Difference between revisions of "2010 AMC 12A Problems/Problem 21"

Revision as of 20:33, 22 June 2022

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?

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

Solutions

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 the $pq+pr+qr$ 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 are 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 $\boxed{\textbf{(A)}\ 4}$ .

Note: One could also multiply $2pq+2pr+2qr = 4$ by 2 and subtract from $p^2+q^2+r^2+4pq+4pr+4qr = 29$ to obtain $p^2+q^2+r^2=21$ 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 $p+q+r = 5$ and $pq + qr + rp = 2$ , we can algebraically derive $pqr$ .

Applying Vieta's formulas on the $x^3$ term yields $2p^2q+2pq^2+2q^2r+2qr^2+2r^2p+2rp^2+8pqr = 4$ .

Notice that $(p+q+r)(pq+qr+rp) = p^2q+pq^2+q^2r+qr^2+r^2p+rp^2+3pqr$ , so $2p^2q+2pq^2+2q^2r+2qr^2+2r^2p+2rp^2+6pqr = 2(p+q+r)(pq+qr+rp) = 20.$

Subtracting this from $2p^2q+2pq^2+2q^2r+2qr^2+2r^2p+2rp^2+8pqr = 4$ yields $2pqr = -16$ , so $pqr = -8$ , which means that $p$ , $q$ , and $r$ are the roots of the cubic $x^3 - 5x^2 + 2x + 8$ , and it is not hard to find that these roots are $-1$ , $2$ , and $4$ . The largest of these values is $\boxed{\textbf{(A)}\ 4}$ .

Solution 3

First, $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0$ has exactly $3$ roots. Therefore, $y = (kx^3+lx^2+mx+n)^2 = 0$ .

So, $k^2x^6+2klx^5+(2km+l^2)x^4+2(kn+lm)x^3+ax^2-bx-c = 0$

By matching the coefficients of the first $4$ terms, we have $k^2 = 1, 2kl = -10, 2km+l^2 = 29, 2kn+2lm = -4$

Solving the equations above, we have $2$ sets of solutions; first set of which is $k = 1, l = -5, m = 2, n = 8$ . Second set of which is $k = -1, l = 5, m = -2, n = -8$ . After squaring both sets, they are the same i.e. $x^3-5x^2+2x+8 = 0$ .

This is equal to $(x-4)(x-2)(x+1) = 0$ . The largest root is $\boxed {\textbf{(A) 4}}$

~Arcticturn

@@ Line 64: / Line 64: @@
 == Solution 3==
+First, <math>y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0</math> has exactly <math>3</math> roots. Therefore, <math>y = (kx^3+lx^2+mx+n)^2 = 0</math>.
+So, <math>k^2x^6+2klx^5+(2km+l^2)x^4+2(kn+lm)x^3+ax^2-bx-c = 0</math>
+By matching the coefficients of the first <math>4</math> terms, we have <math>k^2 = 1, 2kl = -10, 2km+l^2 = 29, 2kn+2lm = -4</math>
+Solving the equations above, we have <math>2</math> sets of solutions; first set of which is <math>k = 1, l = -5, m = 2, n = 8</math>. Second set of which is <math>k = -1, l = 5, m = -2, n = -8</math>. After squaring both sets, they are the same i.e. <math>x^3-5x^2+2x+8 = 0</math>.
+This is equal to <math>(x-4)(x-2)(x+1) = 0</math>. The largest root is <math>\boxed {\textbf{(A) 4}}</math>
+~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

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