Difference between revisions of "2019 AIME I Problems/Problem 12"
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solving the above quadratic yields <math>r=9+\sqrt{221}</math> so our answer is <math>9+221=\boxed{230}</math> | solving the above quadratic yields <math>r=9+\sqrt{221}</math> so our answer is <math>9+221=\boxed{230}</math> | ||
− | ==See | + | ==Solution 3== |
+ | I would like to use a famous method namely coni method. | ||
+ | |||
+ | Statement .If we consider there complex number <math>A,B,C</math> in argand plane then <math>\angle ABC =\arg{\frac{B-C}{B-A}}</math>. | ||
+ | |||
+ | According to the question given, we can assume ,<math>A= f(f(z)),B=f(z),C= x</math> respectively. | ||
+ | |||
+ | WLOG,<math>Z_1= \frac{f(f(z))-f(z)}{z-f(z)}</math>. | ||
+ | According to the question <math>\arg{Z_1}=\frac{\pi}{2}</math>. | ||
+ | |||
+ | So,<math>\Re (Z_1)=0</math>. | ||
+ | |||
+ | Now, <math>Z_1=\frac{z(z-19)(z+1)(z-20)}{-z(z-20)}</math>. | ||
+ | |||
+ | <math>\implies Z_1= -(z^2-18z-19)</math>. | ||
+ | WLOG,<math>z=a+11i</math> .where <math>a=m+\sqrt{n}</math>. | ||
+ | |||
+ | So,<math>\Re (Z_1)= -(a^2-18a-140)</math>. | ||
+ | Solving,<math>a^2-18a-140=0</math> .get , | ||
+ | |||
+ | <math>a=9</math>±<math>\sqrt{221}</math>. | ||
+ | So, possible value of <math>a=9+\sqrt{221}</math>. | ||
+ | |||
+ | <math>\boxed{m+n=230}</math>. | ||
+ | ~ftheftics. | ||
+ | ==See also== | ||
{{AIME box|year=2019|n=I|num-b=11|num-a=13}} | {{AIME box|year=2019|n=I|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:47, 3 January 2020
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
Solution 3
I would like to use a famous method namely coni method.
Statement .If we consider there complex numberin argand plane then
.
According to the question given, we can assume , respectively.
WLOG,.
According to the question
.
So,.
Now, .
.
WLOG,
.where
.
So,.
Solving,
.get ,
±
.
So, possible value of
.
.
~ftheftics.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.