Difference between revisions of "2020 AIME I Problems/Problem 14"
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The requested sum is <math>36+49=85</math>.~TheUltimate123 | The requested sum is <math>36+49=85</math>.~TheUltimate123 | ||
+ | ==Solution 5 (Official MAA)== | ||
+ | Note that because <math>P\big(P(3)\big)=P\big(P(4)\big)= 0</math>, <math>P(3)</math> and <math>P(4)</math> are roots of <math>P(x)</math>. There are two cases. | ||
+ | CASE 1: <math>P(3) = P(4)</math>. Then <math>P(x)</math> is symmetric about <math>x=\tfrac72</math>; that is to say, <math>P(r) = P(7-r)</math> for all <math>r</math>. Thus the remaining two roots must sum to <math>7</math>. Indeed, the polynomials <math>P(x) = \left(x-\frac72\right)^2 + \frac{11}4 \pm i\sqrt3</math> satisfy the conditions. | ||
+ | CASE 2: <math>P(3)\neq P(4)</math>. Then <math>P(3)</math> and <math>P(4)</math> are the two distinct roots of <math>P(x)</math>, so<cmath>P(x) = \big(x-P(3)\big)\big(x-P(4)\big)</cmath>for all <math>x</math>. Note that any solution to <math>P\big(P(x)\big) = 0</math> must satisfy either <math>P(x) = P(3)</math> or <math>P(x) = P(4)</math>. Because <math>P(x)</math> is quadratic, the polynomials <math>P(x) - P(3)</math> and <math>P(x) - P(4)</math> each have the same sum of roots as the polynomial <math>P(x)</math>, which is <math>P(3) + P(4)</math>. Thus the answer in this case is <math>2\big(P(3) + P(4)\big)-7</math>, and so it suffices to compute the value of <math>P(3)+P(4)</math>. | ||
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+ | Let <math>P(3)=u</math> and <math>P(4) = v</math>. Substituting <math>x=3</math> and <math>x=4</math> into the above quadratic polynomial yields the system of equations | ||
+ | <cmath>\begin{align*} | ||
+ | u &= (3-u)(3-v) = 9 - 3u - 3v + uv\ | ||
+ | v &= (4-u)(4-v) = 16 - 4u - 4v + uv. | ||
+ | \end{align*}</cmath>Subtracting the first equation from the second gives <math>v - u = 7 - u - v</math>, yielding <math>v = \frac72.</math> Substituting this value into the second equation gives<cmath>\dfrac72 = \left(4 - u\right)\left(4 - \dfrac72\right),</cmath>yielding <math>u = -3.</math> The sum of the two solutions is <math>2\left(\tfrac72-3\right)-7 = -6</math>. In this case, <math>P(x)= (x+3)\left(x-\frac72\right)</math>. | ||
+ | |||
+ | The requested sum of squares is <math>7^2+(-6)^2 = {85}</math>. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=I|num-b=13|num-a=15}} | {{AIME box|year=2020|n=I|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:11, 25 February 2021
Contents
[hide]Problem
Let be a quadratic polynomial with complex coefficients whose coefficient is Suppose the equation has four distinct solutions, Find the sum of all possible values of
Solution 1
Either or not. We first see that if it's easy to obtain by Vieta's that . Now, take and WLOG . Now, consider the parabola formed by the graph of . It has vertex . Now, say that . We note . Now, we note by plugging in again. Now, it's easy to find that , yielding a value of . Finally, we add . ~awang11, charmander3333
Remark: We know that from .
Solution 2
Let the roots of be and , then we can write . The fact that has solutions implies that some combination of of these are the solution to , and the other are the solution to . It's fairly easy to see there are only possible such groupings: and , or and (Note that are interchangeable, and so are and ). We now casework: If , then so this gives . Next, if , then Subtracting the first part of the first equation from the first part of the second equation gives Hence, , and so . Therefore, the solution is ~ktong
Solution 3
Write . Split the problem into two cases: and .
Case 1: We have . We must have Rearrange and divide through by to obtain Now, note that Now, rearrange to get and thus Substituting this into our equation for yields . Then, it is clear that does not have a double root at , so we must have and or vice versa. This gives and or vice versa, implying that and .
Case 2: We have . Then, we must have . It is clear that (we would otherwise get implying or vice versa), so and .
Thus, our final answer is . ~GeronimoStilton
Solution 4
Let . There are two cases: in the first case, equals (without loss of generality), and thus . By Vieta's formulas .
In the second case, say without loss of generality and . Subtracting gives , so . From this, we have .
Note , so by Vieta's, we have . In this case, .
The requested sum is .~TheUltimate123
Solution 5 (Official MAA)
Note that because , and are roots of . There are two cases. CASE 1: . Then is symmetric about ; that is to say, for all . Thus the remaining two roots must sum to . Indeed, the polynomials satisfy the conditions. CASE 2: . Then and are the two distinct roots of , sofor all . Note that any solution to must satisfy either or . Because is quadratic, the polynomials and each have the same sum of roots as the polynomial , which is . Thus the answer in this case is , and so it suffices to compute the value of .
Let and . Substituting and into the above quadratic polynomial yields the system of equations Subtracting the first equation from the second gives , yielding Substituting this value into the second equation givesyielding The sum of the two solutions is . In this case, .
The requested sum of squares is .
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.