60 degrees

It’s non-soluble without a length as the angle changes dependent on the height of the major triangle.

That’s the bit I don’t get. We know the shape can’t change (it can’t be a taller and thinner for example), so the whole thing can only scale in unison.

You are right, but it is ill-posed. Unless anyone spots an error in my alebra. Didn’t see one. four unknowns but only three independent equations. That means you can find x in terms of a ratio of another angle you can’t find. So you’ll have to draw it.

Ask yourself why it might be on Facebook 😉

Tired, if you think it is ill-posed, you find two solutions and draw them accurately 🙂

Never was much good at geometry, but I’ll have a think about it. I think it must have a neat solution.

At airport waiting for a flight so this is perfect :))

@Tired’s got it with the comment that the solution can be an equation, it doesn’t have to be a single number (“42” and all that 🙂 )

Will double check once on plane where I can write out the various equations. @Tired you kissed the chance to accuse everyone else of going off at a tangent … I’ll get my coat

There’s only one answer.

You can reduce the problem to the one quadrilateral at the top, CDE and the centrepoint.

We know its 4 angles: C=20° (given), D= 180°-40° = 140°; C= 180°-30°= 150° and the bottom 50° (given).

fixed angles = fixed shape, the 4 angles given by the line joining C and D are also fixed.

It is not ill posed I am working through it at the moment but I think the key is that the main triangle is an isosceles triangle and therefore your four unknowns can be reduced to three as you know the two sides are equal.

If it turns out you can produce more than one answer by solving simultaneous equations, one of them is going to be obviously unusable, like a minus angle or something otherwise irrational.

Initially it looks unsolvable, but when everyone started saying the answer was a nice convenient round number, I’ve been scratching my head.

Of course, they may now pipe up and tell me their CAD software was rounding!

The most distubing thing about this is that it looks uncannily like my writing…

your four unknowns can be reduced to three as you know the two sides are equal.

That’s what my 11 year old son said – I made him have a shower so didn’t have to hurt my brain any more.

Lord, there are some folk making this look very hard! The answer has been stated at least twice already!

and since then, there’s been another question:

Can you prove it with a formula, rather than with Autocad or a pen, paper and protractor?

Probably not, but working out the answer didn’t take autocad/pen/paper or protractor – Just the knowledge that the 3 internal angles of a triangle sum to 180.

3 internal angles of a triangle sum to 180.

No way!

Probably not, but working out the answer didn’t take autocad/pen/paper or protractor – Just the knowledge that the 3 internal angles of a triangle sum to 180.

No it doesn’t as the unknowns cancel each other out.

try again

I can work out loads of angles that work that it could be, but no single one for certain.

Show your workings

60 with proof coming up.

Basically, mirror triangle EB(centre) around E. You know the angles at E must sum to 180, so it’s 30 + 30 + 120. x = half 120.

twice.

I made it 60, but can’t be arsed posting my working out….

The answer’s 20.. but I can’t get an elegant solution.

Brute force – define the base as length 1.. and then you can get the lines:
AC => y=x tan80
AE => y=x tan 70

BC => y=(1-x) tan 80
BD => y=(1-x) tan 60

Calculate the intersections to give points D and E as:
x y
D: 0.233955556881022 1.32682789633788
E: 0.67364817766693 1.85083315679665

This gives line DE a slope of 50 degrees to horizontal – so the angle between AE and DE is 20 degrees.

I’m sure there must be a simpler way – but I’m too tired!

Can’t you stick any number in?

My head says you shouldn’t be able to but I’ve tried with x=10 and x=20 and it seems to add up.

Probably just made a mistake though

As I showed, geometry does not work. However trigonometry does:

Four applications of the Sin rule to find the lengths of the inner l1,l2,l3 and l4 from vertices and midpoints to the crossover, then the Cosine rule to find the distance across the triangle and finally the Sin rule again will get you to (excuse the lack of brackets:

Sinx = (Sin60/Sin50)(Sin10/Sin40)Sin50/Sqrt((Sin10Sin60/Sin40Sin50)^2 + (Sin20Sin50/Sin30Sin70)^2 – 2*(Sin10Sin60/Sin40Sin50)*(Sin20Sin50/Sin30Sin70)Cos50)

EDIT:

Here is the proof:

Denote the midpoint F and assume AB is 1

Then:
Sin rule 1: BF/Sin30 = 1/Sin50
Sin rule 2: BF/Sin130 = EF/Sin20 so EF = (Sin30/Sin50) * (Sin20/Sin130)

Repeat:
Sin rule 3: AF/Sin60 = 1/Sin50
Sin rule 4: DF/Sin10 = AF/Sin40 so DF = (Sin50/Sin50) * (Sin10/Sin40)

Now we have two sides of the inner triangle and the angle 50, so it’s cosine rule time

DE^2 = DF^2+EF^2-2*DF*EF*Cos50
that gives us the third side of that triangle so we can use the Sin rule one last time

Sin rule 5: DF/Sin50 = DE/Sinx

Hence Sinx = DE*Sin50/DF

back substitute for DE and DF and you should eventually get the solution above.

Simples?

Markie, you’ve assumed that line DE is at 90° to CB. Not sure how?

TiRed – Lovely. 😕

and then? 🙂

Blinded by maths here, and I did a few courses on projective geometry and linear algebra at uni. 😳

TiRed, I started off on my own equations earlier, thinking I was getting somewhere, then it all started to look very familiar. Looked back up at yours at the top of the page. then packed it in!

Actually there was a small typo in one line (STW is not the place to type algebra)

Here is the proof:

Denote the midpoint F and assume AB is 1

Then:
Sin rule 1: BF/Sin70 = 1/Sin50 so BF = Sin70/Sin50
Sin rule 2: BF/Sin30 = EF/Sin20 so EF = (Sin70/Sin50) * (Sin20/Sin30)

Repeat:
Sin rule 3: AF/Sin60 = 1/Sin50 so AF = Sin60/Sin50
Sin rule 4: DF/Sin10 = AF/Sin40 so DF = (Sin60/Sin50) * (Sin10/Sin40)

Now we have two sides of the inner triangle and the angle 50, so it’s cosine rule time

DE^2 = DF^2+EF^2-2*DF*EF*Cos50
that gives us the third side of that triangle so we can use the Sin rule one last time

Sin rule 5: DE/Sin50 = DF/Sinx

Hence Sinx = DF*Sin50/DE

back substitute for DE and DF and you should eventually get the solution above.

WAIT !!!!

It’s all a trap – if you look really carefully at the original image you can see the faint outline of a treadmill at the bottom

😯

This is a “well known” mathematical problem. I think it is called Langley’s triangle? You can prove it without triginomety by projecting additional points outside the triangle.

EDIT: Google says the original Langley problem was slightly different angles – but same concept. Here is the solution for that: http://www.gogeometry.com/LangleyProblem.html

That proof doesn’t work because the second isosceles triangle does not come about as the angles are different – trig works as follows using excel so I think the answer should be 20 deg.

BF 1.226681597
EF 0.839099631
AF 1.130515875
DF 0.305407289
DF^2 0.093273612
EF^2 0.704088191
DE^2 0.467911114
DE 0.684040287
Sin X 0.342020143
X 19.98060704

But really smart people came up with these

TiRed – your original had typos in two lines and it is simples for someone who said it couldn’t be solved – just kidding I had forgotten the Sine and Cosine rules.

x=35

Angles at plane E are therefore:
115 + 35 + 30 = 180

The other angle within the triangle x is in is 95

Angles at plane D are therefore:
45 + 95 + 40 = 180

The triangle CDE has the angles
C=20 which we already knew
D=45
E=115
Which add up to 180

mefty’s link to classic 80-80-20 triangle problems and variations gave me a little more satisfaction.

I managed to work one of them through to get the x=20 answer, but it was very tricky.

I used this method to find the classical framework (where the bottom angles are 50 and 60):
http://www.cut-the-knot.org/triangle/80-80-20/Classical5.shtml

Then applied this over the top:
http://www.cut-the-knot.org/triangle/80-80-20/60-70Sol1.shtml

It all felt like a leap away from where I was trying on my own though.

Nasty.

http://www.arbelos.co.uk/Papers/Triangle-problem.pdf