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Topic: Perimeter and Inradius (Read 538 times) |
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Sameer
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Perimeter and Inradius
« on: Feb 10th, 2005, 12:55pm » |
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Let h1,h2,h3 be the lengths of the sides of a triangle and let r be the radius of its inscribed circle. Find the minimum of (h1+h2+h3)/r over all triangles
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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Icarus
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Re: Perimeter and Inradius
« Reply #1 on: Feb 10th, 2005, 3:05pm » |
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::6::
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Sameer
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Re: Perimeter and Inradius
« Reply #2 on: Feb 11th, 2005, 6:46am » |
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Method?
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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Grimbal
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Re: Perimeter and Inradius
« Reply #3 on: Feb 11th, 2005, 8:09am » |
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::6[sqrt]3::
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Sir Col
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My native wits tells me that Grimbal is correct, but I can't say why...
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Let the vertices of the triangle be A, B, C.
The incentre, O, is the point of intersection of the angle bisectors of the triangle; call this point O.
Let the point where the circle is tangental to the triangle be P, Q, R.
Triangle AOR is congruent with triangle AOQ.
Let angle AOR = X = angle AOQ.
Let AR = x = AQ; CQ = y = CP; BP = z = BR.
Let radius, OR = r = OQ.
Therefore tanX=x/r
Similarly, we get tanY=y/r and tanZ=z/r.
As (h1+h2+h3)/r = h1/r+h2/r+h3/r = (x/r+y/r)+(y/r+z/r)+(z/r+x/r) = 2(tanX+tanY+tanZ)
As 2(X+Y+Z) is one full rotation, we know that X+Y+Z=pi, but I can't show that this minimal arrangement occurs when the triangle is equilateral.
However, I suspect it is, and when X=Y=Z=pi/3, tan(pi/3)=sqrt(3), so we get (h1+h2+h3)/r = 6sqrt(3).
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| « Last Edit: Feb 11th, 2005, 9:35am by Sir Col » |
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Eigenray
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Re: Perimeter and Inradius
« Reply #5 on: Feb 11th, 2005, 1:16pm » |
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Another way is to convince yourself that the area A of the triangle is given by A=sr, where s = (h1+h2+h3)/2, and reduce it to a previously solved problem.
But as for minimizing tan(X)+tan(Y)+tan(Z), subject to X+Y+Z=[pi],
suppose Z is fixed, and minimize tan(X)+tan(Y), subject to X+Y=[pi]-Z=C.
Since tan is convex on (0,[pi]/2), we have
2tan(C/2) <= tan(X) + tan(C-X).
Moreover, since tan is strictly convex, equality holds only for X=C-X=Y.
A standard argument then shows that tan(X)+tan(Y)+tan(Z) is minimized for X=Y=Z.
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| « Last Edit: Feb 11th, 2005, 1:40pm by Eigenray » |
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JocK
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Re: Perimeter and Inradius
« Reply #6 on: Feb 12th, 2005, 3:57am » |
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Let's denote the area of the triangle as A, the perimeter as p, and the inradius as r.
The question posed (minimise p/r) is equivalent to asking:
For a fixed perimeter, what triangle maximises its inradius?
However, given that for any triangle the area is equal to the semi-perimeter times the inradius:
A = r p/2
maximising the inradius for fixed perimeter is equivalent to maximising the area for given perimeter. So the question of this thread can be posed as:
For a fixed perimeter, what triangle maximises its area?
The optimum arrangement is obviously an equilateral triangle (do I need to prove this?  ) for which p/r = 6[sqrt]3.
[e]-- Sorry... didn't read the hidden text in Eigenray's message before posting above...  --
Btw: if you want to answer the problem using straightforward analysis, it is perhaps easiest to start from r2 = (1-h1)(1-h2)(1-h3) valid for triangles with unit semi-perimeter and maximise r2 subject to h1+h2+h3=2 using Lagrange multipliers. [/e]
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| « Last Edit: Feb 12th, 2005, 4:17am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Icarus
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Re: Perimeter and Inradius
« Reply #7 on: Feb 12th, 2005, 6:46am » |
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It was obvious to me that the minimum had to occur for the equilateral triangle. Symmetry demands it. But I messed up in calculating the inradius.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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