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## Shepard Fairey

Counterpoint in blue: paintings by Alexandre Motte (Dinard)

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## Mathematical Cover-Up

Great mathematician John H. Conway passed away Saturday April 11th, as a result of complications from COVID-19.

He had a slyly bent sense of humour. Together, he and Princeton University colleague Alexander Soifer managed to get the world record for the shortest paper ever published in a serious Math Journal.

The paper was published in the January 2005 issue of the American Mathematical Monthly. Numberphile recalls the event.

The paper asks the question: Can $n^2+ 1$ unit equilateral triangles cover an equilateral triangle of side $> n$, say $n + \epsilon$? then proceeds to give the answer: $n^2+ 2$ can, together with figures showing two different ways of achieving the result1)Figure 1 is easy enough to understand: take the bottom horizontal « strip » of the size-$n$ triangle (comprised of $2n – 1$ triangles); by squeezing it horizontally and adding $2$ triangles, we increase its height until it covers up the $n + \epsilon$ triangle.2)Figure 2 is slightly more difficult: take the bottom horizontal strip of the size-$n$ triangle; squeeze it vertically so as to make it a little wider (to fit an $n + \epsilon$ side); meanwhile its height will decrease a little; do the same with all remaining horizontal strips (but the triangle on top), you end up covering up the $n + \epsilon$ triangle except for a triangular area on top that is slightly larger than a unit triangle; this remaining area can be covered-up using three unit triangles, as figure 2 shows. I believe there is a typo in the figure though: the length on the bottom left should read $1 – {\epsilon \over n}$ instead of $1 – \epsilon$..

The folks at numberphile worry that Conway and Soifer have written nothing about the possibility of an $n^2 + 1$ unit triangles cover-up: Tony Padilla (at 4 min 40): « Yes, so that’s what I think about it. I don’t think they even answered the question. »

Actually, in his book How Does One Cut a Triangle?, Soifer recollects the $n^2 + 1$ case was the easy part (« Area considerations alone show the need for at least $n^2 + 1$ of them »). So they needed not bother to encumber their proof with it.

It could have gone like this:

Consider the « canonic » cover-up of the triangle of side $n + \epsilon$ with $n^2$ small triangles of side $1 + {\epsilon \over n}$.
Consider the combined $n^2 + 2$ vertices of these small triangles.
The distance between any two vertices is greater than or equal to $1 + {\epsilon \over n}$, so no two vertices belong to the same unit triangle.
Consequently a cover-up of the $n + \epsilon$ triangle must be comprised of at least $n^2 + 2$ unit triangles.
$\Box$

Notes   [ + ]

 1 ↑ Figure 1 is easy enough to understand: take the bottom horizontal « strip » of the size-$n$ triangle (comprised of $2n – 1$ triangles); by squeezing it horizontally and adding $2$ triangles, we increase its height until it covers up the $n + \epsilon$ triangle. 2 ↑ Figure 2 is slightly more difficult: take the bottom horizontal strip of the size-$n$ triangle; squeeze it vertically so as to make it a little wider (to fit an $n + \epsilon$ side); meanwhile its height will decrease a little; do the same with all remaining horizontal strips (but the triangle on top), you end up covering up the $n + \epsilon$ triangle except for a triangular area on top that is slightly larger than a unit triangle; this remaining area can be covered-up using three unit triangles, as figure 2 shows. I believe there is a typo in the figure though: the length on the bottom left should read $1 – {\epsilon \over n}$ instead of $1 – \epsilon$.
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## Erugo Purakashi :)

Pino is cute ♥

Fellow citizens! The time is now to consume. Why skimp, when you deserve more? Fellow citizens! Do your part, and make waste. Life is easier when you lighten the load. Fellow citizens! The time is…

City of Romdeau mantra
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## Quasi-cristaux

Une belle conférence de Denis Gratias pour l’université de tous les savoirs.

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## Théorème de Cantor Bernstein

Soient A et B deux ensembles. Supposons l’existence d’une injection de A dans B et d’une injection de B dans A. Alors A et B sont de même cardinalité.

Une démonstration élémentaire est disponible ici.