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[Geometry] Quadrature of the Parabola

2021-11-01 09:39 作者:AoiSTZ23  | 我要投稿

By: Tao Steven Zheng (鄭濤)

【Problem】

In "Quadrature of a Parabola", Archimedes (c. 287 - 212 BC) calculated the area segment of a parabola using the method of exhaustion. He begins by inscribing the largest triangle (triangle ABC) within the parabolic segment. The method proceeds by filling the remaining space with more triangles whereby each stage doubles the number of triangles. Archimedes determined that the total area of the triangles constructed at each stage is 1/4 of the area of the triangles constructed in the previous stage.


Let A_T be the area of the first triangle (pictured in yellow), and A_P be the area of the parabolic segment. Determine the ratio of the two areas A_P%20%3A%20A_T.

【Solution】

Let A_1%20 represent the total area of the blue triangles, %20A_2 be the total area of the light red triangles , A_3 be the total area of the dark red triangles. Each stage is 1/4 the area of the previous. So,

A_1%20%3D%20%5Cfrac%7B1%7D%7B4%7DA_T

%20A_2%20%3D%20%5Cfrac%7B1%7D%7B4%7DA_1%20%3D%20%5Cfrac%7B1%7D%7B16%7DA_T

A_3%20%3D%20%5Cfrac%7B1%7D%7B4%7DA_2%20%3D%20%5Cfrac%7B1%7D%7B64%7DA_T

This process continues indefinitely, generating the infinite geometric series:

A_P%20%3D%20A_T%20%2B%20%5Cfrac%7B1%7D%7B4%7DA_T%20%2B%20%5Cfrac%7B1%7D%7B16%7DA_T%20%2B%20%5Cfrac%7B1%7D%7B64%7DA_T%20%2B%20...

A_P%20%3D%20%5Cleft(1%20%2B%20%5Cfrac%7B1%7D%7B4%7D%20%2B%20%5Cfrac%7B1%7D%7B16%7D%20%2B%20%5Cfrac%7B1%7D%7B64%7D%20%2B%20...%20%5Cright)%20A_T%20


Here, the first term is %20t_1%20%3D%201%20 and the common ratio is %20r%20%3D%201%2F4. Using the infinite geometric formula gives:

%7BS%7D_%7B%5Cinfty%7D%20%3D%20%5Cfrac%7Bt_1%7D%7B1-r%7D

%20%7BS%7D_%7B%5Cinfty%7D%20%3D%20%5Cfrac%7B1%7D%7B1-1%2F4%7D

%7BS%7D_%7B%5Cinfty%7D%20%3D%20%5Cfrac%7B4%7D%7B3%7D

Subsequently,

A_P%20%3D%20%5Cfrac%7B4%7D%7B3%7DA_T

Therefore, the ratio of the two areas is

%20A_P%20%3A%20A_T%20%3D%204%3A3%20


【Historical Note】

Archimedes was not aware of the infinite geometric series formula. He used proof by contradiction to arrive at the result!


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