Let ABC be an isosceles triangle with AB = AC and A'B'C'D' the inscribed square in ABC based on AC (A' on AB, B' on BC, C' on CA near C and D' on CA near A). Let y be the side of the square inscribed in BA'C' based on A'B' and z the side of the square inscribed in the right triangle D'A'A based on AD'.
If y = z, find the angles of the triangle.
Solution:
Let x be the side of the square A'B'C'D' and h the altitude BB*.
we have:
x = bh / (b+h)
y = bh^2 / (b+h)^2 = x^2 / b
z = xAD' / (x + AD') = x^2cotA /(x + xcotA) = xcotA / (1 + cotA)
y = z ==> x / b = cotA / (1 + cotA) ==>
(h/b) / (1 + (h/b)) = cotA / (1 + cotA) ==>
(h/c) / (1 + (h/c)) = cotA / (1 + cotA) ==>
sinA / (1 + sinA) = cotA / (1 + cotA) ==>
sinA = cotA ==> (cosA)^2 + cosA - 1 = 0 ==>
cosA = (-1 + sqrt(5))/2 (since 0 < A < 180) ==> A = Αrccos(1/φ) and B = C = 90 - (A/2)
Note: B* divides AC in golden ratio.
Addendum (13 April):
There is one more solution for A > 90 d.
See Hyacinthos Message 19985
