In a circle with center O,suppose A, P, B are | vedclass

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(A) Let the circle be the unit circle with center $O(0,0)$. Let $A = (1, 0)$,$B = (\cos 	heta, \sin 	heta)$,and $P = (\cos(	heta/2), \sin(	heta/2)

Vedclass · Accepted Answer

$\frac{1}{\sqrt{5}}$

Vedclass · Answer

(A) Let the circle be the unit circle with center $O(0,0)$. Let $A = (1, 0)$,$B = (\cos 	heta, \sin 	heta)$,and $P = (\cos(	heta/2), \sin(	heta/2))$.The area of $	riangle AOB = \frac{1}{2} |x_A(y_O - y_B) + x_O(y_B - y_A) + x_B(y_A - y_O)| = \frac{1}{2} |1(0 - \sin 	heta) + 0 + \cos 	heta(0 - 0)| = \frac{1}{2} \sin 	heta$.The area of $	riangle APB = \frac{1}{2} |x_A(y_P - y_B) + x_P(y_B - y_A) + x_B(y_A - y_P)| = \frac{1}{2} |1(\sin(	heta/2) - \sin 	heta) + \cos(	heta/2)(\sin 	heta - 0) + \cos 	heta(0 - \sin(	heta/2))|$.Using $\sin 	heta = 2\sin(	heta/2)\cos(	heta/2)$ and $\cos 	heta = 1 - 2\sin^2(	heta/2)$:Area of $	riangle APB = \frac{1}{2} |\sin(	heta/2) - 2\sin(	heta/2)\cos(	heta/2) + 2\sin(	heta/2)\cos^2(	heta/2) - \sin(	heta/2)(1 - 2\sin^2(	heta/2))| = \frac{1}{2} |\sin(	heta/2) - 2\sin(	heta/2)\cos(	heta/2) + 2\sin(	heta/2)\cos^2(	heta/2) - \sin(	heta/2) + 2\sin^3(	heta/2)| = \frac{1}{2} |2\sin(	heta/2)(\sin^2(	heta/2) + \cos^2(	heta/2)) - 2\sin(	heta/2)\cos(	heta/2)| = \sin(	heta/2)(1 - \cos(	heta/2))$.Given $\frac{\frac{1}{2} \sin 	heta}{\sin(	heta/2)(1 - \cos(	heta/2))} = \sqrt{5} + 2$.Since $\sin 	heta = 2\sin(	heta/2)\cos(	heta/2)$,the ratio becomes $\frac{\cos(	heta/2)}{1 - \cos(	heta/2)} = \sqrt{5} + 2$.Let $x = \cos(	heta/2)$. Then $x = (\sqrt{5} + 2)(1 - x)$ $\Rightarrow x = \sqrt{5} + 2 - x(\sqrt{5} + 2)$ $\Rightarrow x(1 + \sqrt{5} + 2) = \sqrt{5} + 2$ $\Rightarrow x = \frac{\sqrt{5} + 2}{\sqrt{5} + 3} = \frac{(\sqrt{5} + 2)(\sqrt{5} - 3)}{5 - 9} = \frac{5 - 3\sqrt{5} + 2\sqrt{5} - 6}{-4} = \frac{-1 - \sqrt{5}}{-4} = \frac{\sqrt{5} + 1}{4}$.For $2	heta$,the ratio is $\frac{\cos 	heta}{1 - \cos 	heta}$.Since $\cos 	heta = 2\cos^2(	heta/2) - 1 = 2(\frac{\sqrt{5} + 1}{4})^2 - 1 = 2(\frac{5 + 1 + 2\sqrt{5}}{16}) - 1 = \frac{6 + 2\sqrt{5}}{8} - 1 = \frac{3 + \sqrt{5}}{4} - 1 = \frac{\sqrt{5} - 1}{4}$.Ratio $= \frac{(\sqrt{5} - 1)/4}{1 - (\sqrt{5} - 1)/4} = \frac{\sqrt{5} - 1}{4 - \sqrt{5} + 1} = \frac{\sqrt{5} - 1}{5 - \sqrt{5}} = \frac{\sqrt{5} - 1}{\sqrt{5}(\sqrt{5} - 1)} = \frac{1}{\sqrt{5}}$.

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