# Blue

trigonography.com

• I'm the Trigonographer at the newly-launched trigonography.com, featuring my various "proofs without words" for trigonometric identities and relations.

• I occasionally post to my research and mathematical miscellany blog, "The Bloog!".

• Honk if you love the Function Monkey!

 3h awarded Nice Answer 1d comment Are there theoretical applications of trigonometry?Interestingly, this connection to alternating (aka, "up-down") permutations drives my proof of a geometric interpretation of the power series of secant and tangent. See this answer and my note "Zig-Zag Involutes, Up-Down Permutations, and Secant and Tangent" 1d revised Triangle with a bisected side and a trisected sideRestored better title; minor edits 1d revised Triangle with a bisected side and a trisected sideBetter title; minor edits 1d answered New Golden Ratio Construct: which one of my constructs is superior/simplest--squares & circles or just circles? 1d comment Locus of a point on a fixed-length segment whose endpoints slide along orthogonal lines@Learner: ... and yet, even after I edited my answer to include the general case, you awarded a sizable bounty to an answer that handled only the $xy$-axis case. Go figure. (I'm not bothered about the bounty ---hypergeometric is quite welcome to more internet points--- but I'm disappointed by the inconsistency. I invested (more) time and effort into refining an answer to address a concern that apparently wasn't much of a concern, after all.) 1d awarded Necromancer 2d revised Given point in an angle's interior, find a segment with endpoints on the angle's sides, with the given point as its midpointBetter title, added link 2d comment The Golden Ratio in a Circle, Triangle, and Square: simple geometry/trigonometry construction.BTW: My construction: (1) Build $\square WXYZ$. (2) Rotate $Y$ about $W$ by $30^\circ$, both clockwise & counterclockwise. (3) Draw segments connecting $W$ to the rotated pts. (4) Mark $P$ & $Q$ where these segments meet the edges of the square. ($\triangle WPQ$ is equilateral.) (4) Use the Angle Bisector tool to bisect $\angle WPY$. (5) Mark $K$ where the angle bisector meets diagonal $\overline{WY}$. (6) Drop a perpendicular from $K$ to $\overline{PY}$; mark intersection $L$. (7) Draw circle about $K$ through $L$. (8) Mark endpoints of your target segments on the perpendicular. Done! 2d comment The Golden Ratio in a Circle, Triangle, and Square: simple geometry/trigonometry construction.I concur with @Aretino; the ratio is $1.65242\dots$ . As I've mentioned before: GeoGebra calculations are quite accurate. If your construction is "real" (that is, you're actually constructing tangent circles and perpendicular lines and so forth, not merely dragging elements into what looks like the right place), then you should have little doubt about whether the golden ratio appears. I recommend investing some time into making "real" constructions before posting more conjectures. (You could ask here for advice about how to make such constructions, but the GeoGebra forums might be better.) 2d answered The Golden Ratio in a Circle and Equilateral Triangle. Geomertic/Trigonometric Proof? 2d revised (Elegant) proof of an inequality: $h(x) \geq 1- (1-\frac{x}{1-x})^2$, where $h$ is the binary entropy functionBetter title, added link 2d comment How do squares of non-right triangles relate?See this answer. May 1 revised Locus of a point on a fixed-length segment whose endpoints slide along orthogonal linesGeneralized to arbitrary pairs of perpendicular lines Apr 30 revised Locus of a point on a fixed-length segment whose endpoints slide along orthogonal linesExpanded answer beyond initial hint. Apr 30 revised Solving $\left(\;1-a\cos(\theta-\alpha)\;\right)\left(\;1-b\cos(\theta-\beta)\;\right)=\frac14\left(1-a^2\right)\left(1-b^2\right)$ for $\theta$Better title Apr 29 revised Examples of applications of the Theorems of Pappus and Ménélaüs.Better title; edits for clarity Apr 29 revised Saccheri quadrilaterals: perpendicularity of midpoint segment, and comparative lengths of summit and baseBetter title Apr 29 revised Solving $2\cos\left(2\theta\right) = \sqrt{3}$Better title; minor edits Apr 29 answered Golden Ratio Conjecture in three simple Geogebra shapes--circle, triangle, and square.