Difference between revisions of "2020 AIME I Problems/Problem 14"
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Hence, <math>a+b=2(m+n)-7=2\cdot \frac{1}{2}-7=-6</math>, and so <math>(a+b)^2=(-6)^2=36</math>. | Hence, <math>a+b=2(m+n)-7=2\cdot \frac{1}{2}-7=-6</math>, and so <math>(a+b)^2=(-6)^2=36</math>. | ||
Therefore, the solution is <math>49+36=\boxed{085}</math> ~ktong | Therefore, the solution is <math>49+36=\boxed{085}</math> ~ktong | ||
+ | |||
+ | == Solution 3 == | ||
+ | Write <math>P(x) = x^2+wx+z</math>. Split the problem into two cases: <math>P(3)\ne P(4)</math> and <math>P(3) = P(4)</math>. | ||
+ | |||
+ | Case 1: We have <math>P(3) \ne P(4)</math>. We must have | ||
+ | <cmath>w=-P(3)-P(4) = -(9+3w+z)-(16+4w+z) = -25-7w-2z.</cmath> | ||
+ | Rearrange and divide through by <math>8</math> to obtain | ||
+ | <cmath>w = \frac{-25-2z}{8}.</cmath> | ||
+ | Now, note that | ||
+ | <cmath>z = P(3)P(4) = (9+3w+z)(16+4w+z) = \left(9 + 3\cdot \frac{-25-2z}{8} + z\right)\left(16 + 4 \cdot \frac{-25-2z}{8} + z\right) =</cmath> | ||
+ | <cmath>\left(-\frac{3}{8} + \frac{z}{4}\right)\left(\frac{7}{2}\right) = -\frac{21}{16} + \frac{7z}{8}.</cmath> | ||
+ | Now, rearrange to get | ||
+ | <cmath>\frac{z}{8} = -\frac{21}{16}</cmath> | ||
+ | and thus | ||
+ | <cmath>z = -\frac{21}{2}.</cmath> | ||
+ | Substituting this into our equation for <math>w</math> yields <math>w = -\frac{1}{2}</math>. Then, it is clear that <math>P</math> does not have a double root at <math>P(3)</math>, so we must have <math>P(a) = P(3)</math> and <math>P(b) = P(4)</math> or vice versa. This gives <math>3+a = \frac{1}{2}</math> and <math>4+b = \frac{1}{2}</math> or vice versa, implying that <math>a+b = 1-3-4 = -6</math> and <math>(a+b)^2 = 6</math>. | ||
+ | |||
+ | Case 2: We have <math>P(3) = P(4)</math>. Then, we must have <math>w = -7</math>. It is clear that <math>P(a) = P(b)</math> (we would otherwise get <math>P(a)=P(3)=P(4)</math> implying <math>a \in \{3,4\}</math> or vice versa), so <math>a+b=-w=7</math> and <math>(a+b)^2 = 49</math>. | ||
+ | |||
+ | Thus, our final answer is <math>49+36=\fbox{085)</math>. ~GeronimoStilton | ||
==See Also== | ==See Also== |
Revision as of 11:01, 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 $49+36=\fbox{085)$ (Error compiling LaTeX. Unknown error_msg). ~GeronimoStilton
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.