Difference between revisions of "2020 AIME II Problems/Problem 11"
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Subtracting <math>(3)</math> from <math>(1)</math>, we get <math>r - s = \dfrac{3 + b}{2}</math>. Adding this to <math>(2)</math>, we get <math>2r = 3 \implies r = \dfrac{3}{2}</math>. This gives us that <math>t = \dfrac{-5}{3}</math> from <math>(4)</math>. Substituting these values into <math>(5)</math> and <math>(6)</math>, we get <math>s = \dfrac{c-7}{3}</math> and <math>s = \dfrac{-3c - 6}{10}</math>. Equating these values, we get <math>\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = \dfrac{52}{19} = R(0)</math>. Thus, our answer is <math>52 + 19 = \boxed{071}</math>. ~ TopNotchMath | Subtracting <math>(3)</math> from <math>(1)</math>, we get <math>r - s = \dfrac{3 + b}{2}</math>. Adding this to <math>(2)</math>, we get <math>2r = 3 \implies r = \dfrac{3}{2}</math>. This gives us that <math>t = \dfrac{-5}{3}</math> from <math>(4)</math>. Substituting these values into <math>(5)</math> and <math>(6)</math>, we get <math>s = \dfrac{c-7}{3}</math> and <math>s = \dfrac{-3c - 6}{10}</math>. Equating these values, we get <math>\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = \dfrac{52}{19} = R(0)</math>. Thus, our answer is <math>52 + 19 = \boxed{071}</math>. ~ TopNotchMath | ||
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==Solution 3== | ==Solution 3== |
Revision as of 13:57, 23 July 2020
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 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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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