Difference between revisions of "2010 AIME II Problems/Problem 7"
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== Problem 7 == | == Problem 7 == | ||
− | Let <math>P(z)= | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>P(z)=z^3+az^2+bz+c</math>, where a, b, and c are real. There exists a complex number <math>w</math> such that the three roots of <math>P(z)</math> are <math>w+3i</math>, <math>w+9i</math>, and <math>2w-4</math>, where <math>i^2=-1</math>. Find <math>|a+b+c|</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
− | == Solution == | + | |
+ | == Solution (vieta's) == | ||
Set <math>w=x+yi</math>, so <math>x_1 = x+(y+3)i</math>, <math>x_2 = x+(y+9)i</math>, <math>x_3 = 2x-4+2yi</math>. | Set <math>w=x+yi</math>, so <math>x_1 = x+(y+3)i</math>, <math>x_2 = x+(y+9)i</math>, <math>x_3 = 2x-4+2yi</math>. | ||
− | Since <math>a,b,c\in{R}</math>, the imaginary part of a,b,c must be 0. | + | Since <math>a,b,c\in{R}</math>, the imaginary part of <math>a,b,c</math> must be <math>0</math>. |
+ | |||
+ | Start with a, since it's the easiest one to do: <math>y+3+y+9+2y=0, y=-3</math>, | ||
+ | |||
+ | and therefore: <math>x_1 = x</math>, <math>x_2 = x+6i</math>, <math>x_3 = 2x-4-6i</math>. | ||
+ | |||
+ | Now, do the part where the imaginary part of c is 0 since it's the second easiest one to do: | ||
+ | <math>x(x+6i)(2x-4-6i)</math>. The imaginary part is <math>6x^2-24x</math>, which is 0, and therefore <math>x=4</math>, since <math>x=0</math> doesn't work. | ||
+ | |||
+ | So now, <math>x_1 = 4, x_2 = 4+6i, x_3 = 4-6i</math>, | ||
+ | |||
+ | and therefore: <math>a=-12, b=84, c=-208</math>. Finally, we have <math>|a+b+c|=|-12+84-208|=\boxed{136}</math>. | ||
+ | |||
+ | == Solution 1b == | ||
+ | |||
+ | Same as solution 1 except that when you get to <math>x_1 = x</math>, <math>x_2 = x+6i</math>, <math>x_3 = 2x-4-6i</math>, you don't need to find the imaginary part of <math>c</math>. We know that <math>x_1</math> is a real number, which means that <math>x_2</math> and <math>x_3</math> are complex conjugates. Therefore, <math>x=2x-4</math>. | ||
+ | |||
+ | == Solution 2 (casework) == | ||
+ | |||
+ | Note that at least one of <math>w+3i</math>, <math>w+9i</math>, or <math>2w-4</math> is real by complex conjugate roots. We now separate into casework based on which one. | ||
+ | |||
+ | Let <math>w=x+yi</math>, where <math>x</math> and <math>y</math> are reals. | ||
− | + | Case 1: <math>w+3i</math> is real. This implies that <math>x+yi+3i</math> is real, so by setting the imaginary part equal to zero we get <math>y=-3</math>, so <math>w=x-3i</math>. Now note that since <math>w+3i</math> is real, <math>w+9i</math> and <math>2w-4</math> are complex conjugates. Thus <math>\overline{w+9i}=2w-4</math>, so <math>\overline{x+6i}=2(x-3i)-4</math>, implying that <math>x=4</math>, so <math>w=4-3i</math>. | |
− | + | Case 2: <math>w+9i</math> is real. This means that <math>x+yi+9i</math> is real, so again setting imaginary part to zero we get <math>y=-9</math>, so <math>w=x-9i</math>. Now by the same logic as above <math>w+3i</math> and <math>2w-4</math> are complex conjugates. Thus <math>\overline{w+3i}=2w-4</math>, so <math>\overline{x-6i}=2(x-9i)-4</math>, so <math>x+6i=2x-4-18i</math>, which has no solution as <math>x</math> is real. | |
− | + | Case 3: <math>2w-4</math> is real. Going through the same steps, we get <math>y=0</math>, so <math>w=x</math>. Now <math>w+3i</math> and <math>w+6i</math> are complex conjugates, but <math>w=x</math>, which means that <math>\overline{x+3i}=x+6i</math>, so <math>x-3i=x+6i</math>, which has no solutions. | |
− | <math>x | ||
− | + | Thus case 1 is the only one that works, so <math>w=4-3i</math> and our polynomial is <math>(z-(4))(z-(4+6i))(z-(4-6i))</math>. Note that instead of expanding this, we can save time by realizing that the answer format is <math>|a+b+c|</math>, so we can plug in <math>z=1</math> to our polynomial to get the sum of coefficients, which will give us <math>a+b+c+1</math>. Plugging in <math>z=1</math> into our polynomial, we get <math>(-3)(-3-6i)(-3+6i)</math> which evaluates to <math>-135</math>. Since this is <math>a+b+c+1</math>, we subtract 1 from this to get <math>a+b+c=-136</math>, so <math>|a+b+c|=\boxed{136}</math>. | |
− | + | ~chrisdiamond10 | |
== See also == | == See also == | ||
{{AIME box|year=2010|num-b=6|num-a=8|n=II}} | {{AIME box|year=2010|num-b=6|num-a=8|n=II}} | ||
+ | {{MAA Notice}} |
Latest revision as of 01:17, 12 February 2021
Problem 7
Let , where a, b, and c are real. There exists a complex number such that the three roots of are , , and , where . Find .
Solution (vieta's)
Set , so , , .
Since , the imaginary part of must be .
Start with a, since it's the easiest one to do: ,
and therefore: , , .
Now, do the part where the imaginary part of c is 0 since it's the second easiest one to do: . The imaginary part is , which is 0, and therefore , since doesn't work.
So now, ,
and therefore: . Finally, we have .
Solution 1b
Same as solution 1 except that when you get to , , , you don't need to find the imaginary part of . We know that is a real number, which means that and are complex conjugates. Therefore, .
Solution 2 (casework)
Note that at least one of , , or is real by complex conjugate roots. We now separate into casework based on which one.
Let , where and are reals.
Case 1: is real. This implies that is real, so by setting the imaginary part equal to zero we get , so . Now note that since is real, and are complex conjugates. Thus , so , implying that , so .
Case 2: is real. This means that is real, so again setting imaginary part to zero we get , so . Now by the same logic as above and are complex conjugates. Thus , so , so , which has no solution as is real.
Case 3: is real. Going through the same steps, we get , so . Now and are complex conjugates, but , which means that , so , which has no solutions.
Thus case 1 is the only one that works, so and our polynomial is . Note that instead of expanding this, we can save time by realizing that the answer format is , so we can plug in to our polynomial to get the sum of coefficients, which will give us . Plugging in into our polynomial, we get which evaluates to . Since this is , we subtract 1 from this to get , so .
~chrisdiamond10
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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