1984 AHSME Problems/Problem 10

Problem

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

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

Solution

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 $2-i, \boxed{\text{B}}$ .

1984 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1984 AHSME Problems/Problem 10

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