Difference between revisions of "2010 AMC 12A Problems/Problem 21"
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<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8</math> | <math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8</math> | ||
− | + | == Solutions == | |
− | == Solution 1== | + | === Solution 1 === |
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>. | 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>. | ||
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<cmath>(x^3-ux^2+vx-w)^2</cmath> | <cmath>(x^3-ux^2+vx-w)^2</cmath> | ||
− | < | + | <cmath>= x^6-2ux^5+(u^2+2v)x^4-(2uv+2w)x^3+(2uw+v^2)x^2-2vwx+w^2</cmath> |
[Quick note: Since we don't know <math>a</math>, <math>b</math>, and <math>c</math>, we really don't even need the last 3 terms of the expansion.] | [Quick note: Since we don't know <math>a</math>, <math>b</math>, and <math>c</math>, we really don't even need the last 3 terms of the expansion.] | ||
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&\boxed{\textbf{(A)}\ 4}\end{align*}</cmath> | &\boxed{\textbf{(A)}\ 4}\end{align*}</cmath> | ||
− | == Solution 2 == | + | === Solution 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>. | 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. | 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. | ||
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Notice that squaring the first equation yields <math>p^2+q^2+r^2+2pq+2qr+2pr= 25</math>, which is similar to the second equation. | Notice that squaring the first equation yields <math>p^2+q^2+r^2+2pq+2qr+2pr= 25</math>, which is similar to the second equation. | ||
− | Subtracting this from the second equation, we get <math>2pq+2pr+2qr = 4</math>. Now that we have | + | Subtracting this from the second equation, we get <math>2pq+2pr+2qr = 4</math>. Now that we have the <math>pq+pr+qr</math> term, we can manpulate the equations to |
yield the sum of squares. <math>2(p^2+q^2+r^2+2pq+2qr+2pr)-2pq-2pr-2qr= 25*2-4</math> or <math>2p^2+2q^2+2r^2+2pq+2qr+2pr = 46</math>. We finally reach <math>(p+q)^2+(q+r)^2+(p+r)^2 = 46</math>. | yield the sum of squares. <math>2(p^2+q^2+r^2+2pq+2qr+2pr)-2pq-2pr-2qr= 25*2-4</math> or <math>2p^2+2q^2+2r^2+2pq+2qr+2pr = 46</math>. We finally reach <math>(p+q)^2+(q+r)^2+(p+r)^2 = 46</math>. | ||
Since the answer choices are integers, we can guess and check squares to get <math>\{(p+q)^2, (q+r)^2, (p+r)^2\} = \{1, 9, 36\}</math> in some order. We can check that this works by adding then and seeing <math>2p+2q+2r = 10</math>. We just need to take the lowest value in the set, square root it, and subtract the resulting value from 5 to get <math>\boxed{\textbf{(A)}\ 4}</math>. | Since the answer choices are integers, we can guess and check squares to get <math>\{(p+q)^2, (q+r)^2, (p+r)^2\} = \{1, 9, 36\}</math> in some order. We can check that this works by adding then and seeing <math>2p+2q+2r = 10</math>. We just need to take the lowest value in the set, square root it, and subtract the resulting value from 5 to get <math>\boxed{\textbf{(A)}\ 4}</math>. | ||
− | Alternative method: | + | Note: One could also multiply <math>2pq+2pr+2qr = 4</math> by 2 and subtract from <math>p^2+q^2+r^2+4pq+4pr+4qr = 29</math> to obtain <cmath>p^2+q^2+r^2=21</cmath> 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 <math>p+q+r = 5</math> and <math>pq + qr + rp = 2</math>, we can algebraically derive <math>pqr</math>. | After reaching <math>p+q+r = 5</math> and <math>pq + qr + rp = 2</math>, we can algebraically derive <math>pqr</math>. | ||
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Notice that <math>(p+q+r)(pq+qr+rp) = p^2q+pq^2+q^2r+qr^2+r^2p+rp^2+3pqr</math>, so <cmath>2p^2q+2pq^2+2q^2r+2qr^2+2r^2p+2rp^2+6pqr = 2(p+q+r)(pq+qr+rp) = 20.</cmath> | Notice that <math>(p+q+r)(pq+qr+rp) = p^2q+pq^2+q^2r+qr^2+r^2p+rp^2+3pqr</math>, so <cmath>2p^2q+2pq^2+2q^2r+2qr^2+2r^2p+2rp^2+6pqr = 2(p+q+r)(pq+qr+rp) = 20.</cmath> | ||
− | Subtracting this from <math>2p^2q+2pq^2+2q^2r+2qr^2+2r^2p+2rp^2+8pqr = 4</math> yields <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== | ||
+ | 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 == | == See also == |
Latest revision as of 16:45, 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.