Geometry problem

Plictisit

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Jan 25, 2011
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Re: Homework help

Hello, Plictisit!


\(\displaystyle \text{Draw radii }CO\text{ and }NO.\)

\(\displaystyle \text{Let: }\,r \;=\:AO = BO = CO = NO.\)

\(\displaystyle \text{Let: }\,MN \cap CO \:=\:T.\)

\(\displaystyle \text{Let: }\,a \:=\:pT \,=\,TQ\,=\,CT.\)


\(\displaystyle \text{Then secant }MN\text{ looks like this:}\)

. . \(\displaystyle \begin{array}{cccccccccc} M && P && T && Q && N \\ \circ & ---- & \circ & --- & \circ & -- & \circ & ---- & \circ \\ & \frac{3}{2}a && a && a && \frac{3}{2}a \end{array}\)

\(\displaystyle \text{And: }\:TO \,=\,r-a\)


\(\displaystyle \text{In right triangle }NTO\!:\;\;TO^2 + TN^2 \:=\:ON^2 \quad\Rightarrow\quad (r-a)^2 + \left(\tfrac{5}{2}a\right)^2 \;=\;r^2\)

. . \(\displaystyle \text{which simplifies to: }\;\tfrac{29}{4}a^2 - 2ar \:=\:0 \quad\Rightarrow\quad \tfrac{a}{4}(29a - 8r) \:=\:0 \quad\Rightarrow\quad a \:=\:\tfrac{8}{29}r\)


\(\displaystyle \text{Therefore, point }T\text{ is located so that: }\:CT \:=\:\tfrac{5}{29}r\)

 
Thank you for the answer. I understood what you explained here but somehow i have to represent this and i can't measure 0.810344828. Is there any way to draw the result ? By the way this is for arhitecture class so i have to preety much solve things by drawing them.
 
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