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

(Solution 2)
(The correct answer (according to official answers) was E, not C! The incenter and excenters are all possible points.)
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==Solution==
 
==Solution==
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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>.
 
  
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<asy>
+
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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unitsize(4cm);
 
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draw((0,0)--(0,2)--(2,0)--cycle);
 
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draw((2-sqrt(2),2-sqrt(2))--(0,0));
 
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draw((2-sqrt(2),2-sqrt(2))--(0,2));
 
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draw((2-sqrt(2),2-sqrt(2))--(2,0));
 
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draw((2-sqrt(2),2-sqrt(2))--(0,2-sqrt(2)));
 
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draw((2-sqrt(2),2-sqrt(2))--(2-sqrt(2),0));
 
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draw((2-sqrt(2),2-sqrt(2))--(1,1));
 
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label("$O$",(0,0),WNW);
 
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label("$A$",(0,2),NW);
 
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label("$B$",(2,0),NE);
 
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label("$P$",(2-sqrt(2),2-sqrt(2)),NNE);
 
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label("$C$",(2-sqrt(2),0),S);
 
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label("$D$",(0,2-sqrt(2)),W);
 
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label("$x$",(0,1-sqrt(1/2)),W);
 
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label("$x$",(1-sqrt(1/2)),S);
 
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label("$x$",(2-sqrt(2),1-sqrt(1/2)),E);
 
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label("$x$",(1-sqrt(1/2),2-sqrt(2)),N);
 
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label("$2-x$",(2-sqrt(1/2),0),S);
 
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label("$F$",(1,1),NE);
 
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label("$2-x$",(0,2-sqrt(1/2)),W);
 
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label("$2-x$",(1/2,3/2),NE);
 
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label("$2-x$",(3/2,1/2),NE);
 
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</asy>
 
  
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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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==Solution 2==
 
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We notice that <math> (x, y) </math> is the incenter of the triangle bounded by the x-axis, the y-axis, and the line <math>x+y=2</math>. The problem is asking for <math>x</math>, which is just the inradius. The inradius is <math>\frac{a+b-c}{2}</math>, so the answer is <math>\boxed{\mathrm{(C) \ } 2-\sqrt{2}}</math>
 
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-purplepenguin2
 
 
==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}}

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 $\boxed{(\mathbf{E})\ \mathrm{not\ uniquely\ determined}}$.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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