Geometry Puzzle

Answer

Need a couple of "way back" facts to knock this one out.

Sum of angles in a triangle= X

Opposite angle are equal

I did it quickly, but I'm pretty sure the answer is : the swing on a Nova 3000 lathe X the number of fingers a GOOD shop teacher has on his/her right hand.

Steve

Re: Geometry Puzzle *LINK*

This is a variation of the so-called Langley problem. You can find a nice animated solution at the link below. It takes a little bit more than the basic fact of the sum of the angles in a triangle being equal to 180


http://agutie.homestead.com/FiLEs/LangleyProblem.h

Re: Answer

Sorry for the delay in response. I'm not a turner, and assumed you were right. Looking up that lathe, though, I see that it claims a capacity of 16" over the bed, which I think means it has an 8" swing? If so, your answer is not correct. And the solution is a bit more complex than merely sum of angles = 180.

Re: Geometry Puzzle

Good link. The solution to this one uses similar constructions, but still cannot be solved (by me at least!) using just this construction and basic sum or angles = 180. What answer did you get, and what other geometrical facts did you need?

Trying again...

I got to a point where I just brute forced it with some numbers (and assumed that the drawing is not to scale).

I now get an answer that proves that I remember 50% of the math I've been taught.

I'd love to see how you solve the problem without brute force.

Steven

Does the 50% relate to your first try?

If so, the actuary in me says that on average, you are exactly right!

I'd be interested in your brute force method here. Will post the solution tomorrow if I get no other takers.

BTW, the drawing is to scale, so you can "cheat" to find the answer if my hint above is not clear, and if that will help you find a solution.

Brute force (no computers involved)

Assuming that you know enough geometry to be dangerous, we can start by calculating all of the solvable angles via the sum of angles=180.

Two intersecting lines in the middle form equal angles of 50 and 130 degrees allow me to end up with only four mystery angles.

ADE

AED

DEB

EDC

I know that:

EDC+DEB=130

AED+DEB=140

ADE+EDC=150

AED+ADE=160

Some quick mathematical substitution tells me:

AED=EDC+10

and

ADE=DEB+20

At this point, I knew that I was looking for four numbers that were either ending in 5 or 0, so I guessed 0 and randomly started with 50. I added to it to get 60, which gave AED and EDC. I plugged them into the formula and found the other two were 80 and 100. I then back checked the answers against all of the formulas and it worked...

Brute force.

Steve

Re: Brute force (no computers involved)

Steve: Nothing wrong with brute force, as long as it is testing a finite number of cases rather than pure trial and error.

Speaking of "brute force", I can't figure out how to quote with this forum software, so I am going to just copy your entire post and insert comments set off with asterisks.

Your post:

Assuming that you know enough geometry to be dangerous, we can start by calculating all of the solvable angles via the sum of angles=180.

Two intersecting lines in the middle form equal angles of 50 and 130 degrees allow me to end up with only four mystery angles.

ADE

AED

DEB

EDC

*********************

And knowing any one of these makes the others trivial

*********************

I know that:

EDC+DEB=130

AED+DEB=140

ADE+EDC=150

AED+ADE=160

Some quick mathematical substitution tells me:

AED=EDC+10

and

ADE=DEB+20

********************

I'm with you so far

********************

At this point, I knew that I was looking for four numbers that were either ending in 5 or 0,

***********************

How do you know that? It will probably be obvious once you state it, but I don't see it.

***********************

so I guessed 0 and randomly started with 50. I added to it to get 60, which gave AED and EDC. I plugged them into the formula and found the other two were 80 and 100. I then back checked the answers against all of the formulas and it worked...

************************

But I think it worked because you were testing it with the same formulas you used to determine the other angles once one was set. I think you could "prove" any angle worked with this method!

I assume you are saying AED = 60, EDC = 50, ADE = 100, and DEB = 80, Wouldn't AED = 53, EDC = 43, ADE =107, and DEB = 87 satisfy these conditions just as well? I think there is a condition missing.

Alex

I give up...

You are correct. I didn't try any other numbers since I assumed that there were enough permutations to prevent other combinations from working. I assumed ending in 0 would prevent a 3+3 or 7+7, but clearly your numbers also satisfy my logic.

I'm pretty sure I'm right. Now, can you tell me why? :-)

Steve

Re: I give up...

From drawing to scale I can measure BDC as equal to AED.

Correct.

Both are 30 degrees. But absent measuring, how can you derive this?

Solution *LINK*

AS was mentioned in another post, the answer is 30 degrees. But getting there, without measurement, is non-trivial.

First, we construct some isosceles triangles, with sides equal to the base of ABC:

On Side AB, mark a point K such that CK = BC. BCK is isosceles, with angle BCK = 20

On Side AC, mark a point L such that KL = BC. CKL is isosceles (and equilateral)

On side AB, mark a point M such that LM = BC. KLM is isosceles, with angle KLM = 100 (the other two triangles have 80 and 60 degree angles at K, leaving 40 for this triangle)

Now look at triangle CLM. Angle CLM is 160 (adding angles from the two triangles we constructed at that point). CLM is also isosceles, with sides - BC. Thus, we know that angle LCM is 10 degrees.

ACD is also 10 degrees. So we know that M is the same point as D. I'll refer to it as "D" from here on.

Now consider two triangles sharing side ED, EBD and ELD. Angle EBD = 20 = Angle ELD.

At this point comes a fact I either never knew or had long forgotten, the Inscribed Angle Theorem (see link). It tells us that these four points lie on a circle.

Using the same circle, but a different chord, Angle LDE = angle LBE.

Triangle LCB is isosceles with angle LCB = 80, from which we conclude that LBC is 50 and LBE is 10.

From that point, it is just a couple of steps of sum of angles = 180 to find AED = 30.

The picture below adds the construction lines mentioned above, which may help to follow the discussion.




Inscribed angle theorem