^{2}+(a_{y}-\frac{a_{y}+b_{y}}{2})^{2}}[/tex]

    [tex]B=\sqrt{(b_{x}-\frac{a_{x}+b_{x}}{2})^{2}+(ab_{y}-\frac{a_{y}+b_{y}}{2})^{2}}[/tex]

    Progress?
     
    Last edited: Dec 14, 2007
  6. Dec 14, 2007 #5

    Gib Z

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    Your question didn't say anything about the origin being involved anywhere >.<
     
  7. Dec 14, 2007 #6

    Gib Z

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    Well anyway, what I meant was to use to distance formulas to find actual expressions for the distance between A & X, and B & X, and between A and B, then sub those expressions into your original question.
     
  8. Dec 14, 2007 #7

    HallsofIvy

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    There was no coordinate system in the orginal problem (I think that was what Gib Z meant when he said "Your question didn't say anything about the origin being involved anywhere") but you are welcome to set up a coordinate system. If you set up a coordinate system so that the origin is at the midpoint of AB and the x-axis lies along AB, then you can write A as [itex](x_0, 0)[/itex] and [itex](-x_0,0)[/itex]. For any point X, with coordinates (x,y), then the length of AX is [itex]\sqrt{(x-x_0)^2+ y^2}[/itex], the length of XY is [itex]\sqrt{(x+ x_0)^2+ y^2}[/itex] and the length of AB is [itex]2x_0[/itex]. Put THOSE into [itex]AX^2+ BX^2= AB^2[/itex] and see what equation you.

    You don't need to do this analytically. If you remember the basic geometric propostion that "if the vertex of an angle lies on a circle, and both rays of the angle cut the circle then the measure of the angle is equal to half the measure of the arc cut off the circle by the angle." Here, you know that angle is a right angle so arc cut off a circle is twice that: 180 degrees or half a circle. From that you get another geometric proposition: "The right angle vertex of a right triangle having line segment AB as hypotenuse lies on the circle with diameter AB."

    Finally, as Gib Z says, you need to work on your notation! You started by telling us that A, B, and X are points and then say
    [tex]X=\frac{AB}{\sqrt{A+B}}[/tex]
    Even after Gib Z complained that it makes no sense to talk about [itex]\sqrt{A+ B}[/itex] when A and B are points, you still say:
    [tex]A=\sqrt{(a_{x}-\frac{a_{x}+b_{x}}{2})^{2}+(a_{y}-\frac{a_{y}+b_{y}}{2})^{2}}[/tex]
     
  9. Dec 15, 2007 #8

    npa

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    Hi everybody!
    I would like to excuse in advance, because I'm Russian and haven't studied maths in english, that's why I could have some problems with english terminology.
    Anyway, I'll try to post my solution:
    If [tex]AX^{2}[/tex] + [tex]BX^{2}[/tex] = [tex]AB^{2}[/tex] then triangle ABX is a right triangle, where AB is a hypotenuse and angle AXB=90 degrees. Let's look at a circle with diameter AB and take any point M on this circle. The angle AMB=90 degrees (as an inscribed angle that subtends diameter), then the triangle ABM is a right triangle and we have : [tex]AM^{2}[/tex] + [tex]BM^{2}[/tex] = [tex]AB^{2}[/tex].
    So the set of points X is a circle with a diameter AB.
     
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