Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2012 AIME I Problems/Problem 14"
(→‎Solution 3 (Messy))
Line 27: Line 27:
  
 
<math>m + p + r = 0</math>
 
<math>m + p + r = 0</math>
 +
 
<math>n + q + t = 0</math>
 
<math>n + q + t = 0</math>
  
Line 79: Line 80:
  
 
<math>250 - \frac{3}{2} \cdot (-\frac{250}{3})</math>,  
 
<math>250 - \frac{3}{2} \cdot (-\frac{250}{3})</math>,  
−
Answer = <math>\boxed{375}</math>.  
+
Answer = <math>\boxed{375}</math>.
−
 
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=13|num-a=15}}
 
{{AIME box|year=2012|n=I|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:57, 23 June 2018

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 = $\boxed{375}$.

See also

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
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_14&oldid=95558"