Difference between revisions of "Principle of Insufficient Reason"

m (Introductory)
m (Intermediate)
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1. The fraction,  
 
1. The fraction,  
  
<math>\frac{ab+bc+ac}{(a+b+c)^2}</math>
+
<cmath>\frac{ab+bc+ac}{(a+b+c)^2}</cmath>
  
 
where <math>a,b</math> and <math>c</math> are side lengths of a triangle, lies in the interval <math>(p,q]</math>, where <math>p</math> and <math>q</math> are rational numbers. Then, <math>p+q</math> can be expressed as <math>\frac{r}{s}</math>, where <math>r</math> and <math>s</math> are relatively prime positive integers. Find <math>r+s</math>.
 
where <math>a,b</math> and <math>c</math> are side lengths of a triangle, lies in the interval <math>(p,q]</math>, where <math>p</math> and <math>q</math> are rational numbers. Then, <math>p+q</math> can be expressed as <math>\frac{r}{s}</math>, where <math>r</math> and <math>s</math> are relatively prime positive integers. Find <math>r+s</math>.

Revision as of 19:39, 15 January 2024

Statement

In the absence of any relevant evidence, agents should distribute their credence equally among all the possible outcomes under consideration. I.e., If there is not enough reason for two quantities to be different (i.e. they're indistinguishable), the extrema occurs when the quantities are the same. So basically, a symmetric expression is maximized or minimized when the $2$ quantities are the same. This is what enabled us to say WLOG.

Proof

Problems

Introductory

A hexagon is inscribed in a circle with radius $1$. What is the maximum area of the hexagon? (Solution)

Intermediate

1. The fraction,

\[\frac{ab+bc+ac}{(a+b+c)^2}\]

where $a,b$ and $c$ are side lengths of a triangle, lies in the interval $(p,q]$, where $p$ and $q$ are rational numbers. Then, $p+q$ can be expressed as $\frac{r}{s}$, where $r$ and $s$ are relatively prime positive integers. Find $r+s$.

(Source: I made it. Solution here)

Olympaid

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