Math Is Fun Forum

Hi ktesla39;

No, it's not Python.

Mathematica has its own [proprietary] language, and you need to have the program (not free) to run the code.

I only have a basic knowledge of programming in it, as any Mathematica coder would quickly see if they saw my code.

I'm more familiar with BASIC, and was able to use some of that knowledge in Mathematica, which can run adjusted BASIC-like code, but with reduced efficiency compared to its own language.

Hi Bob;

Thanks for your method...I'll have to look at it tomorrow (bed now).

But before I go, here's my method:

Hi;

Whether I'm logged in or out, I get this message: "You do not have permission to access this page."

Hi Kerim;

Thanks for trying for a numerical proof of my graphical solution!

Xlnt! Excel Solver agrees with me on the AE+BG sum! That sum is also the length of FB, which is √28 ≈5.2915026222 (see my post #17 drawing).

However, your area (1.484614945) is nearly twice the smallest area that I got (≈0.742307489).

When G is anywhere on FB, AE+BG is always at its smallest sum (which I discovered by T&E), meaning that CEG's area can vary between its smallest (≈0.742307489) and its largest (≈6.9282032303).

In my previous post (#17), I showed how to accurately position G on FB at the location where CEG's area is smallest: draw a circle, radius √(12/5.25), on B, intersecting with FB at G. CEG's area grows if G moves along FB from that initial location, as the video in post #13 shows (G moving off FB does the same).

Well, that's odd!

Both images display on my desktop, my laptop, my TV (which has internet connection), and my mobile phone (yes, I finally got one of those clever bricks...sadly, I had to ditch my trusty 3G Samsung GT-B2710 when our 3G network shut down about 6 weeks ago).

Here they are from my Google Drive account, but in 'hide' boxes, because of their huge display size (I could've shrunk them, I suppose):

I hope you can see my Google Drive images!

Btw, can you see my images in posts #2 (3 pics), #5 & #9, and my videos when you click on their links in posts #7, #9 & #13?

I'm sorry, Kerim, but I can't follow much of your explanation.

The main reason for that is my low maths standard. My last year of school was 4th-year high, after which I went straight to work.

However, I enjoy doing maths puzzles, and have learnt some things along the way.

Good work on getting a solution!

Is my 'solution' (the one in the two shaded boxes in my previous post) anywhere near yours?

In the next day or two I'll have a closer look at what you did, and maybe I'll see something there to help me progress towards a solution too.

Thx - Phro

Ah, yes...that makes sense now.

What was also throwing me (apart from the 'overline' and 'm', neither of which I'd seen before), was your 0.5/2. I was using 1/4 (same thing).

Not to mention the multiplication dot, and writing the square longhand instead of with '^2'.

I gave up too quickly with trying to fathom your style out, but then returned to it later because I knew that you knew what you were doing, and that I should persevere.

Hi Bob;

I took some measurements when triangle CEG is smallest (from my observation) that don't occur at other CEG positions, and have shown them in the image below:

Btw, FB is the line along which G travels (from B to F), and is the "constant-slope trajectory" I referred to in post #4.

Here's a link to an animation video I made of CEG travelling its full distance along the x-axis: Triangle area. It uses the image details from post #5.
Note: The two bold-font boxes in the bottom right-hand corner are dynamic, continually updating as CEG moves along its path along the x-axis. The area box changes constantly...but the AE+BG box remains unchanged, confirming the accuracy of the FB trajectory slope.

I've had a good look to see if I could improve on my observation 'solution' technique, but no luck so far.

Edit: I've just realised that if my drawing actually is of the smallest CEG, then it's probably no coincidence that AEC & CGB are right angles, giving CE and CG their shortest route to AE and BG respectively. Any other angles there would stretch the size of CEG larger than what it is in the image.

Hi Bob;

I went 'P&O' discovery cruising again this morning, and saw that AEC and CGB are both right angles for minimum CEG.

I zoomed in at Geogebra's maximum level, adjusted CEG's current position a tiny bit, and the standard curved symbol for both angles magically changed into right-angle symbols.

As you can see from my image, I couldn't quite get the angles to actually be 90°, but that was close enough for Geogebra.

Maybe this info can help with what you're working on...

Hi johntom...and welcome to the forum!

Unless I don't fully understand the problem (or maybe there some info missing), I think I've found the smallest AE+BG...provided that either AE or BG (depending on the case - see images) can be length zero. If not, then the smallest AE+BG would be an infinitesimally small amount larger than zero!

Also, there seems to be an infinite number of areas possible for triangle CEG (all when AE+BG is smallest), ranging from √3 (see edit below) to 4√3, with the difference between adjacent sizes being immeasurably small.

I've drawn the following images based on AE or BG (as the case may be) having zero length. If zero length is not allowed, the change to the next larger size would be indiscernible (that is, if my conclusions are correct)!

The first image shows triangle CEG, with side CB as the base along the x axis.

Triangle CEG, with vertex C remaining on the x axis, then rotates -120° (ie, anticlockwise) at vertex C. 

The second image shows CEG's position at the end of the -120° rotation, with side EC having become the new base along the x axis:

During the -120° rotation, CEG changes size (first shrinking from its initial 2x2x2, then growing to its final 4x4x4), while maintaining constant smallest AE+BG throughout.

EDIT: The smallest CEG area I've found is 0.7423074889 (rounded to 10 decimal places), when CEG is about 41% into its 120° anticlockwise rotation (see image below).

These images could also have been drawn with CEG under the x axis.

Hi Bob,

Let FR extend to U for straight-line FRU, with FR = RU. Then RU = RS, so U is a point on the directrix!

Thanks, Bob...that opened my eyes to a new field for me.

So I've explored that further, and progressed to the drawings in my post #7.

For my method #3, I had initially just used the 1.28x² - 3.84x - 5.12y = -8 equation that appeared in the Algebra window for method #2, in which I used the Parabola tool.

However, that equation was way more horrible than I expected, given that I'd used simple plotting figures: Focus (1.5,2) and Directrix y=0! I've got no idea how to construct that equation, neither does it make any sense to me.

A good online tutorial showed me how to find the much simpler-looking equation (x-1.5)² = 4(y-1), and gave me some understanding on how to make such things (if they're simple enough, that is).

Hi JohnG;

Hope your experimenting is going well!

I did some of my own today, but I've hidden my results (a Geogebra drawing I did + explanatory notes). When you're ready to have a peek, just click on the hide box.

I hope my hidden info helps, but have a go at things yourself first!