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        riddles >> medium >> Smallest Equilateral Triangle
        
(Message started by: ThudanBlunder on Nov 28th, 2009, 2:51am)
Title: Smallest Equilateral Triangle
Post by ThudanBlunder on Nov 28th, 2009, 2:51am
Find the area of the smallest equilateral triangle inscribable in a right triangle of legs a and b.
Title: Re: Smallest Equilateral Triangle
Post by Immanuel_Bonfils on Nov 30th, 2009, 6:39pm

[hide]ABC is the right triangle and Phttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sub1.gifPhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sub2.gifPhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sub3.gifthe inscribed equilateral with vetices respecively in AB, BC and CA, side length x.
Let http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifABC = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif(constant or known: sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif=b/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(ahttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif+ bhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif) ) and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifPhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sub3.gifPhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sub2.gifC=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif(variable or unknown).
Sinus law in triangle Phttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sub1.gifBPhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sub2.gifleads to
x = a.sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif/(sin(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/3+http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif)+coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif.sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif)  (*),
that has a minimum for cot http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif=(sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif+sin(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/3-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif))/cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/3-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif)).

Substitution  in (*) gives us xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subm.gifthat multiplied by (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3)/4 leads to the required area.
[/hide]
Title: Re: Smallest Equilateral Triangle
Post by ThudanBlunder on Dec 1st, 2009, 2:49am
I will bear that in mind, but I was looking for a simple formula f(a,b).
Title: Re: Smallest Equilateral Triangle
Post by SMQ on Dec 1st, 2009, 9:02am
Place the right angle at the origin with leg a along the positive y-axis and leg b along the positive x-axis.  Pick a point X = (x, 0) on side b.  If we pick another point P anywhere on the triangle, there are two points Q and Q' which will form an equilateral triangle with X and P.  The sets of Q and Q' over all P form [hide]images of the triangle rotated by +/-120o around point X[/hide].  To find the inscribed equilateral triangle containing X we can find point Y = (0, y): the [hide]intersection of the image of the hypotenuse (rotated +120o) with leg a, i.e. the y-axis[/hide].  With some algebra, we obtain y = [hide][(b - x)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3]/2 - (b + x)(bhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3 - a)/[2(b + ahttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3)][/hide].

We now seek to minimize XY = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x2 + y2).  Equivalently, and more simply, we can minimize XY2 = x2 + y2.  Setting d/dx (x2 + y2) = 0 gives, following some more algebra, x2 + y2 = [hide]a2b2/(a2 + abhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3 + b2)[/hide].

Finally, the area of the inscribed equilateral triangle = (XY2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3)/4 = [(x2 + y2)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3]/4, so the area of the minimal inscribed equilateral triangle is [hide](a2b2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3)/[4(a2 + abhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3 + b2)][/hide].

--SMQ
Title: Re: Smallest Equilateral Triangle
Post by ThudanBlunder on Dec 1st, 2009, 10:51am
Excellent, SMQ, as always.

I will shortly post a solution that uses a neat little substitution to simplify matters. Perhaps even a diagram, but I am presently using Vista and most of my drawing software doesn't seem to want to work with it. Although Cabri doesn't seem very fussy.
Title: Re: Smallest Equilateral Triangle
Post by ThudanBlunder on Dec 26th, 2009, 2:24pm
Sorry, no diagram.

But with a pencil and paper draw the right-angled triangle ABC, with AC as the hypotenuse.
Let AB = a and BC = b
Now draw the inscribed equilateral triangle DEF with D on AC, E on AB, and F on BC
Let BF = b1 and FC = b2
Let DE = EF = FD = s
Let angle EFB = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif
Let angle DCF = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif

Then angle FDC = 60 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif
and
b = b1 + b2
  = s*coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif + [s*sin(60 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)/sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif]

Now letting  r =  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3  + b/a and p = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3b/a + 1
we obtain 1/s = (rsinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif + pcoshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)/2b

We seek 1/s to be a maximum when we differentiate wrt to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif
and setting d(1/s)/dhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gifto zero we obtain
tanhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif = r/p and minimum value of s2 = a2b2/(a2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3ab + b2)

And required area = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3s2/4 =  (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3/4)a2b2/(a2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3ab + b2)



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