at least one a_i has to be equal to 1/2

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Source: Oliforum Contest V 2017 p8 https://artofproblemsolving.com/community/c2487525_oliforum_contes

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#1 h Sep 29, 2021, 10:03 AM

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Fix $a_1, . . . , a_n \in (0, 1)$ and define $f(I) = \prod_{i \in I} a_i \cdot \prod_{j \notin I} (1 - a_j)$ for each $I \subseteq \{1, . . . , n\}$ . Assuming that $\sum_{I\subseteq \{1,...,n\}, |I| odd} {f(I)} = \frac12,$ show that at least one $a_i$ has to be equal to $\frac12$ .

(Paolo Leonetti)

This post has been edited 1 time. Last edited by parmenides51, Sep 29, 2021, 10:19 AM

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#2 h Sep 29, 2021, 10:38 AM

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parmenides51 wrote:

Fix $a_1, . . . , a_n \in (0, 1)$ and define $f(I) = \prod_{i \in I} a_i \cdot \prod_{j \notin I} (1 - a_j)$ for each $I \subseteq \{1, . . . , n\}$ . Assuming that $\sum_{I\subseteq \{1,...,n\}, |I| odd} {f(I)} = \frac12,$ show that at least one $a_i$ has to be equal to $\frac12$ .

(Paolo Leonetti)

Let $I_n= \{1,...,n\}$ , define $b_i=1-a_i$ for all $i \in I_n$ .

Consider the polynomial $P(X)=\prod_{i=1}^{n}(a_iX-b_i)$ . We want to prove $P(1)=0$ .

Note that the condition is equivalent to $P(-1)-P(1)=(-1)^n$ , but $P(-1)=(-1)^n$ (since $a_i+b_i=1$ ), hence the result

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