wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> medium >> Inscribed Triangle
        
(Message started by: Barukh on Sep 3rd, 2007, 12:55am)
Title: Inscribed Triangle
Post by Barukh on Sep 3rd, 2007, 12:55am
Given an arbitrary triangle, construct another triangle which is inscribed in the first and has the minimal perimeter.
Title: Re: Inscribed Triangle
Post by Eigenray on Sep 3rd, 2007, 4:16pm
How about this.
Title: Re: Inscribed Triangle
Post by Barukh on Sep 4th, 2007, 5:19am

on 09/03/07 at 16:16:06, Eigenray wrote:
How about this.

Bingo!  :D

Have you got a proof? What kind of a triangle will you get?
Title: Re: Inscribed Triangle
Post by Eigenray on Sep 4th, 2007, 7:34pm
Let the triangle be ABC.  I reasoned in 3 steps:

(1) Fix R on side AB, and Q on AC.  How do you find P on BC minimizing QR + PQ + PR?
[hide]Reflect Q across BC to Q', and let P be the intersection of RQ' and BC.[/hide]

(2) Fix R on side AB.  How do you find P on BC and Q on AC minimizing QR + PQ + PR?
[hide]Reflect R across AC to R', and across BC to R''.  Then QR+PQ+PR = R'Q+QP+PR'' http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif R'R'', so let P,Q be the intersection of R'R'' with BC, AC, respectively[/hide]

(3) How do you find P,Q, and R?
[hide]Reflect B across AC to B', and A across BC to A'.  We must find R on AB minimizing R'R'' above.  Subject to |R'B| = |RB| = |R''B|, it is clear(?) geometrically that |R'R''| is minimized when http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifAR'R'' = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifBR''R'.  That is, R'R'' is parallel to the vertical lines BU and B'V in the diagram.  Then |UR'| = |BR''| = |BR| = |R'B'| = |R''V|.  So the desired line R'R'' is midway between BU and B'V.[/hide]

The diagram assumes WLOG http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif > http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2, and that ABC is acute.  (What happens if it's not?)

Can you show the following: the following triangles are all similar

ABC ~ AQR ~ PBR ~ PQC

in the ratio 1 : coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif : coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif : coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif ?

Thus |PQR| = (1 - cos2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif - cos2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif - cos2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif)*|ABC|,

and the minimal perimeter is

PQ+QR+RP = a coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif + b coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif + c coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif
= (a sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif)(b sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif)(c sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif)/ K
= 2 K/R = 8 K2/(abc),

where K = |ABC|, and 2R = a/sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif is the diameter of the circumscribed triangle.

Amusingly, the perimeter of PQR is

2 * sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif * sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif * c,

which is "dual" to the area of the original triangle

1/2 * a * b * sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif.

Is there a deeper reason for this?


on 09/04/07 at 05:19:32, Barukh wrote:
What kind of a triangle will you get?

???
Title: Re: Inscribed Triangle
Post by ecoist on Sep 4th, 2007, 9:23pm
Help me out here, guys.  Shouldn't the inscribed triange be required to touch all three sides of the enclosing triangle?  Otherwise, the answer is a single point with perimeter zero.

If the inscribed triangle is required to touch all three sides of the enclosing triangle, I can construct such a triangle with perimeter equal to [hide]twice the length of a shortest side of the enclosing triangle[/hide], which perimeter is shorter than the one in Eigenray's picture.  What am I missing?
Title: Re: Inscribed Triangle
Post by Eigenray on Sep 4th, 2007, 9:56pm

on 09/04/07 at 21:23:42, ecoist wrote:
Help me out here, guys.  Shouldn't the inscribed triange be required to touch all three sides of the enclosing triangle?  Otherwise, the answer is a single point with perimeter zero.

That's hardly inscribed!


Quote:
If the inscribed triangle is required to touch all three sides of the enclosing triangle, I can construct such a triangle with perimeter equal to [hide]twice the length of a shortest side of the enclosing triangle[/hide], which perimeter is shorter than the one in Eigenray's picture.  What am I missing?

Is this for an obtuse triangle?  It seems my construction goes horribly wrong in this case.  If it is obtuse, maybe the solution is to take the [hide]shortest altitude[/hide]?
Title: Re: Inscribed Triangle
Post by ecoist on Sep 4th, 2007, 10:12pm
Sorry, I'm still lost, Eigenray!  Your inscribed triangle touches all three sides of the enclosing triangle, exactly as I meant.  And yes, my construction is obtuse.  Oops!  You meant the enclosing triangle could be obtuse.  My construction doesn't care; my constructed inscribed triangle is always obtuse.
Title: Re: Inscribed Triangle
Post by Barukh on Sep 5th, 2007, 5:01am
Nice solution, Eigenray! Although I had a different one in mind.


Quote:
???

[hide]Connect the points A, P; B, Q; C, R. What do you get?[/hide] And why?


on 09/04/07 at 22:12:47, ecoist wrote:
Oops!  You meant the enclosing triangle could be obtuse.  My construction doesn't care; my constructed inscribed triangle is always obtuse.

Are you saying you can beat Eigenray’s solution in case the enclosing triangle is acute?
;)


on 09/04/07 at 21:56:31, Eigenray wrote:
Is this for an obtuse triangle?  It seems my construction goes horribly wrong in this case.  If it is obtuse, maybe the solution is to take the [hide]shortest altitude[/hide]?

;)
Title: Re: Inscribed Triangle
Post by ecoist on Sep 5th, 2007, 9:32am

Quote:
Are you saying you can beat Eigenray’s solution in case the enclosing triangle is acute?


I'm saying exactly that unless I miss-understand the meaning of "triangle inscribed in a triangle".
Title: Re: Inscribed Triangle
Post by Barukh on Sep 5th, 2007, 9:50am

on 09/05/07 at 09:32:30, ecoist wrote:
I'm saying exactly that unless I miss-understand the meaning of "triangle inscribed in a triangle".

Could you please provide an example?
Title: Re: Inscribed Triangle
Post by Eigenray on Sep 5th, 2007, 10:05am

on 09/04/07 at 22:12:47, ecoist wrote:
Your inscribed triangle touches all three sides of the enclosing triangle, exactly as I meant.

I meant that civilized people would not call a point an inscribed triangle.


on 09/05/07 at 05:01:15, Barukh wrote:
[hide]Connect the points A, P; B, Q; C, R. What do you get?[/hide]

Well, that was unexpected!  [hide]Drop perpendiculars AP, BQ, CR.  Now, BR = a coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif,  BP = c coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif,  and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifRBP = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif, so RP = b coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif, and ABC is similar to PBR.  Similarly, we get http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifBRP = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifARQ, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifAQR = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif = PQC, and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifRPB = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifQPC.

So a light beam emitted from P towards Q will reflect around and around in triangle PQR.[/hide]  This is the same as the triangle previously constructed.
Title: Re: Inscribed Triangle
Post by ecoist on Sep 5th, 2007, 10:28am

Quote:
Could you please provide an example?


Gladly, Barukh!  Using Eigenray's figure, my triangle is BCP.  Although this triangle is degenerate, it lies within Eigenray's black triangle and touches all three sides of that triangle.  Its perimeter is clearly 2|BC|, which is less than the perimeter of PQR.

Why is this example illegitimate?
Title: Re: Inscribed Triangle
Post by Barukh on Sep 5th, 2007, 11:17am

on 09/05/07 at 10:28:46, ecoist wrote:
Its perimeter is clearly 2|BC|, which is less than the perimeter of PQR.

Why is this example illegitimate?

Leaving  aside the legitimity of this triangle, its perimeter is more than Eigenray's solution, which for example follows from this remark:


on 09/04/07 at 19:34:39, Eigenray wrote:
Amusingly, the perimeter of PQR is

2 * sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif * sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif * c,
Title: Re: Inscribed Triangle
Post by Sameer on Sep 5th, 2007, 1:03pm
From Eigenray's explanation it seems the triangle with smallest perimeter is a [hide] orthic triangle - is that the right term? the triangle formed by the feet of the altitudes [/hide]. This shows that for an obtuse triangle [hide] the minimum permiter triangle i.e. the orthic triangle lies outside the triangle, so there has to be another triangle?? [/hide]. I still don't have my own proof. Working on it!!
Title: Re: Inscribed Triangle
Post by ecoist on Sep 5th, 2007, 4:20pm
How stupid of me!  Thanks, Barukh.  I had a mental cramp that no perimeter of an inscribed triangle could be less than 2|BC|.
Title: Re: Inscribed Triangle
Post by Barukh on Sep 6th, 2007, 4:56am

on 09/05/07 at 10:05:30, Eigenray wrote:
Well, that was unexpected!  

Exactly! This triangle is called [hide]orthic[/hide], as Sameer pointed out rightly.

Eigenray, if we continue your drawing for another 3 triangle, the picture we get is almost self-explanatory (where the solution is drawn is dashed line): segments AB and A’B’ are parallel (why?),  RR’ is a straight line (why?), and any other line connecting corresponding points will have a greater length (solution by H.A.Schwarz).


on 09/05/07 at 13:03:16, Sameer wrote:
This shows that for an obtuse triangle [hide] the minimum permiter triangle i.e. the orthic triangle lies outside the triangle, so there has to be another triangle?? [/hide]. I still don't have my own proof. Working on it!!


;)
Title: Re: Inscribed Triangle
Post by Sameer on Sep 6th, 2007, 9:48pm
Interesting. 5 reflections of the triangle!! How do you start with point R? As in the internal traces - how do you start with it? Is there a way to work out this method based on Eigenray's equation? e.g. equation 1 translates to reflection, etc...
Title: Re: Inscribed Triangle
Post by Barukh on Sep 7th, 2007, 2:00am

on 09/06/07 at 21:48:41, Sameer wrote:
Is there a way to work out this method based on Eigenray's equation? e.g. equation 1 translates to reflection, etc...

In his derivation, Eigenray proved the mirror property of the orthic triangle: its sides form equal angles with the corresponding side of the enclosing triangle. Following this, is it clear that a reflection of orthic triangle in any side will produce a pair of parallel segments. This, in fact, determines uniquely a sequence of reflections, if we want to get a straight line of the sides of the orthic triangle (Eigneray used 2 such reflections). Now, it is left to notice that after the 5th reflection we get a pair of parallel lines of the enclosing triangle (this fact is still to be proved).
Title: Re: Inscribed Triangle
Post by Aryabhatta on Sep 7th, 2007, 3:48pm
I believe this is called as Fagnano's problem which Fagnano had solved using Calculus.
(See the name of the image which Barukh attached  :P)

Schwartz was the first to give the elegant proof by mirroring.

Sorry for adding nothing to this thread  :-/
Title: Re: Inscribed Triangle
Post by Sameer on Sep 7th, 2007, 4:01pm
Wow I didn't know this problem had a name!!  8) I googled and got this famous site!!

http://www.cut-the-knot.org/Curriculum/Geometry/Fagnano.shtml
Title: Re: Inscribed Triangle
Post by ecoist on Sep 7th, 2007, 6:57pm
Sameer wrote:


Quote:
after the 5th reflection we get a pair of parallel lines of the enclosing triangle (this fact is still to be proved)


In the figure, rewrite the middle A and B as A'' and B'', and write the middle C as C'.  Then the figures ABCB''A'' and A'B'C'A''B'' are congruent.  Hence the angles between the lines AB and A''B'' and that between A'B' and A''B'' are equal; whence AB and A'B' are parallel.

This suggests that the choice of R can vary, as long as the dotted red line stays inside the figure.

Oops!  Goofed again!  That dotted red line must pass through R'' on A''B'' as well as R' on A'B'!  Since moving R right/left causes R'' to move left/right, there is only one choice for R.

So what happens if angle ABC is greater than 90 degrees, when there is no R?  Is Eigenray's suggestion, an altitude, the solution for obtuse triangles?
Title: Re: Inscribed Triangle
Post by Barukh on Sep 7th, 2007, 11:07pm

on 09/07/07 at 18:57:42, ecoist wrote:
Oops!  Goofed again!  That dotted red line must pass through R'' on A''B'' as well as R' on A'B'!  Since moving R right/left causes R'' to move left/right, there is only one choice for R.

Yes, you got it right.


Quote:
So what happens if angle ABC is greater than 90 degrees, when there is no R?  Is Eigenray's suggestion, an altitude, the solution for obtuse triangles?

Exactly.
Title: Re: Inscribed Triangle
Post by TenaliRaman on Sep 8th, 2007, 2:30am

on 09/07/07 at 18:57:42, ecoist wrote:
So what happens if angle ABC is greater than 90 degrees, when there is no R?  Is Eigenray's suggestion, an altitude, the solution for obtuse triangles?

From cut-the-knot,

Excerpt :
In obtuse or right triangles, however, no inscribed proper (with distinct vertices) triangle attains the minimum possible perimeter. There exist triangles whose perimeter is arbitarily close to twice the length of the shortest altitude but no proper triangle attains this value.

-- AI
P.S -> So, i guess one has to prove the non-existence here.
Title: Re: Inscribed Triangle
Post by srn347 on Sep 9th, 2007, 6:33pm
For any triangle in that triangle, there is one in that one with smaller perimeter. It would have to have perimeter of zero which wouldn't even be a triangle. So there is no answer.
Title: Re: Inscribed Triangle
Post by Barukh on Sep 9th, 2007, 9:43pm

on 09/09/07 at 18:33:24, srn347 wrote:
For any triangle in that triangle, there is one in that one with smaller perimeter. It would have to have perimeter of zero which wouldn't even be a triangle. So there is no answer.

Let me remind you that "inscribed triangle" means a triangle touching all 3 sides of the original triangle. Therefore, it cannot have a zero perimeter.

If you want to make your argument stronger, please take the figures posted by Eigenray or me and try to show an inscribed triangle with smaller perimeter.
Title: Re: Inscribed Triangle
Post by SWF on Sep 9th, 2007, 11:00pm
Imagine a tiny loop of rubber band much smaller than the triangle that is forced to have frictionless contact at a point on each of the three sides of the triangle, and the contact points allow slip of the band so that tension is the same throughout the rubber band. The rubber band must be stretched to make it touch each side, but will take on the shape of the inscribed triangle of minimum perimeter to minimize the strain energy due to stretching.

At any of the points where the rubber band touches a side of the triangle, for it to be in equilibrium, the angle "incidence" and "reflection" must be the same. Thus, the shape of the minimizing inscribed triangle is the reflecting ray of light path as described by Eigenray.

The same should be true not matter what the shape within which you are trying to inscribe a polygon of minimium perimeter- although there are likely to be several shapes having a local local minimum (or maximum) perimeter. You are likely to get some of the contact points to coalesce. For example, trying to find a minimum permeter quadrilateral might degnerate into a triangle, and there should alway be at least one such shape (otherwise you would have perpetual motion).


Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board