Simple Geometry Problem: Finding the Delta Angle in an Irregular Quadrilateral

[nburo]

"Simple" geometry problem

Hello everyone,

I need your help in this "simple" geometry problem. I need to find the delta angle, knowing everything else given into the image (see attachment). It might even not be possible at all, for all I know. I've tried very hard, but I still have no solution.

Everything is explained in the image. Do not forget that DJ is the bisection of the delta angle. Also, AD and CB are the non orthogonal diagonals of the irregular quadrilateral.

The angles ACE = ABE are easily found, same for the length of AE. There are a few other things that can be easily found too.

Thank you for your help.

 

[Studiot]

What is the purpose of DFHJ and AHEI ?

In other words why are they introduced?

 

[nburo]

I misspelled theta, sorry.

What is the purpose of DFHJ and AHEI ?

In other words why are they introduced?

You might want to ignore them, if they are not necessary (sorry for the confusion). They might become handy later on, when this problem is solved.

The context of this problem is related to pool/billiards. To make long story short, A is the ball you want to pot into a pocket, D is the cue ball, theta is the acceptable angle needed for the object ball to be potted, delta is the unknown angle, which represents the angle needed for the cue ball to hit and pot the object ball. Phi (the angle between the bisection of delta and theta, i.e. the angle between DFHJ and AHEI) represents the cut angle. See this article : http://billiards.colostate.edu/technical_proofs/TP_3-4.pdf

Hope it helps =)

 

[Studiot]

If the quadrilateral ABCD was cyclic then your delta + your theta = 180.

Unfortunately I can see no reason why the quad should be cyclic.

You know the length of two sides and the included angle and the length of one diagonal.

The cosine rule will get you the length of the other diagonal, but, unless the quad is cyclic, you still need one more side length or another angle.

 

[nburo]

Oh well, thank you!

 

[coolul007]

I think delta = tetha + 2*phi

 

[Studiot]

I think delta = tetha + 2*phi

Go on

 

[coolul007]

You mean I have to show you a pretty drawing, it will take a while.. The key for me is that IB =IC = 2r

 

[Studiot]

You don't 'have' to do anything.

But the main purpose of these forums is to help others so if you can contribute I'm sure nburo will welcome your efforts.

 

[coolul007]

You mean I have to show you a pretty drawing, it will take a while.. The key for me is that IB =IC = 2r

I may revise this as I try to formalize this I may have " jumped the gun". It's that "assumption" problem.

 

[nburo]

I must admit I am very interested by your solution. I'm looking forward for more hints you might have.

Thanks to both of you for taking your time. It's really appreciated.

 

[Studiot]

I think there might be a way to perform the calculation, however it is not simple.

There is not enough information to solve either triangle ACD and ABD by themselves, however using the facts AD is common to both and AB = AC might allow the solution of some simultaneous trigonometric equations.

It would also be helpful to redraw you quadrilateral so the letters follow round the quad, rather than having ABDC as you have now.

 

[nburo]

I think there might be a way to perform the calculation, however it is not simple.

There is not enough information to solve either triangle ACD and ABD by themselves, however using the facts AD is common to both and AB = AC might allow the solution of some simultaneous trigonometric equations.

It would also be helpful to redraw you quadrilateral so the letters follow round the quad, rather than having ABDC as you have now.

Here you go.

 

[coolul007]

The "simple" solution exists when IB = IC =2r, on the original drawing. Angle BIC = theta. Then delta = theta - 2 phi.

I think there is a flaw in the problem statement, I do not see how one can know phi without knowing delta.

 

[Studiot]

But why does IB = IC ?

 

[coolul007]

AE meets CB at right angles and AC = AB = 2r, therefore EB = EC and IE is perpendicular to CB. since EC = EB and IE =IE by SAS IB=IC.

 

[nburo]

I get why IB = IC, but not the IB = IC = 2r

 

[coolul007]

It doesn't equal 2r, my comment was if it does then that's the answer. What makes this problem "impossible" is that H or J is not fixed. phi becomes some angle but we don't know where the vertex is. AD can be thought of as a radius going through an angle of zero with AC through theta/2 to AI. Without H or J fixed at some distance this can't be solved. We were given a two stipulation fact: JD bisects delta AND meets AE at angle phi.

 

[nburo]

It doesn't equal 2r, my comment was if it does then that's the answer. What makes this problem "impossible" is that H or J is not fixed. phi becomes some angle but we don't know where the vertex is. AD can be thought of as a radius going through an angle of zero with AC through theta/2 to AI. Without H or J fixed at some distance this can't be solved. We were given a two stipulation fact: JD bisects delta AND meets AE at angle phi.

F, H and J are fixed. I'm studying this particular case.

 

[Studiot]

You have 10 (relevant) variables in a quadrilateral.

4 side lengths
4 vertex angles
2 diagonal lengths

You say you know four of these -one angle, two sides and one diagonal.

Attached are six equations (10 including the knowns) for the 6 unknowns.

These equations apply to a general quadrilateral. They cna be greatly simplified if the vertices can be shown to lie on a circle as I have already mentioned.

I have relabelled them to conform to conventional order. Theta is conventionally used for the intersection angle of the diagonals.

You can solve them at your leisure.

 

[nburo]

You have 10 (relevant) variables in a quadrilateral.

4 side lengths
4 vertex angles
2 diagonal lengths

You say you know four of these -one angle, two sides and one diagonal.

Attached are six equations (10 including the knowns) for the 6 unknowns.

These equations apply to a general quadrilateral. They cna be greatly simplified if the vertices can be shown to lie on a circle as I have already mentioned.

I have relabelled them to conform to conventional order. Theta is conventionally used for the intersection angle of the diagonals.

You can solve them at your leisure.

Thank you, but I think one of your 6 equations might be dependent of the others. (equation 6, I think)