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

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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> OA </math>). Also, <math> OP, BP, </math> and <math> AP </math> are angle bisectors since <math> P </math> is the incenter. <math> OPC\congruent 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\congruent FPB </math> and <math> ADP\con 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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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>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1984|num-b=17|num-a=19}}
 
{{AHSME box|year=1984|num-b=17|num-a=19}}

Revision as of 20:12, 28 October 2011

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

Consider the triangle bound by the x-axis, the y-axis, and the line $x+y=2$. The point equidistant from the vertices of this triangle is the incenter, the point of intersection of the angle bisectors and the center of the inscribed circle. Now, remove the coordinate system. Let the origin be $O$, the y-intercept of the line be $A$, the x-intercept of the line be $B$, and the point be $P$.

[asy] 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 $x$ in the diagram is what we are looking for: the distance from the point to the x-axis ($OB$). Also, $OP, BP,$ and $AP$ are angle bisectors since $P$ is the incenter. $OPC\cong OPD$ by $AAS$, and $PD=OC$, since $PC||OD$, so $OC=CP=PD=DO=x$. Therefore, since $OA=OB=2$, we have $DA=CB=2-x$. Also, $CPB\cong FPB$ and $ADP\cong AFP$ by $AAS$, so $AF=FB=DA=CB=2-x$, and $AB=4-2x$. However, we know from the Pythagorean Theorem that $AB=2\sqrt{2}$. Therefore, $4-2x=2\sqrt{2}\implies x=2-\sqrt{2}, \boxed{\text{C}}$.

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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