Difference between revisions of "1984 AHSME Problems/Problem 10"
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NICE! You can also just graph and use pythag. | NICE! You can also just graph and use pythag. | ||
+ | Or get the solution by sheer obviousness (I'm serious) | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1984|num-b=9|num-a=11}} | {{AHSME box|year=1984|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:01, 29 September 2024
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 .
NICE! You can also just graph and use pythag. Or get the solution by sheer obviousness (I'm serious)
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.