Difference between revisions of "2012 AIME I Problems/Problem 14"
(→Solution 1) |
(→Solution 2) |
||
Line 9: | Line 9: | ||
===Solution 2=== | ===Solution 2=== | ||
− | + | Assume <math>q</math> and <math>r</math> are real, so at least one of <math>a,</math> <math>b,</math> and <math>c</math> must be real, with the remaining roots being pairs of complex conjugates. Without loss of generality, we assume <math>a</math> is real and <math>b</math> and <math>c</math> are <math>x + yi</math> and <math>x - yi</math> respectively. By symmetry, the triangle described by <math>a,</math> <math>b,</math> and <math>c</math> must be isosceles and is thus an isosceles right triangle with hypotenuse <math>\overline{ab}.</math> Now since <math>P(z)</math> has no <math>z^2</math> term, we must have <math>a+b+c = a + (x + yi) + (x - yi) = 0</math> and thus <math>a = -2x.</math> Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, <math>a-x=y</math> and thus <math>y=-3x.</math> We can then solve for <math>x</math>: | |
<cmath> | <cmath> |
Revision as of 20:23, 18 April 2012
Problem 14
Complex numbers
and
are zeros of a polynomial
and
The points corresponding to
and
in the complex plane are the vertices of a right triangle with hypotenuse
Find
Solution
Solution 1
By Vieta's formula, the sum of the roots is equal to 0, or . Therefore,
. 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
and the other leg be
. Without the loss of generality, let
be the hypotenuse. The magnitudes of
,
, and
are just
of the medians because the origin, or the centroid in this case, cuts the median in a ratio of
. So,
because
is two thirds of the median from
. Similarly,
. The median from
is just half the hypotenuse because the hypotenuse of any right triangle is just half the hypotenuse. So,
. Hence,
. Therefore,
.
Solution 2
Assume and
are real, so at least one of
and
must be real, with the remaining roots being pairs of complex conjugates. Without loss of generality, we assume
is real and
and
are
and
respectively. By symmetry, the triangle described by
and
must be isosceles and is thus an isosceles right triangle with hypotenuse
Now since
has no
term, we must have
and thus
Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse,
and thus
We can then solve for
:
Now is the distance between
and
so
and thus
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 |