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Difference between revisions of "2010 AMC 12A Problems/Problem 21"
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== Solution 2 ==
 
== Solution 2 ==
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The <math>x</math> values in which <math>y=x^6-10x^5+29x^4-4x^3+ax^2</math> intersect at <math>y=bx+c</math> are the same as the of <math>y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c</math>.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.
+
The <math>x</math> values in which <math>y=x^6-10x^5+29x^4-4x^3+ax^2</math> intersect at <math>y=bx+c</math> are the same as the zeros of <math>y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c</math>.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 <math>(x-p)^2(x-q)^2(x-r)^2</math>.  
 
Let the function be <math>(x-p)^2(x-q)^2(x-r)^2</math>.  
  

Revision as of 11:37, 20 December 2015

Problem

The graph of $y=x^6-10x^5+29x^4-4x^3+ax^2$ lies above the line $y=bx+c$ except at three values of $x$, where the graph and the line intersect. What is the largest of these values?

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$


Solution 1

The $x$ values in which $y=x^6-10x^5+29x^4-4x^3+ax^2$ intersect at $y=bx+c$ are the same as the zeros of $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c$.

Since there are $3$ zeros and the function is never negative, all $3$ zeros must be double roots because the function's degree is $6$.

Suppose we let $p$, $q$, and $r$ be the roots of this function, and let $x^3-ux^2+vx-w$ be the cubic polynomial with roots $p$, $q$, and $r$.

\begin{align*}(x-p)(x-q)(x-r) &= x^3-ux^2+vx-w\\ (x-p)^2(x-q)^2(x-r)^2 &= x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0\\ \sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} &= x^3-ux^2+vx-w = 0\end{align*}

In order to find $\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c}$ we must first expand out the terms of $(x^3-ux^2+vx-w)^2$.

\[(x^3-ux^2+vx-w)^2\] $= x^6-2ux^5+(u^2+2v)x^4-(2uv+2w)x^3+(2uw+v^2)x^2-2vwx+w^2$

[Quick note: Since we don't know $a$, $b$, and $c$, we really don't even need the last 3 terms of the expansion.]

\begin{align*}&2u = 10\\ u^2+2v &= 29\\ 2uv+2w &= 4\\ u &= 5\\ v &= 2\\ w &= -8\\ &\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} = x^3-5x^2+2x+8\end{align*}

All that's left is to find the largest root of $x^3-5x^2+2x+8$.

\begin{align*}&x^3-5x^2+2x+8 = (x-4)(x-2)(x+1)\\ &\boxed{\textbf{(A)}\ 4}\end{align*}

Solution 2

The $x$ values in which $y=x^6-10x^5+29x^4-4x^3+ax^2$ intersect at $y=bx+c$ are the same as the zeros of $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c$.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 $(x-p)^2(x-q)^2(x-r)^2$.

Applying Vieta's formulas, we get $2p+2q+2r = 10$ or $p+q+r = 5$. Applying it again, we get, after simplification, $p^2+q^2+r^2+4pq+4pr+4qr = 29$.

Notice that squaring the first equation yields $p^2+q^2+r^2+2pq+2qr+2pr= 25$, which is similar to the second equation. Subtracting this from the second equation, we get $2pq+2pr+2qr = 4$. Now that we have to $pq+pr+qr$ term, we can manpulate the equations to yield the sum of squares. $2(p^2+q^2+r^2+2pq+2qr+2pr)-2pq-2pr-2qr= 25*2-4$ or $2p^2+2q^2+2r^2+2pq+2qr+2pr = 46$. We finally reach $(p+q)^2+(q+r)^2+(p+r)^2 = 46$. Since the answer choices a integers, we can guess and check squares to get $\{(p+q)^2, (q+r)^2, (p+r)^2\} = \{1, 9, 36\}$ in some order. We can check that this works by adding then and seeing $2p+2q+2r = 10$. We just need to take the lowest value in the set, square root it, and subtract the resulting value from 5 to get $\boxed{\textbf{(A)}\ 4}$.


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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