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Aug
27
awarded  Nice Answer
Aug
26
awarded  Necromancer
Jul
23
awarded  Yearling
Jul
10
comment How much heigth can a roll of pipe insulation cover?
50 ft² = 7200 in²
Jul
5
comment Writing circles as $|z-a| = \lambda |z-b|$ for the same $a,b$
What have you tried?
May
15
comment Three planes in general position, one point in each, construct sections
May we use a compass and a 1 unit line segment too?
May
2
answered Construct the great circle (geodesic) in spherical or Riemanian geometry
May
1
comment Analytic Geometry - vectors and points
There is the same question and an answer here: math.stackexchange.com/questions/728383/…
Apr
19
comment Cabri 3D - Rotating a triangle
Are $T$ and $T_0$ given and you must find out the line $AB$?
Jan
9
comment Centroids and Harmonic Means
I think you mean one of them is HALF of the harmonic mean of the others. Don't you?
Jan
6
comment In the regular hexagon tell each area,But How find this length with $B_{i}B_{i+1}$?
@HandeBruijn Very good. It's really astonishing to get an exact answer. +1.
Jan
6
comment In the regular hexagon tell each area,But How find this length with $B_{i}B_{i+1}$?
@HandeBruijn Check the value of $x$. $x$ is not $45.27468489$.
Jan
5
comment In the regular hexagon tell each area,But How find this length with $B_{i}B_{i+1}$?
$x$ cannot be more than $2*A_1A_2$.
Dec
27
comment In the regular hexagon tell each area,But How find this length with $B_{i}B_{i+1}$?
If the hexagon were regular, then its total area $A$ should be: $A=6 \frac{L^2 \sqrt{3}}{4}$. As $L= |A_1A_2|=15$ , the area should be $A=\frac{675 \sqrt{3}}{2}$, but adding up all the given areas we get 15210?!
Dec
26
comment Finding Locus of a Midpoint of a Chord
@Pakquebchsoflwty By the converse of Thales Theorem the circle whose diameter is $OY$ is unique and $M$ is a point of this circle.The "lowest" position where $M$ can be is when $C$ is on $A$ or on $B$.In this case the smaller square becomes a point and the bigger square has its center on $Y$. $M$ will be the midpoint of $DC$ or $EC$ and $M$ will be a point of the perpendicular bisector of $YO$. Therefore $M$ is a point of the upper half semicircle. – RicardoCruz 1 hour ago
Dec
26
answered Finding Locus of a Midpoint of a Chord
Dec
23
awarded  Informed
Dec
20
comment Geometry construction problem
@RoryDaulton You're welcome. I'm happy to help.
Dec
20
comment Geometry construction problem
There is something that doesn't fit in the equation you got. I think the second term of the RHS should be $2 \sqrt{r^2-(d-z)^2}$. If you correct it, you will get a quadratic equation.
Dec
19
awarded  Constituent