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

(Added a copy of the problem, and fixed some LaTeX)
(Cleaned up the solution and added more explanation)
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==Solution==
 
==Solution==
  
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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  
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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  
  
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<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>
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<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>
  
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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>
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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 $P$ lies in the same plane as a given square of side $1$. Let the vertices of the square, taken counterclockwise, be $A, B, C$ and $D$. Also, let the distances from $P$ to $A, B$ and $C$, respectively, be $u, v$ and $w$. What is the greatest distance that $P$ can be from $D$ if $u^2 + v^2 = w^2$?

$\textbf{(A)}\ 1 + \sqrt{2}\qquad \textbf{(B)}\ 2\sqrt{2}\qquad \textbf{(C)}\ 2 + \sqrt{2}\qquad \textbf{(D)}\ 3\sqrt{2}\qquad \textbf{(E)}\ 3+\sqrt{2}$

Solution

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

\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*}

Thus we see that $P$ lies 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 $(0, -1-\sqrt{2}),$ in which case $PD$ is $1 - (-1 - \sqrt{2}) = \boxed{\textbf{(C)}\ 2 + \sqrt{2}}.$

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