Difference between revisions of "2020 AIME I Problems/Problem 14"
m (→Solution 1) |
m (→Solution 2) |
||
Line 9: | Line 9: | ||
== Solution 2 == | == Solution 2 == | ||
− | Let the roots of <math>P(x)</math> be <math>m</math> and <math>n</math>, then we can write <math>P(x)=x^2-(m+n)x+mn</math>. The fact that <math>P(P(x))=0</math> has solutions <math>x=3,4,a,b</math> implies that some combination of <math>2</math> of these are the solution to <math>P(x)=m</math>, and the other <math>2</math> are the solution to <math>P(x)=n</math>. It's fairly easy to see there are only <math>2</math> possible such groupings: <math>P(3)=P(4)=m</math> and <math>P(a)=P(b)=n</math>, or <math>P(3)=P(a)=m</math> and <math>P(4)=P(b)=n</math> (Note that <math>a,b</math> are interchangeable, and so are <math>m</math> and <math>n</math>). We now | + | Let the roots of <math>P(x)</math> be <math>m</math> and <math>n</math>, then we can write <math>P(x)=x^2-(m+n)x+mn</math>. The fact that <math>P(P(x))=0</math> has solutions <math>x=3,4,a,b</math> implies that some combination of <math>2</math> of these are the solution to <math>P(x)=m</math>, and the other <math>2</math> are the solution to <math>P(x)=n</math>. It's fairly easy to see there are only <math>2</math> possible such groupings: <math>P(3)=P(4)=m</math> and <math>P(a)=P(b)=n</math>, or <math>P(3)=P(a)=m</math> and <math>P(4)=P(b)=n</math> (Note that <math>a,b</math> are interchangeable, and so are <math>m</math> and <math>n</math>). We now casework: |
If <math>P(3)=P(4)=m</math>, then | If <math>P(3)=P(4)=m</math>, then | ||
<cmath>9-3(m+n)+mn=16-4(m+n)+mn=m \implies m+n=7</cmath> | <cmath>9-3(m+n)+mn=16-4(m+n)+mn=m \implies m+n=7</cmath> |
Revision as of 22:12, 27 May 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
Remark: We know that from
.
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 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.