My Math Forum  

Go Back   My Math Forum > High School Math Forum > Geometry

Geometry Geometry Math Forum


Thanks Tree16Thanks
Reply
 
LinkBack Thread Tools Display Modes
March 2nd, 2017, 09:46 AM   #51
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 2,487
Thanks: 630

I can't believe that this thread has gone on for 50 posts. Obviously Kadomole came up with an equation that matched what he thought was "". He simply does not know what "" is. If he is, say, 12 or 13 years old, I would commend him for his (misguided) attempts. If he is older than that, it is really very sad.
Country Boy is offline  
 
March 2nd, 2017, 10:02 AM   #52
Senior Member
 
Joined: May 2016
From: USA

Posts: 632
Thanks: 257

Quote:
Originally Posted by Country Boy View Post
I can't believe that this thread has gone on for 50 posts. Obviously Kadomole came up with an equation that matched what he thought was "". He simply does not know what "" is. If he is, say, 12 or 13 years old, I would commend him for his (misguided) attempts. If he is older than that, it is really very sad.
If you look at post #48, it is quite obvious that he does know what $\pi$ is.

We are now on post 51, and we still do not have his putative "proof." I strongly suspect that we never shall be given a look at that "proof."
JeffM1 is offline  
March 3rd, 2017, 03:01 AM   #53
Senior Member
 
Joined: Apr 2014
From: Glasgow

Posts: 1,988
Thanks: 646

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Quote:
Originally Posted by topsquark View Post
Okay, I'll play along.
Do as you wish, but personally I wouldn't waste your time with this silliness.
Benit13 is offline  
March 8th, 2017, 11:50 PM   #54
Newbie
 
Joined: Jan 2017
From: mwanza tanzania

Posts: 20
Thanks: 0

Quote:
Originally Posted by topsquark View Post
Okay, I'll play along.

$\displaystyle A_{circ} = \pi r^2 = A_{tri} = \frac{1}{2}b(2r)$

$\displaystyle b = \frac{\pi r^2}{\frac{1}{2}(2r)} = \pi r$

-Dan
You are right; now use these HINTS: {area of triangle=0.5r^2[4+[pi^2]/4]sinx=pi[r^2], sinx={8pi}/[16+[pi]^2}, draw a right-angled triangle with angle x, opposite side = 8pi, hypotenuse side = 16+pi^2, and adjacent side = plus[+] or minus [-] sqrt[[16+pi^2]]^2 - 64pi^2, obtain value of tanx from that triangle, use tanx=[2tanx/2]/[1-[tanx/2]^2], use tanx/2 = pi/4, finish on calculating value of pi. Test your pi value by drawing a right-angled triangle with sides sqrt8, sqrt8 and 4, one angle = pi/4, calculate different trigonometric ratio like sin[pi]/2, tan[pi]/4, sin[pi/4], etc. What do you comment on pi value? kadomole

Last edited by skipjack; March 9th, 2017 at 12:05 AM.
kadomole simon kadomole is offline  
March 9th, 2017, 12:11 AM   #55
Global Moderator
 
Joined: Dec 2006

Posts: 17,172
Thanks: 1285

Quote:
Originally Posted by kadomole simon kadomole View Post
. . . finish on calculating value of pi.
How?
skipjack is offline  
March 9th, 2017, 02:38 PM   #56
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 1,558
Thanks: 599

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by kadomole simon kadomole View Post
You are right; now use these HINTS: {area of triangle=0.5r^2[4+[pi^2]/4]sinx=pi[r^2]
The area of what triangle? The only formula I can come up with is $\displaystyle A = \frac{1}{2} ab ~ \sin(C)$ where a and b are two sides and C is the angle between them. Best guess is that you are talking about a triangle inscribed in the "top half" of your circle and x is the angle between two radii? But then where did the $\displaystyle 4 + \pi ^2 / 4$ and $\displaystyle \pi r^2$ come from?

A diagram would be helpful.

-Dan

Last edited by skipjack; March 9th, 2017 at 03:01 PM.
topsquark is offline  
March 10th, 2017, 01:12 PM   #57
Newbie
 
Joined: Jan 2017
From: mwanza tanzania

Posts: 20
Thanks: 0

Quote:
Originally Posted by topsquark View Post
The area of what triangle? The only formula I can come up with is $\displaystyle A = \frac{1}{2} ab ~ \sin(C)$ where a and b are two sides and C is the angle between them. Best guess is that you are talking about a triangle inscribed in the "top half" of your circle and x is the angle between two radii? But then where did the $\displaystyle 4 + \pi ^2 / 4$ and $\displaystyle \pi r^2$ come from?

A diagram would be helpful.

-Dan
take that half of Isoscles triangle with height=2r,and base=[pi]r/2,tanC/2=tanx/2=[pi]r/2/2r=[pi]/4 ,a=b=Hypotonuse =[r]sqrt[4+pi^2/4] , 0.5[ab]sinC=[pi]r^2kadomole
kadomole simon kadomole is offline  
March 10th, 2017, 01:16 PM   #58
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 9,425
Thanks: 640

This guy is unbelievable...
Denis is offline  
March 10th, 2017, 03:15 PM   #59
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: CA

Posts: 1,238
Thanks: 637

Quote:
Originally Posted by Denis View Post
This guy is unbelievable...
well when you feed the trolls they grow
romsek is offline  
March 10th, 2017, 06:28 PM   #60
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,488
Thanks: 887

Math Focus: Elementary mathematics and beyond
I have chosen to intervene. I ask all other members to refrain from further discussion until the issues addressed below are resolved.

Quote:
Originally Posted by kadomole simon kadomole View Post
It has come to my notice that this equation 3y^4-32y-192=0 gives a value of (pi) a mathematical constant which is approximately to 3.1425.......can this equation be used to give a value of pi to any number of decimal places needed instead of known equation pi=circumference of a circle divide to its radius? kadomole
kadomole simon kadomole,

As previously asked, how is this equation used to determine the value of $\pi$ (pi)? Note that $\pi\ne3.1425...$

Please respond explicitly. Do not refer to previous posts without directly quoting them. Include all relevant details. Failure to do so will result in this thread being closed.
Thanks from MarkFL, Denis and JeffM1
greg1313 is offline  
Reply

  My Math Forum > High School Math Forum > Geometry

Tags
3y432y



Thread Tools
Display Modes






Copyright © 2017 My Math Forum. All rights reserved.