1984 AHSME Problems/Problem 18

Revision as of 18:21, 4 June 2021 by Purplepenguin2 (talk | contribs) (Solution 2)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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

$\mathrm{(A) \ }\sqrt{2}-1 \qquad \mathrm{(B) \ }\frac{1}{2} \qquad \mathrm{(C) \ } 2-\sqrt{2} \qquad \mathrm{(D) \ }1 \qquad \mathrm{(E) \ } \text{Not uniquely determined}$


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 $\boxed{\mathrm{(C) \ } 2-\sqrt{2}}$


See Also

1984 AHSME (ProblemsAnswer KeyResources)
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