Track / Stage 6 / 181 of 400 #1181 of 2000

Problem 1181

National olympiad, first round

Geometry Difficulty 6.2 Prove it

Problem 4. Let $A B C D$ be a trapezoid with $A B \| C D$ and $(A C) \cap(B D)=\{O\}$ . If $M \in(A D)$ and $N \in(B C)$ such that points $M, O$ and $N$ are collinear, then:

a) express the vectors $\overrightarrow{O M}$ and $\overrightarrow{O N}$ in terms of the vectors $\overrightarrow{O C}$ and $\overrightarrow{O D}$

b) prove that $\frac{A B}{C D} \leq \frac{1}{2}\left(\frac{N B}{C N}+\frac{M A}{D M}\right)$ .[^0]

## NATIONAL MATHEMATICS OLYMPIAD Local stage - 5.03. 2016 GRADING GUIDE - 9th Grade

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Problem 4. Let $ABCD$ be a trapezoid with $AB \parallel CD$ and $(AC) \cap (BD) = \{O\}$ . If $M \in (AD)$ and $N \in (BC)$ such that the points $M, O$ , and $N$ are collinear, then:

a) express the vectors $\overrightarrow{OM}$ and $\overrightarrow{ON}$ in terms of the vectors $\overrightarrow{OC}$ and $\overrightarrow{OD}$ ;

b) prove that $\frac{AB}{CD} \leq \frac{1}{2}\left(\frac{NB}{CN} + \frac{MA}{DM}\right)$ .

## Grading Rubric.

(2p) Let $\frac{AB}{CD} = k, \frac{NB}{CN} = q, \frac{MA}{DM} = p$ . From the similarity of triangles $\triangle AOB \sim \triangle COD$ , it follows that $\frac{AB}{CD} = \frac{OA}{OC} = \frac{OB}{OD}$ , hence, $\left\{\begin{array}{l}\overrightarrow{OA} = -k \cdot \overrightarrow{OC} \\ \overrightarrow{OB} = -k \cdot \overrightarrow{OD}\end{array}\right.$

(2p) From $\frac{MA}{DM} = p$ , we obtain $\overrightarrow{OM} = \frac{-k \cdot \overrightarrow{OC} + p \cdot \overrightarrow{OD}}{1 + p}$ , and from $\frac{NB}{CN} = q$ , we obtain $\overrightarrow{ON} = \frac{-k \cdot \overrightarrow{OD} + q \cdot \overrightarrow{OC}}{1 + q}$ .

(1p) Since the vectors $\overrightarrow{OM}$ and $\overrightarrow{ON}$ are collinear, it follows that $\frac{-k}{q} = \frac{p}{-k}$ , i.e., $k^2 = pq$ .

(2p) Therefore, $k = \sqrt{pq} \leq \frac{p + q}{2}$ , from which we obtain $\frac{AB}{CD} \leq \frac{1}{2}\left(\frac{NB}{CN} + \frac{MA}{DM}\right)$ .[^1]

[^0]: ${ }^{1}$ The actual working time is 3 hours;

${ }^{2}$ All problems are mandatory;

${ }^{3}$ Each problem is graded from 0 to 7.

[^1]: ${ }^{1}$ Each grader awards an integer number of points;

${ }^{2}$ Any other correct solution is graded accordingly.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.