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 · 2024-12-03 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 · 2024-12-03 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.

phrontister · 2024-12-05 01:14:37

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...

Last edited by phrontister (2024-12-07 00:20:10)

KerimF · 2024-12-06 21:32:28

I don't think our dear guest, Johntom, can solve his interesting exercise, analytically.

Here is what I did:
A(0.0)
C(4,0)
B(6,0)
E(m,n)
G(p,q)

Therefore:
AE = sqrt(m^2 + n^2)
BG = sqrt[(6-p)^2 + q^2]

EC^2 = (4-m)^2 + n^2
CG^2 = (p-4)^2 + q^2
EG^2 = (p-m)^2 + (q-n)^2

equ_1: EG^2 = EC^2
(p-m)^ + (q-n)^2 = (4-m)^2 + n^2
p^2 - 2pm + m^2 + q^2 - 2qn + n^2 = 16 - 8m + m^2 + n^2
res_1: p^2 - 2pm + q^2 - 2qn = 16 -8m

equ_2: EG^2 = CG^2
(p-m)^ + (q-n)^2 = (p-4)^2 + q^2
p^2 - 2pm + m^2 + q^2 - 2qn + n^2 = p^2 - 8p + 16 + q^2
- 2pm + m^2 - 2qn + n^2 = - 8p + 16
res_2: 2pm - m^2 + 2qn - n^2 = 8p - 16

From res_1 and res_2
equ_3: p^2 + q^2 = 8p - 8m
res_3: m = p - (p^2 + q^2) / 8

From res_1 and equ_3:
p^2 + q^2 = 16 -8m + 2pm + 2qn
8p - 8m = 16 -8m + 2pm + 2qn
8p = 16 + 2pm + 2qn
2pm = 8p - 16 - 2qn
pm = 4p - 8 - qn
res_4: m = (4p - 8 - qn) / p

From res_3 and res_4:
p - (p^2 + q^2) / 8 = (4p - 8 - qn) / p
p - (p^2 + q^2) / 8 = 4 - 8/p - qn/p
qn/p = - p + (p^2 + q^2) / 8 + 4 - 8/p
qn = - p^2 + p*(p^2 + q^2) / 8 + 4p - 8
res_5: n = [ - p^2 + p*(p^2 + q^2) / 8 + 4p - 8 ] / q

So far, if I didn't make mistakes, we have m and n in function of p and q.
So, we need another equation which is actually the trick of this exercise:

Let us add F on EG, so that ACF is a right angle. We have now the angles ECF + FCG = pi/3 (of the equilateral triangle ECG)
FCG = pi/3 - ECF
tan(ECF) = (4 -m)/n
tan(FCG) = (p - 4)/q

We know: tan(A-B) = [tan(A)-tan(B)] / [1+tan(A)*tan(B)]
tan(FCG) = tan(pi/3 - ECF) = [ tan(pi/3) - tan(ECF) ] / [ 1 + tan(pi/3)*tan(ECF) ]
tan(FCG) = [ sqrt(3) - tan(ECF) ] / [ 1 + sqrt(3)*tan(ECF) ]
Therefore:
equ_6: (p - 4)/q = [ sqrt(3) - (4-m)/n ] / [ 1 + sqrt(3)*(4-m)/n ]

If we replace m and n from res_3 and res_5 in equ_6, we get an equation with two unknowns only, p and q.
In other words, we can get p=f(q) or q=f(p)

By replacing p, m and n (which are known in function of q) in the function SUM(q)
SUM(q) = AE + GB = sqrt(m^2 + n^2) + sqrt[(6-p)^2 + q^2]
SUM(q) is minimum when the derivative SUM'(q) = 0

Et Voila.

phrontister · 2024-12-07 01:18:19

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

KerimF · 2024-12-07 03:48:58

phrontister wrote:

Good work on getting a solution!

I didn't get a numerical solution
All the equations I had are non-linear and the best way to solve them numerically is not obvious! I may need to use the 'solver' of Excel.

phrontister wrote:

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

What you did is very good, mainly if AE+BG=5.2915 happens to be the smallest sum to get an equilateral triangle. Perhaps I missed your graphical proof.

phrontister · Yesterday 00:46:32

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!

KerimF · Yesterday 00:52:37

phrontister wrote:

But I'm as close to finding a non-graphical solution as ever!

I liked to prove numerically that your graphical solution is right and your AE+BG=5.2915 is indeed the smallest sum to get an equilateral triangle.

By using the solver of Excel, I got:
AE+BG= 5.291502622
Area = 1.484614945

Please note that on my post #14, equ_3 is wrong. Also, by shifting the origin A to C, the equations become simpler.

But I am not sure if showing here what I did could interest you or anyone else. The last equation happened to be non-linear, so I had to solve it by Excel’s solver while finding the minimum sum by trial and error (thanks to Excel).

Kerim

Note:
I couldn't see your two images:
post #13, https://i.imgur.com/bs1yfhul.jpg
post #17, https://i.imgur.com/4BS9AQTl.jpg

Last edited by KerimF (Yesterday 01:06:00)

phrontister · Yesterday 13:51:16

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?

Last edited by phrontister (Yesterday 23:09:20)

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 2024-12-03 01:28:01

Re: Calculate the area of a triangle

#12 2024-12-03 03:07:45

Re: Calculate the area of a triangle

#13 2024-12-05 01:14:37

Re: Calculate the area of a triangle

#14 2024-12-06 21:32:28

Re: Calculate the area of a triangle

#15 2024-12-07 01:18:19

Re: Calculate the area of a triangle

#16 2024-12-07 03:48:58

Re: Calculate the area of a triangle

#17 Yesterday 00:46:32

Re: Calculate the area of a triangle

#18 Yesterday 00:52:37

Re: Calculate the area of a triangle

#19 Yesterday 13:51:16

Re: Calculate the area of a triangle

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