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May 8th, 2015, 05:21 AM   #1
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Bisector and inner circle. Hard to solve

On the plane xy there are four points: O(0, 0) A(0, 3) B(0, -3), C(4,0).

When point D is the intersections of bisector of angle ABC (Angle B)and the x-axis, then OD : DC = ? : ?.
and the coordinates of the inner center of ABC are (?/2; ?)

I have nooo idea how to solve it I would appreciate your help
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May 8th, 2015, 06:53 AM   #2
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I'm not sure this is the simplest way to do it but a "brute strength" method would be to use the cosine law to find cosine of angle B:
The side opposite angle B has length 5 and the other two sides are of length 5 and 6:
$b^2= a^2+ c^2- 2ac cos(B)$
25= 25+ 36- 2(30)cos(B)
co(B)= 36/60= 6/10= 0.6

The bisector makes angle B/2 with the y-axis and $cos(B/2)= \sqrt{\frac{cos(B)+ 1}{2}}$. Here that is $cos(B/2)= \sqrt{0.8}$ so $sin(B/2)= \sqrt{1- 0.8}= \sqrt{0.2}$ and $tan(B/2)= \frac{\sqrt{.2}}{\sqrt{.8}}= \sqrt{\frac{1}{4}}= \frac{1}{2}$.

Since $tan(B/2)= \frac{OD}{OB}$, $\frac{1}{2}= \frac{OD}{3}$ so OD= 3/2.

Last edited by Country Boy; May 8th, 2015 at 06:59 AM.
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May 8th, 2015, 07:44 AM   #3
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Since the intersection of the angle bisectors of a triangle is the center of its incircle, OD = r and DC = 4 - r, where r is the inradius.

To compute the inradius use A = Sr, where A is the area of the triangle and S is the semiperimeter (half the perimeter).
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