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Topic: Inscribed circle (Read 3862 times) |
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Barukh
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Given an equilateral triangle with side 1, and point C, it's center. Find the radius of the inscribed circle.
This simple problem has many different solutions. The problem is to find as simple solution as possible.
It is not easy to give a definition of "the simplest solution". But the following example explains the point: Solution using trigonometry is not considered simple in this case.
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| « Last Edit: Jul 25th, 2014, 4:10am by Barukh » |
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towr
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Re: Inscribed circle
« Reply #1 on: Jul 25th, 2014, 5:28am » |
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The triangle neatly fits around a triangle of six circles.
Which would make the side of the triangle equal to two circle diameters, plus [e]twice[/e] the side of an equilateral triangle circumscribed by the circle (which is sqrt(3) times the radius).
So R = 1/[4+sqrt(3)] [e]R = 1/[4+2 sqrt(3)][/e]
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| « Last Edit: Jul 25th, 2014, 6:07am by towr » |
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Barukh
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Re: Inscribed circle
« Reply #2 on: Jul 25th, 2014, 5:53am » |
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Nice solution, towr! But the answer is a bit off. 
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towr
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Re: Inscribed circle
« Reply #3 on: Jul 25th, 2014, 6:06am » |
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Ah.. Yes..
I should have added another side of triangle.
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rloginunix
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Re: Inscribed circle
« Reply #4 on: Jul 25th, 2014, 4:04pm » |
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(I haven't looked at towr's suggestion, so if mine is a duplicate - let me know and I will delete this post).
Not sure what the metric of simplicity here is. In ruler and compass constructions it is usually the number of steps required to accomplish a given construction, step being defined as a construction of a line or a circle.
In this problem as given I would construct a perpendicular through C to the base to locate E (3 steps) and then bisect the given angle CAE (4 steps) to locate F and lock in r (7 steps total):
The only thing that's left to do then is to use Euclid's B6.P3 (bisector cuts the base proportionally) to come up with the proportion:
r/CF = AE/AC
Use the fact that CF = CE - r and that the medians intersect each other in the ratio 2:1. The rest follows from Pythagoras:
AE = 1/2, AC = sqrt(3)/3, CE = sqrt(3)/6 and after simple manipulations you get:
r = 1 - sqrt(3)/2
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rloginunix
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Re: Inscribed circle
« Reply #5 on: Jul 25th, 2014, 4:05pm » |
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If. If you are willing to accept the literal interpretation of the above result you could construct the answer in just two steps:
1). Extend BC till it intersects the opposite side at G.
2). Cir(B, BG) till it intersects AB at H such that AH = r:
If (in reverse) that can be counted as a standalone solution then, I guess, Euclid's propositions based proof will be more laborious.
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Barukh
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Re: Inscribed circle
« Reply #6 on: Jul 26th, 2014, 3:51am » |
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First of all, rloginunix, you solution is also very nice.
on Jul 25th, 2014, 4:04pm, rloginunix wrote:| Not sure what the metric of simplicity here is. |
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As I said, the notion of simplicity is not well defined. But you suggested an interesting idea here: count the number of propositions in Euclid's "Elements"! In this respect, your solution explicitly contains 2 propositions - the one in the link, and the other concerning the intersection of the medians.
Quote:| If you are willing to accept the literal interpretation of the above result you could construct the answer in just two steps: |
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Yes. And this gives an idea of the most elegant solution I've seen so far. Can you think of it?
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towr
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You can add a copy of the drawing rotate 30 degrees around A, and arrive quite easily at that rather nice side - height conclusion.
Though the way I did it is probably still more work than necessary.
But I think I could successfully explain it to a twelve-year old. (To name a measure of simplicity.)
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rloginunix
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Re: Inscribed circle
« Reply #8 on: Jul 26th, 2014, 12:41pm » |
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on Jul 26th, 2014, 3:51am, Barukh wrote:
I'm trying to but failing at the moment.
With Euclid completely out of the picture you can find r via square areas:
The answer you get is:
sqrt(3)/(6 + 4*sqrt(3))
which numerically is the same as above. Not literal interpretation and not elegant.
Tried using B3.P36 but one distance there is awkward.
In this thread I learned from SWF that tessellations can be used in problem solving. Formations of multiple well known objects that yield some interesting results. But I can't see how it can be used here at this instance.
(Sorry, towr. I didn't read your reply while composing this message because I wanted to put my thoughts down as they happened. Now I'll read it).
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Barukh
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Re: Inscribed circle
« Reply #9 on: Jul 27th, 2014, 8:56am » |
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I will use rloginunix's notation, and denote D the remaining vertex of the original triangle.
Reflect the original configuration in line AB. Let F' be the image of F. Then, triangle ADF' is isosceles.
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towr
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Re: Inscribed circle
« Reply #10 on: Jul 27th, 2014, 9:16am » |
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Ah, yes.
Easy enough to proof by some arithmetic on known angles. that does make it rather simple.
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dudiobugtron
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Re: Inscribed circle
« Reply #11 on: Jul 27th, 2014, 5:17pm » |
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Call the triangle ABD, then consider the triangles ACB, ACD and BCD. These are each 1/3 of the total area of the main triangle, and have the same base, so their height must be 1/3 of the height of ABD as well. But their height is also the radius of the inscribed circle.
So to find the radius, just find the height of the triangle (eg: by pythagoras, or known triangles).
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rloginunix
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Re: Inscribed circle
« Reply #12 on: Jul 27th, 2014, 7:42pm » |
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(off: sorry for sporadic replies. Circumstances beyond my control made this past weekend very busy for me, with virtually no leisure and computer access time.)
My literal interpretation take. Using the original formula we can say that "side = height + radius". If we reflect the point E about F to locate G then (after trivial reasoning) we get two similar isosceles 30-75-75 triangles: ADG and GAE:
Hence, their corresponding sides must be in the same proportion:
EG/AG = AG/AD
or
2r = AE^2
where AE is the angular bisector of the triangle AFC and can be calculated via a known formula:
AE^2 = AF*AC(1 - CF^2/(AF + AC)^2)
Numeric results come out as above.
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rloginunix
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Re: Inscribed circle
« Reply #13 on: Jul 27th, 2014, 8:27pm » |
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on Jul 27th, 2014, 5:17pm, dudiobugtron wrote:Call the triangle ABD, then consider the triangles ACB, ACD and BCD. These are each 1/3 of the total area of the main triangle, and have the same base, so their height must be 1/3 of the height of ABD as well. But their height is also the radius of the inscribed circle.
So to find the radius, just find the height of the triangle (eg: by pythagoras, or known triangles). |
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I'm sorry to intervene here, dudiobugtron, but what Barukh is really saying in the problem statement is "Find the radius of the inscribed circle." ... shown in the picture below.
It looks like you're referring to the inscribed circle of the entire equilateral triangle while the objective is to find the radius of the circle inscribed into the smaller isosceles triangles you are referring to (these are two different radii).
Sorry again. I mean it in a good way.
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SWF
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Based on tesselation of hexagons, squares, and triangles, make circles of radius R centered at each vertex (and an extra circle at center of each hexagon. Each side of the triangle shown is hypotenuse of right triangle with sides of (2+sqrt(3))R and (3+2*sqrt(3))R. Sum of the squares of those two lengths must be 1, which leads to an expression for R.
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towr
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Re: Inscribed circle
« Reply #15 on: Jul 27th, 2014, 10:43pm » |
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Dudiobugtron's idea does offer another approach, because we can split ABC into three triangles with height r. Combined these have a base of 1 + 2/3 sqrt(3), and an area 1/3rd of ABD, so r = [1/3 * 1 * 1/2 sqrt(3)]/[1 + 2/3 sqrt(3)] = 1 - 1/2 sqrt(3) as before.
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| « Last Edit: Jul 27th, 2014, 10:46pm by towr » |
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dudiobugtron
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Re: Inscribed circle
« Reply #16 on: Jul 27th, 2014, 11:48pm » |
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Thanks for the pointer rloginunix! (and thanks for salvaging my idea towr!)
I had my phone set to not download images, but it didn't show a broken link like it usually does, so I had no idea there were even photos! I wondered why noone else's answers made any sense; I figured you were all talking in some triangle-related code I had yet to learn...
(I even noticed that rloginunix' pictures thread in the general forum had been updated, and wondered why since I hadn't seen pictures anywhere!)
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rloginunix
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Re: Inscribed circle
« Reply #17 on: Jul 28th, 2014, 7:20pm » |
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Another tessellation gem! Am I the only one who feels like a kid in the candy store?
I've missed my opportunity last time so may be I'll ask this time. Are you willing to share with us how you come up with these things? I mean there many different tessellation types (I've looked them up on wiki) so how do you decide which type to use for a given problem?
Also, are there any good tutorials or books on the subject?
Lastly, when I got home I simply had to do it myself so that I understand it better. I'm posting this rendering here for everyone to enjoy but if you don't want me to I will delete it:
(just to be clear, the picture below is 100% SWF's idea, not mine, I simply filled in a few details):
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SWF
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Re: Inscribed circle
« Reply #18 on: Jul 28th, 2014, 10:32pm » |
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That is a very nice picture, rloginunix! I was hoping you would improve on my hastily made figure. It would be helpful to show lines of length equal to legs of the right triagles with hypotenuse 1.
To continue the calculation...
1=R2(2+sqrt(3))2+R2(3+2*sqrt(3))2= R2(28+ 16*sqrt(3))=R2(4+2*sqrt(3))2
R=1/(4+2*sqrt(3))=1-sqrt(3)/2
I will try to answer you questions tomorrow when I have more time.
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rmsgrey
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Re: Inscribed circle
« Reply #19 on: Jul 29th, 2014, 7:46am » |
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The asymmetry of the diagram bothers me - it's not immediately obvious that the green triangle must be equilateral.
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towr
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Well, the idea I used in my first solution gives a nice symmetrical tiling, if you prefer.
It's actually pretty much the same, except I rotated the hexagons, and choose the centers of the hexagons to make the triangle.
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| « Last Edit: Jul 29th, 2014, 10:09am by towr » |
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rloginunix
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Re: Inscribed circle
« Reply #21 on: Jul 29th, 2014, 12:10pm » |
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That's another beautiful arrangement, towr. And you kept it from us for that long.
rmsgrey, your concern with asymmetry in SWF's drawing is easy to alleviate. I'm at work right now and can't do it but once I get home I'll post the drawing.
Meanwhile, what you do is in your mind's eye imagine the green triangle hanging freely on the top vertex with the bottom two vertexes unbuckled from the plane.
Simply swing the entire green triangle right (or left) 30 degrees (keeping the top vertex where it is). The remaining two vertexes will snap into new positions such that the "distances" between all three vertexes will be "2 sqaures + 1 hexagon" making it visually equilateral.
(I hope SWF is OK with me intervening.)
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towr
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Re: Inscribed circle
« Reply #22 on: Jul 29th, 2014, 12:27pm » |
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on Jul 29th, 2014, 12:10pm, rloginunix wrote:| That's another beautiful arrangement, towr. And you kept it from us for that long. |
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Well, I only had the inside of the triangle when I gave my first solution. I hadn't considered how to tile the plane with it.
But it has some nice properties, I think. You can even compare the height and width of the large triangle easily in terms of the heights and widths of the small triangles and squares. (Maybe I should also just have drawn the hexagons as 6 red triangles)
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| « Last Edit: Jul 29th, 2014, 12:29pm by towr » |
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rloginunix
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Re: Inscribed circle
« Reply #23 on: Jul 29th, 2014, 3:34pm » |
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Addressing rmsgrey's asymmetric look and feel of the original drawing concern: swing the green triangle about the top vertex 30 degrees counterclockwise. It should be easily seen that the distance between any two vertexes of the green triangle is "square + hexagon + square" (and that all the boundary circles are either cut nicely through their centers or touched):
I've highlighted the heights of the parent triangle so that the sides of the right triangles can be easily seen and calculated. The short side (for the bottom left vertex moving towards the right vertex): R + R + Pythagoras = R(2 + sqrt(3)). The long side (moving towards the top vertex): R + Pythagoras + R + R + Pythagoras = R(3 + 2*sqrt(3)).
In my previous comment I've mentioned that you can swing the green triangle 30 degrees clockwise. When you do so you'd have to move the top vertex of the green triangle one "polygon side" length (2R) over to the left.
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SWF
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Re: Inscribed circle
« Reply #24 on: Jul 29th, 2014, 9:19pm » |
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That last figure is a nice improvement in that it is clear the triangle is equilateral and the side length is R(4+2sqrt(3)). No need for the right triangles with this. No matter where you move the triangle, it should work out, but of course (now that I think about it more) the most symmetrical has it centered on one of the small triangles of the pattern.
I picked this tessellation because the angles in the problem were the same as in a dodecagon, but I was thinking more in terms of the arrangement of circles into a dodecagon. towr's tessellation (if colored differently) is the same as the tesselation of dodecagons with triangles. There is also one with dodecagons, hexagons, and squares, that would be worth trying. Other patterns might work too, such as the Archimedian tessalation of squares and triangles, as long as the pattern invoves the key angles.
I recall seeing a book on tessellations for quiltmakers. Martin Gardner probably wrote an article on them, and of course, there is Escher.
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