Difference between revisions of "2022 AIME I Problems/Problem 1"
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We need to cancel <math>a</math> and <math>c.</math> Since <math>\operatorname{lcm}(16,20)=80,</math> we subtract <math>4\cdot[(3)+(4)]</math> from <math>5\cdot[(1)+(2)]</math> to get <cmath>b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.</cmath> | We need to cancel <math>a</math> and <math>c.</math> Since <math>\operatorname{lcm}(16,20)=80,</math> we subtract <math>4\cdot[(3)+(4)]</math> from <math>5\cdot[(1)+(2)]</math> to get <cmath>b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.</cmath> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
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==Solution 2.5 (Similar to Sol 2)== | ==Solution 2.5 (Similar to Sol 2)== |
Revision as of 03:46, 4 September 2022
Contents
Problem
Quadratic polynomials and
have leading coefficients
and
respectively. The graphs of both polynomials pass through the two points
and
Find
Solution 1 (Linear Polynomials)
Let Since the
-terms of
and
cancel, we conclude that
is a linear polynomial.
Note that
so the slope of
is
It follows that the equation of is
for some constant
and we wish to find
We substitute into this equation to get
from which
~MRENTHUSIASM
Solution 2 (Quadratic Polynomials)
Let
for some constants
and
We are given that
and we wish to find
We need to cancel
and
Since
we subtract
from
to get
~MRENTHUSIASM
Solution 2.5 (Similar to Sol 2)
Like Solution 2, we can begin by setting and
to the quadratic above, giving us
We can first add and
to obtain
Next, we can add and
to obtain
By subtracting these two equations, we find that
so substituting this into equation
we know that
so therefore
~jessiewang28
Solution 3 (Pure Brute Force)
Let
By substitutes and
into these equations, we can get:
Hence,
and
.
Similarly,
Hence,
and
.
Notice that and
.
Therefore
~Littlemouse
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=sUfbEBCQ6RY
Video Solution by MRENTHUSIASM (English & Chinese)
https://www.youtube.com/watch?v=XcS5qcqsRyw&ab_channel=MRENTHUSIASM
~MRENTHUSIASM
Video Solution
https://youtu.be/MJ_M-xvwHLk?t=7
~ThePuzzlr
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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.