Difference between revisions of "2012 AIME I Problems/Problem 14"
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<math>\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}</math> | <math>\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}</math> | ||
− | Note that we are trying to find <math>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} \ | + | Note that we are trying to find <math>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}</math>\newline<math> = \Vec{a} \cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - 2\Vec{a}\cdot \Vec{c}</math>. |
<math>\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})</math> | <math>\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})</math> |
Revision as of 23:08, 4 March 2020
Contents
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 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 median 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
Solution 3 (Messy)
Let the roots ,
, and
each be represented by complex numbers
,
, and
.
By Vieta's formulas, their sum is 0. Breaking into real and imaginary components, we get:
And, we know that the sum of the squares of the magnitudes of each is 250, so
Given the complex plane, we set each of these complex numbers to points: ,
,
.
WLOG let
be the vertex opposite the hypotenuse.
If the three points form a right triangle, the vectors from to
and
's dot product is 0.
Substituting and likewise, simplifying:
Rearranging we get:
The answer is the distance from to
=
.
Substituting the equation equal to 250,
Taking our original equations summing to 0, and squaring each we get:
Adding, we get:
Substituting again we obtain:
Substituting the equivalence of :
Solving for , we find it equal to
.
Substituting this value into our answer expression, we get:
,
Answer =
.
Solution 4 (clean)
As noted in the previous solutions, .
Let
,
,
and we have
.
Then the given
translates to
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
so
and we may conclude.
~ rzlng
Solution 5 (vectors)
As shown in the other solutions, .
Without loss of generality, let be the complex number opposite the hypotenuse.
Note that there is an isomorphism between under
and
under
.
Let ,
, and
be the corresponding vectors to
,
, and
.
Thus
Now implies that
Also note that because there is a right angle at ,
and
are perpendicular.
Note that we are trying to find \newline
.
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.