Let $A(2,2)$ and $B(7,7)$ be the points in the plane,Define $R$ as the region in the first quadrant consisting of those points $C$ such that $ABC$ is a acute triangle.Find area of region $R$.
For the triangle to be acute angled triangle,angles $A,B,C$ should be all acute angles.But i cant figure out the boundary of the region $R$ in which $C$ can lie.
Some hint/suggestion is needed.Please help me.
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diya
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4$\angle A$ is acute iff $C$ is lying above line $x+y = 4$. $\angle B$ is acute iff $C$ is lying below the line $x+y = 14$.$\angle C$ is acute iff $C$ is lying outside the circle having $AB$ as diameter. – achille hui Dec 29 '15 at 06:09
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Draw lines perpendicular to $AB$ at $A$ and at $B$, then you'll see that in the region $R_1 = \{(x,y)\in\Bbb R^2:4\le x+y \le 14\}$, $\angle A$ and $\angle B$ are acute, and the area of $R_1$ is $\dfrac{14^2-4^2}{2}=90$. For $\angle C < 90^\circ$, we need to substract the circle with $AB$ as the diameter, so the answer is $90-\pi(\dfrac{5}{\sqrt{2}})^2=90-\dfrac{25\pi}{2}$
Please see the graph.
GNUSupporter 8964民主女神 地下教會
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If $C$ is in this region, then angles $A$ and $B$ are guaranteed to be acute, but angle $C$ might not. – Joey Zou Dec 29 '15 at 06:04
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@JoeyZou Thanks for your comment. Corrected the answer. – GNUSupporter 8964民主女神 地下教會 Dec 29 '15 at 06:10