# Trigonometry Question

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• September 2nd 2012, 11:40 AM
DonManolo
Trigonometry Question
Being http://forum2011.obmep.org.br/latexr...d23525e932.gif a triangle and http://forum2011.obmep.org.br/latexr...0c9a0fdaaa.gif a point inside http://forum2011.obmep.org.br/latexr...d23525e932.gif.
Find the coordinates of the points http://forum2011.obmep.org.br/latexr...319e541aee.gif and http://forum2011.obmep.org.br/latexr...0caf1283d6.gif on the sides of http://forum2011.obmep.org.br/latexr...d23525e932.gif, in function of the coordinates of the points http://forum2011.obmep.org.br/latexr...c130d45b94.gif and http://forum2011.obmep.org.br/latexr...0c9a0fdaaa.gif, in a way that http://forum2011.obmep.org.br/latexr...0c9a0fdaaa.gif divides the line segment http://forum2011.obmep.org.br/latexr...bc048a646a.gif in half.
(Tip: Consider the point http://forum2011.obmep.org.br/latexr...b72eacbe29.gifto be on the origin of the coordinates system and point http://forum2011.obmep.org.br/latexr...957afab571.gif to be on the http://forum2011.obmep.org.br/latexr...3038fdd89b.gif axis.)
• September 3rd 2012, 09:30 PM
kalyanram
1 Attachment(s)
Re: Trigonometry Question
Hey Don,
Refer to the figure attached. Choose x-axis along $AB$ and y-axis along $AC$ with the origin at $A$.

$Q,R$ can be interchanged.

Case 1.
$QR$ on sides $AB$ and $AC$ we have $Q=(2\alpha,0)$, $R=(0,2\beta)$. With the additional condition that $0 \le \alpha \le \frac{b}{2}$, $0 \le \beta \le \frac{c}{2}$

Case 2.
$QR$ on sides $AC$ and $BC$ we have the abscissa of $Q$ is $0$ and hence abscissa of $R$ has to be $2\alpha$. Let the ordinate of $Q$ be $\gamma$ then we have ordinate of $R$ has to be $2\beta-\gamma$ and it has to satisfy the line equation $\frac{y}{x-b} = -\frac{c}{b} \implies \gamma = -\frac{c(2\alpha - b)}{b}$ with the additional bound constraints $0 \le \alpha \le \frac{b}{2}$ and $0 \le \gamma \le c$

Case 3.
Can be done on similar lines of Case2.

Now use the rotational matrix to translate to the Rectangular Cartesian Co-ordinates.

~Kalyan.
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