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Math Help - Hard math problem

  1. #1
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    Hard math problem

    My friend gave me a good question today so I want to see if you smart geniuses can answer it

    Consider a fixed line AB=4. Also consider a line CD, such that
    AB=CD=4, and that AB is the perpendicular bisector of CD AND CD is the perpendicular bisector of AB.
    Now, consider the set of all points P on CD (which there are inifineitely many). For every unique point P, there is a unique point Q such that QBPA is cyclic (all four points lie on a circle).
    The set of all possible points Q creates a familiar figure.
    What is that figure and what is the area of that figure?
    Prove this.

    I said it was a cirlce with area 4pi, but he said it was wrong?
    What other familar shape can it possibly be? If i draw it it seems like a circle (i just plotted a lot of points)
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  2. #2
    Moo
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    Hello,

    For every unique point P, there is a unique point Q such that QBPA is cyclic
    Is there some information missing here ?
    Because for a point P, there are infinitely many points Q such that Q,B,P,A lie on a circle

    Or am I missing something ?
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  3. #3
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    Sorry, same user, i forogt user name

    anyways, you are right

    I forgot one more condition (forogt what he tole me):

    QP is bisected by AB. Thus there are two unique points Q for every P.

    Anyways, inifintiely many points P will create a unique figure for P.

    What is the figure and what is that area
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  4. #4
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    Quote Originally Posted by mluo View Post
    Consider a fixed line segment AB=4. Also consider a line segment CD, such that
    AB=CD=4, and that AB is the perpendicular bisector of CD AND CD is the perpendicular bisector of AB.
    Now, consider the set of all points P on CD (which there are inifineitely many). For every unique point P, there are two unique points Q such that QBPA is cyclic (all four points lie on a circle) and that QP is bisected by AB.
    The set of all possible points Q creates a familiar figure.
    What is that figure and what is the area of that figure?
    Prove this.

    I said it was a cirlce with area 4pi, but he said it was wrong?
    What other familar shape can it possibly be? If i draw it it seems like a circle (i just plotted a lot of points)
    1. Please don't double post. It's against the rules and it wastes everybodies time.

    2. See attachment
    Attached Thumbnails Attached Thumbnails Hard math problem-cycl_viereck.png  
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  5. #5
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    It is an ellipse, with foci at A and B and eccentricity 1/√2. So its area is 4√2π.

    Now prove it.
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  6. #6
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    Isnt the proof just Angle Bisector or am I wrong?
    This is because the chords AP and BP equal.
    Do you have a good solution?
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  7. #7
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    Quote Originally Posted by person8901 View Post
    Isnt the proof just Angle Bisector or am I wrong?
    This is because the chords AP and BP equal.
    Do you have a good solution?
    I have a solution, though I wouldn't call it a "good" solution. (I would prefer to have an argument using synthetic geometry to show that AQ+QB = const.)

    Take a coordinate system in which A and B are the points (±a,0), and C and D are (0,±a). (We are told that a=2, but I prefer to work with letters rather than numbers.)

    If a circle contains A, B and P, then its centre must be on the y-axis, say at the point (0,t). Then the radius of the circle is \sqrt{t^2+a^2}, and its equation is x^2 + (y-t)^2 = t^2+a^2. Thus P is the point (0,t-\sqrt{t^2+a^2}).

    The condition that PQ is bisected by AB tells you that the y-coordinate of Q is the negative of the y-coordinate of P. So if Q=(x,y) then y =  \sqrt{t^2+a^2} - t. Putting those coordinates into the equation of the circle, and simplifying, we get the equation x^2 = 4t(\sqrt{t^2+a^2}-t) = 4ty. Thus t = \frac{x^2}{4y}. Substitute that value of t into the equation y = \sqrt{t^2+a^2} - t, and you get y = \sqrt{\frac{x^4}{16y^2} + a^2} - \frac{x^2}{4y}. Write this as 4y^2+x^2 = \sqrt{x^4 + 16a^2y^2}, square both sides, simplify, and the result comes out as \frac{x^2}{2a^2} + \frac{y^2}{a^2} = 1, the standard equation of an ellipse with semi-axes √2a and a.
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