Difference between revisions of "2021 AMC 12B Problems/Problem 18"
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Let <math>z</math> be a complex number satisfying <math>12|z|^2=2|z+2|^2+|z^2+1|^2+31.</math> What is the value of <math>z+\frac 6z?</math> | Let <math>z</math> be a complex number satisfying <math>12|z|^2=2|z+2|^2+|z^2+1|^2+31.</math> What is the value of <math>z+\frac 6z?</math> | ||
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===Solution 1=== | ===Solution 1=== | ||
The answer being in the form <math>z+\frac 6z</math> means that there are two solutions, some complex number and its complex conjugate. <cmath>a+bi = \frac{6}{a-bi}</cmath> <cmath>a^2+b^2=6</cmath> We should then be able to test out some ordered pairs of <math>(a, b)</math>. After testing it out, we get the ordered pairs of <math>(-1, \sqrt{5})</math> and its conjugate <math>(-1, -\sqrt{5})</math>. Plugging this into answer format gives us <math>\boxed{\textbf{(A) }-2}</math> ~Lopkiloinm | The answer being in the form <math>z+\frac 6z</math> means that there are two solutions, some complex number and its complex conjugate. <cmath>a+bi = \frac{6}{a-bi}</cmath> <cmath>a^2+b^2=6</cmath> We should then be able to test out some ordered pairs of <math>(a, b)</math>. After testing it out, we get the ordered pairs of <math>(-1, \sqrt{5})</math> and its conjugate <math>(-1, -\sqrt{5})</math>. Plugging this into answer format gives us <math>\boxed{\textbf{(A) }-2}</math> ~Lopkiloinm | ||
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+ | ==See Also== | ||
+ | {{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} |
Revision as of 19:36, 11 February 2021
Contents
[hide]Problem
Let be a complex number satisfying What is the value of
Solution
Solution 1
The answer being in the form means that there are two solutions, some complex number and its complex conjugate. We should then be able to test out some ordered pairs of . After testing it out, we get the ordered pairs of and its conjugate . Plugging this into answer format gives us ~Lopkiloinm
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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