MHB A center of mass determined triangle: Find the angles of the triangle P_1P_2P_3

lfdahl
Gold Member
MHB
Messages
747
Reaction score
0
Masses $m, 2m$ and $\sqrt{3}m$ are located at points $P_1, P_2$ and $P_3$ on a circle $C$
so that their center of mass coincides with the center of $C$.
Find the angles of the triangle $P_1P_2P_3$.
 
Mathematics news on Phys.org
Here´s the suggested solution:

Let $C$ be the unit circle of the $xy$-plane, with mass $m$ at the point $(1; 0)$.

If \[\angle P_1OP_2 = \alpha, \: \: \: \angle P_1OP_3 = \beta\] – then\[m\cdot 0+2m\sin \alpha +\sqrt{3}m\sin \beta = 0\: \: \: \: (1). \\ m\cdot 1 +2m\cos \alpha +\sqrt{3}m\cos \beta = 0\: \: \: \: (2).\]From $(1)$: \[\frac{\sin \alpha }{\sin \beta }= -\frac{1}{2}\sqrt{3}\]substitute in $(2)$ and obtain\[1 + 2 \cos \alpha +\sqrt{3}\sqrt{1-\frac{4}{3}\sin^2\alpha } = 0,\]whence\[\cos \alpha = -\frac{1}{2}, \: \: \alpha = 120^{\circ}.\]so that\[\sin \beta = -1, \: \: \beta = 270^{\circ}.\]Thus, the angles of $\bigtriangleup P_1P_2P_3$ are: $45^{\circ}, 60^{\circ}, 75^{\circ}$.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top