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Do you study about computer science?
No, not at all.
I just love doing number puzzles, with or without computer aided software.
For me, tackling a challenging number puzzle is simply a nice way to relax.
Hi ktesla39;
I revised my code to be much more Mathematica-like:
Code:
Length[Cases[Table[Divisors[i++],{i,999}],{_,_,_}]]
Output:
11
Hi ktesla39;
By the way, which language did you use in that program? It looks like python, is it?
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 ktesla39,
I'll try python or JS automation to solve the question by dividing all integers from 1-999.
Good idea...I tried it with Mathematica, and here's my code and solution:
It tested all integers in that range, and for each integer that had only 3 divisors it printed the solution number, the integer number, and the divisors. When done, it gave the total number of solutions.
Edit:
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 Bob;
Yes, I got that answer too...with a 27-character Mathematica formula using Sum & Mod that also runs in WolframAlpha.
Hi;
Whether I'm logged in or out, I get this message: "You do not have permission to access this page."
Hi Kerim;
...there are more than one value for the triangle area that satisfies the given condition.
Reading the question on the OP, one has the impression that there is just one.
Yes, I have the same impression.
That's why I wrote these:
Post #2 - "or maybe there's some info missing"
Post #4 - "the OP's problem doesn't say anything about the sought area size"
Post #7 - "I would've thought that, ideally, there'd only be one location for the smallest AE+BG, and that we'd only need to report the area of CEG at that location.
So maybe the OP's wording in post #1 is missing some information we need?"
The only satisfactory goal (for when AE+BG is smallest) has to be to find either:
(a) the largest CEG area; or
(b) the smallest CEG area.
The number of different CEG areas in between (a) & (b) is infinite.
I'm opting for (b), as (a) is too easy. But maybe it's supposed to be an easy puzzle?!
Unfortunately, the OP hasn't returned to the thread to clarify...
Hi Kerim;
Thanks for trying for a numerical proof of my graphical solution!
By using the solver of Excel, I got:
AE+BG= 5.291502622
Area = 1.484614945
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).
I couldn't see your two images:
post #13, https://i.imgur.com/bs1yfhul.jpg
post #17, https://i.imgur.com/4BS9AQTl.jpg
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?
Here's an update to my post #9, showing the interesting length relationships a bit differently.
It includes a formula I hadn't thought of before:
That makes it easy to position G accurately on FB: ie, draw a circle on B, with radius a, and use the Intersect tool on FB and the circle to create G on FB at the exact right spot (so no more moving G along FB in search of the exact right spot!)
And hey presto, everything falls easily into place: the right angles, some other angles, the GB-length relationships, the smallest AE+BG and smallest CEG area...!
But I'm as close to finding a non-graphical solution as ever!
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
Hi Bob;
Here's an update to my post #9...this time showing formulas:
Video link: Triangle area (smallest) - formulas
The figures in the shaded boxes are governed by the formulas.
But...I still can't prove
1. that the smallest AE+BG only occurs with G on FB (which I think it does); and
2. that the smallest area of triangle CEG only occurs as shown in the image (which I think it does).
Edit: Just spotted something: triangles FGC and AEC are congruent. I wonder if that helps with anything...
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;
Still trying to work out what you did...
Have worked it out...but it looks like you were testing me by deliberately misplacing the decimal point!
Here's my formula (with a = AE):
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 Kerim;
Although I can't follow you...
Here's a link to a little video I just made that I hope will help: Triangle area.
It shows one of the paths that equilateral triangle CEG can take. It's the path that I've described in my posts, and as far as I can tell is the only one where the required smallest AE+BG occurs.
...I just wonder if this problem may have more than one solution.
The path I've described maintains the smallest AE+BG for the whole path, not just particular locations of the triangle along the path. That gives innumerable solutions differing in size!
In my post #4, I wrote:
Btw, the OP's problem doesn't say anything about the sought area size. We're trying for the smallest atm, and I've mentioned that there's a myriad of sizes between the smallest and largest, but maybe the OP wants the largest?
I would've thought that, ideally, there'd only be one location for the smallest AE+BG, and that we'd only need to report the area of CEG at that location.
So maybe the OP's wording in post #1 is missing some information we need?
Or...I've simply misunderstood/overlooked/misconstrued something or other!
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...
I'm assuming you used Geogebra and moved a point around until you reached a minimum.
True!
I couldn't come up with anything I'd call the proper approach, so I used the 'play and observe' (P&O) technique that works for me sometimes.
So...I parked the small 2x2x2 CEG on CB, and moved G up-left from its starting position at B, along a constant-slope trajectory that produced a constant smallest AE+BG during CEG's 120° anticlockwise rotation at C.
CEG shrank immediately after G departed B, and then grew into 4x4x4 when E reached A.
I found my minimum CEG by moving G slowly along the trajectory line...on a very zoomed-in screen.
Btw, the OP's problem doesn't say anything about the sought area size. We're trying for the smallest atm, and I've mentioned that there's a myriad of sizes between the smallest and largest, but maybe the OP wants the largest? Well, I think the largest is this (as also shown in my second image in post #2):
That would turn this puzzle into quite an easy one!
Still trying to work out what you did...
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,
Left part of diagram...
...So R is unique and FRS is perpendicular to the directrix.
Second part of diagram...
...So FRS is a line of symmetry for the parabola.
Redrawing the diagram so that FRS is straight......places F, T and U in their correct positions.
Here's a slow-motion video of that happening: FRS straightening.
Hi Bob,
Let T be the midpoint of FS. Then FT = TS so T is a point on the parabola
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!
I really like the first part, the section prior to your further exploration.
The logic is easy to understand, and, unlike the online approach that I used (that results in the same equation as yours), you've used Pythagoras, which I already know.
The online method (though simple enough), requires learning & memorising some additional techniques...which I can do without!
I'll explore the other part later, maybe tomorrow.
Phro. y = x^2 is a simple example of a parabola.
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 John;
Does your parabola show up in the Algebra window? If it does, then I don't understand why Geogebra doesn't recognise it.
The point on the directrix then drove a Locus which gave me a parabola.
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):
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!