Difference between revisions of "2015 AIME I Problems/Problem 10"
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which must hold for all <math>x</math>. Since <math>a\neq 0</math> we have that (1) <math>t+u+v=q+r+s</math>, (2) <math>tu+uv+tv=qr+qs+rs</math> and (3) <math>a(tuv-qrs)=24</math>. Since <math>q+r+s+t+u+v</math> is the sum of 1,2,3,5,6, and 7, we have <math>q+r+s+t+u+v=24</math> so that by (1) we have <math>q+r+s=12</math> and <math>t+u+v=12</math>. We must partition 1,2,3,5,6,7 into 2 sets each with a sum of 12. Consider the set that contains 7. It can't contain 6 or 5 because the sum of that set would already be <math>\geq 12</math> with only 2 elements. If 1 is in that set, the other element must be 4 which is impossible. Hence the two sets must be <math>\{2,3,7\}</math> and <math>\{1,5,6\}</math>. Note that each of these sets happily satisfy (2). By (3), since the sets have products 42 and 30 we have that <math>|a|=\frac{24}{|tuv-qrs|}=\frac{24}{12}=2</math>. Since <math>a</math> is the leading coefficient of <math>f(x)</math>, the leading coefficient of <math>(f(x))^2</math> is <math>a^2=|a|^2=2^2=4</math>. Thus the leading coefficient of <math>g(x)</math> is 4, i.e. <math>k=4</math>. Then from earlier, <math>|f(0)|=\sqrt{g(0)+144}=\sqrt{1260k+144}=\sqrt{1260\cdot4+144}=\sqrt{5184}=72</math> so that the answer is <math>\boxed{072}</math>. | which must hold for all <math>x</math>. Since <math>a\neq 0</math> we have that (1) <math>t+u+v=q+r+s</math>, (2) <math>tu+uv+tv=qr+qs+rs</math> and (3) <math>a(tuv-qrs)=24</math>. Since <math>q+r+s+t+u+v</math> is the sum of 1,2,3,5,6, and 7, we have <math>q+r+s+t+u+v=24</math> so that by (1) we have <math>q+r+s=12</math> and <math>t+u+v=12</math>. We must partition 1,2,3,5,6,7 into 2 sets each with a sum of 12. Consider the set that contains 7. It can't contain 6 or 5 because the sum of that set would already be <math>\geq 12</math> with only 2 elements. If 1 is in that set, the other element must be 4 which is impossible. Hence the two sets must be <math>\{2,3,7\}</math> and <math>\{1,5,6\}</math>. Note that each of these sets happily satisfy (2). By (3), since the sets have products 42 and 30 we have that <math>|a|=\frac{24}{|tuv-qrs|}=\frac{24}{12}=2</math>. Since <math>a</math> is the leading coefficient of <math>f(x)</math>, the leading coefficient of <math>(f(x))^2</math> is <math>a^2=|a|^2=2^2=4</math>. Thus the leading coefficient of <math>g(x)</math> is 4, i.e. <math>k=4</math>. Then from earlier, <math>|f(0)|=\sqrt{g(0)+144}=\sqrt{1260k+144}=\sqrt{1260\cdot4+144}=\sqrt{5184}=72</math> so that the answer is <math>\boxed{072}</math>. | ||
+ | ==Solution 5== | ||
+ | By drawing the function, WLOG let <math>f(1)=f(5)=f(6)=12</math>. Then, <math>f(2)=f(3)=f(7)</math>. Realize that if we shift <math>f(x)</math> down 12, then this function <math>f(x)-12</math> has roots <math>1,5,6</math> with leading coefficient <math>-2</math> because <math>f(2)-12=-24=-2(1)(-3)(-4)</math>. Therefore <math>f(x)=-2(x-1)(x-5)(x-6)+12</math>, and then <math>|f(0)|=60+12=\boxed{72}</math>. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2015|n=I|num-b=9|num-a=11}} | {{AIME box|year=2015|n=I|num-b=9|num-a=11}} |
Revision as of 18:14, 7 October 2017
Problem
Let be a third-degree polynomial with real coefficients satisfying Find .
Solution
Let = . Since is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing 12 and -12, it is easy to see that f(1)=f(5)=f(6), and f(2)=f(3)=f(7); otherwise more bends would be required in the graph. Since only the absolute value of f(0) is required, there is no loss of generalization by stating that f(1)=12, and f(2)=-12. This provides the following system of equations. Using any four of these functions as a system of equations yields
Solution 2
Express in terms of powers of : By the same argument as in the first Solution, we see that is an odd function about the line , so its coefficients and are 0. From there it is relatively simple to solve (as in the above solution, but with a smaller system of equations): and
Solution 3
Without loss of generality, let . (If , then take as the polynomial, which leaves unchanged.) Because is third-degree, write where clearly must be a permutation of from the given condition. Thus However, subtracting the two equations gives , so comparing coefficients gives and thus both values equal to . As a result, . As a result, and so . Now, we easily deduce that and so removing the without loss of generality gives , which is our answer.
Solution 4
The following solution is similar to solution 3, but assumes nothing. Let . Since has degree 3, has degree 6 and has roots 1,2,3,5,6, and 7. Therefore, for some . Hence . Note that . Since has degree 3, so do and ; and both have the same leading coefficient. Hence and for some (else is not cubic) where is the same as the set . Subtracting the second equation from the first, expanding, and collecting like terms, we have that which must hold for all . Since we have that (1) , (2) and (3) . Since is the sum of 1,2,3,5,6, and 7, we have so that by (1) we have and . We must partition 1,2,3,5,6,7 into 2 sets each with a sum of 12. Consider the set that contains 7. It can't contain 6 or 5 because the sum of that set would already be with only 2 elements. If 1 is in that set, the other element must be 4 which is impossible. Hence the two sets must be and . Note that each of these sets happily satisfy (2). By (3), since the sets have products 42 and 30 we have that . Since is the leading coefficient of , the leading coefficient of is . Thus the leading coefficient of is 4, i.e. . Then from earlier, so that the answer is .
Solution 5
By drawing the function, WLOG let . Then, . Realize that if we shift down 12, then this function has roots with leading coefficient because . Therefore , and then .
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.