10.5 What is the most non-isosceles triangle? 319
If we assume the regular pentagon inscribed in a unit circle, the side-
lengths of these squares are respectively
√
10
8
(√
5+1
)
and √
5
2
(√
10 − 2
√
5 −
√
5+1
)
.
The first square exceeds the second one by1
8(
20 + 5
√
2+
√
10 − 4
√
5 − 4
√
50 − 10
√
5
)
≈ 0. 0324786 ···.
One of you [GC] suggested an “easy” construction of the square as
follows.
This indeed is not exact, but it gives an approximate rectangle with
base
1
2
√
5(2− 24
√
5+
√
5),
which is smaller only by 0 .08%!
As for the triangles, the first one has length√
5
2(
4+
√
5 −
√
15 + 6
√
5
)
,
and the second one has shorter length
√10(√5+1)
(
√
5 −1)√
5 −
√
5+2
√
6. The difference
between the two is approx 0. 0328 ···.