Puzzle challenge...

[(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:

No more guessing while factoring quadratics! Po-Shen Loh has been thinking about how to explain school math concepts in more thoughtful and interesting ways. Professor Loh creates inspiring lessons for Daily Challenge, the first and only online math class that captivates: live.poshenloh.com/