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This oval is made up of 4 arcs, 2 on the left and right sides of radius 1 and 2 on top and bottom of radius $R$. Given that the the oval fits in a $4 \times 8$ rectangle, is it possible to find $R$ ?

oval in rectangle

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Wait, what radius? Ellipses don't have radii... –  J. M. Dec 6 '11 at 15:09
    
@J.M.: The question is about ovals, not ellipses. Ovals are actually constructed from pairs of arcs. –  Heike Dec 6 '11 at 15:21
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@Heike: So the title has deceived me, I presume... I have no love for titles not agreeing with post bodies. –  J. M. Dec 6 '11 at 15:26
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If Heike's interpretation of the question is the intended one, you should replace "ellipse" by "oval" and "arcs" by "circular arcs". Also, if I understand Heike's solution correctly, it assumes continuous tangents at the transition points. If this is supposed to be part of the problem statement, you should explicate it. –  joriki Dec 6 '11 at 15:27
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@Heike: The curve isn't smooth in the technical sense, only $C^1$. –  joriki Dec 6 '11 at 15:41

1 Answer 1

Let $L$ be the length of the box, and $H$ be its height. By Pythagoras' theorem we have $(R-1)^2=((L-2)/2)^2+(R-H/2)^2$ from which it follows that $$R=\frac{(L-2)^2+H^2-4}{4(H-2)}$$ For this particular example we have $L=8$, $H=4$, and $R=6$.

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Its hard to understand the solution without a diagram. Your value of $R$ is correct though. –  schooler Dec 7 '11 at 12:02
    
@schooler: The solution assumes continuous tangents. That implies that the centres of the circles must lie in the same direction from the transition points. So the centres of the big circles lie at the intersections of the lines through the centres of the small circles and the transition points. The triangle that Heike applied Pythagoras' theorem to is one formed by a centre of a small circle, a centre of a big circle and the centre of the rectangle. –  joriki Dec 7 '11 at 17:35

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