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Difference between revisions of "1984 AHSME Problems/Problem 18"

(The correct answer (according to official answers) was E, not C! The incenter and excenters are all possible points.)
m (→‎Solution)
 
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The <math>x</math>-axis and <math>y</math>-axis intersect at <math>(0,0)</math>, while the line <math>x+y=2</math> intersects the <math>x</math>-axis at <math>(2,0)</math> and the <math>y</math>-axis at <math>(0,2)</math>.  Let <math>A (0,2)</math>, <math>B (2,0)</math>, and <math>C (0,0)</math> be these three points.  Then the incenter and the three excenters of <math>\triangle ABC</math> must all be equidistant from all three of these lines.  Since the <math>B</math>-excenter has a negative <math>x</math>-coordinate while the incenter and other two excenters have positive <math>x</math>-coordinates, <math>x</math> is <math>\boxed{(\mathbf{E})\ \mathrm{not\ uniquely\ determined}}</math>.
 
The <math>x</math>-axis and <math>y</math>-axis intersect at <math>(0,0)</math>, while the line <math>x+y=2</math> intersects the <math>x</math>-axis at <math>(2,0)</math> and the <math>y</math>-axis at <math>(0,2)</math>.  Let <math>A (0,2)</math>, <math>B (2,0)</math>, and <math>C (0,0)</math> be these three points.  Then the incenter and the three excenters of <math>\triangle ABC</math> must all be equidistant from all three of these lines.  Since the <math>B</math>-excenter has a negative <math>x</math>-coordinate while the incenter and other two excenters have positive <math>x</math>-coordinates, <math>x</math> is <math>\boxed{(\mathbf{E})\ \mathrm{not\ uniquely\ determined}}</math>.
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-j314andrews
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1984|num-b=17|num-a=19}}
 
{{AHSME box|year=1984|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:11, 3 July 2025

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

See Also

1984 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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