2012 AIME I Problem 14

Attempt Problem 14 of the 2012 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AIME I solutions, or check the answer key.

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14.

Complex numbers a,a, b,b, and cc are the zeros of a polynomial P(z)=z3+qz+r,P(z) = z^3 + qz + r, and ∣a∣2+∣b∣2+∣c∣2=250.|a|^2 + |b|^2 + |c|^2 = 250. The points corresponding to a,a, b,b, and cc in the complex plane are the vertices of a right triangle with hypotenuse h.h. Find h2.h^2.

Answer: 375
Concepts:complex numberVieta’s Formulasright triangle
Difficulty rating: 3060
Small Hint:

There is no z2z^2 term, so a+b+c=0;a + b + c = 0; if the right angle is at b,b, then b=−(a+c)b = -(a + c)

Big Hint:

The hypotenuse midpoint a+c2\frac{a + c}{2} is equidistant from all three vertices; also ∣a∣2+∣c∣2=∣a−c∣2+∣a+c∣22|a|^2 + |c|^2 = \frac{|a - c|^2 + |a + c|^2}{2}

Solution:

Since P(z)P(z) has no z2z^2 term, a+b+c=0.a + b + c = 0. Say the right angle is at b;b; then the hypotenuse joins aa and c,c, so h=∣a−c∣,h = |a - c|, and b=−(a+c).b = -(a + c). The midpoint d=a+c2d = \frac{a + c}{2} of the hypotenuse is the circumcenter of the right triangle, so ∣b−d∣=h2.|b - d| = \frac{h}{2}. Since b−d=−32(a+c),b - d = -\frac{3}{2}(a + c), this gives ∣a−c∣=3 ∣a+c∣.|a - c| = 3\,|a + c|.

By the parallelogram law, ∣a∣2+∣c∣2=∣a−c∣2+∣a+c∣22,|a|^2 + |c|^2 = \frac{|a - c|^2 + |a + c|^2}{2}, and ∣b∣2=∣a+c∣2,|b|^2 = |a + c|^2, so 250=9 ∣a+c∣2+∣a+c∣22+∣a+c∣2=6 ∣a+c∣2. \begin{aligned} 250 &= \frac{9\,|a + c|^2 + |a + c|^2}{2} \\ &\quad {}+ |a + c|^2 \\ &= 6\,|a + c|^2. \end{aligned}

Therefore h2=∣a−c∣2=9 ∣a+c∣2h^2 = |a - c|^2 = 9\,|a + c|^2 =9⋅2506= \frac{9 \cdot 250}{6} =375.= 375.

Problem 13#13
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