Difference between revisions of "1984 AHSME Problems/Problem 10"
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Revision as of 11:49, 5 July 2013
Problem
Four complex numbers lie at the vertices of a square in the complex plane. Three of the numbers are 
, and 
. The fourth number is
Solution
Perhaps the easiest way to attack this is to transfer this to the Cartesian plane. The points then would be 
and 
, assuming the real axis was horizontal. Let these points be 
and 
, respectively. The remaining point is then the intersection of the following perpendicular lines:
The one perpenicular to 
and passing through 
and the one perpendicular to 
and passing through .
The slope of the first line is the negative reciprocal of the slope of the line through , which, using the slope formula, is 
, so the slope of the perpendicular line is 
. It passes through 
, so the equation of the line in point slope form is 
, or 
. Similarly, the slope of the second line is 
, and, since it passes through 
, its equation is 
, or 
. To find the intersection, we have 
, and solving for 
yields 
. Plugging this back into the equation yields 
, so the remaining point in the Cartesian plane is 
, and in the complex plane is 
.
See Also
| 1984 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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. 
