## Acute Geometry Problem

I saw a plea for help on Twitter from Astronomer Bryan Gaensler who is stuck with his son’s homework.

So please give him a hand by solving this to find a, b and c.

Your time starts now.

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December 3, 2018 at 2:08 pm

a=1, b=15, c=22.5

December 3, 2018 at 2:24 pm

To see that there is not a unique answer, imagine a rod of length 7.5 units, attached by a hinge to a rod of length 2 units, which is itself attached at its other end by a hinge to a rod of length 3 units. These hinges are such that the three rods are constrained to a plane. Suppose that the first-mentioned hinge is at the apex of the diagram, and the plane of the rods is the plane of the paper without loss of generality. Then the free end of the 3-unit rod can be slid along the 7.5-unit rod as the hinge angles vary, while a line can be drawn through the free end of the 7.5-unit rod parallel to the 3-unit rod. This line is side c of the major triangle.

December 3, 2018 at 9:07 pm

I think there are infinitely many solutions with 1<a<5.

December 3, 2018 at 9:28 pm

I agree, a can be anything that the basic triangle inequality allows. I think it’s actually a nice question, assuming it was intentional.

December 3, 2018 at 5:10 pm

The c and the 3 are parallel, so 7.5/a=b/2=c/3. 2 equations and 3 unknowns so not suluble. However, if the angle at the top is a right angle (which it appears to be but is not so labelled), Pythagoras comes to our aid; a=root(5); b= 15/root(5); c=22.5/root(5)

December 4, 2018 at 7:45 am

Yeah, surely it’s intended to be a right angle.

December 4, 2018 at 9:56 am

I don’t think it can be taken as obvious that the angle at the top is exactly a right angle. But I do think it’s clear that the angle between b and c is to be taken as less than a right angle. In that case, the cosine rule allows 1 < a < sqrt(13), combined with b=15/a and c=3b/2.

December 4, 2018 at 6:58 pm

We could print it and measure the angles experimentally… The angle at the top is clearly larger than 90 degrees. Ah, theorists…!

December 4, 2018 at 8:38 pm

It looks like you have to solve a cubic expression. Do it numerically.

December 5, 2018 at 2:02 pm

I quickly did it numerically and got a=2.02183658 b=7.41899723 and c= 11.12849584.

Seems a tricky problem to give in school, though.

December 5, 2018 at 2:22 pm

If anyone cares, in Python the solution is:

def f(x):

return np.array([x[1]**2-x[3]**2-7.5**2+(x[2]-x[3])**2,2**2-x[4]**2-x[0]**2+(3.-x[4])**2,x[3]/x[4]-(x[2]-x[3])/(3.-x[4]),7.5/x[0]-x[1]/2.,x[3]/x[4]-x[1]/2.])

guess=np.array([3.,3.,5.,1.,1.])

roots_solve=scipy.optimize.fsolve(f,guess)

print(roots_solve)

[ 2.02183658 7.41899723 11.12849584 5.50995127 1.48536281]

December 5, 2018 at 5:08 pm

That would be Monty Python?

December 6, 2018 at 9:43 am

There is actually a Monte-Carlo code using python which is called Monte Python. ðŸ™‚

December 6, 2018 at 9:47 am

Yes, the name does come from Monty Python.

December 6, 2018 at 11:02 am

Just to be clear, the programming language is named after

Monty Python’s Flying Circus.December 6, 2018 at 11:56 am

Which also provided the inspiration for Monty Python’s Flying C++

December 6, 2018 at 12:22 pm

Actually, C++ code is far more absurd than anything in Monty Python’s Flying Circus. ðŸ˜€

December 6, 2018 at 12:23 pm

Most of my code would be graded C–