Enunciado do desafio :: Challenge statement
Os arcos de circunferência a grosso verdes e azuis têm o mesmo comprimento. A soma das áreas delimitadas pelos arcos e linhas a grosso azuis é igual à soma das áreas delimitadas pelos verdes? Justifique.
Nota: os diâmetros são perpendiculares.
The thick green and blue circle arcs have the same lengths. Does the sum of the areas limited by the thick blue arcs and lines equal the sum of the areas limited by the green ones? Justify.
Remark: the diameters are perpendicular.
Solution by Jacques Glorieux:
Let’s denote by 
The area of the green trapezium (trapezoid) 
![\dfrac{1}{2}[R\sin (a+b)+R\sin (b)][R\cos (b)-R\cos (a+b)]=](https://s0.wp.com/latex.php?latex=%5Cdfrac%7B1%7D%7B2%7D%5BR%5Csin+%28a%2Bb%29%2BR%5Csin+%28b%29%5D%5BR%5Ccos+%28b%29-R%5Ccos+%28a%2Bb%29%5D%3D&bg=ffffff&fg=000000&s=0 "\dfrac{1}{2}[R\sin (a+b)+R\sin (b)][R\cos (b)-R\cos (a+b)]=")
![\dfrac{1}{2}R^{2}[\sin (a+b)+\sin (b)][\cos (b)-\cos (a+b)]\qquad (1)](https://s0.wp.com/latex.php?latex=%5Cdfrac%7B1%7D%7B2%7DR%5E%7B2%7D%5B%5Csin+%28a%2Bb%29%2B%5Csin+%28b%29%5D%5B%5Ccos+%28b%29-%5Ccos+%28a%2Bb%29%5D%5Cqquad+%281%29&bg=ffffff&fg=000000&s=0 "\dfrac{1}{2}R^{2}[\sin (a+b)+\sin (b)][\cos (b)-\cos (a+b)]\qquad (1)")
The area of the green trapezium (trapezoid) 
![\dfrac{1}{2}R^{2}[\cos (b)+\cos (a+b)][\sin (a+b)-\sin (b)]\qquad (2)](https://s0.wp.com/latex.php?latex=%5Cdfrac%7B1%7D%7B2%7DR%5E%7B2%7D%5B%5Ccos+%28b%29%2B%5Ccos+%28a%2Bb%29%5D%5B%5Csin+%28a%2Bb%29-%5Csin+%28b%29%5D%5Cqquad+%282%29&bg=ffffff&fg=000000&s=0 "\dfrac{1}{2}R^{2}[\cos (b)+\cos (a+b)][\sin (a+b)-\sin (b)]\qquad (2)")
By summing 
If we look at the original drawing, we see that the blue figure is just a particular case of the green one when 
The sum of the areas of the two blue trapezia (trapezoids) is thus the same as the sum of the areas of the two green trapezes.
As the total green (blue) area is equal to the sum of the areas of the two green (blue) trapezia (trapezoids) plus twice the area under one of the chord (i.e. 
[Correction of May 12, 2010: trapezium (trapezoid), trapezia (trapezoids) instead of trapeze, trapezes A. Tavares]
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Resolução de Jacques Glorieux:
Designemos por
A área do trapézio verde
A área do trapézio verde
Somando
Se observarmos o desenho original, vemos que a figura azul é o caso particular da verde para
A soma das áreas dos dois trapézios azuis é por isso a mesma que a da soma das áreas dos dois trapézios verdes.
Como o total das áreas verdes (azuis) é igual à soma das áreas dos dois trapézios verdes (azuis) mais o dobro da área sob uma corda (i.e.
