Difference between revisions of "1977 AHSME Problems/Problem 5"
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| + | == Problem 5 == | ||
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| + | The set of all points <math>P</math> such that the sum of the (undirected) distances from <math>P</math> to two fixed points <math>A</math> and <math>B</math> equals the distance between <math>A</math> and <math>B</math> is | ||
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| + | <math>\textbf{(A) }\text{the line segment from }A\text{ to }B\qquad | ||
| + | \textbf{(B) }\text{the line passing through }A\text{ and }B\qquad\\ | ||
| + | \textbf{(C) }\text{the perpendicular bisector of the line segment from }A\text{ to }B\qquad\\ | ||
| + | \textbf{(D) }\text{an ellipse having positive area}\qquad | ||
| + | \textbf{(E) }\text{a parabola} </math> | ||
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==Solution== | ==Solution== | ||
Solution by e_power_pi_times_i | Solution by e_power_pi_times_i | ||
The answer is <math>\textbf{(A)}</math> because <math>P</math> has to be on the line segment <math>AB</math> in order to satisfy <math>PA+PB=AB</math>. | The answer is <math>\textbf{(A)}</math> because <math>P</math> has to be on the line segment <math>AB</math> in order to satisfy <math>PA+PB=AB</math>. | ||
Latest revision as of 11:27, 21 November 2016
Problem 5
The set of all points 
such that the sum of the (undirected) distances from to two fixed points 
and 
equals the distance between and is
Solution
Solution by e_power_pi_times_i
The answer is 
because has to be on the line segment 
in order to satisfy 
.
