# Minimising L1 and L2: Official Solution

*English translation by SOTA – AI Community of the Russian original. Organisers who would like this translation removed can email sota.ai.community@gmail.com.*

*Task 3 of the municipal stage (Moscow) of the All-Russian School Olympiad (VsOSh) 2025/26 in artificial intelligence, grades 9–11: official answer and solution (the statement is in a separate file). Original: [ans-ai-9-11-mun-msk-25-26.pdf](https://vos.olimpiada.ru/upload/files/Arhive_tasks/2025-26/mun/ai/ans-ai-9-11-mun-msk-25-26.pdf).*

**a)** Find the value $c_1$ that minimises $L_1(c)$. If there are several optimal values, give the smallest of them as your answer.

**Answer:** 3

**Scoring criterion:** exact match of the answer — 50 points

**b)** Find the value $c_2$ that minimises $L_2(c)$. If there are several optimal values, give the smallest of them as your answer.

**Answer:** 35/6

**Scoring criterion:** exact match of the answer — 50 points

**Maximum score for the task — 100**

**Solution.** Let us sort the values: 1, 2, 3, 9, 10, 10.

**a)** *Minimum of $L_1$.*

$L_1(c)$ is minimised at the *median* of the sample. With an even number of elements, the set of optima is the whole segment between the two middle values. Here the middle values are 3 and 9, so the optima are $c \in [3, 9]$. By the rule of the task, we take the smallest:

$$
c_1 = 3.
$$

**b)** *Minimum of $L_2$.*

$L_2(c)$ is a quadratic parabola; the minimum is attained at the *mean*:

$$
c_2 = \frac{1 + 2 + 3 + 9 + 10 + 10}{6} = \frac{35}{6}.
$$

(The optimum is unique, so the instruction to take the "smallest" value changes nothing here.)