7
$\begingroup$

Note: Although this may seem maths-related, I did do some research on meta and it seems that questions which take an 'elegant' solution are on-topic.

There are many ways to tell if four points are cyclic (i.e lie on a circle). However, given four points and angles and lengths as shown below, what is the condition that they are cyclic (in terms of a,b,c,x and y)?

enter image description here

Improve this question
$\endgroup$

3 Answers 3

Reset to default
6
$\begingroup$

The condition is:

$c\sin x+a\sin y=b\sin (x+y)$.

Proof:

Name the points $O,A,B,C$ so that $OA=a,OB=b,OC=c$. Perform an inversion with center $O$ and radius $b$; clearly this transformation fixes $B$. Say, this sends the points $A,B,C$ to $P,Q,R$ respectively (we already know that $Q\equiv B$), and define $p,q,r$ by $OP=p$ and so on. The properties of inversion imply that $O,A,B,C \text{ are concyclic} \iff P,Q,R \text{ are collinear}.$ Then taking $OQ=OB$ as the $x-$axis, the coordinates of these points are $P(p\cos x,p\sin x),Q(q,0),R(r\cos y,-r\sin y).$ So the condition $P,Q,R$ being collinear translates to $$\det\left| \begin{array}{ccc} q & 0 & 1\\ p\cos x & p\sin x &1\\ r\cos y & -r\sin y & 1\end{array}\right|= 0$$ and after some simple calculations, this reduces to $q(p\sin x+r\sin y)=pr \sin(x+y)$. Now note that by the definition of inversion, $p=\frac{b^2}{a},q=b,r=\frac{b^2}{c}$; so subbing these into the previous equation yields the desired result : $$c\sin x+a\sin y=b\sin (x+y).$$ $\blacksquare$

Improve this answer
$\endgroup$
2
$\begingroup$

If my understanding of the problem is correct, to find out if the following image is true,

enter image description here
We need to make sure that X+Y+N = Z+M = 180.
I will spare you the basic math, but all the angles can be found by using basic sin/cos functions.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Yes, this will work. However, cosine rule and sine rule and square roots... ugh. There is a nicer method, which is why this is on puzzling SE not maths SE. I haven't checked if doing it this way reduces to the 'nice' condition easily though. $\endgroup$
    – Wen1now
    Commented May 22, 2017 at 1:56
  • $\begingroup$ @Wen1now soooo, you accept the answer that required using sin cos after all??? Ank's answer is very good, but it is very complicated and math based and uses sin and cos, which are the 3 things you said you didn't want. $\endgroup$
    – stack reader
    Commented May 22, 2017 at 4:49
  • $\begingroup$ The problem isn't cosine and sine, it is that the cosine rule and the sine rule can get potentially very messy, with inverse sine and other stuff. I mean Ankoganit's solution did not require multiple sine/cosine rule iterations. The method I had envisaged was far more elementary but a solution is a solution. $\endgroup$
    – Wen1now
    Commented May 22, 2017 at 5:17
  • $\begingroup$ @Wen1now While "a solution is a solution" is a solution, you needn't leave it as the final word. You can post your own answer, noting it was the intended solution, and still give the Accept to Ankoganit's worthy answer. Or conversely you can follow humn's recent example (see the corollary) and raise the question again, this time with enough specificity to get the originally intended answer. $\endgroup$
    – Rubio ♦
    Commented May 22, 2017 at 8:45
  • $\begingroup$ @Rubio I have done that with previous questions, but I think it seems slightly... cheap? Not sure if that's the right word. However, if you guys want my approach, I would be happy to post it. $\endgroup$
    – Wen1now
    Commented May 22, 2017 at 9:24
2
$\begingroup$

Elementarier solution

Let's prove the reverse direction first. Assuming they are cyclic, we can let the radius of the circle be R.

enter image description here

Then we know that, by the extended sine rule, $d=2R\sin(x)$, $e=2R\sin(x+y)$ and $f=2R\sin(y)$. Then using Ptolemy's (which was on the wikipedia link I provided in the question), we know that $af+cd=be$. Thus after dividing through by $2R$ we have $a\sin(y)+c\sin(x)=b\sin(x+y)$.

For the reverse direction, suppose that the equality is true. Note that this equality uniquely defines b (it cannot lie on the other side of the line as the angles would then be wrong). Then the four points are cyclic.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Ah, of course ( * facepalm *). Nice solution. :) $\endgroup$
    – Ankoganit
    Commented May 23, 2017 at 2:03

Not the answer you're looking for? Browse other questions tagged or ask your own question.