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

Wiggle Wam (talk | contribs) (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...") |
Sevenoptimus (talk | contribs) (Added a copy of the problem, and fixed some LaTeX) |
||

Line 1: | Line 1: | ||

+ | ==Problem== | ||

+ | |||

+ | A point <math>P</math> lies in the same plane as a given square of side <math>1</math>. Let the vertices of the square, | ||

+ | taken counterclockwise, be <math>A, B, C</math> and <math>D</math>. Also, let the distances from <math>P</math> to <math>A, B</math> and <math>C</math>, respectively, be <math>u, v</math> and <math>w</math>. | ||

+ | What is the greatest distance that <math>P</math> can be from <math>D</math> if <math>u^2 + v^2 = w^2</math>? | ||

+ | |||

+ | <math>\textbf{(A)}\ 1 + \sqrt{2}\qquad | ||

+ | \textbf{(B)}\ 2\sqrt{2}\qquad | ||

+ | \textbf{(C)}\ 2 + \sqrt{2}\qquad | ||

+ | \textbf{(D)}\ 3\sqrt{2}\qquad | ||

+ | \textbf{(E)}\ 3+\sqrt{2} </math> | ||

+ | |||

+ | ==Solution== | ||

+ | |||

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^2+PB^2=PC^2,</math> so | 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^2+PB^2=PC^2,</math> so | ||

− | \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>\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> |

Thus, we see that <math>P</math> is 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> so <math>PD</math> is <math>\boxed{2+\sqrt{2}}.</math> | Thus, we see that <math>P</math> is 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> so <math>PD</math> is <math>\boxed{2+\sqrt{2}}.</math> |

## Revision as of 18:53, 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 on the coordinate plane with as the origin. (This means that and We are given that so

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