2020 AIME II Problems/Problem 11
Contents
Problem
Let , and let
and
be two quadratic polynomials also with the coefficient of
equal to
. David computes each of the three sums
,
, and
and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If
, then
, where
and
are relatively prime positive integers. Find
.
Solution 1
Let and
. We can write the following:
Let the common root of
be
;
be
; and
be
. We then have that the roots of
are
, the roots of
are
, and the roots of
are
.
By Vieta's, we have:
Subtracting from
, we get
. Adding this to
, we get
. This gives us that
from
. Substituting these values into
and
, we get
and
. Equating these values, we get
. Thus, our answer is
. ~ TopNotchMath
Solution 2
Let have shared root
,
have shared root
, and the last pair having shared root
. We will now set
, and
. We wish to find
, and now we compute
.
From here, we equate coefficients. This means
. Now,
. Finally, we know that
Solution 3
We know that .
Since , the constant term in
is
. Let
.
Finally, let .
. Let its roots be
and
.
Let its roots be
and
.
. Let its roots be
and
.
By vietas,
We could work out the system of equations, but it's pretty easy to see that .
~quacker88
Solution 4 (Official MAA)
Let the common root of and
be
, the common root of
and
be
, and the common root of
and
be
. Because
and
are both roots of
and
has leading coefficient
, it follows that
Similarly,
and
. Adding these three equations together and dividing by
yields
so
Similarly,
Comparing the
coefficients yields
, and comparing the constant coefficients yields
. The fact that
implies that
. Adding these two equations yields
, and so substituting back in to solve for
gives
. Finally,
The requested sum is
. Note that
and
.
Video Solution
https://youtu.be/BQlab3vjjxw ~ CNCM
Another one: https://www.youtube.com/watch?v=AXN9x51KzNI
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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