Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1984 AHSME Problems/Problem 18"
(Solution 2)
 
Line 34: Line 34:
  
 
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>.
 
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>.
 +
==Solution 2==
 +
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>
  
 +
-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}}

Latest revision as of 18:21, 4 June 2021

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, $4-2x=2\sqrt{2}\implies x=2-\sqrt{2}, \boxed{\text{C}}$.

Solution 2

We notice that $(x, y)$ is the incenter of the triangle bounded by the x-axis, the y-axis, and the line $x+y=2$. The problem is asking for $x$, which is just the inradius. The inradius is $\frac{a+b-c}{2}$, so the answer is $\boxed{\mathrm{(C) \ } 2-\sqrt{2}}$

-purplepenguin2

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1984_AHSME_Problems/Problem_18&oldid=155235"