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Solve geometric quantities using GeometricScene

Xiangyang Zhou
Posted 2 days ago

I constructed the following GeometricScene with parameters {s,r}. The scene consists of a circle of radius r, and an equilateral triangle {a,b,c} inscribed in the circle. s being the length of one side of the triangle. I want to calculate the value of s, dependent on the value of r.

scene=GeometricScene[{{a,b,c,o},{s,r}},{CircleThrough[{a,b,c},o,r],GeometricAssertion[Triangle[{a,b,c}],"Equilateral"],s==TriangleMeasurement[{a,b,c},"Perimeter"]}];
RandomInstance[scene]
s/.%["Quantities"]

It returns 4.41219 as the value of s. I'm guessing behind the scenes, Mathematica chose a value for r and use that value for s. Is there any way to get an analytical solution of s dependent on r?
POSTED BY: Xiangyang Zhou
8 Replies
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"Quantities" gave the values for the particular instance of the scene found by RandomInstance. You can use GeometricSolveValues on the abstract scene instead:

scene = GeometricScene[{{a, b, c, o}, {s, 
     r}}, {CircleThrough[{a, b, c}, o, r], 
    GeometricAssertion[Triangle[{a, b, c}], "Equilateral"], 
    s == TriangleMeasurement[{a, b, c}, "Perimeter"]}];

GeometricSolveValues[scene, s]

{ConditionalExpression[s, 
  Sqrt[(Indexed[a, {1}] - Indexed[b, {1}])^2 + (Indexed[a, {2}] - 
        Indexed[b, {2}])^2] + 
     Sqrt[(Indexed[a, {1}] - Indexed[c, {1}])^2 + (Indexed[a, {2}] - 
        Indexed[c, {2}])^2] + 
     Sqrt[(Indexed[b, {1}] - Indexed[c, {1}])^2 + (Indexed[b, {2}] - 
        Indexed[c, {2}])^2] == 
    s && (Indexed[a, {1}] - Indexed[b, {1}])^2 + (Indexed[a, {2}] - 
       Indexed[b, {2}])^2 == (Indexed[b, {1}] - 
       Indexed[c, {1}])^2 + (Indexed[b, {2}] - 
       Indexed[c, {2}])^2 == (Indexed[a, {1}] - 
       Indexed[c, {1}])^2 + (Indexed[a, {2}] - 
       Indexed[c, {2}])^2 && (Indexed[a, {1}] - 
       Indexed[o, {1}])^2 + (Indexed[a, {2}] - 
       Indexed[o, {2}])^2 == (Indexed[b, {1}] - 
       Indexed[o, {1}])^2 + (Indexed[b, {2}] - 
       Indexed[o, {2}])^2 == (Indexed[c, {1}] - 
       Indexed[o, {1}])^2 + (Indexed[c, {2}] - Indexed[o, {2}])^2 == 
    r^2 && r > 0]}

GeometricSolveValues output for abstract scene
POSTED BY: Ian Ford

Hi Ian,

Thanks for helping out here. If you Eliminate all the indexed variables from the returned equations as Gianluca suggested, you'll get this equation: 81 r^4 s^6 - 30 r^2 s^8 + s^10 == 0, and solve for s will result in the following solutions:

{{s->0},{s->0},{s->0},{s->0},{s->0},{s->0},{s->-3 SqrtBox["3"] r},{s->-SqrtBox["3"] r},{s->SqrtBox["3"] r},{s->3 SqrtBox["3"] r}}

It returns two non-zero positive solutions: 3 Sqrt[3] r and Sqrt[3] r. What puzzles me is the solution of Sqrt[3] r. It is a solution to the equation obtained from Eliminate, but it isn't a solution to the geometric problem. I expect 3 Sqrt[3] r to be the only non-negative solution. What're your thoughts?
POSTED BY: Xiangyang Zhou

I am no expert, but maybe the culprit is the first equation, that contains three square roots. Perhaps Eliminate transforms it into a polynomial equation, and in doing so it introduces spurious solutions.
POSTED BY: Gianluca Gorni

You can check that s == r*Sqrt[3] is not a solution this way:

sols = Reduce[Most[eqs] && s == r*Sqrt[3]] // Reduce
Implies[sols, r == 0] // Reduce
POSTED BY: Gianluca Gorni

r*Sqrt[3] is a solution of the system returned by Eliminate[Most[eqs], Cases[eqs, _Indexed, All]]. This is what puzzles me, it is a solution to the eqs from the scene, yet it is NOT a solution to the scene itself. The expected solution is 3 Sqrt[3] r.

POSTED BY: Xiangyang Zhou

Eliminate gives necessary conditions, not always sufficient. Try this:

GeometricScene[{{a, b, c, o}, {s, r}}, {CircleThrough[{a, b, c}, o, r],
    GeometricAssertion[Triangle[{a, b, c}], "Equilateral"], 
   s == TriangleMeasurement[{a, b, c}, "Perimeter"], 
   s != 3 Sqrt[3] r}]["Conclusions"]
POSTED BY: Gianluca Gorni

This certainly got me to the next step. However, it gives the following solutions,

{{s->0},{s->0},{s->0},{s->0},{s->0},{s->0},{s->-3 SqrtBox["3"] r},{s->-SqrtBox["3"] r},{s->SqrtBox["3"] r},{s->3 SqrtBox["3"] r}}

What puzzels me is that Sqrt[3]*r is not a solution. If I add that to the scene, Mathematica will have a hard time to find an instance.

scene=GeometricScene[{{a,b,c,o},{s,r}},{CircleThrough[{a,b,c},o,r],GeometricAssertion[Triangle[{a,b,c}],"Equilateral"],s==TriangleMeasurement[{a,b,c},"Perimeter"],s==Sqrt[3] r}];
RandomInstance[scene]

Was it a bug or something's wrong in my scene? Thank you!
POSTED BY: Xiangyang Zhou

This may be a start:

eqs = scene["AlgebraicFormulation"]
Eliminate[Most[eqs], Cases[eqs, _Indexed, All]]
Solve[%, r]
POSTED BY: Gianluca Gorni
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