The mathematician formerly known as DayLateDon

Oct
17
comment What is the equation of the reflections of a fixed point across all the tangents to a fixed circle?
Eliminating the trig terms gives this equation for the curve $$( x^2 + y^2 - a^2 )^2 = 4 r^2\;( ( x - a )^2 + y^2 )$$ where I've taken the circle's radius to be $r$. Compare to the cardioid equation $$( x^2 + y^2 - a^2 )^2 = 4 a^2\;( ( x - a )^2 + y^2 )$$
Oct
17
comment What is the equation of the reflections of a fixed point across all the tangents to a fixed circle?
You can write the final result more-simply as $$(2\cos t-a \cos 2t, 2 \sin t - a\sin 2 t)$$ When $a=1$ (or when the radius of the circle is $a$), the curve is a cardioid. So, a cardioid is the locus of reflections of a particular point on a circle through all tangents to that circle. Neat!
Oct
17
comment Stereographic projection proof that is geometrical.
(+1) Very nice!
Oct
13
revised Problem about right triangles.
added 71 characters in body
Oct
13
answered Problem about right triangles.
Oct
12
comment Number of Distinct Regular n-gons, given n
Look-up the concept of "relatively prime" numbers.
Oct
6
comment Area of the intersection of four circles of equal radius
Duplicate?
Oct
5
comment Prove that point is on the perimeter of circle
The other circle is very large, and lives to the upper-left. Its center ---and its point of tangency with $C_2$--- is on the perpendicular to $\rho$ at $O_2$.
Oct
5
comment Prove that point is on the perimeter of circle
Interestingly, the result is also true for the "other" circle tangent to $C_1$, $C_2$, and $\lambda$: if the center of that circle is $O_5$, then $\overleftrightarrow{O_2O_5}$ meets $\lambda$ at point $J$ on $C_4$ (diametrically opposite $I$ on that circle).
Oct
5
comment Prove that point is on the perimeter of circle
Let $r$ be the radius of $\bigcirc O_3$. Let $P$ be the foot of the perpendicular from $O_3$ to $\kappa$, so that $|PO_4| = r$. Then $\triangle O_2PO_3\sim \triangle O_2O_4I$ implies $$\frac{|O_2I|}{|O_2O_3|} = \frac{|O_2O_4|}{|O_2P|} \quad\to\quad \frac{|O_2I|}{2+r}=\frac{2}{2-r}\quad\to\quad |O_2I| =\frac{2(2+r)}{2-r}$$ Knowing that $r = 6 - 4\sqrt{2}$ gives $|O_2I| = 2\sqrt{2}$, and we're done. While we might not be allowed to assume that value of $r$, we can deduce it from Descartes' Theorem. There's probably a more-clever solution.
Oct
2
awarded Nice Answer
Oct
1
answered A Triangle Determinant
Oct
1
comment Line parallel to parallel sides of a trapezium dividing it into equal areas
Duplicate? math.stackexchange.com/q/638862/409
Sep
30
awarded Explainer
Sep
29
awarded Good Answer
Sep
28
comment A Regular Tetrahedron is a cool Polyhedron.
"I take it that this regular triangle is a face". Then you're right: every pair of vertices determine an edge, so that the skeleton is the complete graph on $n$ vertices ($K_n$), which is only a polyhedron for $n=4$ (since $K_n$ for $n\geq 5$ is non-planar). But that's too easy! Note that in the extreme (non-serious) case of a spherical "infini-hedron", any two "vertices" admit a third such that the (non-edge) chords joining them form an equilateral triangle; it seems to me that the real challenge here is to determine whether this can happen with "finite" polyhedra.
Sep
28
comment A Regular Tetrahedron is a cool Polyhedron.
I don't see (3) as being obvious. How do you justify it?
Sep
27
comment How is $A\sin\theta +B\cos\theta = C\sin(\theta + \phi)$ derived?
See my picture proof answer to this (duplicate?) question.
Sep
24
awarded Autobiographer
Sep
24
awarded Autobiographer
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