Difference between revisions of "2020 AIME I Problems/Problem 14"
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Let <math>P(x)</math> be a quadratic polynomial with complex coefficients whose <math>x^2</math> coefficient is <math>1.</math> Suppose the equation <math>P(P(x))=0</math> has four distinct solutions, <math>x=3,4,a,b.</math> Find the sum of all possible values of <math>(a+b)^2.</math> | Let <math>P(x)</math> be a quadratic polynomial with complex coefficients whose <math>x^2</math> coefficient is <math>1.</math> Suppose the equation <math>P(P(x))=0</math> has four distinct solutions, <math>x=3,4,a,b.</math> Find the sum of all possible values of <math>(a+b)^2.</math> | ||
− | == Solution == | + | == Solution 1 == |
− | Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)(4 - 4a + \frac{8a - 1}{2}) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333 | + | Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)\left(4 - 4a + \frac{8a - 1}{2}\right) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333 |
+ | |||
+ | <b>Remark</b>: We know that <math>c=\frac{8a-1}{2}</math> from <math>P(3)+P(4)=3+a</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | Let the roots of <math>P(x)</math> be <math>m</math> and <math>n</math>, then we can write <math>P(x)=x^2-(m+n)x+mn</math>. The fact that <math>P(P(x))=0</math> has solutions <math>x=3,4,a,b</math> implies that some combination of <math>2</math> of these are the solution to <math>P(x)=m</math>, and the other <math>2</math> are the solution to <math>P(x)=n</math>. It's fairly easy to see there are only <math>2</math> possible such groupings: <math>P(3)=P(4)=m</math> and <math>P(a)=P(b)=n</math>, or <math>P(3)=P(a)=m</math> and <math>P(4)=P(b)=n</math> (Note that <math>a,b</math> are interchangeable, and so are <math>m</math> and <math>n</math>). We now casework: | ||
+ | If <math>P(3)=P(4)=m</math>, then | ||
+ | <cmath>9-3(m+n)+mn=16-4(m+n)+mn=m \implies m+n=7</cmath> | ||
+ | <cmath>a^2-a(m+n)+mn=b^2-b(m+n)+mn=n \implies a+b=m+n=7</cmath> | ||
+ | so this gives <math>(a+b)^2=7^2=49</math>. | ||
+ | Next, if <math>P(3)=P(a)=m</math>, then | ||
+ | <cmath>9-3(m+n)+mn=a^2-a(m+n)+mn=m \implies a+3=m+n</cmath> | ||
+ | <cmath>16-4(m+n)+mn=b^2-b(m+n)+mn=n \implies b+4=m+n</cmath> | ||
+ | Subtracting the first part of the first equation from the first part of the second equation gives | ||
+ | <cmath>7-(m+n)=n-m \implies 2n=7 \implies n=\frac{7}{2} \implies m=-3</cmath> | ||
+ | Hence, <math>a+b=2(m+n)-7=2\cdot \frac{1}{2}-7=-6</math>, and so <math>(a+b)^2=(-6)^2=36</math>. | ||
+ | Therefore, the solution is <math>49+36=\boxed{085}</math> ~ktong | ||
+ | |||
+ | == Solution 3 == | ||
+ | Write <math>P(x) = x^2+wx+z</math>. Split the problem into two cases: <math>P(3)\ne P(4)</math> and <math>P(3) = P(4)</math>. | ||
+ | |||
+ | Case 1: We have <math>P(3) \ne P(4)</math>. We must have | ||
+ | <cmath>w=-P(3)-P(4) = -(9+3w+z)-(16+4w+z) = -25-7w-2z.</cmath> | ||
+ | Rearrange and divide through by <math>8</math> to obtain | ||
+ | <cmath>w = \frac{-25-2z}{8}.</cmath> | ||
+ | Now, note that | ||
+ | <cmath>z = P(3)P(4) = (9+3w+z)(16+4w+z) = \left(9 + 3\cdot \frac{-25-2z}{8} + z\right)\left(16 + 4 \cdot \frac{-25-2z}{8} + z\right) =</cmath> | ||
+ | <cmath>\left(-\frac{3}{8} + \frac{z}{4}\right)\left(\frac{7}{2}\right) = -\frac{21}{16} + \frac{7z}{8}.</cmath> | ||
+ | Now, rearrange to get | ||
+ | <cmath>\frac{z}{8} = -\frac{21}{16}</cmath> | ||
+ | and thus | ||
+ | <cmath>z = -\frac{21}{2}.</cmath> | ||
+ | Substituting this into our equation for <math>w</math> yields <math>w = -\frac{1}{2}</math>. Then, it is clear that <math>P</math> does not have a double root at <math>P(3)</math>, so we must have <math>P(a) = P(3)</math> and <math>P(b) = P(4)</math> or vice versa. This gives <math>3+a = \frac{1}{2}</math> and <math>4+b = \frac{1}{2}</math> or vice versa, implying that <math>a+b = 1-3-4 = -6</math> and <math>(a+b)^2 = 6</math>. | ||
+ | |||
+ | Case 2: We have <math>P(3) = P(4)</math>. Then, we must have <math>w = -7</math>. It is clear that <math>P(a) = P(b)</math> (we would otherwise get <math>P(a)=P(3)=P(4)</math> implying <math>a \in \{3,4\}</math> or vice versa), so <math>a+b=-w=7</math> and <math>(a+b)^2 = 49</math>. | ||
+ | |||
+ | Thus, our final answer is <math>49+36=\boxed{085}</math>. ~GeronimoStilton | ||
+ | |||
+ | ==Solution 4== | ||
+ | Let <math>P(x)=(x-r)(x-s)</math>. There are two cases: in the first case, <math>(3-r)(3-s)=(4-r)(4-s)</math> equals <math>r</math> (without loss of generality), and thus <math>(a-r)(a-s)=(b-r)(b-s)=s</math>. By Vieta's formulas <math>a+b=r+s=3+4=7</math>. | ||
+ | |||
+ | In the second case, say without loss of generality <math>(3-r)(3-s)=r</math> and <math>(4-r)(4-s)=s</math>. Subtracting gives <math>-7+r+s=r-s</math>, so <math>s=7/2</math>. From this, we have <math>r=-3</math>. | ||
+ | |||
+ | Note <math>r+s=1/2</math>, so by Vieta's, we have <math>\{a,b\}=\{1/2-3,1/2-4\}=\{-5/2,-7/2\}</math>. In this case, <math>a+b=-6</math>. | ||
+ | |||
+ | The requested sum is <math>36+49=85</math>.~TheUltimate123 | ||
==See Also== | ==See Also== |
Latest revision as of 13:07, 3 June 2020
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
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 |
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