Naked Science Forum
Non Life Sciences => Geek Speak => Topic started by: syhprum on 04/06/2013 14:24:00

This puzzle appears on Facebook adverts as part of a scam with the comment that 97% of the viewers cannot work out the angles a and c for the moment this includes me.
I am assuming the sides AB and AC are equal
I feel that simultaneous equations are needed or is there a more simple way ?

... I feel that simultaneous equations are needed or is there a more simple way ?
I think it is down to simultaneous equations (https://en.wikipedia.org/wiki/System_of_linear_equations#Elimination_of_variables).
I am assuming the sides AB and AC are equal
IMO the fact the outer triangle is isosceles isn't helpful.
The values below seem to fit numerically ...
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a = 120^{o}
c = 100^{o}
x = 60^{o}
y = 20^{o}
but they don't match the geometry of the diagram where angle y>x

I omitted to label the angles that you have labelled x and y which I should have done, the assumption that it is Isosceles makes the calculations of the angles at B and C possible.
I am tempted to cheat and gun up Mathematica on simultaneous equations
PS Y+X must equal 80°

Just measured the angles on the diagram and they are quite accurate : within +/ 2^{o},
So the solution must have y>x , (x approx 10^{o}), (so my answer, which has x>y, must be wrong [:I] )

It's a trick.
The three angles at A, B and C have to add to 180 degrees
B is 50, C is 50 so A must be 80
i.e. it's 180(50+50)
But x+y = 80 and y =80 so x = zero.
there is no second triangle a, c, C.

Y=70° X=10° c= 150° a=70 all done by guess work not proper maths it would not get marks in an exam bring on a maths expert and let us have a formal solution.

Facebook, may in fact be right.
97% of the viewers may be CORRECT!!!
and 3% of the viewers may be WRONG!!!
I came up with 4 equations, 4 unknowns.
1) a+c = 360  140 = 220 (360° around the central part)
2) x+c = 180  20 = 160 (180° around right most triangle)
3) y+a = 180  40 = 140 (180° around left most triangle)
4) x+y = 80 (80°is given for angle A, also can calculate with A+B+C = 180)
Can anybody see a fifth independent equation?
So, solving my equations:
1) a+c=220
2) x+c=160
3) y+a=140
4) x+y=80
Subtract (4) from (2), and one gets:
1) a+c=220
24) cy=80
3) y+a=140
Add (24) + (3), and one gets:
1) a+c=220
24+3) a+c=220
SO,
My 4 equations and 4 unknowns were not independent. And trying to solve them, I end up with 1 equation and 2 unknowns.
All angles have to be positive.
So, if you take the last equation (4), x+y = 80°
You can choose any x & y < 80°, and solve for a & c.
Unless there is another equation that I'm missing.
For example:
x = 1
y = 79
c = 159
a = 61
or
y = 1
x = 79
c = 81
a = 139
And, anything in between.
Thus, there is no unique answer.

BC,
I think you misinterpreted the original diagram.
Note the semicircle going from lines AB to line AC, thus indicating the entire angle A = 80°,
and x+y=80°
x must be nonzero, otherwise, one could not have the 20° component at C.

Whew, I'm rusty on my trig.
I should be able to figure out the lengths of sides for the lower triangle.
Then, from that determine the lengths of sides of the other two triangles.

I think that calculation of the length of lines or distance of the central point from the long sides should pin down an unique solution.
as you point out x cannot be zero.
I should have kept my school Tech drawing kit!

I started off thinking it was meant to be a tetrahedron, but no  it has to be flat. If you name the point where c and a meet as point D, you can then draw in a vertical line down there to a point E on BC. Armed with that, you can start to assign lengths for all sides of the triangle BCD based on simple soh cah toa stuff, and then you can apply more to the triangle ABC in the same manner. That will show you if BA and BD are the same length. Whatever the case though, it will certainly force a single answer out.
Edit:
Starting point: BD=1
BE = cos 10
DE = sin 10
EC = DE * tan 60
BC = EC + BE
BF = 0.5 * BC (F is the point half way along BC)
BA = BF / cos 40
So BA appears to be about 0.85 the length of BD, if I've crunched the numbers correctly. Have I got it right thus far?

Ok,
I pounded a bit of brute force through it.
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The outer triangle is an isosceles triangle, so one can set the outer edges to length 1.
From this one can calculate the length of the base.
Then I made the inner triangles into right triangles, and worked from the little triangle bottom right (306090 triangle), and worked back around counter clockwise, up to where I could solve for the triangle with Y in it.
I ended up with y=70° (to the numerical precision of my calculations).
Then, from my equations above, one can solve the rest.
y=70°
x=10°
a=70°
c=150°
These angles seem a little bit bizarre, as if there is some geometric voodoo that I have forgotten.
All the numbers seem plausible based on the diagram, but I could always have made a mistake somewhere.

Really hearty congratulations on a formal solution these are the same values that I produced by
successive approximation guesswork.
According to the scam you are now eligible to pay out £4.50 a week to receive more puzzles

Ah, found my silly error  the last bit should have been BA = BF / cos 50 and not BA = BF / cos 40. Yes, BA=BD.

Ok,
Sorry about editing my diagram a couple of times, I had managed to let a layer shift, then added some inconsequential parenthesis to one of the equations.
Looking back at my diagram above.
I calculated that the length of d/sin(10) = `1 (length from B to middle).
This made my left most triangle into an isosceles triangle.
Thus, the angles a = y, and one easily calculates the angle to be 70°
Anyway, that simplifies the calculations a bit, but I've still got some thinking on why this all worked out like it did.
No idea why I need to pay £4.50 for new puzzles, when all I need is Syhprum to scramble my brain, and remind me that I'm no longer taking 11^{th} grade math!!! [xx(]

No SOH CAH TOA for us we had to memorise Peter Has (perpendicular over hypotenuse) Been Here (base over hypotenuse) Playing Billiards (perpendicular over base) which seems to have worked Knowing it after 70+ years

For me, the SOH, CAH, TOA mnemonic has been sufficient to remember my sin, cos, and tan relations longterm.
But, it is infrequent enough that I do trig problems like this that I have to work a bit to remember how to use all the equations. And, it never hurts to write my mnemonic down as I'm pulling out the equations.
I am still struggling a bit on whether or not one could have solved this "puzzle" without using the sin/cos/tan trig functions, or perhaps using them only superficially.
As far as math stuff. I do wonder about future generations. There is so much that one can look up online. I missed the sliderules, and did very little table lookup and interpolation, but landed early in the calculator generation.
We didn't have fancy graphing calculators, or calculators that could symbolically solve/simplify equations. So, one learned things like a quadratic makes a U, a 3^{rd} order makes a sideways S, a 4^{th }order makes a W, etc.
Does having a calculator do it all for you make it easier, or harder? I suppose half of the battle would be learning how to use the calculator.

I grew up in the era when we had a complex money system that made domestic financial calculations difficult and the base unit the pound was very large about half my fathers weekly wage when I was a boy.
I did not like slide rules but found log tables very useful for interest calculations etc.
I do not think this calculation could be done without the use of trigonometry as the key to the whole thing is the bottom triangle about which we have complete information.
What I like about this modern era are spell checkers which have made me much more confident about writing previously being rather dyslectic.

In my opinion there is a fairly simple solution. It involves use of the sin rule  Y/sin y = Z/sin z
To make it explicit:
Triangle ABC is isosceles, so AB = AC
In the triangle ABX  AX/sin 40° = AB/sin a
In the triangle ACX  AX/sin 20° = AC/sin c
therefore  sin a/sin c = sin 40°/sin 20° = 2 cos 20° (double angle formula)
moreover a + c = 220° (angles in a full circle)
This provides 2 equations in 2 unknowns.

The last few lines of my last post left me with a bit of work to do:
therefore  sin a/sin c = sin 40°/sin 20° = 2 cos 20° (double angle formula)
moreover a + c = 220° (angles in a full circle)
This provides 2 equations in 2 unknowns.
Substituting (220° c) for a leads to
(sin 220° cos c cos 220°sin c) /sin c = 2 cos 20°
cot c sin 220° cos 220° = 2 cos 20°
cot c = (cos 220° + 2 cos 20°)/sin 220° = (cos 40° 2 cos 20°)/sin 40°
From tables I find that cos 40° = 0.76604444 cos 20° = 0.93969262 sin 40° = 0.64278761
Therefore cot c = (0.76604444 2 * 0.93969262) / 0.64278761 = 1.7320508
And therefore c = 150° and a = 70°
*****
The fact that the solutions prove to be such convenient round numbers convinces me that there should be an even simpler way of tackling this with geometric constructions rather than trig equations. More anon maybe?

Whew,
That was much simpler than what I had ended up with using the SOH, CAH, TOA & building right triangles.
As with problems in many math books, the solutions often work out to be whole numbers. While the left most triangle is an isosceles triangle, the bottom one, and the right most ones are not.
In my solution, I could have stopped once I discovered that it was an isosceles triangle, but proving that without some advanced trig functions might have been illusive.
And, of course, a general solution would need to be independent from such constructions. Perhaps the next step would be to scramble the triangles so that the given angles are whole numbers, and the calculated angles are not.

Update: I have tried to generalize the problem to find out if there might be a geometric method.
The result is that the angles a & c are not in whole number of degrees for any of the other angle inputs (which are in whole numbers of degrees) for this particular generalization.
The conclusion is that there is no particular geometric construction that could help to solve this problem except possibly for the particular case.
Finding the appropriate geometric construction may not be possible, and would certainly involve more time and labour than my trigonometric solution, especially because the trigonometric solution
can be applied to more general modifications of the problem
is well adapted to Excel spreadsheet