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Title: A parabola and a triangle Post by pex on Oct 26th, 2012, 5:26am 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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Title: Re: A parabola and a triangle Post by Noke Lieu on Oct 29th, 2012, 11:20pm [hide] 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[/hide]... 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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Title: Re: A parabola and a triangle Post by pex on Oct 30th, 2012, 1:19am on 10/29/12 at 23:20:01, Noke Lieu wrote:
... yes. Yes, there is ;D Of all possible parameterizations, you seem to have chosen the least convenient one... |
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Title: Re: A parabola and a triangle Post by Noke Lieu on Oct 30th, 2012, 4:11am played for and got! Needed to reflect my current project... |
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Title: Re: A parabola and a triangle Post by Grimbal on Nov 1st, 2012, 9:18am on 10/29/12 at 23:20:01, Noke Lieu wrote:
Note that you can satisfy that with a linear transform. And such a transform preserves the ratio of surfaces. |
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Title: Re: A parabola and a triangle Post by Immanuel_Bonfils on Nov 15th, 2012, 3:37pm The ratio, already known by Archimedes, is 3/4 . |
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Title: Re: A parabola and a triangle Post by pex on Nov 16th, 2012, 1:08pm on 11/15/12 at 15:37:17, Immanuel_Bonfils wrote:
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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