2012 AIME I Problems/Problem 14

Problem 14

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

Solution 1

By Vieta's formula, the sum of the roots is equal to 0, or $a+b+c=0$ . Therefore, $\frac{(a+b+c)}{3}=0$ . Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be $x$ and the other leg be $y$ . Without the loss of generality, let $\overline{ac}$ be the hypotenuse. The magnitudes of $a$ , $b$ , and $c$ are just $\frac{2}{3}$ of the medians because the origin, or the centroid in this case, cuts the median in a ratio of $2:1$ . So, $|a|^2=\frac{4}{9}\cdot((\frac{x}{2})^2+y^2)=\frac{x^2}{9}+\frac{4y^2}{9}$ because $|a|$ is two thirds of the median from $a$ . Similarly, $|c|^2=\frac{4}{9}\cdot(x^2+(\frac{y}{2})^2)=\frac{4x^2}{9}+\frac{y^2}{9}$ . The median from $b$ is just half the hypotenuse because the median of any right triangle is just half the hypotenuse. So, $|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}$ . Hence, $|a|^2+|b|^2+|c|^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250$ . Therefore, $h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}$ .

Solution 2

Assume $q$ and $r$ are real, so at least one of $a,$ $b,$ and $c$ must be real, with the remaining roots being pairs of complex conjugates. Without loss of generality, we assume $a$ is real and $b$ and $c$ are $x + yi$ and $x - yi$ respectively. By symmetry, the triangle described by $a,$ $b,$ and $c$ must be isosceles and is thus an isosceles right triangle with hypotenuse $\overline{bc}.$ Now since $P(z)$ has no $z^2$ term, we must have $a+b+c = a + (x + yi) + (x - yi) = 0$ and thus $a = -2x.$ Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, $a-x=y$ and thus $y=-3x.$ We can then solve for $x$ :

$\begin{align*} |a|^2 + |b|^2 + |c|^2 &= 250\\ |-2x|^2 + |x-3xi|^2 + |x+3xi|^2 &= 250\\ 4x^2 + (x^2 + 9x^2) + (x^2 + 9x^2) &= 250\\ x^2 &= \frac{250}{24} \end{align*}$

Now $h$ is the distance between $b$ and $c,$ so $h = 2y = -6x$ and thus $h^2 = 36x^2 = 36 \cdot \frac{250}{24} = \boxed{375.}$

Solution 3 (Messy)

Let the roots $a$ , $b$ , and $c$ each be represented by complex numbers $m + ni$ , $p + qi$ , and $r + ti$ . By Vieta's formulas, their sum is 0. Breaking into real and imaginary components, we get:

$m + p + r = 0$

$n + q + t = 0$

And, we know that the sum of the squares of the magnitudes of each is 250, so

$m^2 + n^2 + p^2 + q^2 + r^2 + t^2 = 250$

Given the complex plane, we set each of these complex numbers to points: $(m, n)$ , $(p, q)$ , $(r, t)$ . WLOG let $(r, t)$ be the vertex opposite the hypotenuse.

If the three points form a right triangle, the vectors from $(r, t)$ to $(m, m)$ and $(p, q)$ 's dot product is 0. $mp + r^2 - r(m + p) + nq + t^2 - t(n + q) = 0$

Substituting $m + p + r = 0$ and likewise, simplifying: $mp + 2r^2 + nq + 2t^2 = 0$

Rearranging we get: $r^2 + t^2 = -\frac{mp + nq}{2}$

The answer is the distance from $(m, n)$ to $(p, q)$ = $m^2 + n^2 + p^2 + q^2 - 2(mp + nq)$ . Substituting the equation equal to 250,

$= 250 - r^2 - t^2 - 2(mp + nq)$ $= 250 + \frac{mp + nq}{2} - 2(mp + nq)$ $= 250 - \frac{3}{2} \cdot (mp + nq)$

Taking our original equations summing to 0, and squaring each we get:

$n + q = -t$ $m + p = -r$

$n^2 + 2nq + q^2 = t^2$ $m^2 + 2mp + p^2 = r^2$

Adding, we get:

$m^2 + n^2 + p^2 + q^2 + 2(mp + nq) = r^2 + t^2$

Substituting again we obtain:

$250 - r^2 - t^2 + 2(mp + nq) = r^2 + t^2$ $2(r^2 + t^2) = 250 + 2(mp + nq)$ $r^2 + t^2 = 125 + (mp + nq)$

Substituting the equivalence of $r^2 + t^2$ :

$-\frac{mp + nq}{2} = 125 + (mp + nq)$

Solving for $mp + nq$ , we find it equal to $-\frac{250}{3}$ .

Substituting this value into our answer expression, we get:

$250 - \frac{3}{2} \cdot (-\frac{250}{3})$ , Answer = $\boxed{375}$ .

Solution 4 (clean)

As noted in the previous solutions, $a+b+c = 0$ . Let $a = a_1+a_2 i$ , $b = b_1+b_2 i$ , $c = c_1+c_2 i$ and we have $a_1 + b_1 + c_1 = a_2 + b_2 + c_2 = 0$ . Then the given $|a|^2 + |b|^2 + |c|^2 = 250$ translates to $\sum_{} ( {a_1}^2 + {a_2}^2 ) = 250.$ Note that in a right triangle, the sum of the squares of the three sides is equal to two times the square of the hypotenuse, by the pythagorean theorem. Thus, we have $2h^2 = (a_1 - b_1)^2 + (a_2 - b_2)^2 + (b_1 - c_1)^2 + (b_2 - c_2)^2 + (a_1 - c_1)^2 + (a_2 - c_2)^2$ $= 2 \left( \sum_{} ( {a_1}^2 + {a_2}^2 ) \right) - 2 \left( \sum_{cyc} a_1 b_1 + \sum_{cyc} a_2 b_2 \right)$ $= 500 - \left( (a_1 + b_1 + c_1)^2 + (a_2 + b_2 + c_2)^2 - \sum_{} ( {a_1}^2 + {a_2}^2 ) \right)$ $= 500 - (0^2 + 0^2 - 250)$ so $h^2 = \boxed{375}$ and we may conclude. ~ rzlng

Solution 5 (vectors)

As shown in the other solutions, $a+b+c = 0$ .\newline Without loss of generality, let $b$ be the complex number opposite the hypotenuse.\newline Note that there is an isomorphism between $\mathbb{C}$ under $+$ and $\mathbb{R}^2$ under $+$ .\newline Let $\Vec{a}$ , $\Vec{b}$ , and $\Vec{c}$ be the corresponding vectors to $a$ , $b$ , and $c$ .\newline Thus $\Vec{a} + \Vec{b} + \Vec{c} = \Vec{0}$ \newline $\Rightarrow 0 = \Vec{0}\cdot \Vec{0} = (\Vec{a} + \Vec{b} + \Vec{c})\cdot (\Vec{a} + \Vec{b} + \Vec{c}) = \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} + 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})$ \newline Now $|a|^2 + |b|^2 + |c|^2 = 250$ implies that $\lVert \Vec{a}\rVert^2 + \lVert \Vec{b}\rVert^2 + \lVert \Vec{c}\rVert^2 = 250$ \newline $\Rightarrow \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} = \lVert \Vec{a}\rVert^2 + \lVert \Vec{b}\rVert^2 + \lVert \Vec{c}\rVert^2 = 250$ \newline Also note that because there is a right angle at $b$ , $\Vec{a} - \Vec{b}$ and $\Vec{c} - \Vec{b}$ are perpendicular.\newline $\Rightarrow (\Vec{a} - \Vec{b})\cdot (\Vec{c} - \Vec{b}) = 0$ \newline $\Rightarrow 0 = (\Vec{a} - \Vec{b})\cdot (\Vec{c} - \Vec{b}) = \Vec{a}\cdot \Vec{c} + \Vec{b} \cdot \Vec{b} - \Vec{a} \cdot \Vec{b} - \Vec{b} \cdot \Vec{c}$ \newline Note that we are trying to find $h^2 = |a-c|^2 = \lVert \Vec{a} - \Vec{c} \rVert^2 = (\Vec{a} - \Vec{c})\cdot (\Vec{a} - \Vec{c}) = \Vec{a} \cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - \Vec{a}\cdot \Vec{c} - \Vec{a}\cdot \Vec{c} = \Vec{a} \cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - 2\Vec{a}\cdot \Vec{c}$ .\newline $\Rightarrow 0 = \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} + 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}) = 250 + 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})$ \newline $\Rightarrow -250 = 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})$ \newline $\Rightarrow -125 = \Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}$ \newline $\Rightarrow -125 = -125 + 0 = (\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}) + (\Vec{a}\cdot \Vec{c} + \Vec{b} \cdot \Vec{b} - \Vec{a} \cdot \Vec{b} - \Vec{b} \cdot \Vec{c}) = 2\Vec{a}\cdot \Vec{c} + \Vec{b} \cdot \Vec{b}$ \newline $\Rightarrow 125 = - \Vec{b} \cdot \Vec{b} - 2\Vec{a}\cdot \Vec{c}$ \newline $\Rightarrow 375 = 250 + 125 = \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} - \Vec{b} \cdot \Vec{b} - 2\Vec{a}\cdot \Vec{c} = \Vec{a}\cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - 2\Vec{a}\cdot \Vec{c} = h^2$ \newline $\Rightarrow h^2 = \boxed{375}$

2012 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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