Difference between revisions of "2021 AIME II Problems/Problem 7"
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~amcrunner | ~amcrunner | ||
==Solution 5== | ==Solution 5== | ||
− | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>. | + | Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>. |
Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math> | Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math> |
Revision as of 22:51, 29 January 2024
Contents
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
. Since you already know
and
, so you can solve for
and
pretty easily and see that
. So the answer is
.
~math31415926535 ~minor edit by Mathkiddie
Solution 2
Note that can be rewritten as
. Hence,
.
Rewriting , we get
.
Substitute
and solving, we get
We refer to this as Equation 1.
Note that gives
. So,
, which implies
or
We refer to this as Equation 2.
Substituting Equation 2 into Equation 1 gives, .
Solving this quadratic yields that .
Now we just try these two cases:
For substituting in Equation 1 gives a quadratic in
which has roots
.
Again trying cases, by letting , we get
, Hence
.
We know that
, Solving these we get
or
(doesn't matter due to symmetry in
).
So, this case yields solutions .
Similarly trying other three cases, we get no more solutions, Hence this is the solution for .
Finally, .
Therefore, .
~Arnav Nigam
Solution 3
For simplicity purposes, we number the given equations and
in that order.
Rearranging and solving for
we have
Substituting
into
and solving for
we get
Substituting
and
into
and simplifying, we rewrite the left side of
in terms of
and
only:
Let
from which
Multiplying both sides by
rearranging, and factoring give
Substituting back and completing the squares produce
If
then combining this with
we know that
and
are the solutions of the quadratic
Since the discriminant is negative, neither
nor
is a real number.
If then combining this with
we know that
and
are the solutions of the quadratic
or
from which
Substituting
into
and
we obtain
and
respectively. Together, we have
so the answer is
~MRENTHUSIASM
Solution 4 (way too long)
Let the four equations from top to bottom be listed 1 through 4 respectively. We factor equation 3 like so:
Then we plug in equation 2 to receive
. By equation 4 we get
. Plugging in, we get
. Multiply by
on both sides to get the quadratic equation
. Solving using the quadratic equation, we receive
. So, we have to test which one is correct. We repeat a similar process as we did above for equations 1 and 2. We factor equation 2 to get
After plugging in equation 1, we get
. Now we convert it into a quadratic to receive
. The value of
will depend on
. So we obtain the discriminant
.
Let d = -5.
Then
, so
, discriminant is
, which makes this a dead end. Thus
For
, making
. This means the discriminant is just
, so we obtain two values for
as well. We get either
or
. So, we must AGAIN test which one is correct.
We know
, and
, so we use these values for testing.
Let
.
Then
, so
. We thus get
, which leads to the quadratic
. The discriminant for this is
. That means this value of
is wrong, so
. Thus we get polynomial
. The discriminant this time is
, so we get two values for
. Through simple inspection, you may see they are interchangeable, as if you take the value
, you get
. If you take the value
, you get
. So it doesn't matter. That means the sum of all their squares is
so the answer is
~amcrunner
Solution 5
Let the four equations from top to bottom be listed through
respectively. Multiplying both sides of
by
and factoring some terms gives us
. Substituting using equations
and
gives us
, and solving gives us
or
. Plugging this back into
gives us
, or using the other solution for
gives us
. Solving both of these equations gives us
when
and
when
.
Multiplying both sides of by
and factoring some terms gives us
. Testing
will give us an imaginary solution for
, so therefore
and
. This gets us
. Solving for
gives us
or
. With a bit of testing, we can see that the correct value of
is
. Now we know
and
,
, and it is obvious that
and
or the other way around, and therefore,
, giving us the answer
~hihitherethere
Video Solution
https://www.youtube.com/watch?v=2rrX1G7iZqg
Video Solution by Interstigation
~Interstigation
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.