Difference between revisions of "2020 AIME I Problems/Problem 14"
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== Problem == | == Problem == | ||
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 <cmath>P(x)=x^2-(m+n)x+mn</cmath> 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 = 36</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 | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | 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 = \boxed{085}</math>. | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | Let <math>P(x) = (x-c)^2 - d</math> for some <math>c</math>, <math>d</math>. | ||
+ | |||
+ | Then, we can write <math>P(P(x)) = ((x-c)^2 - d - c)^2 - d</math>. Setting the expression equal to <math>0</math> and solving for <math>x</math> gives: | ||
+ | |||
+ | <math>x = \pm \sqrt{ \pm \sqrt{d} + d + c} + c</math> | ||
+ | |||
+ | Therefore, we have that <math>x</math> takes on the four values <math>\sqrt{\sqrt{d} + d + c} + c</math>, <math>-\sqrt{\sqrt{d} + d + c} + c</math>, <math>\sqrt{-\sqrt{d} + d + c} + c</math>, and <math>-\sqrt{-\sqrt{d} + d + c} + c</math>. Two of these values are <math>3</math> and <math>4</math>, and the other two are <math>a</math> and <math>b</math>. | ||
+ | |||
+ | We can split these four values into two "groups" based on the radicand in the expression - for example, the first group consists of the first two values listed above, and the second group consists of the other two values. | ||
+ | |||
+ | <math>\textbf{Case 1}</math>: Both the 3 and 4 values are from the same group. | ||
+ | |||
+ | In this case, the <math>a</math> and <math>b</math> values are both from the other group. The sum of this is just <math>2c</math> because the radical cancels out. Because of this, we can see that <math>c</math> is just the average of <math>3</math> and <math>4</math>, so we have <math>2c = 3 + 4 = 7</math>, so <math>(a+b)^2 = 7^2 = 49</math>. | ||
+ | |||
+ | <math>\textbf{Case 2}</math>: The 3 and 4 values come from different groups. | ||
+ | |||
+ | It is easy to see that all possibilities in this case are basically symmetric and yield the same value for <math>(a+b)^2</math>. Without loss of generality, assume that <math>\sqrt{\sqrt{d} + d + c} + c = 4</math> and <math>\sqrt{-\sqrt{d} + d + c} + c = 3</math>. Note that we can't switch the values of these two expressions since the first one is guaranteed to be larger. | ||
+ | |||
+ | We can write <math>\sqrt{\sqrt{d} + d + c} + c = 1 + \sqrt{-\sqrt{d} + d + c} + c</math>. | ||
+ | |||
+ | Moving most terms to the left side and simplifying gives <math>\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} = 1</math>. | ||
+ | |||
+ | We can square both sides and simplify: | ||
+ | |||
+ | <math>\sqrt{d} + d + c - \sqrt{d} + d + c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1</math> | ||
+ | |||
+ | <math>2d + 2c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1</math> | ||
+ | |||
+ | <math>\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = (d+c) - \frac{1}{2}</math> | ||
+ | |||
+ | <math>\sqrt{(d+c)^2 - (\sqrt{d})^2} = (d+c) - \frac{1}{2}</math> | ||
+ | |||
+ | <math>\sqrt{d^2 + 2dc + c^2 - d} = (d+c) - \frac{1}{2}</math> | ||
+ | |||
+ | Squaring both sides again gives the following: | ||
+ | |||
+ | <math>d^2 + 2dc + c^2 - d = d^2 + 2dc + c^2 - d - c + \frac{1}{4}</math> | ||
+ | |||
+ | Nearly all terms cancel out, yielding <math>c = \frac{1}{4}</math>. | ||
+ | |||
+ | By substituting this back in, we obtain <math>\sqrt{\sqrt{d} + d + c} = \frac{15}{4}</math> and <math>\sqrt{-\sqrt{d} + d + c} = \frac{11}{4}</math>. | ||
+ | |||
+ | The sum of <math>a</math> and <math>b</math> is equal to <math>-\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} + 2c = -\frac{15}{4} - \frac{11}{4} + \frac{1}{2} = -6</math>, so <math>(a+b)^2 = 36</math>. | ||
+ | |||
+ | Adding up both values gives <math>49 + 36 = \boxed{085}</math> as our final answer. | ||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/_Iji1DW7QaY?si=t6Qbn2XYAfknnIxr | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
==See Also== | ==See Also== |
Latest revision as of 19:34, 17 September 2023
Contents
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
, so
for 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 gives
yielding
The sum of the two solutions is
. In this case,
.
The requested sum of squares is .
Solution 6
Let for some
,
.
Then, we can write . Setting the expression equal to
and solving for
gives:
Therefore, we have that takes on the four values
,
,
, and
. Two of these values are
and
, and the other two are
and
.
We can split these four values into two "groups" based on the radicand in the expression - for example, the first group consists of the first two values listed above, and the second group consists of the other two values.
: Both the 3 and 4 values are from the same group.
In this case, the and
values are both from the other group. The sum of this is just
because the radical cancels out. Because of this, we can see that
is just the average of
and
, so we have
, so
.
: The 3 and 4 values come from different groups.
It is easy to see that all possibilities in this case are basically symmetric and yield the same value for . Without loss of generality, assume that
and
. Note that we can't switch the values of these two expressions since the first one is guaranteed to be larger.
We can write .
Moving most terms to the left side and simplifying gives .
We can square both sides and simplify:
Squaring both sides again gives the following:
Nearly all terms cancel out, yielding .
By substituting this back in, we obtain and
.
The sum of and
is equal to
, so
.
Adding up both values gives as our final answer.
Video Solution
https://youtu.be/_Iji1DW7QaY?si=t6Qbn2XYAfknnIxr
~MathProblemSolvingSkills.com
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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