Difference between revisions of "1983 AHSME Problems/Problem 29"
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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> | + | 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>\left(0, -1-\sqrt{2}\right),</math> in which case <math>PD</math> is <math>1 - \left(-1 - \sqrt{2}\right) = \boxed{\textbf{(C)}\ 2 + \sqrt{2}}.</math> |
Revision as of 17:58, 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