Difference between revisions of "1983 AHSME Problems/Problem 29"
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− | + | ==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>? | ||
− | Thus | + | <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 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\\ \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 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> | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME box|year=1983|num-b=28|num-a=30}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 01:11, 20 February 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
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Problem 30 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.