Math Is Fun Forum

Hi John;

Does your parabola show up in the Algebra window? If it does, then I don't understand why Geogebra doesn't recognise it.

How did you 'drive' it?

I can think of 3 methods (no doubt there are more) of drawing a parabola in Geogebra:

Method 1. Activate Geogebra's 'Show trace' feature on the apex of the isosceles triangle, and then draw the parabola by moving the triangle's point A along the directrix. However, no parabola entry appears in the Algebra window, indicating that Geogebra doesn't recognise it for interaction. The Intersect tool doesn't work on the parabola, and I think the trace drawing is just a non-interactive image.

This is what I got with the 'Show trace' parabola feature:

Method 2. Create the parabola with Geogebra's Parabola tool.

Method 3. Create the parabola by entering this equation into the Input box: (x-1.5)² = 4(y-1). Equation created from Focus (1.5,2), Directrix y = 0.

For methods 2 & 3, a parabola entry appears in the Algebra window (indicating interaction capability), and I used the Intersect tool to create intersection points F & G on the parabola (see below):

Oops! Must've been my turn to bumble!!

Of course n=1 has only 1 white triangle!

New Excel image below, updated to 100% accuracy!!

Hi Bob;

To me that means for Step 1, n = 1 (and for Step 2, n = 2; Step 3, n = 3; ...etc).

Looks like I got the Part c formula wrong in post #32, so here's my revised set of answers:

I drew it all up in Excel (see image), and it shows that the revised set works (I think!)

Btw, I'd expected to find the above equilateral triangle area formula on MIF, here - Area of Triangles - but it's not there.

Hi Bob;

Btw, I can get exactly the same image with my 'old' (2021) Geogebra Classic 5 (Classic 6 is 2023).

My earliest use of Geogebra was in 2012, but I don't know which version or how capable it was.

Hi Bob;

Expressing my method isn't as easy as I'd hoped, but here goes (and I'm sorry, I don't know how to do non-LaTeX subscripts here on MIF):

Rule: Term 1 (t1) = 4, and subsequent terms (t2, t3, etc) follow the [+1] incrementing form { t2 = t1 + t2's term position }, with arithmetic operators for the respective terms being assigned from the repeating sequence { + / * – } (starting with '+' for t2).

Here's my demonstration worksheet for the first 9 terms (note that t6's operator begins the first repeat of the operator sequence):

As you can see from the formulas in column H, calculations for terms after t1 are done entirely in column G. The other columns explain the calculations.

I thought the same about my initial efforts, but then started again with an Excel spreadsheet to help organise & lay out my thoughts.

Light dawned while working out relationships between the adjacent given terms in column G and the various possible ways in which to arrive at each next term. That led to discovering the use of the 4 arithmetic operators, followed by recognising that the group of 4 operators was a repeating sequence. 

I'd really like to know the rule to get that answer!

Hi Bob;

Here's my method...but with a different result from mathenjoyer's:

#    Term

The bold numbers are a sequence of 6 numbers from 2 to 7, and the operators used are the repeating sequence + ÷ × – .   

I say 'repeating sequence', as the 5th term repeats the '+' used in the 1st term....from which it follows that the 6th term would use the '÷' from the 2nd term.

Oculus8596 has missed the whole point of tackling these wonderfully challenging Project Euler Problems!

I've tidied it up a fair bit, but I'm sure it can still be improved further (or changed altogether!).

Haven't tried #45 yet, but succeeded with #40 after several clumsy goes at it.

Solved it in M, starting with some awful code that I didn't like, improved it a bit (still didn't like it), and eventually got something I'm quite pleased with.

Yes...I agree with your last comment.

Always learning!

I suspect that rounding might come into this...

I've never heard of error intervals, but rounding is uppermost in my mind atm because I'm working on a tricky problem where rounding is critical in finding solutions.