Difference between revisions of "2020 AIME I Problems/Problem 14"
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== Solution 1 == | == Solution 1 == | ||
Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)(4 - 4a + \frac{8a - 1}{2}) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333 | Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)(4 - 4a + \frac{8a - 1}{2}) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333 | ||
+ | |||
+ | <b>Remark</b>: We know that <math>c=\frac{8a-1}{2}</math> from <math>P(3)+P(4)=3+a</math>. | ||
== Solution 2 == | == Solution 2 == |
Revision as of 16:37, 29 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
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 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 |
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