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Topic: A parabola and a triangle (Read 5705 times) |
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pex
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A parabola and a triangle
« on: Oct 26th, 2012, 5:26am » |
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Given are a parabola, and a line intersecting this parabola in two distinct points. Let S be the region bounded by the parabola and the line, and let T be the largest triangle that can be inscribed in S.
Show that the ratio (area of T) : (area of S) is independent of the choice of parabola and line.
(Yes, this is a classic, and probably googlable. I found it interesting regardless.)
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Noke Lieu
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Re: A parabola and a triangle
« Reply #1 on: Oct 29th, 2012, 11:20pm » |
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ax2+bx+c
area under that = (ax3/3) + (bx2/2)+cx
Area bounded by line and x-axis (mx2/2) + nx
Difference between these two areas = area bounded by parabola and line
(-ax3/3) + ((m-b)x2/2)+(n-c)x
Line and parabola intersect at
ax2+bx+c = mx+n
0=(m-b)x+(n-c)-ax2
the roots of which are (-(m-b)+_sqrt((m-b)2-4a(n-c)) )/2a
Sooo...
Find the difference between root1 and root2 for the horizontal component of the base; plug them back into y=mx+c, and find the differences to work out hte vertical component, then use Pythagoras' theorem to work out the base length...
Find the mean of the two roots to determine the hieght of the triangle... and notice how there has to be a a more elegant (or, indeed, accurate) way... 
It's easier with the constraint that the line's horizontal to the 'vertical' parabola...
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a shade of wit and the art of farce.
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pex
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Re: A parabola and a triangle
« Reply #2 on: Oct 30th, 2012, 1:19am » |
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on Oct 29th, 2012, 11:20pm, Noke Lieu wrote:| ... and notice how there has to be a a more elegant (or, indeed, accurate) way... |
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... yes. Yes, there is  Of all possible parameterizations, you seem to have chosen the least convenient one...
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Noke Lieu
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Re: A parabola and a triangle
« Reply #3 on: Oct 30th, 2012, 4:11am » |
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played for and got!
Needed to reflect my current project...
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Grimbal
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Re: A parabola and a triangle
« Reply #4 on: Nov 1st, 2012, 9:18am » |
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on Oct 29th, 2012, 11:20pm, Noke Lieu wrote:| It's easier with the constraint that the line's horizontal to the 'vertical' parabola... |
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Note that you can satisfy that with a linear transform. And such a transform preserves the ratio of surfaces.
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Immanuel_Bonfils
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Re: A parabola and a triangle
« Reply #5 on: Nov 15th, 2012, 3:37pm » |
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The ratio, already known by Archimedes, is 3/4 .
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pex
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on Nov 15th, 2012, 3:37pm, Immanuel_Bonfils wrote:| The ratio, already known by Archimedes, is 3/4 . |
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Well, I did say it was a classic...  My solution follows below.
The first thing to realize is that "all parabolas are equal"; that is, we may choose a coordinate system in which the given parabola is described by y = x2.
The second thing to realize is that this problem becomes much easier if we don't parameterize the line as y = Ax + B, but instead in terms of its intersection points with the parabola. Let's say they are (L, L2) and (R, R2), with L < R. It follows that the line has the equation y = (L+R)x - LR.
The area of S is simply integral(x = L..R) [(L+R)x - LR - x2] dx
= [(L+R)x2/2 - LRx - x3/3](x = L..R)
= (L+R)(R2-L2)/2 - LR(R-L) - (R3-L3)/3
= (R-L)/6 ( 3(L+R)2 - 6LR - 2(R2+LR+L2) )
= (R-L)/6 (L2 - 2LR + R2)
= (R-L)3/6.
Now, let us introduce T, the largest triangle that can be inscribed in S. It is obvious that its vertices will be (L, L2), (R, R2), and (M, M2), for some number M with L < M < R.
Refer to the attached figure. We can find the areas of U and V in the same way as we found that of S:
area of U = (M-L)3/6
area of V = (R-M)3/6
and hence, area of T = (area of S) - (area of U) - (area of V)
= (1/6) ((R-L)3 - (M-L)3 - (R-M)3)
= (3/6) (-LR2+L2R +LM2-L2M +MR2-M2R)
= (1/2) (R-L) (-LR-M2+LM+MR)
= (R-L)(M-L)(R-M)/2.
The maximum of this expression is attained at M = (L+R)/2, so that
area of T = (R-L)3/8
and indeed, (area of T) / (area of S) = 3/4.
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| « Last Edit: Nov 16th, 2012, 4:07pm by pex » |
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