Some random (non-usual) geometric length spamming ratios (Storage)
by kamatadu, May 21, 2023, 8:34 AM
![\[OH = \sqrt{ 9R^2 - (a^2+b^2+c^2)}.\]](http://latex.artofproblemsolving.com/9/9/c/99c45cec0a669721a7679dde26ce40ccbeb6679a.png)
![\[OH^2 = R^2 (1 - 8 \cos A \cos B \cos C).\]](http://latex.artofproblemsolving.com/d/1/3/d13e52f3a2c59d2a8f91d3206c3170e39231f823.png)
![\[\sin\left(\dfrac{A}{2}\right)\sin\left(\dfrac{B}{2}\right)\sin\left(\dfrac{C}{2}\right)=\dfrac{r}{4R}.\]](http://latex.artofproblemsolving.com/1/a/f/1af79948d59a2c6e55e68a440cdc8914257db8db.png)
![\[\tan A + \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C. \]](http://latex.artofproblemsolving.com/3/e/f/3effaf199e2b4a3b2968a04b36c786e257e9e509.png)
Now a random Lemma which I typed but later noticed that I didn't use it in the proof
If 
is a circle, and 
is another circle not passing through the center of . If 
is the image of under an Inversion w.r.t. , then 
share a common radical axis.
Let 
denote the radical axis of 
. Now 
denote the intersection point of the line joining the centers of and . Now construct the circle with center and which is orthogonal to , and name it 
. Note that as lies on the radical axis, we get that 
which thus implies that is also orthogonal to .Now perform an Inversion
. Note that this Inversion fixes . Moreover, as Inversion preserves the angle between clines (it is defined as the angle between tangents the intersection points of the circles in case of two circles, and as the angle between the intersecting line and the tangent to the circles at the point of intersection in case of a circle and a line), we get that the angle between and is also 
. This simply means that the circles and are also orthogonal thus finishing the proof. 
If is a circle, and is another circle passing through the center of . If is the image of under an Inversion w.r.t. , then is the radical axis of . 
denote the center of . Let the line passing through the centers of and intersect at 
, 
; and at , ; and at 
respectively. Note that is simply the inverse of . Now we know that as Inversion induces harmonic bundles, we get 
. Moreover, is the midpoint of 
. This implies that 
which simply means 
thus proving our corollary.
Bruh, here is another lemma... This one is a bit useful as a nice concept though. One is a dumb proof (hidden below) of mine (for the case when the Involution is on a line), and the other is a proof communicated to me by i3435.
If 
and 
are two reciprocal pairs under some Involution. Then there exists a unique Involution swapping these pairs of points. For the proof, take any arbitrary point and let one of the Involution swap 
and the other Involution swap 
. Then we get that,![\[(A,B;X,P)=(B,A;Y,P'),\]](http://latex.artofproblemsolving.com/2/d/0/2d0c8163f4549f2f0997b77d1a0d4dc7a81c93ef.png)
from the first Involution and also that,![\[(A,B;X,P)=(B,A;Y,P''),\]](http://latex.artofproblemsolving.com/8/e/7/8e7db280faa9010fbde9ca88e34cd55881f04a57.png)
from the second Involution and thus 
.
If 
and 
are two Involutions on a line such that under both these Involutions, 
)
.
be the exact ordering of the points on the line. Then let 
denote the center of the Inversion of 
denote the midpoint of 
(as 
and 
denote the circles with diameters 
which gives that 
(where the last equality is due to the ordering we defined earlier). This gives that 
which is a contradiction as we had 
and 
denote the centers of the Inversions respectively 
)
lie to the left of 
(we can do this since Involutions just swap the reciprocal pairs). Here is a diagram (not accurate) for easier reading.![[asy]
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
/*
A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings.
*/
pair A = (-10,0);
pair X = (-7,0);
pair O1 = (-1,0);
pair O2 = (1,0);
pair Y = (7,0);
pair B = (10,0);
import graph;
size(15cm);
pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps);
real xmin = -11, xmax = 11, ymin = -1, ymax = 2;
pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw(A--B, linewidth(0.5));
dot("$A$", A, N);
dot("$X$", X, N);
dot("$O_{1}$", O1, N);
dot("$O_{2}$", O2, N);
dot("$Y$", Y, N);
dot("$B$", B, N);
[/asy]](http://latex.artofproblemsolving.com/6/6/f/66fcda8a1cf51f0faff84963647e27a53a650aa4.png)
and 
.
This however contradicts the fact that we picked 
would either be positive or negative, but would never equal 
.This post has been edited 5 times. Last edited by kamatadu, Jun 12, 2024, 6:47 PM
March 2025
February 2025