1984 AHSME Problems/Problem 10

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Four complex numbers lie at the vertices of a square in the complex plane. Three of the numbers are $1+2i, -2+i$, and $-1-2i$. The fourth number is

$\mathrm{(A) \ }2+i \qquad \mathrm{(B) \ }2-i \qquad \mathrm{(C) \ } 1-2i \qquad \mathrm{(D) \ }-1+2i \qquad \mathrm{(E) \ } -2-i$


Perhaps the easiest way to attack this is to transfer this to the Cartesian plane. The points then would be $(1, 2), (-2, 1),$ and $(-1, -2)$, assuming the real axis was horizontal. Let these points be $A, B$ and $C$, respectively. The remaining point is then the intersection of the following perpendicular lines: The one perpenicular to $AB$ and passing through $A$ and the one perpendicular to $BC$ and passing through $C$. The slope of the first line is the negative reciprocal of the slope of the line through $AB$, which, using the slope formula, is $\frac{1}{3}$, so the slope of the perpendicular line is $-3$. It passes through $(1,2)$, so the equation of the line in point slope form is $y-2=-3(x-1)$, or $y=-3x+5$. Similarly, the slope of the second line is $\frac{-1}{-3}=\frac{1}{3}$, and, since it passes through $(-1,-2)$, its equation is $y+2=\frac{1}{3}(x+1)$, or $y=\frac{1}{3}x-\frac{5}{3}$. To find the intersection, we have $-3x+5=y=\frac{1}{3}x-\frac{5}{3}$, and solving for $x$ yields $x=2$. Plugging this back into the equation yields $y=-1$, so the remaining point in the Cartesian plane is $(2, -1)$, and in the complex plane is $2-i, \boxed{\text{B}}$.

NICE! You can also just graph and use pythag.

See Also

1984 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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