# Blog

### Thinking Out Loud – EGMO 2013 Problem 1

EGMO 2013, problem 1. The side $$\textrm{BC}$$ of the triangle $$\textrm{ABC}$$ is extended beyond $$\textrm{C}$$ to $$\textrm{D}$$ so that $$\textrm{CD} = \textrm{BC}$$. The side $$\textrm{CA}$$ is extended beyond $$\textrm{A}$$ to $$\textrm{E}$$ so that $$\textrm{AE} = 2\textrm{CA}$$. Prove that if $$\textrm{AD} = \textrm{BE}$$, then the triangle $$\textrm{ABC}$$ is right-angled.

### EGMO Countries – Germany

by/di Luigi Amedeo Bianchi

This is the first in a series of posts about the teams taking part in EGMO 2018. Today you can join us in finding out more about Germany with Susanne Armbruster, the Team Leader.

Questo è il primo di una serie di post che ci faranno scoprire qualcosa di più sulle squadre presenti alle EGMO 2018. Oggi parliamo della Germania con Susanne Armbruster, la Team Leader.

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### Thinking Out Loud – EGMO 2013 Problem 2

EGMO 2013, problem 2. Determine all integers $$m$$ for which the $$m\times m$$ square can be dissected into five rectangles, the side lengths of which are the integers $$1, 2, 3, \dotsc, 10$$ in some order.

### Women in Mathematics beyond Stereotypes — Claire Voisin

by/di Emanuele Tron

In this series of posts we would like to talk about women in mathematics who are not as well known as, say, Hypatia or Emmy Noether, and are at the same time quite different from the usual mathematicians' stereotypes. Our mathematician of the day is Claire Voisin, who managed to balance having a family with five children and doing high level research in Hodge theory (earning a CNRS Gold Medal in the process).

In questa serie di post vorremmo parlare di matematiche meno famose di, ad esempio, Ipazia o Emmy Noether e che, allo stesso tempo, non corrispondono ai consueti stereotipi sui matematici. La matematica di oggi è Claire Voisin, che è riuscita a bilanciare le esigenze di una famiglia con cinque figli e la ricerca nell'ambito della teoria di Hodge (vincendo tra l'altro una medaglia d'oro del CNRS).

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### Thinking Out Loud – EGMO 2013 Problem 3

EGMO 2013, problem 3. Let $$n$$ be a positive integer.

1. Prove that there exists a set $$S$$ of $$6n$$ pairwise different positive integers, such that the least common multiple of any two elements of $$S$$ is no larger than $$32n^2$$.
2. Prove that every set $$T$$ of $$6n$$ pairwise different positive integers contains two elements the least common multiple of which is larger than $$9n^2$$.