Difference between revisions of "2020 AIME I Problems/Problem 14"
m |
(→Solution 3) |
||
Line 39: | Line 39: | ||
Thus, our final answer is <math>49+36=\boxed{085}</math>. ~GeronimoStilton | Thus, our final answer is <math>49+36=\boxed{085}</math>. ~GeronimoStilton | ||
+ | |||
+ | ==Solution 4== | ||
+ | Let <math>P(x)=(x-r)(x-s)</math>. There are two cases: in the first case, <math>(3-r)(3-s)=(4-r)(4-s)</math> equals <math>r</math> (without loss of generality), and thus <math>(a-r)(a-s)=(b-r)(b-s)=s</math>. By Vieta's formulas <math>a+b=r+s=3+4=7</math>. | ||
+ | |||
+ | In the second case, say without loss of generality <math>(3-r)(3-s)=r</math> and <math>(4-r)(4-s)=s</math>. Subtracting gives <math>-7+r+s=r-s</math>, so <math>s=7/2</math>. From this, we have <math>r=-3</math>. | ||
+ | |||
+ | Note <math>r+s=1/2</math>, so by Vieta's, we have <math>\{a,b\}=\{1/2-3,1/2-4\}=\{-5/2,-7/2\}</math>. In this case, <math>a+b=-6</math>. | ||
+ | |||
+ | The requested sum is <math>36+49=85</math>.~TheUltimate123 | ||
==See Also== | ==See Also== |
Revision as of 17:48, 13 March 2020
Problem
Let be a quadratic polynomial with complex coefficients whose
coefficient is
Suppose the equation
has four distinct solutions,
Find the sum of all possible values of
Solution 1
Either or not. We first see that if
it's easy to obtain by Vieta's that
. Now, take
and WLOG
. Now, consider the parabola formed by the graph of
. It has vertex
. Now, say that
. We note
. Now, we note
by plugging in again. Now, it's easy to find that
, yielding a value of
. Finally, we add
. ~awang11, charmander3333
Solution 2
Let the roots of be
and
, then we can write
. The fact that
has solutions
implies that some combination of
of these are the solution to
, and the other
are the solution to
. It's fairly easy to see there are only
possible such groupings:
and
, or
and
(Note that
are interchangeable, and so are
and
). We now to casework:
If
, then
so this gives
.
Next, if
, then
Subtracting the first part of the first equation from the first part of the second equation gives
Hence,
, and so
.
Therefore, the solution is
~ktong
Solution 3
Write . Split the problem into two cases:
and
.
Case 1: We have . We must have
Rearrange and divide through by
to obtain
Now, note that
Now, rearrange to get
and thus
Substituting this into our equation for
yields
. Then, it is clear that
does not have a double root at
, so we must have
and
or vice versa. This gives
and
or vice versa, implying that
and
.
Case 2: We have . Then, we must have
. It is clear that
(we would otherwise get
implying
or vice versa), so
and
.
Thus, our final answer is . ~GeronimoStilton
Solution 4
Let . There are two cases: in the first case,
equals
(without loss of generality), and thus
. By Vieta's formulas
.
In the second case, say without loss of generality and
. Subtracting gives
, so
. From this, we have
.
Note , so by Vieta's, we have
. In this case,
.
The requested sum is .~TheUltimate123
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.