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I have a situation where I know the cartesian coordinates of the 2 vertices of a triangle that form its base, hence I know the length of the base and this is fixed.

I also know the angle opposite the base and this is also fixed.

Now what I want to do is figure out how to compute all possible positions for the third vertex.

My maths is rusty, I reverted to drawing lots of pictures and with the help of some tracing paper I believe that the set of all possible vertices that satisfies the fixed base and opposite angle prescribes a circle or possibly some sort of ellipse, my drawings are too rough to discern which.

I started with a simple case of an equilateral triangle, with a base length of two, i.e. the 3rd vertex is directly above the x origin, 0, base runs from -1 to 1 along the x-axis

then i started drawing other triangles that had that same base, -1 to 1 and the same opposite angle of 60 degrees or pi/3 depending on your taste

now i need to take it to the next step and compute the x and y coordinates for all possible positions of that opposite vertex.

struggling with the maths, do i use the sin rule, i.e. sin a / A = sin b/ B and so on, or do I need to break it down into right angle triangles and then just use something along the lines of a^2 + b^2 = c^2

ultimately, I intend to plot the line that represents all the possible vertex positions but i have to figure out the mathematical relationship between that and the facts, namely,

base is fixed running from (-1,0) to (1,0)
angle opposite the base is 60 deg

I then need to extend to arbitrary bases and opposite angles, but thought starting with a nice simple one might be a good stepping stone.

apologies if my formatting is poor - first post on math stack exchange - i am more a stack overflow sort of guy..

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    $\begingroup$ The locus of the vertex is a part circle, due to the theorem " angles in the same segment are equal" $\endgroup$ Aug 4, 2015 at 16:39
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    $\begingroup$ If the angle opposite is $\pi\over 2$, you have a semi-circle; you can compare the behavior of other angles to this based on whether they are greater or less than $\pi\over 2$... $\endgroup$
    – abiessu
    Aug 4, 2015 at 16:46
  • $\begingroup$ @David - brilliant - i think that is the fact that will allow me to solve my problem, and certainly give me the necessary search term for a whole load of relevant googling - happy to accept this as the answer if you want to present it as one. $\endgroup$
    – bph
    Aug 4, 2015 at 16:47
  • $\begingroup$ Glad to be of help $\endgroup$ Aug 4, 2015 at 16:52

4 Answers 4

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A general solution to the problem:

Let the triangle have sides $a,b,c$ and angles opposite to these sides as $A,B,C$ respectively. We fix $a$ between points $(x_1,y_1)$ and $(x_2,y_2)$. For a fixed $A$, we need to find the mathematical equation of the locus of third point, $(x_3,y_3)$.

Using sine rule, we have: $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k $$ where $k$ is easily found since we know $a$ and $A$.

In the triangle, angle $C = 180 - (B+A)$. We also see that: $$ b = k \cdot \sin B $$ Similarly, $$ c = k \cdot \sin (180-(B+A)) = k \cdot \sin (B+A) $$ For different values of $B$, we will have corresponding values of $b$ and $c$. Knowing $a,b,c$ fixes the triangle.

From basic geometry, we have: $b= \sqrt{(x_3-x_2)^2+(y_3-y_2)^2}$ and $c= \sqrt{(x_3-x_1)^2+(y_3-y_1)^2}$.

The locus of $(x_3,y_3)$ is given by simultaneously solving: $$ (x_3-x_1)^2+(y_3-y_1)^2 = k^2 \cdot \sin ^2 (B+A) $$ $$ (x_3-x_2)^2+(y_3-y_2)^2 = k^2 \cdot \sin ^2 (B) $$ when $B$ varies from $0$ to $(180-A)$ degrees. The locus will be the arc of the circle. The base $a$ serves as a chord.

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  • $\begingroup$ very elegant algebraic manipulation - i had initially ventured down this route but not got so far - thanks for filling in the gaps although i will have to refresh my knowledge on solving simultaneous equations. I think this compliments the more geometric approach of using central angle theorem and same angles for a given segment very nicely $\endgroup$
    – bph
    Aug 5, 2015 at 9:33
  • $\begingroup$ also how do you format the maths so nicely - is there someway i can see the markup you used? $\endgroup$
    – bph
    Aug 5, 2015 at 11:09
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    $\begingroup$ @bph I am not sure how to send you the markup for the above answer, as stackexchange doesn't allow messages. Have a look at: math.stackexchange.com/editing-help to help you understand the formatting. The equations are written using the same markup as one would use in LaTeX. Math Stackexchange uses mathjax.org The LaTeX syntax for math is not too hard I would say - should take you an hour or so to get the basics and anything else that you need you would be able to look it up online! :) $\endgroup$
    – Mihir
    Aug 5, 2015 at 11:31
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    $\begingroup$ @bph Just realized that you can see the markup for my answer by clicking on edit, just below it. $\endgroup$
    – Mihir
    Aug 5, 2015 at 11:33
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    $\begingroup$ @bph: Another way to see the markup is to highlight the equation and then right-click on it. A context menu should pop up. Select "Show Math As > TeX commands" $\endgroup$
    – PM 2Ring
    Aug 5, 2015 at 12:00
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Vertices of the triangles sought form two circle's arcs.

See Inscribed angle — Theorem on Wikipedia.

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The locus of the vertex is a part circle, due to the theorem " angles in the same segment are equal"

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  • $\begingroup$ That's not true. $\endgroup$
    – CiaPan
    Aug 1, 2018 at 15:52
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Here is the fleshing out of my geometric approach, inspired by David Quinn's input utilising the 'angles for the same segment are equal' theorem and the 'central angle theorem'

enter image description here

for case where chord is horizontal (parallel to x-axis)

(and any chord can be made to be parallel by rotating the circle appropriately)

$$ a = \frac{x_2-x_1}{2} $$ $$ r = \frac{a}{\sin(\theta)} $$ $$ r = \frac{\left(\frac{x_2-x_1}{2}\right)}{\sin(\theta)} $$ $$ b = \frac{a}{\tan(\theta)} $$ $$ b = \frac{\left(\frac{x_2-x_1}{2}\right)}{\tan(\theta)} $$ therefore if the origin, $o$ has cartesian coordinates $[o_x,o_y]$ $$ o_x = x_1 + \frac{x_2-x_1}{2} $$ $$ o_x = \frac{x_1+x2}{2} $$ $$ o_y = y_1 + \frac{\left(\frac{x_1+x2}{2}\right)}{\tan(\theta)} $$ so we know the origin, $[o_x,o_y]$ and radius, $r$, of the circle prescribing the locus of all solutions

so if we want to convert back from polar coordinates to cartesian coordinates we use the following:

$$ x = r\cos(\phi) $$ $$ y = r\sin(\phi) $$ $\phi$ is the angle as we travel round the circle, $r$ the radius

this assumes the centre of the circle is at the origin so that has to be corrected, e.g. $$ x = r\cos(\theta) + \frac{x_1+x2}{2} $$ $$ y = r\sin(\phi) + y_1 + \frac{\left(\frac{x_1+x2}{2}\right)}{\tan(\theta)} $$

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  • $\begingroup$ got there in the end with the latex formatting.. would like to use some square brackets round a \frac though $\endgroup$
    – bph
    Aug 5, 2015 at 11:57
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    $\begingroup$ As an alternative LaTeX / MathJax reference, you may find the Wikipedia Help:Displaying a formula article useful. However, not everything mentioned in that article will work here. $\endgroup$
    – PM 2Ring
    Aug 5, 2015 at 12:05
  • $\begingroup$ thats a really useful link - has enabled me to stick some brackets round fractions, went round rather than square - looks ok! $\endgroup$
    – bph
    Aug 5, 2015 at 12:13

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