Difference between revisions of "2021 AIME II Problems/Problem 7"
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==Solution 2== | ==Solution 2== | ||
+ | <math>ab + bc + ca = -4</math> can be rewritten as <math>ab + c(a+b) = -4</math> | ||
+ | Hence, <math>ab = 3c - 4</math> | ||
+ | |||
+ | Rewriting <math>abc+bcd+cda+dab = 14</math>, we get <math>ab(c+d) + cd(a+b) = 14</math> | ||
+ | Substitute <math>ab = 3c - 4</math> and solving, we get | ||
+ | <math>3c^{2} - 4c - 4d - 14 = 0</math> | ||
==See also== | ==See also== | ||
{{AIME box|year=2021|n=II|num-b=6|num-a=8}} | {{AIME box|year=2021|n=II|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:54, 22 March 2021
Contents
[hide]Problem
Let and
be real numbers that satisfy the system of equations
There exist relatively prime positive integers
and
such that
Find
.
Solution 1
From the fourth equation we get substitute this into the third equation and you get
. Hence
. Solving we get
or
. From the first and second equation we get
, if
, substituting we get
. If you try solving this you see that this does not have real solutions in
, so
must be
. So
. Since
,
or
. If
, then the system
and
does not give you real solutions. So
. From here you already know
and
, so you can solve for
and
pretty easily and see that
. So the answer is
.
~ math31415926535
Solution 2
can be rewritten as
Hence,
Rewriting , we get
Substitute
and solving, we get
See also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.