1984 AHSME Problems/Problem 18

Problem

A point $(x, y)$ is to be chosen in the coordinate plane so that it is equally distant from the x-axis, the y-axis, and the line $x+y=2$ . Then $x$ is

$\mathrm{(A) \ }\sqrt{2}-1 \qquad \mathrm{(B) \ }\frac{1}{2} \qquad \mathrm{(C) \ } 2-\sqrt{2} \qquad \mathrm{(D) \ }1 \qquad \mathrm{(E) \ } \text{Not uniquely determined}$

Solution

The $x$ -axis and $y$ -axis intersect at $(0,0)$ , while the line $x+y=2$ intersects the $x$ -axis at $(2,0)$ and the $y$ -axis at $(0,2)$ . Let $A (0,2)$ , $B (2,0)$ , and $C (0,0)$ be these three points. Then the incenter and the three excenters of $\triangle ABC$ must all be equidistant from all three of these lines. Since the $B$ -excenter has a negative $x$ -coordinate while the incenter and other two excenters have positive $x$ -coordinates, $x$ is $\boxed{(\mathbf{E})\ \mathrm{not\ uniquely\ determined}}$ .

-j314andrews

1984 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1984 AHSME Problems/Problem 18

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