1984 AHSME Problems/Problem 18
Problem
A point 
is to be chosen in the coordinate plane so that it is equally distant from the x-axis, the y-axis, and the line 
. Then 
is
Solution
Consider the triangle bound by the x-axis, the y-axis, and the line . 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 
, the y-intercept of the line be 
, the x-intercept of the line be 
, and the point be 
.
![[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]](http://latex.artofproblemsolving.com/3/3/8/338db661a59c7385704518e59aca471d5bd27a7d.png)
Notice that in the diagram is what we are looking for: the distance from the point to the x-axis (
). Also, 
and 
are angle bisectors since is the incenter. 
by 
, and 
, since 
, so 
. Therefore, since 
, we have 
. Also, 
and 
by , so 
, and 
. However, we know from the Pythagorean Theorem that 
. Therefore, 
.
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 | ||
