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The radius of the circle you've constructed is close to $3/4$ of the length of the starting segments, but not exactly. For ease of calculation let's assume the starting segments are $2$ units long. Apply Pythagoras' theorem to find that the yellow segment has length $\sqrt 3$ and the blue segment has length $b:=\frac12(\sqrt{15}-\sqrt 3)$. The ratio of these ...
Here's a short coordinate proof: Suppose the triangle has side length $2$, one vertex lies at the origin, and the right half of the $x$-axis bisects the triangle. The circle has radius $\sqrt{2}$, so its Cartesian equation is $x^{2} + y^{2} = 2$. The yellow/blue chord lies on $x = \frac{1}{2}\sqrt{3}$, which intersects the circle at $\frac{1}{2}(\sqrt{3}, ... 2 It does imply that everything is in the plane, otherwise it wouldn't say "figure" or talk about two rays "supporting" it, it would be a cone in 3D. Diameter is a good suggestion due to the Thales' theorem but it won't work because the condition is violated at the endpoints of the diameter. One obvious solution is a concentric circle of smaller radius. ... 2 The conjecture already fails at$3$-dimension. For$3$-dimension, consider the square anti-prism with vertices at $$\left( \pm \sqrt{3}, \pm \sqrt{3}, \sqrt{2} \right),\quad ( \pm \sqrt{6}, 0, -\sqrt{2} )\quad\text{ and }\quad( 0, \pm \sqrt{6}, -\sqrt{2})$$ It has volume$V = 16(\sqrt{2}+1)$and diameter$d = \sqrt{20 + 6\sqrt{2}}$. Its "compactness" ... 1 There are infinitely many figures like that. Let$p(\theta)$be any smooth periodic function of period$2\pi$. It is known that the necessary and sufficient condition for such a function$p(\theta)$to be the support function of a convex set$K$is $$p(\theta) + p''(\theta) > 0 \quad\text{ for all }\;\theta$$ The boundary of the convex set$K$is given ... 1 The first construction can be shown to given the golden ratio by considering ($a$is the length of the sides of the square): $$\phi=\frac{a}{\sqrt{(a/2)^2+a^2}-a/2}=\frac{1+\sqrt{5}}{2}$$ The second construction is exactly the same as the first. I will leave it to you to think about. As for the third, it looks to be the same as the first two as well, but ... 1 By the "secant-tangent" aspect of the Power of a Point theorem, we have $$|\overline{AB}||\overline{AC}| = |\overline{AD}|^2 \quad\to\quad q\cdot(p+q) = p^2 \quad\to\quad \frac{p}{q} = \frac{p+q}{p} = \frac{|\overline{BC}|}{|\overline{AB}|}$$ The relation between$p$and$q$says exactly that$\frac{p}{q}$is the golden ratio,$1.618\dots$. (Compare ... 1 Here is a diagram of the situation. Let us calculate the relevant lengths, assuming that the radius of the circle is$1$so that we don't have to deal with scaling. One should note that$\angle CAB$is$\pi/6$, equal to$\angle CDE$. Thus, the length of$AC$is$\frac{1}{\cos(\pi/6)}$. The length of$CE$and$EF$separately is$\tan(\pi/6)$. Noting that ... 1 $$\frac{2t+s}{2r-t} = \frac{2r\tan 30^\circ+r\sec 30^\circ}{2r-r\tan 30^\circ}$$ So far as the conjecture goes, we can stop here, since there's clearly no chance of introducing$\sqrt 5, and thus no appearance of the golden ratio. For completeness, though, we can evaluate the ratio and get ... $$\frac{4}{11}\;\left(\;1 + 2 \sqrt{3}\;\right) = ... 1 We know that ab = cd by the power of the point P, and also that a + b = c + d because the lengths of the chords AB and CD are equal. So$$ b = \frac{cd}{a} \\ \implies a + \frac{cd}{a} = c + d \\ \implies a^2 + cd = ac + ad \\ \implies a^2 - ac + cd - ad = 0 \\ \implies (a-c)(a-d) = 0 \\ \implies a = c\quad \text{OR}\quad a = d $$We obtain that ... 1 So applying cosine rule I got$$\small PI^2=4R^2+16R^2\sin^2\frac{B}{2}\sin^2\frac{C}{2} -16R^2\cos A\sin\frac{B}{2}\sin\frac{C}{2}\Bigg(\cos\frac{B}{2}\cos\frac{C}{2}+\sin\frac{B}{2}\sin\frac{C}{2}\Bigg)$$I think that you have a typo (the red part) :$$\begin{align}&\small PI^2=4R^2\color{red}{\cos^2A}+16R^2\sin^2\frac{B}{2}\sin^2\frac{C}{2} ... 1 Suppose by contradiction there exists a linea$passing through a single point$A$. By axiom 3 there exist two other points$B$and$C$and by axiom 1 they belong to a line$r$parallel to$a$. By axiom 1 there exists a line$s$passing through$A$and$B$and by axiom 2 there exists a line$t$passing through$C$and parallel to$s$. Line$t$does not ... 1 That's from Kedlaya I believe? First of all, let's mention a useful theorem: "For any two triples of non-collinear points, we can find a unique affine transformation sending the points of the first triple to then corresponding points of the second triple". That's a known theorem. So, in our case we apply the affine transformation that sends$A,C,D$(which ... 1 First I'll talk about Hilbert axioms. Angles themselves are not an undefined term. Angle is defined with the notion of line and betweenness. Namely an angle is an unordered pair$\{A,B\}$where$A$and$B$are halflines having the same origin (we also exclude the situation when$A$and$B$are contained in the same line) and halfline is a set ... 1 Observe that$DXY$is similar to$YOZ$Implying$DX/XY=YO/OX$i.e$(a+b)/a=a/bOY$is constructed similar to$OX$1 By the "chord-chord" aspect of the Power of a Point Theorem, we have $$|\overline{AX}||\overline{XB}| = |\overline{CX}||\overline{XD}|\quad\to\quad p\cdot p = (p+q)\cdot q \quad\to\quad\frac{p}{q} = \frac{p+q}{p} = \frac{|\overline{CX}|}{|\overline{XE}|}$$ The relation between$p$and$q$says exactly that$\frac{p}{q}$is the golden ratio,$1.618\dots$. ... 1 Perhaps a way to rewrite this definition is as follows (I am not checking correctness, just changing the language). The ordered list$\{P_0,P_1,\cdots,P_{n+1}\}\subseteq\mathbb{E}^n$is called an Euclidean frame if the ordered list$\{Q_1,\cdots,Q_n\}$where$Q_i=P_i-P_0$has the property that for all$i\not=j$,$Q_i$is orthogonal to$Q_j$and$\|Q_i\|=1$. ... 1 Since there's a trigonometry tag, here's such an approach, which yields Blue's answer: From the OP's picture, I assume that the two triangles are congruent. Then, the sine law implies that in$\triangle BED\$: $$\sin \frac{2\pi}{3}:\sin \beta = DE:DB = 2.$$ Thus $$\sin \beta=\frac{\sqrt3}{4},\cos \beta=\frac{\sqrt{13}}4.\tag{1}$$ On the other hand, ...