Thanks to Lucas from Brazil for the suggestion! This problem is adapted from a Kangaroo Maths competition in 2020 for grades 9-10 (around 15 years old).

A circle has diameter AB. A line starts at A and zig-zags exactly 4 times between the diameter and the circumference until it ends at B, as shown below. If each of the angles the line makes with the diameter has the same measure α, what is α equal to?

As usual, watch the video for a solution.

Zig-Zag Line Puzzle

Or keep reading.

.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
P
U
Z
Z
L
E
.
.
.
.
Answer To Zig-Zag Line Puzzle

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

There are many ways to solve the problem, but I will explain one method using the inscribed angle theorem. This states if an inscribed angle has measure x, the arc it subtends is equal to 2x. (For a proof, see Khan Academy.)

Label the other congruent angles as α and let β be the angle the line makes when “bouncing” off the circumference. We will calculate the measure of arc BC in two different ways. First angle CAB is an inscribed angle equal to α that subtends arc BC, so the arc BC has twice the measure 2α.

But we also have:

arc(BC) = arc(BD) + arc(DC)
arc(BC) = 2(angle DFB) + 2(angle CED)
arc(BC) = 2β + 2β
arc(BC) = 4β

We thus have:

arc(BC) = 4β = 2α
⇒ β = α/2

We also have α, α, β are three angles of a triangle, so the angles will sum to 180°. We then substitute for β to solve for α:

2α + β = 180°
2α + α/2 = 180°
(5/2)α = 180°
α = 72°

Thus angle α = 72 degrees.

References 