1220. A sangaku

    Pepe Chapuzas has become fond of geometric sangakus. I think he even invents them... If he finds out some difficult sangaku, he proposes it to his classmates as a challenge. This is the last one:

    A circular sector has a radius of 1 foot and it can be opened and closed as a fan. When drawing the chord of the arc of the sector, this is cut into two pieces: a circular segment (blue) and an isosceles triangle (green). If we inscribe a circle (red) in each part as shown in the drawing..., what should be the angle of the sector so that the circles are equal?

    Nina Guindilla began...

    Dear Teacher:
    If the angle to be calculated measures  2x , then the diameter of the upper circle is the sagitta of the circular segment and measures (in feet):

Diameter  =  1 − cos x

    The lower circle is inscribed in the isosceles triangle, so its diameter can be calculated from the area and the perimeter of the triangle as follows: 

Diameter  =  4 · Area / Perimeter  =  2 · sin x · cos x / (1 + sin x)

    If both diameters are equal we have a trigonometric equation:

1 − cos x  =  2 · sin x · cos x / (1 + sin x)

    Or better yet...

(1 + sin x) · (1 − cos x)  =  2 · sin x · cos x

    Here Nina got stuck... She tried it in several ways... unsuccessfully... This was one of those magical moments that only we teachers can taste: I told her about the change of variables: 

tan (x / 2)  =  t

    Hence:

sin x  =  2t / (1 + t²)

cos x  =  (1 − t²) / (1 + t²)

    Now she could go on ...

(1 + 2t / (1 + t²)) · (1 − (1 − t²) / (1 + t²))  =  2 · 2t / (1 + t²) · (1 − t²) / (1 + t²)

(1 + t² + 2t) / (1 + t²) · (1 + t² − 1 + t²) / (1 + t²)  =  4t / (1 + t²) · (1 − t²) / (1 + t²)

 

    Multiplying twice by  1 + t²  (≠ 0)

(1 + t² + 2t) · 2t²  =  4t · (1 − t²)

    Dividing by  2t  (≠ 0)

(1 + t² + 2t) · t  =  2 · (1 − t²)

t + t³ + 2t²  =  2 − 2t²

t³ + 4t² + t − 2  =  0

    Dividing by  t + 1  (≠ 0)

t² + 3t − 2  =  0

t  =  (− 3 + √17) / 2

t  ≠  (− 3 − √17) / 2

    And substituting back...

x / 2  =  arctan t  =  arctan ((− 3 + √17) / 2)

    The angle sought is...

2x  =  4 · arctan ((− 3 + √17) / 2)  =  117° 15' 58"

 

    Okay!

    Nina Guindilla won good grades...