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Problem. Show that the sum of the areas of the white regions is equal to the sum of the areas of the grey regions. All the angles between consecutive chords are $45^\circ$.

A solution (not totally Euclidean) can be obtained as follows: Say that two perpendicularly intersected chords are split in pieces of sizes $a,b$ and $c,d$, resp. Then, it is not hard to show that $$ a^2+b^2+c^2+d^2=4R^2 $$ where $R$ is the radius of the circle. (Proposition 11 from Archimedes' book of Lemmas.) Suppose now that we rotate all the four chords by angle $\vartheta$, then it can be readily shown that $$ Grey(\vartheta+\Delta\vartheta)=Grey(\vartheta)+{\mathcal O}(\Delta\vartheta^2), $$ and hence $Grey'(\vartheta)=0$.

Can we produce a solution which is suitable for High-Schools?

EDIT. The new figure corresponds to the Proof without words by L. Carter & S. Wagon (1994a), "Proof without Words: Fair Allocation of a Pizza", Mathematics Magazine 67 (4): 267.

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    $\begingroup$ This is "the pizza theorem", see en.wikipedia.org/wiki/Pizza_theorem $\endgroup$ – Gerry Myerson Jun 16 '16 at 12:53
  • $\begingroup$ @YiorgosSSmyrlis The dissection proof might suit you: en.wikipedia.org/wiki/Pizza_theorem#/media/… $\endgroup$ – almagest Jun 16 '16 at 13:10
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    $\begingroup$ The combination of a calculus approach with proposition 11 from Archimedes' book of Lemmas is absolutely stunning, why do you want another proof? Work a bit harder, and explain yours! $\endgroup$ – Jack D'Aurizio Jun 16 '16 at 20:50
  • $\begingroup$ If you want to use the dissection proof, then I think you should read carefully pp391-2 of this journal article by Hans Humenberger, particularly his Figure 4. There is a pdf math.umt.edu/tmme/vol12no1thru3/28_TMEvol12_Humenberger.pdf $\endgroup$ – almagest Jun 17 '16 at 6:44
  • $\begingroup$ This problem was solved by math contestants(ryoji hiraguchi so on) in "1993". Some known facts(dissection proof) were already given by those high school students in 1993. $\endgroup$ – Takahiro Waki Oct 25 '18 at 4:47
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This is an attempt to give some preliminary lemmas, in order to propose to your students a minor variation of your wonderful proof. The parametric equation of a translated circle is given by: $$(a+\cos\varphi,b+\sin\varphi) \tag{0}$$ hence the polar equation is given by: $$ \rho(\theta)^2 = a^2+b^2+1+2a\cos\varphi+2b\sin\varphi,\qquad \tan\theta=\frac{b+\sin\varphi}{a+\cos\varphi} \tag{1}$$ and since the area in polar coordinates is simply given by $\frac{1}{2}\int \rho(\theta)^2\,d\theta$ (that is the hardest part to prove, but I believe that Cavalieri's principle and triangulations should do the job nicely), the pizza theorem boils down to showing that $$\large\scriptstyle \left(\int_{0}^{\pi/4}+\int_{\pi/2}^{3\pi/4}+\int_{\pi}^{5\pi/4}+\int_{3\pi/2}^{7\pi/4}\right)\, \rho^2(\theta)\,d\theta = \left(\int_{\pi/4}^{\pi/2}+\int_{3\pi/4}^{\pi/2}+\int_{5\pi/4}^{3\pi/2}+\int_{7\pi/4}^{2\pi}\right)\, \rho^2(\theta)\,d\theta \tag{2}$$

Now proposition $11$ in Archimedes' book of lemmas can be read as: $$ \forall\theta,\quad \sum_{k=0}^{3}\rho^2\!\!\left(\theta+\frac{k\pi}{2}\right)=4, \tag{3}$$ hence both the RHS and the LHS of $(2)$ equal $\color{red}{\pi}$.

In principle, we may just use $(1)$ to prove that, by naming $g(a,b)$ the $LHS$ of $(2)$, $$\nabla g = 0 \tag{4}$$ holds, but that involves some nasty changes of variables and differentiation under the integral sign, and it is probably not suited for high school students.

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