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Thread: Constructible angles and circles/trigonometry

  1. #1
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    Constructible angles and circles/trigonometry

    I was playing about with compass and straightedge construction, namely with this site here:
    https://sciencevsmagic.net/geo/#

    I've made an observation, and I don't understand why it's happening. In my quest to try and understand it myself, I've also found that I can't do basic trigonometry - I'll call that the subproblem.

    Subproblem:
    Consider a right angle triangle with an unknown hypotenuse, two sides of length 1 and two angles of 45 degrees. Pythagoras' theorem states that the hypotenuse is root 2, about 1.41 . Ok, great, fine.

    Now, the sine rules states a/sin(A) = b/sin(B) = c/sin(C). Therefore, 1/sin(45) = hypotenuse/sin(90). So, naturally, I punched in 1/sin(45) and multiplied the answer by sin(90) to get the hypotenuse, which was about *drum roll* ... 1.05. Hang on, two theorems giving two values for the same measurement? Very confused.

    Main observation I'm actually trying to understand:
    Consider a circle centre A. Points B and C lie on the circumference. Let angle BAC be "A", line BC be "a" and arc BC be "arca".
    For any value "n":
    The perpendicular to the nth divide of "a" will intersect the nth divide of "A" at a point between "a" and "arca".
    More intriguingly, it seems only to work when the nth divide of A is constructible with a compass and straightedge. So, it works when "A" = 90 degrees and n = 3 or 4. It works when "A" = 30 and n = 4, but not n = 3.

    Furthermore, let the point of intersect be "D", the nth divide of a be "E" and where the perpendicular to E meets arca be "F". It appears that, when it does work, the ratio of distances ED/DF remains constant regardless of the angle and value for n.

    Any insight into why all this is happening (and some help with my appalling trigonometry) would be great.
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  2. #2
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    Re: Constructible angles and circles/trigonometry

    Quote Originally Posted by Cametre View Post
    I was playing about with compass and straightedge construction, namely with this site here:
    https://sciencevsmagic.net/geo/#

    I've made an observation, and I don't understand why it's happening. In my quest to try and understand it myself, I've also found that I can't do basic trigonometry - I'll call that the subproblem.

    Subproblem:
    Consider a right angle triangle with an unknown hypotenuse, two sides of length 1 and two angles of 45 degrees. Pythagoras' theorem states that the hypotenuse is root 2, about 1.41 . Ok, great, fine.

    Now, the sine rules states a/sin(A) = b/sin(B) = c/sin(C). Therefore, 1/sin(45) = hypotenuse/sin(90). So, naturally, I punched in 1/sin(45) and multiplied the answer by sin(90) to get the hypotenuse, which was about *drum roll* ... 1.05. Hang on, two theorems giving two values for the same measurement? Very confused.
    Yes, you are! You are working in degrees but have your calculator in radians. You should, even without using a calculator know that sin(90)= 1 and sin(45)= \sqrt{2}/2.

    Main observation I'm actually trying to understand:
    Consider a circle centre A. Points B and C lie on the circumference. Let angle BAC be "A", line BC be "a" and arc BC be "arca".
    For any value "n":
    The perpendicular to the nth divide of "a" will intersect the nth divide of "A" at a point between "a" and "arca".
    What is "a"? Do you mean that the center of the circle is "a" rather than "A" and that "A" is the angle?

    More intriguingly, it seems only to work when the nth divide of A is constructible with a compass and straightedge. So, it works when "A" = 90 degrees and n = 3 or 4. It works when "A" = 30 and n = 4, but not n = 3.
    What does "it" refer to here?

    [quote[Furthermore, let the point of intersect be "D", the nth divide of a be "E" and where the perpendicular to E meets arca be "F". It appears that, when it does work, the ratio of distances ED/DF remains constant regardless of the angle and value for n.

    Any insight into why all this is happening (and some help with my appalling trigonometry) would be great.[/QUOTE]
    You are asking about when "it" works but haven't said what "it" is!
    Thanks from Cametre
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    Re: Constructible angles and circles/trigonometry

    Quote Originally Posted by HallsofIvy View Post
    Yes, you are! You are working in degrees but have your calculator in radians. You should, even without using a calculator know that sin(90)= 1 and sin(45)= \sqrt{2}/2.
    ....... thank you.



    What is "a"? Do you mean that the center of the circle is "a" rather than "A" and that "A" is the angle?
    A is the centre of the circle, "a" is the line formed between point B and C (in other words, "a" is the chord of the angle at A, therefore line a is opposite angle A).


    What does "it" refer to here?
    "It" is the observation that the perpendicular to the nth divide of a intersects the nth divide of A between the chord and the arc.
    An example:
    A = 90 degrees (for simplicity) and n = 3. In this case, the nth divide of line a means a point 1/3 of the way along that line, and the nth divide of angle A is 30degrees. If you draw a line perpendicular to line a at the point described, it intersects the 30degree line between line a and arca.

    I'll pin up a diagram of this example, to make it clearer what A, a, arca and all the rest refer to.

    Constructible angles and circles/trigonometry-circles2.png
    Last edited by Cametre; Nov 21st 2016 at 01:47 PM.
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  4. #4
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    Re: Constructible angles and circles/trigonometry

    Let M be the midpoint of CB and N the point where DE meets AC

    \frac{\text{CE}}{\text{CB}}=\frac{1}{n}, n>2

    angles \text{NAD}=\frac{A}{n}, \text{AND}=180{}^{\circ}-\frac{A}{2}, \text{ADN}=\frac{A}{2}-\frac{A}{n}

    we will compute AD and show that it is less than 1

    use the law of sines in triangle ADN to get

    \frac{\text{AD}}{\text{AN}}= \frac{\sin (A/2)}{\sin (A/2-A/n)}

    we need AN.

    Triangles CNE and CAM are similar gives \text{CN}=\frac{2}{n} and so \text{AN}=1-\frac{2}{n}

    \text{AD}= \frac{(1-2/n)\sin (A/2)}{\sin (A/2-A/n)}

    \text{AD}= \frac{g(A/2)}{g(A/2-A/n)} where g(x)=\frac{\sin  x}{x}

    since g is decreasing over (0,\pi ), we see that AD<1 so the point D is inside the circle
    Thanks from Cametre
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    Re: Constructible angles and circles/trigonometry

    Now this is starting to clear up! All my high-school maths is rushing back to me - very satisfying.

    Also, I retried it with A/n being non-constructible (both A=60, n=3 and A=30, n=3) and the intersection still comes between the chord and the arc, so my earlier point about it only working with constructible angles was mistaken.

    There's only one part I still don't get:

    Quote Originally Posted by Idea View Post
    use the law of sines in triangle ADN to get

    \frac{\text{AD}}{\text{AN}}= \frac{\sin (A/2)}{\sin (A/2-A/n)}
    Where did the "180 - " go? I worked it out like this:

    \frac{\text{AD}}{\sin (AND)}= \frac{\text {AN}}{\sin(ADN)}

    Which is

    \frac{\text{AD}}{\sin (180 - A/2)}= \frac{\text {AN}}{\sin (A/2-A/n)}

    Times sin(180 - a/2)

    \text{AD}= \frac{\text {AN}\sin (180 - A/2)}{\sin (A/2-A/n)}

    Divide AN

    \frac{\text{AD}}{\text{AN}}= \frac{\sin (180 - A/2)}{\sin (A/2-A/n)}

    Thank you both for your help so far. There's hope for me yet!
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