Calculate the area of a triangle

johntom · 2024-11-25 21:59:23

AB=6, point C is on AB, AC=4, make an equilateral triangle CEG with C as the vertex, connect AE and BG, when AE+BG is the smallest, find the area of triangle CEG.

phrontister · 2024-11-28 00:34:40

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.

Last edited by phrontister (2024-11-28 21:58:01)

Bob · 2024-11-28 21:49:46

I'm assuming you used Geogebra and moved a point around until you reached a minimum. I did that using Sketchpad and got this:

I feel that there ought to be an analytical way to do this, so I'm working on that.

Bob

phrontister · 2024-11-29 02:16:39

Bob wrote:

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...

Last edited by phrontister (2024-12-02 12:58:02)

phrontister · 2024-11-29 11:58:39

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...

KerimF · 2024-11-29 12:33:14

Although I can't follow you, I just wonder if this problem may have more than one solution.

phrontister · 2024-11-29 18:53:58

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!

KerimF · 2024-11-29 20:28:39

Thank you, Phrontister, for the interesting explanation.

phrontister · 2024-11-30 01:08:44

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.

Last edited by phrontister (2024-12-02 10:33:21)

phrontister · 2024-12-02 13:46:28

Hi Bob;

phrontister (post #4) wrote:

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!

should be

Here's my formula (with a = AE):

Last edited by phrontister (2024-12-02 15:59:14)

Bob · Yesterday 01:28:01

Actually it's worse than that. I should have calculated the area of CEG. So that calculation is completely wrong.

It should be:

0.5 times EC^2 times root(3)/2 which for my diagram comes out as 0.88

Still working on an analytic solution.

Bob

phrontister · Yesterday 03:07:45

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.

Math Is Fun Forum

#1 2024-11-25 21:59:23

Calculate the area of a triangle

#2 2024-11-28 00:34:40

Re: Calculate the area of a triangle

#3 2024-11-28 21:49:46

Re: Calculate the area of a triangle

#4 2024-11-29 02:16:39

Re: Calculate the area of a triangle

#5 2024-11-29 11:58:39

Re: Calculate the area of a triangle

#6 2024-11-29 12:33:14

Re: Calculate the area of a triangle

#7 2024-11-29 18:53:58

Re: Calculate the area of a triangle

#8 2024-11-29 20:28:39

Re: Calculate the area of a triangle

#9 2024-11-30 01:08:44

Re: Calculate the area of a triangle

#10 2024-12-02 13:46:28

Re: Calculate the area of a triangle

#11 Yesterday 01:28:01

Re: Calculate the area of a triangle

#12 Yesterday 03:07:45

Re: Calculate the area of a triangle

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