# geometry

• Dec 13th 2007, 09:50 AM
jenjen
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.
• Dec 13th 2007, 09:54 AM
topsquark
Quote:

Originally Posted by jenjen
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
• Dec 13th 2007, 11:50 AM
Soroban
Hello, jenjen!

Quote:

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: . $y \;=\;\frac{x\sqrt{x^2+h^2} - x^2}{h}$