I would appreciate if somebody could help me with the following problem
Q: Find $S(\theta)$ ?
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Hint: $OQ\cdot OP=OA^2$ since $OA$ is the tangent line of the circle – Shuchang Aug 31 '13 at 11:55
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That's a nice picture. How did you make it? – Stefan Smith Aug 31 '13 at 12:50
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using smartdraw7 – Young Aug 31 '13 at 13:30
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3Young: Any reason why you did not accept the answer in your question here? http://math.stackexchange.com/questions/320068/how-to-find-int-01h-nxdx/320140#320140 In fact, out of 58 questions asked, you have accepted only 2. High standards indeed. – Ron Gordon Aug 31 '13 at 13:35
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Note that $A(\Delta QPA) = A(\Delta OPF) - A(\Delta OQA) - A(\Delta APF)$, where $F$ is the foot of the perpendicular drawn from $P$ to the $X$ axis.
Note that
$$A(\Delta OPF) = \frac12 OP^2 \sin{\theta} \cos{\theta}$$ $$A(\Delta OQA) = \frac12 OQ \sin{\theta}$$ $$A(\Delta APF) = \frac12 OP \sin{\theta} (OP \cos{\theta}-1)$$
After a little algebra, I get that
$$A(\Delta QPA) = \frac12 (OP-OQ) \sin{\theta} $$
We find $P$ and $Q$ from the intersection of the circle $(x-1)^2+(y-1)^2=1$ and the line $y=\tan{\theta} x$. After some algebra (which I will leave for the reader), I find that
$$OP = \cos{\theta} + \sin{\theta} + \sqrt{2 \sin{\theta} \cos{\theta}}$$ $$OQ = \cos{\theta} + \sin{\theta} - \sqrt{2 \sin{\theta} \cos{\theta}}$$
and therefore
$$A(\Delta QPA) = \sin{\theta} \sqrt{\sin{2 \theta}}$$
Note that $OP \cdot OQ = 1$ as Shuchang pointed out.
Ron Gordon
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