As we know, the measures of central angles of the arcs of a circle is proportional to the lengths of the arcs  How to prove that those ratios are the same?
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How can I prove that:     The length of the arc is proportional to the size of the angle.   Every book use this fact in explaining radians and the fundamental arc length equation $s = r\theta$. However no book proofs this fact.  Is this fact some axiom, some natural law like $\pi$ and the triangle side proportions?  Can I proof the above? Or is it something that you should just accept?
46852
Let ABC be a triangle in which $AB = AC$ and $\measuredangle CAB = 90^{\circ}$. Suppose $M$ and $N$ are points on the hypotenuse $BC$ such that $BM^2 + CN^2 = MN^2$. How do I  prove that $\measuredangle MAN = 45^{\circ}$.
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I think the title of the question says it all. I unfortunately did not seem to conclude anything. Some ideas I had:  It is easy to show that (given $T$ is the rotation) $\{T^n(\theta)\}$ is a set of distinct points. Furthermore, given that the circle is a compact metric, it must have a limit point $x$. By continuity of the rotation function, $T^n(x)$ is a limit as well since taking $T$ of every term yields the same sequence (with only the first term removed). By induction, we have infinitely many distinct limit points $\{T^n(x)\}$.  That's all I could come up with! It was also easy to show that the orbit is infinite. I still don't seem to be able to get close to the required result however.
111319
On the train home, I thought I would try to prove $\pi$ is irrational. I needed a definition, so I used:  $\pi$ is the area of the unit circle.  But what is a circle?  A circle is the set of tuples $(x,y)$ satisfying "$x^2 + y^2 \leq 1$".  That seems a little awkward - is there a more natural way to say it?  The standard distance metric on $\mathbb{R}^2$ is "$\sqrt{(x1-x2)^2 + (y1-y2)^2}$". A circle is the unit ball of this metric, and $\pi$ is its area.  But why do we use this metric in the first place? Why not use the Taxicab metric - it's linear so it'd seem to satisfy even more nice mathematical properties.  Because of the Pythagorean theorem.  And what property of $\mathbb{R}^2$ lets us prove the Pythagorean theorem?  This one took a little more thought. I think it comes about because we want to associate lengths to lines that are equivalent under translations and rotations. Translations seem pretty natural - but I'm having trouble defining rotations without being circular:   If you multiply a coordinate tuple by the appropriate matrix, you get a rotation - but the matrix's entries involve $\sin$ and $\cos$, which I would define as the coordinates on the unit circle(which is circular). I can define the set of all infinite lines that go through the origin. Except for the vertical line, you might assume they all have the form $L_m = \{(x,y)|y = mx\}$ for some $m \in \mathbb{R}$. Given an $m \in \mathbb{R}$, I can't find an obvious way to get the point on the unit circle corresponding to it. The complex number route just gives $\sin$ and $\cos$. Circular, again.  We can determine whether two vectors are orthogonal using the dot product. That reduces defining the entire unit circle to defining just a single quadrant(if we also assume the length of $-1 * x$ is the same as $x$).  A quick Google search shows circles were a primitive notion in Euclid's Elements. There is another question/answer given here, but I was still left confused as to what a rotation actually is in $\mathbb{R}^2$:    So my questions are:   Out of all the possible metrics, why choose the Euclidean metric as the natural one? I think it's rotation that's the root cause, but that might end up being a red herring. How do we define a natural equivalence of points up to rotation from first principles?  This is related to #1. If we define the set of infinite lines through the origin, you can choose a representative element from each one and form a perimeter curve(assuming a continuous choice function). What properties does the circle have that would make it a natural choice?
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Let $H$ be the orthocentre of an acute triangle $ABC$. Consider the circumcenters of triangle $ABH,BCH,CAH$ .Prove that they are the vertices of a triangle that is congruent to $ABC$.   My Work:   Observe that $AH$ is perpendicular to $BC$ &amp; $O_BO_C$. (Let $O_B$ be the circumcentre of $AHC$ and the others are defined same).  Clearly,$AH\perp BC$ &amp; $AH\perp O_BO_C$. Therefore $BC||O_BO_C$  Similarly $AC||O_AO_C$ &amp; $AB||O_AO_B$  But, I failed to show $O_BO_C=BC$  You may also show that  $AO_C=AO_B$
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It would be convenient sometimes to be able to draw an arc in tikz by specifying   the center of the corresponding circle its radius the initial/final angle   i.e., the "natural" way an arc is defined, instead of the "first point of the arc".  Is there a way to do it?
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Let $x_1, \ldots ,x_n$ be distinct points in $\Bbb{R}^d$. Prove that there is a permutation $\sigma$ of $[n]$ such that the $\angle x_{\sigma(i−1)}x_{\sigma(i)}x_{\sigma(i+1)} &lt; \pi/2$ for any $i = 2,\ldots, n−1$.  So I have got this question, and I am not able to understand what &quot;a permutation $\sigma$ of $[n]$&quot; mean? Can someone explain to me this question and provide some hints to approach it? P.S. I have edited out some parts as it was a bit irrelevant.
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I was given a question which was like:     Suppose that a circle, square, rectangle and triangle have same perimeter. How area their areas related??   My work:  I broke the question in parts and tried to prove it seperately:  STEP $1$. Suppose the given perimeter is $P$.So the radius of circle will be ${r=P/2\pi}$.  And hence area of the circle will be:   ${P^2/4\pi}$.  Side of square will be $P/4$ and hence area of square will be $P^2/16$.   It is obvious that ${P^2/4\pi}$  >  $P^2/16$.   So Area of circle is more than area of square.  STEP $2$: Whether by using $AM&gt;GM$ inequality or by using a bit of differentiation, we can get that product off two quantities is maximum when they are equal.  So, Area of square is more than Area of rectangle.  Here, the hardest part come (at least for me). I can't relate the area of triangle to any of Circle, Square or Rectangle. It would have been easy if the triangle was Equilateral, Right angle but the question is talking about a general triangle.  However, when I considered the situation with different examples (Assuming the perimeter and calculating the area, I got the relation:  Circle > Square > Rectangle > Triangle.  I shall be highly thankful if you guys can establish the relation in purely mathematical way.  EDIT: This question does not ask that whether circle have the largest area among all 2-D figures or not. The OP just want to relate the areas of Circle, Square, Rectangle, Triangle in a pure mathematical way.
97304
Consider two unshaded circles $C_r$ and $C_s$ with radii $r&gt;s$ that touch at the origin of the complex plane. The shaded circles $C_1,C_2...C_7$ (labeled in counterclockwise direction sequentially) all touch $C_r$ internally and $C_s$ externally. $C_1$ also touches the real axis and $C_i$ and $C_{ i+1}$ touch for $i=1...6.$  Let $r_i$ denote the radius of $C_i$. Then show that for $i=1,2...,$  $$r_i^{-1} + 3r_{i+2}^{-1} = 3r_{i+1}^{-1} + r_{i+3}^{-1}$$  A picture is attached for clarity.    Attempt:  Under inversion, $C_s$ and $C_r$ are mapped to lines and $C_i, i \in \left\{1,...,6\right\}$ are mapped to circles. The map $f(z) = 1/z$ is conformal, so the angles are preserved wherever $f'(z) \neq 0$. Each circle has $4$ tangent points, (except $C_7$) so under the transformation these $\pi/2$ angles are preserved.  The resulting image I have after the transformation is that the $C_i$ are mapped to circles that lie within the two lines and touch the sides.   The only way I see just now to preserve the angles is to have all the circles of the same radii. If $s$ is the radius of $C_s$ and $r$ the radius of $C_r$ then the centre between these two lines is $\frac{1}{2}\left(1/s - 1/2r\right) $. (But I do not think this makes sense.)  Many thanks
120180
I apologize the poor image quality and that I don't know how to rotate the image, please see image below.    In this tetrahedron, angle α and β are known and the following are true: AC=BC; AO=BO; OC is perpendicular to plane ABC  As a result, the following are true: angles: CAB=CBA=(180-α)/2; OAB=OBA=(180-β)/2  The question is: what is the length of OC?  (In terms of other lengths and angles)
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Let us say you are in an inertial reference frame with a circular planar disk. If you take your meter measuring rods (or perhaps tape measure) you can find the diameter and circumference of the disk. If you divide the circumference by the diameter, you will get exactly $\pi$. Now you start rotating the disk. A book I have claims that now the ratio will be different (lets call it $\pi_\circ$ to avoid ambiguity.) This has caused to reevaluate my understanding of special relativity (this is a set up to general relativity, but the problem itself only requires the special theory.) My problem is that, sure, the measuring rods (or tape) would contract on the circumference, but wouldn't the circumference contract a similar amount, therefore giving a measurement of $\pi$ in every reference frame according to your own measuring rods. Of course, this reasoning goes on to make me question how I understood all of special relativity up this point.
61878
Let $ABCD$ be a convex quadrilateral such that the length of the segment connecting midpoints of the two opposite sides $AB$ and $CD$ equals $\frac{AD+BC}{2}$ . Prove that AD is parallel to BC. I assume that AD is not parallel to BC but I can't find any contradiction.
101833
Consider a $2 \times 1$ rectangle split by a diagonal.  Then the two angles at a corner are ArcTan(2) and ArcTan(1/2), which are about $63.4^\circ$ and $26.6^\circ$. Of course the sum of these angles is $90^\circ = \pi/2$.  I would like to know if these angles are rational multiples of $\pi$. It doesn't appear that they are, e.g., $(\tan^{-1} 2 )/\pi$ is computed as     0.35241638234956672582459892377525947404886547611308210540007768713728\   85232139736632682857010522101960   to 100 decimal places by Mathematica.  But is there a theorem that could be applied here to prove that these angles are irrational multiples of $\pi$?  Thanks for ideas and/or pointers!  (This question arose thinking about Dehn invariants.)
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Consider the circle $$|z-6i|=3;$$ its centre is $(0,6)$ and its radius is $3$.  I want to find the greatest value of $\arg z$ achieved on this circle. My idea is that the tangents to the circle from the origin create right angles with the radius. Then use the Pythagorean Theorem and trigonometry to find the angle $\theta$ between the tangent and the imaginaery axis.  Then the formula for finding the two values of $\arg z$ in the form $p\pi$ where $-1\lt{p}\le1$ are: $$\frac{\pi-2\theta}{2}=argz$$ $$\frac{\pi-2\theta}{2}+2\theta=argz$$
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Since we’re introduced to trigonometric ratios in terms of opposite, perpendicular and hypotenuse, all of which are a part of a right angled triangle. This defines the ratios for angles greater than 0 and less than 90.  For angles greater than 90, we drop a perpendicular from the end-point of the moving arm of the angle and we take the acute angle made by the end-point, the pivot point and the point on the x axis where the perpendicular is dropped.   My question being; The way we calculate angles greater than 90, is it defined or derived? If it is defined, then what is the intuition behind defining it like that and if it is derived , proof (and how it’s connected to 0 to 90 part).  I do realise that these ratios were never really meant to be specific to the triangle.But then again, they were introduced that way. Also can you please recommend some books which explain trigonometry in great detail?  Thank you!
71364
What theorems can I use to prove this?
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Possible Duplicate:          How to prove $$\arctan(1)+\arctan(2)+ \arctan(3)=\pi$$