a) ... 81°
b) ... 90°
c) ... 96°
Pepe Chapuzas solved this problem (these three problems). He found the natural solutions and the other solutions...
Dear Teacher:
The sum of the four internal angles of a quadrilateral equals 360°
alpha + beta + gamma + delta = 360°
If R is the common ratio of the geometric progression...
gamma/R2 + gamma/R + gamma + gamma·R = 360
gamma + gamma·R + (gamma−360)·R2 + gamma·R3 = 0
Then...a)
81R3 − 279R2 + 81R + 81 = 0
R = 3
alpha = 81°/9 = 9°
alpha = 81°/9 = 9°
beta = 81°/3 = 27°
delta = 3 · 81° = 243°
b)
90R3 − 270R2 + 90R + 90 = 0
R = 1
alpha = beta = gamma = delta = 90°
alpha = beta = gamma = delta = 90°
The progression is constant and the quadrilateral is a square.
c)
96R3 − 264R2 + 96R + 96 = 0
R = 2
alpha = 96°/4 = 24°
alpha = 96°/4 = 24°
beta = 96°/2 = 48°
delta = 2 · 96° = 192°
The other solutions are irrational (even negative).
a)
b)
c)
Good ole Pepe - he usually gets there 😏
ResponderEliminar