Using optimization to setup a problem quickly

You are given a square with a point in it. It is situated at a distance of 1, 2 and 3 units respectively from vertices , and i.e. , and . What is angle ?

Instead of tediously calculating the values of and , we can use optimization to find a good enough approximation as a starting point.

Initial setup

These commands set up the loss function to minimize. We want the distances from to the vertices to match 1, 2, and 3. We do not know the length of each side of the square or the exact location of the point . We start with an initial guess:

• Initial guess for square side
• Initial guess for point p
• Create square with side length

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Draw lines from to each vertex:

• Join to
• Show distances
• Show angle

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Using gradient-descent

We can use autograd and gradient-descent to find the optimal values for and .

Calculate the loss

• Target distances
• Calculate distances from p to vertices
• Difference from target
• Square the differences
• Sum of squared differences
• Root mean square error

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Finally, define number of iterations and train using minimize command:

• Minimize the loss

In less than a second, it finds close enough values for and point .

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Verifying the result

Now let's highlight the key elements:

• Show distances

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Moving on to the solution

The key intent of this article is to showcase a good application of gradient descent to quickly set up the premise of a geometric problem. It is not to provide the exact solution to it (which can be found here). So instead of providing a complete solution, we will skip over a few steps.

Create a copy of points and and rotate them

• p-copy = \copy p
• p0-copy = \copy p0
• Create a triangle with these points and p1

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• Rotate the triangle about by
  • Rotate
  • Remove the triangle to show the lines
• Join and

It can be shown that these points are collinear, which leads to the solution: this angle is equal to .

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Numerical verification

• Show angle

It should be close enough to .