Difference between revisions of "2019 AIME I Problems/Problem 12"
(Note: I put my solution as Solution 1 as I think it is more elegant.) |
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Given <math>f(z) = z^2-19z</math>, there are complex numbers <math>z</math> with the property that <math>z</math>, <math>f(z)</math>, and <math>f(f(z))</math> are the vertices of a right triangle in the complex plane with a right angle at <math>f(z)</math>. There are positive integers <math>m</math> and <math>n</math> such that one such value of <math>z</math> is <math>m+\sqrt{n}+11i</math>. Find <math>m+n</math>. | Given <math>f(z) = z^2-19z</math>, there are complex numbers <math>z</math> with the property that <math>z</math>, <math>f(z)</math>, and <math>f(f(z))</math> are the vertices of a right triangle in the complex plane with a right angle at <math>f(z)</math>. There are positive integers <math>m</math> and <math>n</math> such that one such value of <math>z</math> is <math>m+\sqrt{n}+11i</math>. Find <math>m+n</math>. | ||
− | ==Solution== | + | ==Solution 1== |
+ | Notice that we must have <cmath>\frac{f(f(z))-f(z)}{f(z)-z}=-\frac{f(f(z))-f(z)}{z-f(z)}\in\mathbb I.</cmath>However, <math>f(t)=t(t-20)</math>, so | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z^2-19z)(z^2-19z-20)}{z(z-20)}\\ | ||
+ | &=\frac{z(z-19)(z-20)(z+1)}{z(z-20)}\\ | ||
+ | &=(z-19)(z+1)\\ | ||
+ | &=(z-9)^2-100. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Then, the real part of <math>(z-9)^2</math> is <math>100</math>. Since <math>\text{Im}(z-9)=\text{Im}(z)=11</math>, let <math>z-9=a+11i</math>. Then, <cmath>100=\text{Re}((a+11i)^2)=a^2-121\implies a=\pm\sqrt{221}.</cmath>It follows that <math>z=9+\sqrt{221}+11i</math>, and the requested sum is <math>9+221=\boxed{230}</math>. | ||
+ | |||
+ | (Solution by TheUltimate123) | ||
+ | |||
+ | ==Solution 2== | ||
We will use the fact that segments <math>AB</math> and <math>BC</math> are perpendicular in the complex plane if and only if <math>\frac{a-b}{b-c}\in i\mathbb{R}</math>. To prove this, when dividing two complex numbers you subtract the angle of one from the other, and if the two are perpendicular, subtracting these angles will yield an imaginary number with no real part. | We will use the fact that segments <math>AB</math> and <math>BC</math> are perpendicular in the complex plane if and only if <math>\frac{a-b}{b-c}\in i\mathbb{R}</math>. To prove this, when dividing two complex numbers you subtract the angle of one from the other, and if the two are perpendicular, subtracting these angles will yield an imaginary number with no real part. |
Revision as of 19:01, 15 March 2019
Contents
Problem 12
Given , there are complex numbers with the property that , , and are the vertices of a right triangle in the complex plane with a right angle at . There are positive integers and such that one such value of is . Find .
Solution 1
Notice that we must have However, , so Then, the real part of is . Since , let . Then, It follows that , and the requested sum is .
(Solution by TheUltimate123)
Solution 2
We will use the fact that segments and are perpendicular in the complex plane if and only if . To prove this, when dividing two complex numbers you subtract the angle of one from the other, and if the two are perpendicular, subtracting these angles will yield an imaginary number with no real part.
Now to apply this:
The factorization of the nasty denominator above is made easier with the intuition that must be a divisor for the problem to lead anywhere. Now we know so using the fact that the imaginary part of is and calling the real part r,
solving the above quadratic yields so our answer is
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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