New answers tagged intuition
5
votes
Examples of eventual counterexamples
One could reasonably conjecture that there are no positive integers $a,b,c$ satisfying
$$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 4.$$
I say "reasonably", because the smallest ...
Community wiki
2
votes
Examples of eventual counterexamples
$n$ is sufficiently large for $P(a)=T$ $\forall a\in\mathbb N$ to be a 'reasonable' conjecture to make.
$\ldots$
where 'reasonable' is open to interpretation
I won't be too surprised if a ...
Community wiki
0
votes
Examples of eventual counterexamples
The word "eventually" connotes a very long sequence of positive examples before the first counterexample. Gerry Myerson points out that no polynomial of the form $x^n-1,$ when factored over $...
Community wiki
5
votes
Examples of eventual counterexamples
Assertion: Every integer greater than 1 can be written as the sum of
a prime number and a perfect power of a nonnegative integer.
The smallest (and maybe only?) counterexample to this assertion is
$11^...
Community wiki
1
vote
The vertices of a triangle are three random points on a unit circle. The side lengths are, in random order, $a,b,c$. Show that $P(ab>c)=\frac12$
Here is an elementary proof using basic trigonometry for the more general formula $p(ab<rc)=\frac{2 \arctan(r)}{\pi}.$
Fixing one point on a circle of radius $r$ and placing the other two at angles ...
6
votes
Accepted
The vertices of a triangle are three random points on a unit circle. The side lengths are, in random order, $a,b,c$. Show that $P(ab>c)=\frac12$
One can use basic probability theory to prove that
$$P(ab<kc)=\frac{2}{\pi}\arctan k,\qquad k>0.$$
Without loss of generality, the vertices opposite the sides $a,b,c$ are
$$A=e^{2i\beta},\qquad ...
4
votes
The vertices of a triangle are three random points on a unit circle. The side lengths are, in random order, $a,b,c$. Show that $P(ab>c)=\frac12$
A proof from pure geometry
On a unit circle with centre $O$, draw parallel chords $PQ$ and $P'Q'$ such that $PQ'\perp P'Q$. Chord $MN$ is parallel to $PQ$ and passes through $R$, the intersection of $...
6
votes
The vertices of a triangle are three random points on a unit circle. The side lengths are, in random order, $a,b,c$. Show that $P(ab>c)=\frac12$
The following argument is less or more the same as that of Iosif Pinelis, but with less computations and the symmetry is rather explicit. It may be explained without complex numbers, but the ...
2
votes
The vertices of a triangle are three random points on a unit circle. The side lengths are, in random order, $a,b,c$. Show that $P(ab>c)=\frac12$
Denoting the angles corresponding to edges of side lengths $a,b,c$ with $A,B,C$ respectively, by the sine law:
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2.$$
Thus $c>ab$ amounts to $\sin(...
0
votes
The vertices of a triangle are three random points on a unit circle. The side lengths are, in random order, $a,b,c$. Show that $P(ab>c)=\frac12$
We use the formula (from en.wikipedia.org/wiki/Circumcircle#Other_properties ):
diameter = a * b * c / ( 2 * area )
In our case diameter = 2 and we get:
4 * area = a * b * c
If h is the height of ...
5
votes
The vertices of a triangle are three random points on a unit circle. The side lengths are, in random order, $a,b,c$. Show that $P(ab>c)=\frac12$
Without loss of generality, the three points on the unit circle are $1,e^{is},e^{it}$, where $s$ and $t$ are uniformly and independently distributed on the interval $[0,2\pi]$, so that $a=|e^{is}-1|$, ...
3
votes
Densities, pseudoforms, absolute differential forms and measures, differential forms, etc
I am not the greatest expert on the details of this stuff, but since nobody else tried so far, let me have an attempt:
Prelude: Measures
Since you mention measures, I start with that, though this is ...
3
votes
What is so geometric about symplectic geometry?
My take on this is fairly simple-minded.
A metric is a non-degenerate symmetric bilinear form on a vector space. This yields a notion of distance, angle and volume. Dually, we can define a 'cometric' ...
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