2005 AMC 12B Problems/Problem 18

Problem

Let $A(2,2)$ and $B(7,7)$ be points in the plane. Define $R$ as the region in the first quadrant consisting of those points $C$ such that $\triangle ABC$ is an acute triangle. What is the closest integer to the area of the region $R$ ?

$\mathrm{(A)}\ 25 \qquad \mathrm{(B)}\ 39 \qquad \mathrm{(C)}\ 51 \qquad \mathrm{(D)}\ 60 \qquad \mathrm{(E)}\ 80$

Solution

For angle $A$ and $B$ to be acute, $C$ must be between the two lines that are perpendicular to $\overline{AB}$ and contain points $A$ and $B$ . For angle $C$ to be acute, first draw a $45-45-90$ triangle with $\bar{AB}$ as the hypotenuse. Note $C$ cannot be inside this triangle's circumscribed circle or else $\angle C > 90^\circ$ . Hence, the area of $R$ is $\frac{1}{2}$ $14^2$

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2005 AMC 12B Problems/Problem 18

Problem

Solution

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