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Source: Own. Proposed for Romanian National Olympiad 2025.

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#1 h Apr 2, 2025, 5:42 PM • 5 Y

Y by AndreiVila, Ciobi_, RobertRogo, watery, Miquel-point

(a) Let $A,B,C$ be collinear points (in order) and $D$ a point in plane. Consider the disc $\mathcal{D}$ of center $D$ and radius $kBD$ , for some $k\in(0,1)$ . Prove that $\mathcal{D}\cap [AC]$ is either the empty set or a segment of length at most $2kAC$ .
(b) Let $n$ be a positive integer and $P(X)\in\mathbb{C}[X]$ be a polynomial of degree $n$ . Prove that $\sup_{x\in[0,1]}|P(x)|\le(2n+1)^{n+1}\int\limits_{0}^{1}|P(x)|\mathrm{d}x.$

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#3 h Yesterday at 1:29 PM • 2 Y

Y by MS_asdfgzxcvb, chirita.andrei

Probably not the author's intent, but it's a great "continuation" of Gheorghe Țițeica 2025/12.3. I would have much rather preferred this on the national instead of P3.

(a) Since $B$ lies outside the disk, we may assume that $\mathcal{D}\cap[AC]$ is included in the ray $(BC$ . By linearity, it's enough to analyze the case when $\mathcal{D}\cap[AC]=[PC]$ , with $B,P,C$ collinear in this order, and $A=B$ . Call $\theta=\angle BCD$ and $\varphi=\angle BDC$ . In this case, $PC=2kBD\cos\theta=2kBC\frac{BD}{BC}\cos\theta=2kBC\frac{\sin\theta}{\sin\varphi}\cos\theta.$ We are thus left with proving that $\sin 2\theta\leq 2\sin\varphi$ . However, $\varphi\in[\pi-2\theta ,\pi-\theta]$ , and the function $\sin x$ is concave on $[0,\pi]$ , needing to only prove the inequality at the endpoints of the interval, which is trivial.

(b) Let $P(x)=A(x-z_1)\dots (x-z_n)$ and fix a point $a\in[0,1]$ . For each $i=\overline{1,n}$ , select some $k_i\in(0,1)$ and let $\mathcal{D}_i=\{z\mid |z-z_i|=k_i|a-z_i|\}.$ Applying part (a), we must have that $|x-z_i|\geq k_i|a-z_i|$ for all $x\in [0,1]\,\backslash\,\mathcal{D}_i$ , and $\mu([0,1]\,\backslash\,\mathcal{D}_i)\geq 1-2k_i$ . Thus, taking into account only the region $\mathcal{R}= [0,1]\,\backslash\,\bigcup_{i=1}^n\mathcal{D}_i$ , we arrive at $\int_0^1 |P(x)|\, dx \geq \int_{\mathcal{R}} |A||x-z_1|\dots |x-z_n|\, dx \geq k_1\dots k_n\int_{\mathcal{R}} |A||a-z_1|\dots |a-z_n|\, dx \geq k_1\dots k_n \left(1-2\sum_{i=1}^n k_i\right)|P(a)|.$ Notice that this inequality takes place for all $k_i\in(0,1)$ and $a\in[0,1]$ . By AM-GM, the maximum value that $k_1\dots k_n\left(1-2\sum_{i=1}^n k_i\right)$ attains is $\frac{1}{2^n(n+1)^{n+1}}$ for $k_i=\frac{1}{2(n+1)}$ , and thus $\int_0^1 |P(x)|\, dx\geq \frac{1}{2^n(n+1)^{n+1}}\sup_{x\in[0,1]}|P(x)|,$ which is a better bound than the one required.

This post has been edited 1 time. Last edited by AndreiVila, Yesterday at 3:07 PM

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