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

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==Solution==
 
==Solution==
Consider the [[triangle]] bound by the x-axis, the y-axis, and the line <math> x+y=2 </math>. The point [[equidistant]] from the [[vertices]] of this triangle is the [[incenter]], the [[Concurrency|point of intersection]] of the [[Angle bisector|angle bisectors]] and the center of the inscribed [[circle]]. Now, remove the coordinate system. Let the origin be <math> O </math>, the y-intercept of the line be <math> A </math>, the x-intercept of the line be <math> B </math>, and the point be <math> P </math>.
 
  
<asy>
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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>.
unitsize(4cm);
 
draw((0,0)--(0,2)--(2,0)--cycle);
 
draw((2-sqrt(2),2-sqrt(2))--(0,0));
 
draw((2-sqrt(2),2-sqrt(2))--(0,2));
 
draw((2-sqrt(2),2-sqrt(2))--(2,0));
 
draw((2-sqrt(2),2-sqrt(2))--(0,2-sqrt(2)));
 
draw((2-sqrt(2),2-sqrt(2))--(2-sqrt(2),0));
 
draw((2-sqrt(2),2-sqrt(2))--(1,1));
 
label("$O$",(0,0),WNW);
 
label("$A$",(0,2),NW);
 
label("$B$",(2,0),NE);
 
label("$P$",(2-sqrt(2),2-sqrt(2)),NNE);
 
label("$C$",(2-sqrt(2),0),S);
 
label("$D$",(0,2-sqrt(2)),W);
 
label("$x$",(0,1-sqrt(1/2)),W);
 
label("$x$",(1-sqrt(1/2)),S);
 
label("$x$",(2-sqrt(2),1-sqrt(1/2)),E);
 
label("$x$",(1-sqrt(1/2),2-sqrt(2)),N);
 
label("$2-x$",(2-sqrt(1/2),0),S);
 
label("$F$",(1,1),NE);
 
label("$2-x$",(0,2-sqrt(1/2)),W);
 
label("$2-x$",(1/2,3/2),NE);
 
label("$2-x$",(3/2,1/2),NE);
 
</asy>
 
  
Notice that <math> x </math> in the diagram is what we are looking for: the [[distance]] from the point to the x-axis (<math> OB </math>). Also, <math> OP, BP, </math> and <math> AP </math> are angle bisectors since <math> P </math> is the incenter. <math> OPC\cong OPD </math> by <math> AAS </math>, and <math> PD=OC </math>, since <math> PC||OD </math>, so <math> OC=CP=PD=DO=x </math>. Therefore, since <math> OA=OB=2 </math>, we have <math> DA=CB=2-x </math>. Also, <math> CPB\cong FPB </math> and <math> ADP\cong AFP </math> by <math> AAS </math>, so <math> AF=FB=DA=CB=2-x </math>, and <math> AB=4-2x </math>. However, we know from the [[Pythagorean Theorem]] that <math> AB=2\sqrt{2} </math>. Therefore, <math> 4-2x=2\sqrt{2}\implies x=2-\sqrt{2}, \boxed{\text{C}} </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 (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AHSME Problems and Solutions

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