6
$\begingroup$

I have a quadrilateral ABCD. I want to find all the points x inside ABCD such that $$angle(A,x,B)=angle(C,x,D)$$

Is there a known formula that gives these points ?

Example:

ABCD is a rectangle. Let $x_1=mid[A,D]$ and $x_2=mid[B,C]$. The points x are those lying on the line that passes through $x_1$ and $x_2$.

But I want a formula for arbitrary quadrilaterals.

Thank you.

  • 0
    Do you know the properties of inscribed angles?2011-05-10
  • 0
    Or the expression for the cosinus using the inner product?2011-05-10
  • 0
    @user9325: Sorry I don't understand your question. You mean if I have a constraint about the angles (A,x,B) and (C,x,D) ? Nothing appart from them being equal.2011-05-10
  • 0
    I'm having a hard time coming up with something that would work even for "arrowheads"...2011-05-10
  • 0
    @user3749: I want to know what you know about geometry. Do you know the concepts that I have mentioned?2011-05-10
  • 0
    One of the points of angle equality is the intersection of the diagonals $AC$ and $BD$. Apart from that, the set of points seem to lie on difficult curves (many of which contain the vertices, especially if the curves are allowed outside the quadrilateral).2011-05-10
  • 0
    @user9325: Inscribed angles yes, cosinus using the inner product no (will look for it in the Internet).2011-05-10
  • 0
    @Henry: Yes: there are two other possible points that may be "easy" to find. We draw a line between AD, and we look for some $x$ in this line such that angle(AxB)=angle(CxD). We can do the same by drawing a line BC. I hoped that the problem is already known in geometry.2011-05-10
  • 0
    My first thought is to take the known result for a rectangle and try applying a transformation that preserves lines and angles to map the rectangle onto another quadrilateral. I suspect, though, that such transformations are limited (I think they only have 4 degrees of freedom, so would be determined by the images of 2 given points) and won't get from a rectangle to any general quadrilateral.2011-05-10

2 Answers 2

5

If you understand $A,B,C,D,x$ as complex numbers then your condition is $$\frac{x-A}{x-B}/\frac{x-C}{x-D}\in\mathbb{R}.$$ Let us denote that real number $t$, i.e. you have equation $$(x-A)(x-D)=t(x-B)(x-C).$$ For any given $t$ it is a quadratic equation for $x$, so we can solve it; the solution doesn't look very pretty: $$x = \frac{\pm\sqrt{(-A+B t+C t-D)^2-4 (1-t) (A D-B C t)}-A+B t+C t-D}{2 (t-1)}.$$ Anyway, this gives you the points you're looking for (parametrized by $t\in\mathbb{R}$).

5

For a "formula" we would first have to discuss what constitutes an answer, but I made a picture to make it clear that the condition "inside" is not a very natural one.

enter image description here

I used geogebra. Note that when the curve crosses the line CD or AB , you do not have equal angles anymore, instead the smaller angles sum to 180 degrees, but it is fine again when the curve crosses again.

Furthermore, note that if you move the vertex A slightly, the part of the curve that passes through $A$ and $B$ becomes detached and formes a little oval curve.

If you want to see an equation: $$\frac{((a1 - x) (b1 - x) + (a2 - y) (b2 - y))}{\sqrt{((a1 - x)^2 + (a2 - y)^2) ((b1 - x)^2 + (b2 - y)^2)}} = \frac{((c1 - x) (d1 - x) + (c2 - y) (d2 - y))}{\sqrt{((c1 - x)^2 + (c2 - y)^2) ((d1 - x)^2 + (d2 - y)^2)}}$$

It does not get better if you square it.

Non-convex quadrilaterals do not look different, they also can pass from an S-curve to a little oval plus another branch.

enter image description here

  • 0
    What is a formula? http://math.stackexchange.com/questions/38155/what-is-the-difference-between-equation-and-formula :-)2011-05-10
  • 1
    @lhf: Well, yes, so the point is that it is not too hard to find a polynomial equation for the curve (using coordinates and equality of cosines expressed with inner products), but the degree would be quite high and the question of being inside even more intractable than the question of an explicit formula.2011-05-10
  • 0
    My visualization skills are failing me now, but it looks as if the "arrowhead" case might be an even more perverse configuration...2011-05-10
  • 0
    If you told me what an "arrowhead" is, I could just move the vertices in my file.2011-05-10
  • 0
    @user9325, an arrowhead is probably a non-convex quadrilateral like the one in http://www.tutornext.com/system/files/u26/Arrowhead.png2011-05-10
  • 0
    Interesting, so the arrowhead case is neither more nor less difficult than a convex quadrilateral...2011-05-10
  • 0
    @user9325, what feature of geogebra did you use to draw that curve?2011-05-10
  • 1
    @I.J. Kennedy: Geogebra has a function that draws the locus of points when a certain point moves along a line. In my first picture you can see the point $E$ that moves along the bisector of $AB$. $E$ was the center of a circle through $A$ and then I drew a circle with corresponding angles through $CD$, then I defined the intersection of the two circles as $I,J$ and let geogebra draw the locus of $I,J$ when $E$ moves along the bisector.2011-05-10
  • 0
    @user9325: Very nice, thank you. I don't understand how the case of ABCD being a rectangle is a special case of your curve ? in this case, the points x are a line and they do not passe through the points A,B,C,D.2011-05-10
  • 0
    @user9325: Thank you for the formula. If you have time, please put how you obtained it.2011-05-10
  • 0
    @user3749 The curve does pass through A,B,C,D. You have one branch that is a horizontal line and the circle through the four vertices. Now if you wiggle the vertices a bit the intersections of circle and line break up to form an S curve.2011-05-10
  • 1
    @user3749 I want to point out that you have voted neither for my answer nor for the other answer yet. You *said* you would be looking up the connection of cosinus and inner product and this is exactly what I used. Angles equal implies cosinus equal. And in any case, I have to go to bed.2011-05-10