Geometry Puzzle... Simple Solution?

[(X)]

I had a dream of a geometry puzzle that I thought might be solvable "in my head".

Now, I am not so sure.  Can you think of a solution to this puzzle?

Demo Graph Puzzle.G32 (1.28 KB)

[dragonjim]

Sadly, geometry (and mechanics) were not my strong suit at school - how I passed the exam is still a mystery...

However, from what I can still remember from all those years ago, you may want to treat the problem as a mechanics problem:

You have two objects, one travelling in a circular trajectory - (x^2 + y^2) = r^2 assuming the centre of the circle is (0,0) and, because your radius is 1, you can simplify this further to x^2 + y^2 = 1 - and the other travelling along a straight line y = 1 - 2x (the line you have drawn): where will they meet?

So if y = 1 - 2x, then replacing x for y in the circular trajectory would give x^2 + (1 - 2x)^2 = 1 which gives you..

=> x^2 + (1 - 2x - 2x + 4x^2) = 1

=> 5x^2 - 4x + 1 = 1

=> 5x^2 - 4x = 0

=> x(5x - 4) = 0

So, where 5x - 4 = 0, x must be 4/5 or 0.8; put that back into y = 1 - 2x then that gives you y = 1 - 1.6 => y = 0.6

Assuming that what I have done above is correct, the difficult bit will now be translating that into code...

Edit: And from that equation, the other cross over point will be where x = 0, so where y = 1.

[(X)]

Apr 19, 2024 10:41:58 GMT 1 dragonjim said:

Sadly, geometry (and mechanics) were not my strong suit at school - how I passed the exam is still a mystery...

However, from what I can still remember from all those years ago, you may want to treat the problem as a mechanics problem:

You have two objects, one travelling in a circular trajectory - (x^2 + y^2) = r^2 assuming the centre of the circle is (0,0) and, because your radius is 1, you can simplify this further to x^2 + y^2 = 1 - and the other travelling along a straight line y = 1 - 2x (the line you have drawn): where will they meet?

So if y = 1 - 2x, then replacing x for y in the circular trajectory would give x^2 + (1 - 2x)^2 = 1 which gives you..

=> x^2 + (1 - 2x - 2x + 4x^2) = 1

=> 5x^2 - 4x + 1 = 1

=> 5x^2 - 4x = 0

=> x(5x - 4) = 0

So, where 5x - 4 = 0, x must be 4/5 or 0.8; put that back into y = 1 - 2x then that gives you y = 1 - 1.6 => y = 0.6

Assuming that what I have done above is correct, the difficult bit will now be translating that into code...

Edit: And from that equation, the other cross over point will be where x = 0, so where y = 1.

This is great!  A small missing minus sign typo: y = 1 - 1.6 => y = -0.6

[(X)]

This equation solver website seems to agree:

www.symbolab.com/solver/simultaneous-equations-calculator/x%5E%7B2%7D%2By%5E%7B2%7D%3D1%2Cy%3D-2x%2B1?or=input

This is amazing to me since the graph looks a lot like the one in my example.

[(X)]

Here is version 2 of the Graph Puzzle Demo which uses a function that employs the Quadratic formula to return the roots of the equation of the form: ax² + bx + c = 0

Demo Graph Puzzle V2.G32 (4.38 KB)

Function SolveQuadratic(a As Double, b As Double, c As Double) As Bool
 '
 ' For an equation of the form:
 '
 '     ax² + bx + c = 0
 '
 ' There are 2 possible solutions or roots for x as noted in: root(1) and root(2).
 ' It is worth noting that we must allow for the possiblility that the discriminant
 ' may be negative, thus a complex number
 '
 Try
   Global roots(1 To 2) As Double
   Dim discriminant As Double
   
   ' Calculate the discriminant
   discriminant = b ^ 2 - 4 * a * c
   
   ' Check if the roots are real or complex
   If discriminant >= 0 Then
     ' Real roots
     roots(1) = (-b + Sqr(discriminant)) / (2 * a)
     roots(2) = (-b - Sqr(discriminant)) / (2 * a)
   Else
     ' Complex roots
     roots(1) = -b / (2 * a)
     roots(2) = Sqr(-discriminant) / (2 * a)
   End If
   Return True
 Catch
   Return False
 EndCatch
End Function

[dragonjim]

This is great!  A small missing minus sign typo: y = 1 - 1.6 => y = -0.6

Now you see why it is a miracle I passed the exam...

[(X)]

Isn't it surprising that the result should be an exact fractions 0.6 = 3/5 = 6/10 and 0.8 = 4/5 = 8/10 ?
I was rather expecting some irrational number like Pi or the Golden ratio to pop-up.
An exact number... Huh!

[(X)]

This video was a suggested as a link for me to look at when I started Mozilla browser...

    youtu.be/ZBalWWHYFQc

A Different Way to Solve Quadratic Equations:

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