The All-In-One Equation
lo g 2 ( ∣ ∣ sin 2 θ + 2 cos θ + 1 ∣ ∣ ) + lo g 1 6 ( ⌊ cos 2 θ + 2 sin θ + 1 ⌋ 2 ) = lo g 4 6
Determine the sum of all solutions in degrees for 0 ∘ < θ < 3 6 0 ∘ .
Notations:
- ⌊ ⋅ ⌋ denotes the floor function .
- ∣ ⋅ ∣ denotes the absolute value function .
The answer is 90.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
To make it easy for some:
l
o
g
b
n
x
n
=
l
o
g
b
x
.
S
o
l
o
g
b
x
=
l
o
g
b
x
a
n
d
l
o
g
b
n
x
n
=
l
o
g
b
x
.
S
o
l
o
g
4
x
=
l
o
g
4
x
=
l
o
g
1
6
x
2
.
b
=
4
h
e
r
e
a
n
d
x
=
f
(
θ
)
.
Log in to reply
Nice :) Though the point of posting the problem on Brilliant.org is for people to solve it without graphing...
Log in to reply
Thank you. Mine was a comment not a solution.
The equation can be simplified to ∣ sin ( θ ) 2 + 2 cos ( θ ) + 1 ∣ ∗ ⌊ cos ( θ ) 2 + 2 sin ( θ ) + 1 ⌋ = 6 . Because ⌊ cos ( θ ) 2 + 2 sin ( θ ) + 1 ⌋ is an integer, and the maximum value of ∣ sin ( θ ) 2 + 2 cos ( θ ) + 1 ∣ is 3 , then ∣ sin ( θ ) 2 + 2 cos ( θ ) + 1 ∣ is also an integer. If we let x = sin ( θ ) and sketch the graph of f ( x ) = ∣ 1 − x 2 + 2 x + 1 ∣ , we notice that the integral values f ( x ) are f ( x ) = 0 , 1 , 2 , 3 for − 1 < x < 1 . If we set f ( x ) to those values, we see that f ( x ) = 2 and f ( x ) = 3 are the only ones that work in the original equation. These are achieved when x = 0 ° and x = 9 0 ° , so 0 + 9 0 = 9 0 which is the final answer.