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  1. #1
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    geometry

    plz help...



    Suppose we are given an isosceles triangle ABC in the coordinate plane R^2 whose vertices are given by A = (o, h), B= (-x, 0), and C= (x, 0). Then the point J where the angle bisectors meet lies on the y-axis. Find its
    y-coordinate.




    I tried to approach this problem by taking the square of a, letting a = tan beta x. is that the correct way? Thank you so much for the help.
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  2. #2
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    Quote Originally Posted by jenjen View Post
    plz help...



    Suppose we are given an isosceles triangle ABC in the coordinate plane R^2 whose vertices are given by A = (o, h), B= (-x, 0), and C= (x, 0). Then the point J where the angle bisectors meet lies on the y-axis. Find its
    y-coordinate.




    I tried to approach this problem by taking the square of a, letting a = tan beta x. is that the correct way? Thank you so much for the help.
    Maybe. What's a?

    I would do this as follows:

    Find one of the base angles (I think you were calling this \beta) by
    h = x~tan(\beta)

    Then the y coordinate of the point of intersection of the two bisectors will be determined by
    y = x~tan \left ( \frac{\beta}{2} \right )

    -Dan
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  3. #3
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    Hello, jenjen!

    Given an isosceles triangle ABC with vertices: . A (0,h),\;B(-x,0),\;(x, 0)
    Then point J where the angle bisectors meet lies on the y-axis.
    Find its y-coordinate.
    Code:
                    A|
                     *(0,h)
                    /|\
                   / | \
                  /  |  \
                 /   |   \
                /    | J  \
               /     *     \
              / θ *  |y     \
           B / * θ   |       \ C
        - - * - - - -+- - - - * - -
         (-x,0)      D      (x,0)

    Let \theta = \angle JBC = \angle ABJ

    We know that: . \tan2\theta \:=\:\frac{2\tan\theta}{1-\tan^2\!\theta} \:=\:\frac{h}{x}

    . . and we see that: . \tan\theta \:=\:\frac{y}{x}

    So we have: . \frac{2\tan\theta}{1-\tan^2\!\theta} \:=\:\frac{h}{x}\quad\Rightarrow\quad \frac{2\left(\frac{y}{x}\right)}{1-\left(\frac{y}{x}\right)^2} \:=\:\frac{h}{x}

    This simplifies to a quadratic in y\!:\;\;hy^2 + 2x^2y - hx^2\:=\:0

    Quadratic Formula: . y \;=\;\frac{-2x^2\pm\sqrt{4x^4 + 4h^2x^2}}{2h} \;=\;\frac{-x^2\pm x\sqrt{x^2+h^2}}{h}

    . . and the positive root is: . y \;=\;\frac{x\sqrt{x^2+h^2} - x^2}{h}

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