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

Sevenoptimus (talk | contribs) (Added a copy of the problem, and fixed some LaTeX) |
Sevenoptimus (talk | contribs) (Cleaned up the solution and added more explanation) |
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==Solution== | ==Solution== | ||

− | Place the square | + | Place the square in the <math>xy</math>-plane with <math>A</math> as the origin, so that <math>B=(1,0), C=(1,1),</math> and <math>D=(0,1).</math> We are given that <math>PA^2+PB^2=PC^2,</math> so |

− | <cmath>\begin{align*}&(x^2+y^2)+((x-1)^2+y^2)=(x-1)^2+(y-1)^2\\ &2x^2+2y^2-2x+1=x^2+y^2-2x-2y+2\\ &x^2+y^2=-2y+1\\ &x^2+y^2+2y-1=0\\ &x^2+(y+1)^2=2\end{align*}</cmath> | + | <cmath>\begin{align*}&(x^2+y^2)+((x-1)^2+y^2)=(x-1)^2+(y-1)^2\\ \Rightarrow \quad &2x^2+2y^2-2x+1=x^2+y^2-2x-2y+2\\ \Rightarrow \quad &x^2+y^2=-2y+1\\ \Rightarrow \quad &x^2+y^2+2y-1=0\\ \Rightarrow \quad &x^2+(y+1)^2=2.\end{align*}</cmath> |

− | Thus | + | Thus we see that <math>P</math> lies on a circle centered at <math>(0,-1)</math> with radius <math>\sqrt{2}.</math> The farthest point from <math>D</math> on this circle is at the bottom of the circle, at <math>(0, -1-\sqrt{2}),</math> in which case <math>PD</math> is <math>1 - (-1 - \sqrt{2}) = \boxed{\textbf{(C)}\ 2 + \sqrt{2}}.</math> |

## Revision as of 18:58, 27 January 2019

## Problem

A point lies in the same plane as a given square of side . Let the vertices of the square, taken counterclockwise, be and . Also, let the distances from to and , respectively, be and . What is the greatest distance that can be from if ?

## Solution

Place the square in the -plane with as the origin, so that and We are given that so

Thus we see that lies on a circle centered at with radius The farthest point from on this circle is at the bottom of the circle, at in which case is