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3h
revised Given point in an angle's interior, find a segment with endpoints on the angle's sides, with the given point as its midpoint
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4h
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!
4h
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.)
4h
answered The Golden Ratio in a Circle and Equilateral Triangle. Geomertic/Trigonometric Proof?
12h
revised (Elegant) proof of an inequality: $h(x) \geq 1- (1-\frac{x}{1-x})^2$, where $h$ is the binary entropy function
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12h
comment How do squares of non-right triangles relate?
See this answer.
1d
revised Locus of a point on a fixed-length segment whose endpoints slide along orthogonal lines
Generalized to arbitrary pairs of perpendicular lines
2d
revised Locus of a point on a fixed-length segment whose endpoints slide along orthogonal lines
Expanded answer beyond initial hint.
2d
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 $\theta$
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2d
revised Examples of applications of the Theorems of Pappus and Ménélaüs.
Better title; edits for clarity
2d
revised Saccheri quadrilaterals: perpendicularity of midpoint segment, and comparative lengths of summit and base
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2d
revised Solving $2\cos\left(2\theta\right) = \sqrt{3}$
Better title; minor edits
2d
answered Golden Ratio Conjecture in three simple Geogebra shapes--circle, triangle, and square.
2d
comment Golden Ratio Conjecture in three simple Geogebra shapes--circle, triangle, and square.
Perhaps the easiest construction approach would be to start with the circle, and construct a point on it that makes a $30^\circ$ angle with the horizontal. (GeoGebra will let you create an angle of any size you like; alternatively, you can rotate a point about the circle's center.) Then construct the tangent to that point, which gives a line containing a side of the triangle. Use tangents at the top and bottom of the circle to determine where to cut off that side, then go on to complete the triangle and square. Anyway ... I get that the target ratio is $1.6233\dots$. No gold here.
2d
revised Is there a trig identity to help solve this equation? $A \cdot \cos\theta = B+ \sin\theta$
Better title; minor editing
2d
revised Solving $\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$
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2d
comment PHI Conjecture: Fibonacci Number Line Segments & Simple Golden Ratio Puzzle?
I'll also recommend that you grab a copy of GeoGebra (it's free!) and embark on your quests for the Golden Ratio therein (with "rounding" set to, say, 5 decimal places). GeoGebra constructions and calculations are quite accurate, and they'll help reveal conjecture failures easily so that you can quickly move on to ones that have a better chance of being true. Good luck!
2d
comment PHI Conjecture: Fibonacci Number Line Segments & Simple Golden Ratio Puzzle?
You should label these golden ratio puzzles of yours as conjectures, to avoid giving the impression that the results are known to be true. This is your third incorrect claim. (The others are here and here.) In this case, the resulting ratio from lengths $2$-$3$-$5$ is $1.5914\dots$; from lengths $13$-$21$-$34$, it's $1.65637\dots$; and from $8$-$13$-$12$, it's $1.6614\dots$.
2d
comment Two Equilateral Triangles and the Golden Ratio: Simple Geomtric Proof
@N.S.JOHN: Working-through the problem with arbitrary side lengths, say $|\overline{AB}| = p$ and $|\overline{DE}| = q$, the desired ratio is $$\frac{3p+\sqrt{16q^2-3p^2}}{4p}$$ One can determine that, in order to achieve OP's proposed outcome (the golden ratio between the specified lengths), we must have $$\frac{q}{p}\;=\;\frac{1}{2}\;\sqrt{6-\sqrt{5}} \;\approx\; 0.97$$ It seems likely that experimentation with congruent triangles led OP to conjecture the involvement of the golden ratio.
Apr
29
revised Solving $\arcsin(\sqrt{1-x^2}) +\arccos(x) = \text{arccot} \left(\frac{\sqrt{1-x^2}}{x}\right) - \arcsin( x)$
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