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I want to prove that there is no maximum perimeter trapezoid inscribed in a circle.


Edit: the way my textbook defined a trapezoid does not allow for a square to fit in the trapezoid definition. If your definition is different than that, this problem is the equivalent of proving that the maximum perimeter trapezoid inscribed in a circle is a square.

What I have already figured:

I know (and know how to prove) that an inscribed trapezoid must be an isosceles trapezoid.

I also know that the maximum perimeter quadrilateral inscribed in a circle is a square (but don't know how to prove it).

I know that the inscribed trapezoid can get as close to being a square as we like, but it can never be a square. That is why there is no maximum perimeter inscribed trapezoid. But I do not know how to prove that the square is the maximum perimeter quadrilateral inscribed in a circle.

I think that I should approximate the inscribed square by an inscribed trapezoid, but when I tried that I couldn't prove that the perimeter of the square was greater than that of the trapezoid.

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  • $\begingroup$ A square is a trapezoid. A very special case of a trapezoid, maybe, but still definitely a trapezoid. The same way as, for example, a square is a quadrilateral. $\endgroup$ – dxiv Oct 30 '16 at 4:07
  • $\begingroup$ That depends on your definition. My text book doesn't consider a square to be a trapezoid. I will edit the question to add this information. Thanks for pointing that out. $\endgroup$ – Wheepy Oct 30 '16 at 4:10
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    $\begingroup$ That must be one odd book. A trapezoid is defined everywhere as a quadrilateral with (at least) two parallel sides. Following the same (broken) logic your book would argue that the trapezoid is not a quadrilateral. $\endgroup$ – dxiv Oct 30 '16 at 4:26
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    $\begingroup$ It's certainly an odd way of defining it for someone who has never seen it, but this way is chosen by some authors for teaching preferences. That way, when they say "trapezoid", they mean "a trapezoid that is not a square", when they say "rectangle" they mean "a rectangle that is not a square", etc. Both these ways of defining are possible and lead to the same geometry, apart from names. There is, however, no way of defining a trapezoid without fitting the definition of a quadrilateral. Saying "a trapezoid that is not a quadrilateral" doesn't make any sense in any possible definition. $\endgroup$ – Wheepy Oct 30 '16 at 4:39
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    $\begingroup$ @dxiv I hang out in elementary school classrooms and see this definition problem a lot. Is a square a rectangle, or does the definition of rectangle exclude all sides equal? The latter is sometimes better "for teaching preferences" but it's a disaster kids will have to unlearn later when they do more mathematics. And you could define quadrilateral to be "a four sided polygon with no symmetries". That would be consistent with "a square is not a rectangle". But ugly! $\endgroup$ – Ethan Bolker Oct 30 '16 at 14:00
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I shall prove the statement in the title, given the book's definition of a trapezoid. We may assume the trapezoid $T$ inscribed in the unit circle, with horizontal parallel sides $y=a$ and $y=b$, whereby $-1<a<b<1$, and $b>0$. If $a>0$ we can enlarge the perimeter of $T$ by replacing $a$ by $-a$. The four vertices of $T$ can then be written as $$(\pm \cos\psi,-\sin\psi), \quad(\pm\cos\phi,\sin\phi)$$ with $0\leq\psi<{\pi\over2}$ and $0<\phi<{\pi\over2}$. The perimeter $L$ of $T$ computes to $$L(\phi,\psi)=4\sin{\phi+\psi\over2}+2\cos\phi+2\cos\psi\ ,$$ such that $$L_\phi=2\cos{\phi+\psi\over2}-2\sin\phi,\qquad L_\psi=2\cos{\phi+\psi\over2}-2\sin\psi\ .$$ If $\psi=0$ one has $L_\psi=2\cos{\phi\over2}>0$, hence increasing $\psi$ will increase $L$. If both $\phi$ and $\psi$ are $>0$ we can locally increase $L$ unless $L_\phi=L_\psi=0$. The latter would lead to $\psi=\phi$ and then to $\cos\phi=\sin\phi$, hence $\psi=\phi=45^\circ$, which is not feasible here.

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A trapezoid is a quadrilateral with at least one pair of opposite sides being parallel. A square fits this definition. Any theorem applying to a trapezoid also applies to a square.

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  • $\begingroup$ That depends on your definition. My text book doesn't consider a square to be a trapezoid. I will edit the question to add this information. Thanks for pointing that out. $\endgroup$ – Wheepy Oct 30 '16 at 4:11
  • $\begingroup$ @Robert. A trapezoid is a quadrilateral with exactly one pair of opposite sides being parallel. A square fits the definition of a rectangle, not of a trapezoid. $\endgroup$ – imranfat Oct 30 '16 at 4:34
  • $\begingroup$ @imranfat Would you please provide a reference. That would be a very unusual definition in my books. $\endgroup$ – dxiv Oct 30 '16 at 4:38
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    $\begingroup$ The first line on that page: "a convex quadrilateral with at least one pair of parallel sides is referred to as a trapezoid". I'll leave it at that, and if your book defines it otherwise then I'd seriously recommend a different book. $\endgroup$ – dxiv Oct 30 '16 at 4:45
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    $\begingroup$ So if you prove some arbitrary property for trapezoids, then you'd have to prove it again for parallelograms, rectangles, squares etc. It's technically fine to have your own non-standard definitions as long as you spell them out (and +1 you did that in the edit to your question), but that will invite miscommunications down the road. That's (just) my opinion, and this is my last comment here. $\endgroup$ – dxiv Oct 30 '16 at 4:50
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Hint: assume the convex cyclic quadrilateral $ABCD$ includes the center $O$ of its circumscribed circle or radius $r$ (otherwise find a suitable one with larger perimeter). Then each angle $\widehat{AOB}\le \pi$ and $\sum_{cyc}\widehat{AOB} = 2 \pi$. Given that the perimeter is $P = \sum_{cyc} 2 r \sin \widehat{AOB}/2$ and $\sin x$ is concave on $(0,\pi)$ it follows by Jensen's inequality that:

$$P/2r = \sum_{cyc} \sin \frac{\widehat{AOB}}{2} \le 4 \sin\left(\frac{1}{4} \sum_{cyc} \frac{\widehat{AOB}}{2}\right) = 4 \sin \frac{\pi}{4} = 2 \sqrt{2}$$

with equality iff all angles $\widehat{AOB}/2$ are equal i.e. the quadrilateral is a square.

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Note the symmetries of the circle, you should be able to convince yourself that the maximum peremiter quadrilateral is one that maximizes symmetries (if there was a way to make the perimeter larger on one side of a line of symmetry the same action can always be taken by the other side) this leaves the square and the the crossed square as possible shapes, the crossed square has greater perimeter then the square but it is not the limit of expanding the symmetries of a trapezoid, therefore a square has the maximum perimeter of not self intersecting quadrilaterals.

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  • $\begingroup$ That's good intuition, but not a proof. At face value, the exact same argument would work to prove that the square is the inscribed quadrilateral of minimum permieter. $\endgroup$ – dxiv Oct 30 '16 at 5:53
  • $\begingroup$ yeah but i didn't want to give a full proof just a starting point. one can show that the square is fixed at either max or min then simply check against a non square rectangle. @dxiv $\endgroup$ – shai horowitz Oct 30 '16 at 5:55
  • $\begingroup$ you can also prove the shortest perimeter of an inscribedshape is a rectangle with the 2 lengths approaching 0 yielding a min perimeter of 2d. $\endgroup$ – shai horowitz Oct 30 '16 at 5:58
  • $\begingroup$ this minimization fixed point becomes important in proving that the square is the quadrilateral of most area with min perimeter. $\endgroup$ – shai horowitz Oct 30 '16 at 6:09

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