Putnam 2019 A3

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#1 h Dec 10, 2019, 1:24 AM • 2 Y

Y by Adventure10, Rounak_iitr

Given real numbers $b_0,b_1,\ldots, b_{2019}$ with $b_{2019}\neq 0$ , let $z_1,z_2,\ldots, z_{2019}$ be the roots in the complex plane of the polynomial
$P(z) = \sum_{k=0}^{2019}b_kz^k.$ Let $\mu = (|z_1|+ \cdots + |z_{2019}|)/2019$ be the average of the distances from $z_1,z_2,\ldots, z_{2019}$ to the origin. Determine the largest constant $M$ such that $\mu\geq M$ for all choices of $b_0,b_1,\ldots, b_{2019}$ that satisfy
$1\leq b_0 < b_1 < b_2 < \cdots < b_{2019} \leq 2019.$

This post has been edited 1 time. Last edited by djmathman, Sep 14, 2020, 1:20 PM
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awesomemathlete

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awesomemathlete

#2 h Dec 10, 2019, 1:24 AM • 3 Y

Y by Rebel_1, Akaishuichi9598, Adventure10

By AM-GM we have that $\frac{|z_1|+|z_2|+\dots+|z_{2019}|}{2019}\ge(|z_1|\cdot|z_2|\cdots|z_{2019}|)^{\frac{1}{2019}}=\frac{b_0}{b_{2019}}^{\frac{1}{2019}}\ge\boxed{\frac{1}{2019}^{\frac{1}{2019}}}=M$ . Equality is achieved when all the roots have magnitude $M$ . For example the $2020^{\text{th}}$ roots of unity excluding $1$ multiplied by $M$ are the roots of: $2019\cdot\frac{x^{2020}-M^{2020}}{x-M}=2019(x^{2019}+Mx^{2018}+\dots+M^{2019})$ which satisfies the desired constraints on the coefficients.
$\square$

This post has been edited 1 time. Last edited by awesomemathlete, Dec 10, 2019, 1:25 AM

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yayups

#4 h Dec 10, 2019, 5:11 AM • 2 Y

Y by Adventure10, Mango247

We claim the answer is $\boxed{2019^{-2019^{-1}}}$ . The solution follows from the following two claims.

Claim: We always have $\mu\ge 2019^{-2019^{-1}}$ .

Proof: By Vieta's formulas, we have $z_1z_2\cdots z_{2019}=(-1)^{2019}\frac{b_0}{b_{2019}},$ so $|z_1|\cdots|z_{2019}|=\frac{b_0}{b_{2019}}\ge\frac{1}{2019}.$ Thus, by AM-GM, we have $\mu\ge\sqrt[2019]{\frac{1}{2019}}=2019^{-2019^{-1}}$ . $\blacksquare$

Claim: There exists a polynomial $P$ satisfying the conditions of the problem such that $\mu=2019^{-2019^{-1}}$ .

Proof: For convenience, denote $\ell:=2019^{2019^{-1}}$ . We claim that $P(z):=1+\ell z+\ell^2z^2+\cdots+\ell^{2019}z^{2019}$ works. First note that $b_i=\ell^i$ , and since $\ell>1$ , we have $1=b_0<b_1<b_2<\cdots<b_{2019}=2019.$ Now, note that $P(z)(\ell z-1)=(\ell z)^{2020}-1,$ so any root $z_i$ of $P$ satisfies $(\ell z_i)^{2020}=1$ , so $|\ell z_i|=1$ . Thus, $|z_i|=1/\ell$ , so the average of the absolute value of the roots $\mu$ is just $1/\ell=2019^{-2019^{-1}}$ , as desired.

Remark: It is straightforward to motivate this proof by looking the equality case of the previous claim.

This post has been edited 1 time. Last edited by yayups, Dec 10, 2019, 5:12 AM

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panto

#5 h Oct 3, 2022, 1:40 PM • 3 Y

Y by Mango247, Mango247, Mango247

To all solutions above,

How do we know that there is no better bound than $\frac{1}{2019}^{\frac{1}{2019}}$ ?

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4184 posts

#8 h Dec 14, 2022, 6:59 PM

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letsgo i solved a recent putnam #3

We claim that the answer is $2019^{-1/2019}$ . Let $c=2019^{1/2019}$ . This is achieved by the polynomial $(cx)^{2019}+(cx)^{2018}\cdots +cx+1$ since its roots are the 2020th roots of unity other than 1 but divided by $c$ .

Now, by AM-GM, we have $\mu\geq \sqrt[2019]{|z_{0}z_{1}\cdots z_{2019}|}=\sqrt[2019]{|\frac{b_0}{b_{2019}}|}\geq \sqrt[2019]{1/2019},$ so we are done.

remark: solution was motivated by playing around with only quadratics, first trying roots of unity but then seeing that i could divided that by sqrt2, and then the equality case of all roots of unity scaled by a constant motivated the amgm

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#9 h Feb 24, 2023, 2:15 AM • 1 Y

Y by centslordm

We claim the answer is $M=\left(\frac{1}{2019}\right)^{\frac{1}{2019}}$ which is possible when $P(z)=\sum_{k=0}^{2019}(M^{-1}z)^k$ which yields roots $M\exp(2\pi i k/2020)$ for $k=1,2,\dots,2019$ . For the bound, by AM-GM and Vieta's we see $\frac{|z_1|+\dots+|z_{2019}|}{2019}\ge\sqrt[2019]{|z_1\dots z_{2019}|}=\sqrt[2019]{\frac{b_0}{b_{2019}}}\ge\sqrt[2019]{\frac{1}{2019}}$ as desired. $\square$

This post has been edited 1 time. Last edited by Mogmog8, Feb 24, 2023, 9:05 PM

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#10 h Feb 24, 2023, 2:27 AM

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this problem is very nice and almost requires elegant solutions.

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8857 posts

#11 h Mar 12, 2023, 11:17 PM

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Notice that $\mu \leq \sqrt[2019]{|z_1z_2z_3\cdots z_{2019}|} = \sqrt[2019]{\frac{b_0}{b_{2019}}} \leq \sqrt[2019]{\frac 1{2019}}.$ Equality occurs when we set all the $z_i$ to be the 2019th roots of unity scaled by $\sqrt[2019]{\frac 1{2019}}$ , which yields $b_i$ that satisfy the condition.

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#12 h Nov 14, 2023, 11:41 PM

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solution

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#13 h Dec 22, 2023, 8:53 PM

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We have

$\mu = \frac{|z_1|+\dots+|z_{2019}|}{2019} \ge \sqrt[2019]{|z_1z_2z_3\cdots z_{2019}|} = \sqrt[2019]{\frac{b_0}{b_{2019}}} \ge \boxed{\sqrt[2019]{\frac 1{2019}}}.$
Equality occurs when $|z_1|=|z_2| = \dots = |z_{2019}|$ , which can be achieved with a scaled roots of unity formation.

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#14 h Jan 6, 2024, 11:56 AM

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By AM-GM we have
$\frac{|z_1|+ \dotsb + |z_{2019}|}{2019} \geq \sqrt[2019]{|z_1| \cdot \dotsb \cdot |z_{2019}|}$ The RHS is equal to $|\sqrt[2019]{\frac{b_0}{b_{2019}}}| = \sqrt[2019]{\frac{b_0}{b_{2019}}}$ by Vieta's(because magnitudes are multiplicative).
$\newline$
Then, it follows that the minimum of $\mu$ is $\frac{1}{2019^{2019}} = M$ .

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#15 h Apr 13, 2024, 2:27 PM

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The construction was cute :love:

:love:

Clearly mean of finitely many values gets minimized when the values are equal. So $\mu \geq \left(\prod_1^{2019}|z_i|\right)^{1/2019} = \left(\frac{b_0}{b_{2019}}\right)^{1/2019} \geq \left(\frac{1}{2019}\right)^{1/2019}=M$
For the equality case, we note that $|z_i|=M$ which motivates for the roots of $z^{2020}=M^{2020}$ except for the root $M$ . The polynomial $P(z)=\sum_{i=0}^{2019}2019M^{2019-i}z^i$ suffices. Note that $b_0=1,b_{2019}=2019$ and $b_i$ is strictly increasing. $\blacksquare$

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chakrabortyahan

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chakrabortyahan

#16 h Apr 25, 2024, 3:09 PM

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jaah etao e post kore dieche...tobuo ami korbo ...
This did not feel like an A3 though
The bound is trivial by AM-GM .
$\mu = \frac{\mid z_1\mid+ \mid z_2\mid +....+\mid z_{2019}\mid }{2019}\ge \mid z_1z_2...z_{2019}\mid ^{\frac{1}{2019}} = \Large\mid \frac{b_{2019}}{b_0}\Large\mid^{\frac{1}{2019}}\ge \left[\frac{1}{2019}\right]^{\frac{1}{2019}} = M$
Now for equality we have all $\mid z_i\mid$ s equal to $M$
now we first consider the equation $x^{2019}+x^{2018}+....+1 = 0$ the roots of this equation are all with unit modulus and are 2020 th root of unity and no root is $1$ . Now we take $y=Mx$ the equation is $(y/M)^{2019}+...+1 = 0 \\ \implies 2019 y^{2019}+2019My^{2018}+...M^{2019} = 0 \\ \implies 2019 y^{2019}+2019My^{2018}+...+1 = 0.....(*)$
Now the roots of $(*)$ are all with modulus $M$
and note that $M\le 1 \implies M^k\le 1\implies 2019M^k\le 2019$ and $M\le 1 \implies M^{2019}\le M^k \,\forall \, k \le 2019 \\ \implies 1\le 2019M^k$
Hence all the coefficients of $(*)$ are between 1 and 2019 and the roots are with equal modulus $M$ . Thus we get such a polynomial where the equality holds.
$\blacksquare\smiley$

This post has been edited 2 times. Last edited by chakrabortyahan, May 1, 2024, 7:06 AM

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lifeismathematics

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lifeismathematics

#17 h May 8, 2024, 9:57 AM

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Consider the roots of $P(z)$ to be $z_{1}, z_{2}, \cdots , z-{n}$ .

$\frac{|z_{1}|+|z_{2}|+\cdots+|z_{n}|}{n} \geqslant \sqrt[n]{|z_1||z_2|\cdots \cdot |z_n|}$ , so we have $\mu \geqslant \sqrt[n]{\frac{b_0}{b_n}} \geqslant \frac{1}{\sqrt[n]{n}}$ .

So the maxima such $M=\frac{1}{\sqrt[n]{n}}$ . Now we present a construction for which it is indeed a working solution.

Consider $P(z):=nz^{n}+nMz^{n-1}+nM^{2}z^{n-2}+\cdots+nM^{n}$ , clearly $\mu=\frac{1}{\sqrt[n]{n}}$ , now we have $b_{k}=M^{n-k}$ .

notice $b_{k+i}-b_{k}= n^{\frac{k}{n}}\cdot \left(n^{\frac{i}{n}}-1\right)$ , for each $i>k$ we have $b_{k+i}-b_{k}>0$ , and also $b_{0}=1$ , $b_{n}=n$ , hence $1 \geqslant b_{0}<b_{1}<b_{2} \cdots < b_{n} \leqslant n$ . Now just plug $n=2019$ , we are done. $\square$

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