Hey Brilliantinians, here I'm posting the most beautiful problem I've ever seen. The idea is that everyone can share the most interesting/beautiful/awesome problem he/she has ever faced to. The problem needs not to come from Brilliant. Solutions to the problems are also accepted in the comments

Here's mine. Let \(p\geq 3\) be a prime number. We divide every side of a triangle in \(p\) equal parts and every point of division is connected to the opposite vertex. Compute the maximum number of pairwise disjoint parts in which the triangle is divided in funciton of \(p\).

Enjoy!!!

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestWell, if \(p\) is an odd prime, then the maximum such number must be \(3{ p }^{ 2 }-3p+1\)

so that for \(p={3,5,7,11,13,19,...}\), we have \({19,61,127,331,469,817,...}\)

We first note that the triangle is divided into \({ p }^{ 2 }\) disjoint parts by the first \(2\) sets of rays from \(2\) vertices. Then each ray from the last vertex intersects those rays \(2(p-1)\) times, thereby increasing the number of disjoint parts by \(2(p-1)+1\), and there are \(p-1\) such rays from that last vertex. Hence, the total is

\({ p }^{ 2 }+(p-1)(2(p-1)+1)=3{ p }^{ 2 }-3p+1\)

This derivation depends on the fact that no point is shared by \(3\) such rays, hence the primes only.

This graphic shows the case where \(p=5\), so one can see how this is done. Affine transforms of this triangle leaves the number of parts unaffected.

Log in to reply