Difference between revisions of "2010 AMC 12A Problems/Problem 21"
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Subtracting this from <math>2p^2q+2pq^2+2q^2r+2qr^2+2r^2p+2rp^2+8pqr = 4</math> yields <math>2pqr = -16</math>, so <math>pqr = -8</math>, which means that <math>p</math>, <math>q</math>, and <math>r</math> are the roots of the cubic <math>x^3 - 5x^2 + 2x + 8</math>, and it is not hard to find that these roots are <math>-1</math>, <math>2</math>, and <math>4</math>. The largest of these values is <math>\boxed{\textbf{(A)}\ 4}</math>. | Subtracting this from <math>2p^2q+2pq^2+2q^2r+2qr^2+2r^2p+2rp^2+8pqr = 4</math> yields <math>2pqr = -16</math>, so <math>pqr = -8</math>, which means that <math>p</math>, <math>q</math>, and <math>r</math> are the roots of the cubic <math>x^3 - 5x^2 + 2x + 8</math>, and it is not hard to find that these roots are <math>-1</math>, <math>2</math>, and <math>4</math>. The largest of these values is <math>\boxed{\textbf{(A)}\ 4}</math>. | ||
+ | |||
+ | Extra note: One can also use the identity | ||
+ | <cmath>(p+q)(q+r)(p+r)=2pqr+\sum_{sym}p^2q</cmath> | ||
+ | to compute <math>pqr</math>. From <math>p^2q+pq^2+q^2r+qr^2+r^2p+rp^2+4pqr = 2</math>, use our identity to get <math>2pqr+(p+q)(q+r)(p+r)=2</math>. Then use <math>p+q+r=5</math> to rewrite as <math>2pqr+(5-p)(5-q)(5-r)=2</math>. Expanding and using <math>pq+qr+pr=2</math> as well gives the result <math>pqr=-8</math>. | ||
+ | |||
+ | ~~ clarkculus | ||
== Solution 3== | == Solution 3== |
Latest revision as of 17:45, 26 August 2024
Contents
Problem
The graph of lies above the line
except at three values of
, where the graph and the line intersect. What is the largest of these values?
Solutions
Solution 1
The values in which
intersect at
are the same as the zeros of
.
Since there are zeros and the function is never negative, all
zeros must be double roots because the function's degree is
.
Suppose we let ,
, and
be the roots of this function, and let
be the cubic polynomial with roots
,
, and
.
In order to find we must first expand out the terms of
.
[Quick note: Since we don't know ,
, and
, we really don't even need the last 3 terms of the expansion.]
All that's left is to find the largest root of .
Solution 2
The values in which
intersect at
are the same as the zeros of
.
We also know that this graph has 3 places tangent to the x-axis, which means that each root has to have a multiplicity of 2.
Let the function be
.
Applying Vieta's formulas, we get or
.
Applying it again, we get, after simplification,
.
Notice that squaring the first equation yields , which is similar to the second equation.
Subtracting this from the second equation, we get . Now that we have the
term, we can manpulate the equations to
yield the sum of squares.
or
. We finally reach
.
Since the answer choices are integers, we can guess and check squares to get in some order. We can check that this works by adding then and seeing
. We just need to take the lowest value in the set, square root it, and subtract the resulting value from 5 to get
.
Note: One could also multiply by 2 and subtract from
to obtain
The ordered triple {16,4,1} sums to 21, and the answer choices are all positive integers, therefore the answer is 4.
Alternative method:
After reaching and
, we can algebraically derive
.
Applying Vieta's formulas on the term yields
.
Notice that , so
Subtracting this from yields
, so
, which means that
,
, and
are the roots of the cubic
, and it is not hard to find that these roots are
,
, and
. The largest of these values is
.
Extra note: One can also use the identity
to compute
. From
, use our identity to get
. Then use
to rewrite as
. Expanding and using
as well gives the result
.
~~ clarkculus
Solution 3
First, has exactly
roots. Therefore,
.
So,
By matching the coefficients of the first terms, we have
Solving the equations above, we have sets of solutions; first set of which is
. Second set of which is
. After squaring both sets, they are the same i.e.
.
This is equal to . The largest root is
~Arcticturn
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.