MHB Equilateral triangle within a circumscribed circle

AI Thread Summary
The discussion focuses on proving that the segments MT and TW are equal in an equilateral triangle inscribed in a circumscribed circle. Participants suggest using geometric properties and the inscribed angle theorem to establish the relationship between the segments. One contributor proposes using coordinate geometry to demonstrate that the lengths are equal, deriving the necessary equations from the triangle's angles and the circle's radius. The conclusion confirms that MT equals TW, validating the initial claim. The thread effectively combines geometric reasoning with algebraic methods to solve the problem.
Yankel
Messages
390
Reaction score
0
Dear all,

In the attached picture there is an equilateral triangle within a circumscribed circle.

MW is a radius of the circle, and I wish to prove that MT = TW, i.e., that the triangle cuts the radius into equal parts. I thought perhaps to draw lines AM and AW and to try and prove that I get two identical triangles, but failed to do so.

Can you kindly assist ?

Thank you !

View attachment 7682
 

Attachments

  • triangle circle.PNG
    triangle circle.PNG
    4.3 KB · Views: 143
Mathematics news on Phys.org
Here's how I would do it. Call the radius of the circle "r". The three angles, AMB, AMC, and BMC are each 360/3= 120 degrees. The angle AMT is half that, 60 degrees. The right triangle AMT has hypotenuse of length r, angle at M 60 degrees so the length of MT is r cos(60)= r/2.
 
Is it given that $\overline{MW}$ is perpendicular to $\overline{AB}$?
 
Greg, yes it is !
 
By the inscribed angle theorem $\angle{WAB}=\angle{WCB}=30^\circ$. It shouldn't be too difficult to finish up from there. :)
 
I would use coordinate geometry. Begin by orienting the circle such that its center is at the origin, and WLOG, give it a radius of 1 unit:

$$x^2+y^2=1$$

Now, the inscribed triangle has $\overline{AB}$ in quadrants I and IV as a vertical line. The line segment $\overline{MB}$ lines along:

$$y=\tan\left(60^{\circ}\right)x=\sqrt{3}x$$

Hence:

$$x^2+\left(\sqrt{3}x\right)^2=1$$

$$x^2=\frac{1}{4}$$

As $x$ must be positive, there results:

$$x=\frac{1}{2}$$

And so we conclude:

$$\overline{MT}=\overline{TW}$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Replies
9
Views
2K
Replies
1
Views
3K
Replies
4
Views
1K
Replies
2
Views
2K
Replies
6
Views
3K
Replies
1
Views
2K
Replies
16
Views
3K
Back
Top