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        riddles >> easy >> Find the Limit
        
(Message started by: ThudanBlunder on Jan 22^nd, 2009, 7:14am)

Title: Find the Limit
Post by ThudanBlunder on Jan 22^nd, 2009, 7:14am

The triangle ABC below is isosceles with AB = BC. AX bisects angle CAB and X is the point of intersection of the angle bisector and the side CB. Let OC = 1 and let B move towards the side AC of the triangle along the perpendicular bisector BO.

1) What is the limiting length of OX as
a) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif -> 0?
b) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif -> http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/4?

2) What is the locus of X for 0 < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif//4?

Title: Re: Find the Limit
Post by SMQ on Jan 22^nd, 2009, 8:54am

Place O at the origin, and let X have coordinates (x, y). From line AX we have [hide]y = (1 + x) tan http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif[/hide], and from line CB we have [hide]y = (1 - x) tan 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif[/hide]. Setting these equal and using the [hide]double angle formula for tan[/hide] we find [hide]x = (1 + tan² http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif) / (2 - tan² http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif), y = 4 tan http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif/ (3 - tan² http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif)[/hide]. We can now evaluate 1 a) and 1 b) directly as [hide]1/3 and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5[/hide] respectively.

--SMQ

Title: Re: Find the Limit
Post by Eigenray on Jan 22^nd, 2009, 12:36pm

...which gives the [hide]hyperbola (3x+1)² - 3y² = 4[/hide].
Interestingly, [hide]one of the foci is the point C. The other is at (-5/3,0)[/hide] :-/

Title: Re: Find the Limit
Post by Immanuel_Bonfils on Jan 27^th, 2009, 7:52am

No need of Analytic...
[hide]From the bisector theorem we get |CX| = (2 sec http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif)/ (2 + sec http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif) where http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif= 2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif . Then the cosine law on OCX triangle gives x² = 1 + (4 sec ² http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif) / (2 + sec http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif)² - 4 / ( 2 + sec http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif), where x = |OX|.
Then comes traight the limits for http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif->0 giving x ->1/3 and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif-> http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2 giving x->http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif 5 [/hide]