Question

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 $R?$

$25$

$39$

$51$

$60$

$80$

Solution:

Line $AB$ has slope $1.$ For $\angle A$ to be acute, $C$ must lie beyond the line through $A$ perpendicular to $AB;$ in the first quadrant that line runs between $P(4, 0)$ and $Q(0, 4).$ For $\angle B$ to be acute, $C$ must lie before the line through $B$ perpendicular to $AB,$ between $S(14, 0)$ and $T(0, 14).$

For $\angle C$ to be acute, $C$ must lie outside the circle $U$ with diameter $AB,$ whose radius is $\dfrac{AB}{2} = \dfrac{5\sqrt2}{2}.$

The region is the large right triangle $OST$ minus the small right triangle $OPQ$ minus the semicircle-equivalent area of $U$ inside the strip: $\dfrac12 \cdot 14^2 - \dfrac12 \cdot 4^2 - \pi\left(\dfrac{5\sqrt2}{2}\right)^2 = 98 - 8 - \dfrac{25\pi}{2} = 90 - \dfrac{25\pi}{2} \approx 51.$

Thus, the correct answer is C.

2005 AMC 12B Problem 18

Solution:

Problem 18 in Other Years