# A problem with an inscribed oval

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 $R$ ?

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Wait, what radius? Ellipses don't have radii... –  Guess who it is. 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
@Heike: So the title has deceived me, I presume... I have no love for titles not agreeing with post bodies. –  Guess who it is. Dec 6 '11 at 15:26
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
@Heike: The curve isn't smooth in the technical sense, only $C^1$. –  joriki Dec 6 '11 at 15:41

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 $R=6$.
Its hard to understand the solution without a diagram. Your value of $R$ is correct though. –  schooler Dec 7 '11 at 12:02