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Difference between revisions of "1983 AHSME Problems/Problem 29"

(Created page with "Place the square on the coordinate plane with <math>A</math> as the origin. (This means that <math>B=(1,0), C=(1,1),</math> and <math>D=(0,1).</math>We are given that <math>PA...")
(No difference)

Revision as of 11:05, 7 June 2016

Place the square on the coordinate plane with $A$ as the origin. (This means that $B=(1,0), C=(1,1),$ and $D=(0,1).$We are given that $PA^2+PB^2=PC^2,$ so

(x2+y2)+((x−1)2+y2)=(x−1)2+(y−1)22x2+2y2−2x+1=x2+y2−2x−2y+2x2+y2=−2y+1x2+y2+2y−1=0x2+(y+1)2=2

Thus, we see that $P$ is on a circle centered at $(0,-1)$ with radius $\sqrt{2}.$ The farthest point from $D$ on this circle is at the bottom of the circle, at $(0, -1-\sqrt{2}),$ so $PD$ is $\boxed{2+\sqrt{2}}.$

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