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>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | Let <math>f(x) = x^6 - 10x^5 + 29x^4 - 4x^3 + ax^2 - bx - c</math>. Then <math>f(x) \ge 0</math> for all <math>x</math>, and <math>f(x) = 0</math> for three values of <math>x</math>, say <math>x = p</math>, <math>q</math>, and <math>r</math>. Then we can write | ||
+ | <cmath>f(x) = (x - p)^2 (x - q)^2 (x - r)^2 = [(x - p)(x - q)(x - r)]^2.</cmath> | ||
+ | To make this expression easier to work with, let <math>A = p + q + r</math>, <math>B = pq + pr + qr</math>, and <math>C = pqr</math>. Then <math>(x - p)(x - q)(x - r) = x^3 - Ax^2 + Bx - C</math>, so | ||
+ | <cmath>f(x) = (x^3 - Ax^2 + Bx - C)^2.</cmath> | ||
+ | |||
+ | Expanding, we get | ||
+ | \begin{align*} | ||
+ | f(x) &= (x^3 - Ax^2 + Bx - C)^2 \\ | ||
+ | &= x^6 - 2Ax^5 + (A^2 + 2B) x^4 - (2AB + 2C) x^3 + \dotsb. | ||
+ | \end{align*} | ||
+ | Equating coefficients, we obtain the system of equations | ||
+ | \begin{align*} | ||
+ | -2A &= -10, \\ | ||
+ | A^2 + 2B &= 29, \\ | ||
+ | -2AB - 2C &= -4. | ||
+ | \end{align*} | ||
+ | |||
+ | From the first equation, <math>A = (-10)/(-2) = 5</math>. Then from the second equation, <math>B = (29 - A^2)/2 = (29 - 25)/2 = 2</math>. Finally, from the third equation, <math>C = (4 - 2AB)/2 = (4 - 2 \cdot 5 \cdot 2)/2 = -8</math>. Hence, | ||
+ | <cmath>(x - p)(x - q)(x - r) = x^3 - 5x^2 + 2x + 8.</cmath> | ||
+ | Trying different values of <math>x</math>, we find that this cubic equation factors as | ||
+ | <cmath>(x - p)(x - q)(x - r) = (x - 2)(x - 4)(x + 1).</cmath> | ||
+ | Therefore, the largest such value is <math>\boxed{4}</math>. The answer is (A). | ||
== See also == | == See also == |
Revision as of 22:27, 23 November 2020
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?
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
.
Solution 3
Let . Then
for all
, and
for three values of
, say
,
, and
. Then we can write
To make this expression easier to work with, let
,
, and
. Then
, so
Expanding, we get \begin{align*} f(x) &= (x^3 - Ax^2 + Bx - C)^2 \\ &= x^6 - 2Ax^5 + (A^2 + 2B) x^4 - (2AB + 2C) x^3 + \dotsb. \end{align*} Equating coefficients, we obtain the system of equations \begin{align*} -2A &= -10, \\ A^2 + 2B &= 29, \\ -2AB - 2C &= -4. \end{align*}
From the first equation, . Then from the second equation,
. Finally, from the third equation,
. Hence,
Trying different values of
, we find that this cubic equation factors as
Therefore, the largest such value is
. The answer is (A).
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 |
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