Difference between revisions of "2012 AIME I Problems/Problem 14"
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− | ==Problem | + | ==Problem== |
Complex numbers <math>a,</math> <math>b,</math> and <math>c</math> are zeros of a polynomial <math>P(z) = z^3 + qz + r,</math> and <math>|a|^2 + |b|^2 + |c|^2 = 250.</math> The points corresponding to <math>a,</math> <math>b,</math> and <math>c</math> in the complex plane are the vertices of a right triangle with hypotenuse <math>h.</math> Find <math>h^2.</math> | Complex numbers <math>a,</math> <math>b,</math> and <math>c</math> are zeros of a polynomial <math>P(z) = z^3 + qz + r,</math> and <math>|a|^2 + |b|^2 + |c|^2 = 250.</math> The points corresponding to <math>a,</math> <math>b,</math> and <math>c</math> in the complex plane are the vertices of a right triangle with hypotenuse <math>h.</math> Find <math>h^2.</math> | ||
Line 82: | Line 82: | ||
Answer = <math>\boxed{375}</math>. | Answer = <math>\boxed{375}</math>. | ||
− | == Solution 4 (clean | + | == Solution 4 (clean) == |
As noted in the previous solutions, <math>a+b+c = 0</math>. | As noted in the previous solutions, <math>a+b+c = 0</math>. | ||
Let <math>a = a_1+a_2 i</math>, <math>b = b_1+b_2 i</math>, <math>c = c_1+c_2 i</math> and we have <math>a_1 + b_1 + c_1 = a_2 + b_2 + c_2 = 0</math>. | Let <math>a = a_1+a_2 i</math>, <math>b = b_1+b_2 i</math>, <math>c = c_1+c_2 i</math> and we have <math>a_1 + b_1 + c_1 = a_2 + b_2 + c_2 = 0</math>. | ||
− | Then the given <math>|a|^2 + |b|^2 + |c|^2 = 250</math> translates to <math>\sum_{ | + | Then the given <math>|a|^2 + |b|^2 + |c|^2 = 250</math> translates to <math>\sum_{} ( {a_1}^2 + {a_2}^2 ) = 250.</math> |
− | + | 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 <cmath>(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) | + | Thus, we have <cmath>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 </cmath> |
− | <cmath>2 \left( \sum_{ | + | <cmath>= 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)</cmath> |
− | <cmath>500 - \left( (a_1 + b_1 + c_1)^2 + (a_2 + b_2 + c_2)^2 - \sum_{ | + | <cmath>= 500 - \left( (a_1 + b_1 + c_1)^2 + (a_2 + b_2 + c_2)^2 - \sum_{cyc} ( {a_1}^2 + {a_2}^2 ) \right)</cmath> |
− | <cmath>500 - (0^2 + 0^2 - 250) | + | <cmath>= 500 - (0^2 + 0^2 - 250)</cmath> |
so <math>h^2 = \boxed{375}</math> and we may conclude. | so <math>h^2 = \boxed{375}</math> and we may conclude. | ||
~ rzlng | ~ rzlng | ||
+ | |||
+ | ==Solution 5 (dirty because of god intuition + wishful thinking)== | ||
+ | |||
+ | First, note that the roots of this cubic will be <math>a, b</math> and <math>-(a+b)</math> due to Vieta's, which means that the sum of the roots are 0. | ||
+ | |||
+ | Now, we use some god level wishful thinking. It would really be nice if one of these roots was on the real axis, and it was a right isosceles triangle, because that would be very convenient and easy to work with. The neat part is that it actually works | ||
+ | |||
+ | Set one of the roots as <math>r</math>, where <math>r</math> is any real number. WLOG, assume that this is the right angle. With symmetry to respect of the x axis (because symmetry makes the imaginary parts of the other 2 roots cancel out, besides the fact that complex conjugate root theorem forces it). This way, we can set the other 2 roots as <math>\frac{r}{2}+ni</math> and <math>\frac{r}{2}-ni</math>, making the roots add up to 0. | ||
+ | |||
+ | Now, as we want the roots to satisfy the original condition (right triangle) we are going to have to set an equation to find <math>n</math> out. We use the fact that it is an isosceles right triangle to find that <math>\frac{3r}{2}=n</math>, which means that the 2 other roots are now <math>\frac{r}{2}+\frac{3r}{2}i</math> and <math>\frac{r}{2}-\frac{3r}{2}i</math> | ||
+ | |||
+ | Now we use the fact that <math>|a|^2+|b|^2+|c|^2=250</math>. Clearly one of these is <math>r</math> away from the origin, so that gets <math>r^2</math>, and then we get <math>2*\frac{r}{2}^2+\frac{3r}{2}^2</math> which gets us <math>5r^2</math>, getting <math>r^2+5r^2=250</math>, so <math>r=\sqrt{\frac{250}{6}}</math>. So the final answer comes out to | ||
+ | <cmath>(\frac{250}{6}*\frac{9}{4}*2)^2=\boxed{\boxed{375}}</cmath> | ||
+ | |||
+ | -dragoon | ||
+ | |||
+ | P.S.: The main propose, saying that one root is on the real axis and the right triangle is a right isosceles triangle is not actually only a wishful thinking that came out by luck, but actually is something that must be true due to the complex conjugate root theorem. | ||
+ | -SuperDolphin | ||
+ | |||
+ | - in this specific problem this logic is incorrect, as the coefficients of the polynomial can be complex. If a question specified real roots, then this would be appropriate. In this case, it is just wishful thinking as stated above | ||
+ | |||
+ | ==Solution 5 (vectors)== | ||
+ | As shown in the other solutions, <math>a+b+c = 0</math>. | ||
+ | |||
+ | Without loss of generality, let <math>b</math> be the complex number opposite the hypotenuse. | ||
+ | |||
+ | Note that there is an isomorphism between <math>\mathbb{C}</math> under <math>+</math> and <math>\mathbb{R}^2</math> under <math>+</math>. | ||
+ | |||
+ | Let <math>\Vec{a}</math>, <math>\Vec{b}</math>, and <math>\Vec{c}</math> be the corresponding vectors to <math>a</math>, <math>b</math>, and <math>c</math>. | ||
+ | |||
+ | Thus <math>\Vec{a} + \Vec{b} + \Vec{c} = \Vec{0}</math> | ||
+ | |||
+ | <math>\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})</math> | ||
+ | |||
+ | Now <math>|a|^2 + |b|^2 + |c|^2 = 250</math> implies that <math>\lVert \Vec{a}\rVert^2 + \lVert \Vec{b}\rVert^2 + \lVert \Vec{c}\rVert^2 = 250</math> | ||
+ | |||
+ | <math>\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</math> | ||
+ | |||
+ | Also note that because there is a right angle at <math>b</math>, <math>\Vec{a} - \Vec{b}</math> and <math>\Vec{c} - \Vec{b}</math> are perpendicular. | ||
+ | |||
+ | <math>\Rightarrow (\Vec{a} - \Vec{b})\cdot (\Vec{c} - \Vec{b}) = 0</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 <math>h^2 = |a-c|^2</math> | ||
+ | |||
+ | <math>\Rightarrow h^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}</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 -250 = 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})</math> | ||
+ | |||
+ | <math>\Rightarrow -125 = \Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}</math> | ||
+ | |||
+ | <math>\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}</math> | ||
+ | |||
+ | <math>\Rightarrow 125 = - \Vec{b} \cdot \Vec{b} - 2\Vec{a}\cdot \Vec{c}</math> | ||
+ | |||
+ | <math>\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</math> | ||
+ | |||
+ | <math>\Rightarrow h^2 = \boxed{375}</math> | ||
+ | |||
+ | ==Solution 6(similar to solution 4)== | ||
+ | Note that the roots of the polynomial <math>(a,b,c)</math> must sum to <math>0</math> due to the <math>z^2</math> coefficient equaling <math>0</math> because of Vieta's Formulas. This tells us that <math>a+b+c = 0</math> and <math>\overline{a} + \overline{b} + \overline{c} = 0</math> so <math>(a+b+c)(\overline{a+b+c}) = |a|^2 + |b|^2 + |c|^2 + a\overline{b} + \overline{a}b + b\overline{c} + \overline{b}c + c\overline{a} + \overline{c}a = 250 + a\overline{b} + \overline{a}b + b\overline{c} + \overline{b}c + c\overline{a} + \overline{c}a = 0</math> so <math>a\overline{b} + \overline{a}b + b\overline{c} + \overline{b}c + c\overline{a} + \overline{c}a = -250.</math> | ||
+ | |||
+ | Note that terms similar to <math>\overline{a}b + \overline{b}a</math> appear in <math>|a-b|^2</math> so we decide to sum <math>|a-b|^2 + |b-c|^2 + |c-a|^2</math> out of intuition. Note that this corresponds to the sum of the squares of the sidelengths of the right triangle and if WLOG the <math>|c-a|^2</math> side is the hypotenuse then our sum is equal to <math>2|c-a|^2 = 2h^2.</math> | ||
+ | |||
+ | <cmath>2|a|^2 + 2|b|^2 + 2|c|^2 -a\overline{b} - \overline{a}b - b\overline{c} - \overline{b}c - c\overline{a} - \overline{c}a = 500 + 250 = 2h^2</cmath> | ||
+ | |||
+ | As a result we know that <math>h^2 = \boxed{375}</math> | ||
+ | ~SailS | ||
+ | |||
+ | == Video Solution by Richard Rusczyk == | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2012aimei/354 | ||
+ | |||
+ | ~ dolphin7 | ||
== 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}} |
Latest revision as of 23:59, 20 January 2024
Contents
Problem
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 (dirty because of god intuition + wishful thinking)
First, note that the roots of this cubic will be and
due to Vieta's, which means that the sum of the roots are 0.
Now, we use some god level wishful thinking. It would really be nice if one of these roots was on the real axis, and it was a right isosceles triangle, because that would be very convenient and easy to work with. The neat part is that it actually works
Set one of the roots as , where
is any real number. WLOG, assume that this is the right angle. With symmetry to respect of the x axis (because symmetry makes the imaginary parts of the other 2 roots cancel out, besides the fact that complex conjugate root theorem forces it). This way, we can set the other 2 roots as
and
, making the roots add up to 0.
Now, as we want the roots to satisfy the original condition (right triangle) we are going to have to set an equation to find out. We use the fact that it is an isosceles right triangle to find that
, which means that the 2 other roots are now
and
Now we use the fact that . Clearly one of these is
away from the origin, so that gets
, and then we get
which gets us
, getting
, so
. So the final answer comes out to
-dragoon
P.S.: The main propose, saying that one root is on the real axis and the right triangle is a right isosceles triangle is not actually only a wishful thinking that came out by luck, but actually is something that must be true due to the complex conjugate root theorem. -SuperDolphin
- in this specific problem this logic is incorrect, as the coefficients of the polynomial can be complex. If a question specified real roots, then this would be appropriate. In this case, it is just wishful thinking as stated above
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
.
Solution 6(similar to solution 4)
Note that the roots of the polynomial must sum to
due to the
coefficient equaling
because of Vieta's Formulas. This tells us that
and
so
so
Note that terms similar to appear in
so we decide to sum
out of intuition. Note that this corresponds to the sum of the squares of the sidelengths of the right triangle and if WLOG the
side is the hypotenuse then our sum is equal to
As a result we know that
~SailS
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012aimei/354
~ dolphin7
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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