Quintright inequality.
Home.
Let Θn stand for one of the three relations greater-than, equal-to, or less-than. The aim is to compare a quintright to zero, using only integer (in other words, exact) calculations:
A Θ1 0
or, in expanded notation:
2 Aq cos (18°) + 2 Ar cos (36°) + 2 As cos (54°) + 2 At cos (72°) Θ1 0
and to decide which of the three relations Θ1 must stand for.

To reduce notational congestion on this page, Aq will hereafter be abbreviated as q, Ar as r, et cetera.

In the original expression, substitute radical forms for the cosines and multiply both members by two:
q √(10 + √20) + r (√5 + 1) + s √(10 − √20) + t (√5 − 1) vs1 0
and rearrange to give a convenient equivalent form:
q √(10 + √20) + s √(10 − √20) Θ1 − r (√5 + 1) − t (√5 − 1) [#1]

Squaring both members of an inequality is often a useful technique to eliminate square roots, but it possibly reverses the sense of an inequality if one member is negative, and certainly reverses it if both members are negative. To get the right answer, the problem must be broken into numerous cases according to the signs of members.

Left Member The question here is to determine Θ2 in:
q √(10 + √20) + s √(10 − √20) Θ2 0 [#2]
Observe that 10 + √20 and 10 − √20 are themselves positive, and so are their square roots.

Θ2 in #2 s < 0 s = 0 s > 0
q < 0 Θ2 is less-thanΘ2 is less-thansee case alpha
q = 0 Θ2 is less-thanΘ2 is equal-toΘ2 is greater-than
q > 0 see case betaΘ2 is greater-thanΘ2 is greater-than

Case Alpha Equivalent to #2 is:
s √(10 − √20) Θ2 −q √(10 + √20)
Both members are positive, so square it without exchange:
s2(10 − √20) Θ2 q2(10 + √20)
Rearrange:
10(s2 − q2) Θ2 √20(s2 + q2) [#3]

Θ2 in #3 s2 + q2 = 0 s2 + q2 > 0
s2 − q2 < 0 Θ2 is less-thanΘ2 is less-than
s2 − q2 = 0 Θ2 is equal-toΘ2 is less-than
s2 − q2 > 0 Θ2 is greater-thansee below

When in #3 both members are positive, square it without exchange:

100(s2 − q2)2 Θ2 20(s2 + q2)2
which is all integers, and thus can easily be evaluated exactly.
Case Beta Recall:
s √(10 − √20) Θ2 −q √(10 + √20)
Both members are negative, so square it with exchange:
q2(10 + √20) Θ2 s2(10 − √20)
Rearrange:
√20(s2 + q2) Θ2 10(s2 − q2) [#4]

Θ2 in #4 s2 − q2 < 0 s2 − q2 = 0 s2 − q2 > 0
s2 + q2 = 0 Θ2 is greater-thanΘ2 is equal-toΘ2 is less-than
s2 + q2 > 0 Θ2 is greater-thanΘ2 is greater-thansee below

When in #4 both members are positive, square it without exchange:

20(s2 + q2)2 Θ2 100(s2 − q2)2
which is all integers.
Right Member The question here is to determine Θ3 in:
− r (√5 + 1) − t (√5 − 1) Θ3 0
or equivalently
(t − r) Θ3 √5(t + r) [#5]

Θ3 in #5 t + r < 0 t + r = 0 t + r > 0
t − r < 0 see case gamma Θ3 is less-than Θ3 is less-than
t − r = 0 Θ3 is greater-than Θ3 is equal-to Θ3 is less-than
t − r > 0 Θ3 is greater-than Θ3 is greater-than see case delta

Case Gamma In #5 both members are negative, so square it with exchange:
5(t + r)2 Θ3 (t − r)2
which is all integers.
Case Delta In #5 both members are positive, so square it without exchange:
(t − r)2 Θ3 5(t + r)2
which is all integers.
Summary With Θ2 and Θ3 established, resolution of Θ1 can begin. This table handles many cases:

Θ1 in #1 Θ3 is less-than Θ3 is equal-to Θ3 is greater-than
Θ2 is less-than see case epsilonΘ1 is less-thanΘ1 is less-than
Θ2 is equal-to Θ1 is greater-thanΘ1 is equal-toΘ1 is less-than
Θ2 is greater-than Θ1 is greater-thanΘ1 is greater-thansee case zeta

Case Epsilon In #1 both members are negative, so square it with exchange:
r2 (6 + 2 √5) + t2 (6 − 2 √5) + 8 r t Θ1 q2 (10 + √20) + s2 (10 − √20) + 2 q s √80
Move the radicals left, and divide by two:
√5(r2 − t2 − q2 + s2 − 4 q s) Θ1 − 3 r2 − 4 r t + 5 q2 − 3 t2 + 5 s2
Define
X = r2 − t2 − q2 + s2 − 4 q s
and
Y = − 3 r2 − 4 r t + 5 q2 − 3 t2 + 5 s2
This consolidates the problem to
√5X Θ1 Y [#6]

Θ1 in #6 Y < 0 Y = 0 Y > 0
X < 0 same as Y2 Θ1 5X2Θ1 is less-thanΘ1 is less-than
X = 0 Θ1 is greater-thanΘ1 is equal-toΘ1 is less-than
X > 0 Θ1 is greater-thanΘ1 is greater-thansame as 5X2 Θ1 Y2

Case Zeta In #1 both members are positive, so square it without exchanging:
q2 (10 + √20) + s2 (10 − √20) + 2 q s √80 Θ1 r2 (6 + 2 √5) + t2 (6 − 2 √5) + 8 r t
Move the radicals right, and divide by two:
− 3 r2 − 4 r t + 5 q2 − 3 t2 + 5 s2 Θ1 √5(r2 − t2 − q2 + s2 − 4 q s)
Again define
X = r2 − t2 − q2 + s2 − 4 q s
and
Y = − 3 r2 − 4 r t + 5 q2 − 3 t2 + 5 s2
This consolidates the problem to
Y Θ1 √5X [#7]

Θ1 in #7 X < 0 X = 0 X > 0
Y < 0 same as 5X2 Θ1 Y2Θ1 is less-thanΘ1 is less-than
Y = 0 Θ1 is greater-thanΘ1 is equal-toΘ1 is less-than
Y > 0 Θ1 is greater-thanΘ1 is greater-thansame as Y2 Θ1 5X2