The key to the question is "Normal"

Today I went for a normal walk

A "normal" in geometry is not something "common" or "ordinary". A geometrical "normal" is always perpendicular to something else.

Let us say that this something is...

...a line that extends from a fixed point, to the walker.

If the walker then walks along the "normal" of this thing, they will...

...always walk in a circle. The point is the center of the circle. The line in question is the radius of the circle. Walking along the normal of a radius of a circle means you walk along the circle itself.

If you walk this way, one foot will move further than the other.

Now let us do the Maths (sorry, Spoiler and MathML cannot be combined):

Let us assume that the distance between the walker's feet are, well, one foot, 30 cm, or 0.3 meters. Let us assume they went $x$ number of laps around the circle. So the distances walked is $(r + 0) \cdot x \cdot 2\pi = 3000$ for the inner foot, and $(r + 0.3) \cdot x \cdot 2\pi = 3100$ for the outer.

Divide one by the other and we get:

$\frac{(r + 0) \cdot x \cdot 2\pi}{(r + 0.3) \cdot x \cdot 2\pi} = \frac{3000}{3100} \Rightarrow$

$x\cdot 2\pi$ cannot be zero so we can cancel that factor out

$\frac{r}{r + 0.3} = \frac{30}{31} \Rightarrow$

$31\cdot r = 30\cdot r + 9$

$r = 9$

Hence...

...they walked along a circle that had a radius of 9 meters, with the inner foot directly on the circle. I leave it to the reader to figure out how many laps it was.

The key to the question is "Normal"

Today I went for a normal walk

A "normal" in geometry is not something "common" or "ordinary". A geometrical "normal" is always perpendicular to something else.

Let us say that this something is...

...a line that extends from a fixed point, to the walker.

If the walker then walks along the "normal" of this thing, they will...

...always walk in a circle. The point is the center of the circle. The line in question is the radius of the circle. Walking along the normal of a radius of a circle means you walk along the circle itself.

If you walk this way, one foot will move further than the other.

Now let us do the Maths:

Let us assume that the distance between the walker's feet are, well, one foot, 30 cm, or 0.3 meters. Let us assume they went $x$ number of laps around the circle. So the distances walked is $(r + 0) \cdot x \cdot 2\pi = 3000$ for the inner foot, and $(r + 0.3) \cdot x \cdot 2\pi = 3100$ for the outer.

Divide one by the other and we get:

$\frac{(r + 0) \cdot x \cdot 2\pi}{(r + 0.3) \cdot x \cdot 2\pi} = \frac{3000}{3100} \Rightarrow$

$x\cdot 2\pi$ cannot be zero so we can cancel that factor out

$\frac{r}{r + 0.3} = \frac{30}{31} \Rightarrow$

$31\cdot r = 30\cdot r + 9$

$r = 9$

Hence...

...they walked along a circle that had a radius of 9 meters, with the inner foot directly on the circle. I leave it to the reader to figure out how many laps it was.

The key to the question is "Normal"

Today I went for a normal walk

A "Normal" in geometry is not something "common" or "ordinary". A geometrical "normal" is always perpendicular to something else.

Let us say that this something is...

...a line that extends from a fixed point, to the walker.

If the walker then walks along the "normal" of this thing, they will...

...always walk in a circle. The point is the center of the circle. The line in question is the radius of the circle. Walking along the normal of a radius of a circle means you walk along the circle itself.

If you walk this way, one foot will move further than the other.

Now let us do the Maths (sorry, Spoiler and MathML cannot be combined):

Let us assume that the distance between the walker's feet are, well, one foot, 30 cm, or 0.3 meters. Let us assume they went $x$ number of laps around the circle. So the distances walked is $(r + 0) \cdot x \cdot 2\pi = 3000$ for the inner foot, and $(r + 0.3) \cdot x \cdot 2\pi = 3100$ for the outer.

Divide one by the other and we get:

$\frac{(r + 0) \cdot x \cdot 2\pi}{(r + 0.3) \cdot x \cdot 2\pi} = \frac{3000m}{3100m} \Rightarrow$

$x\cdot 2\pi$ cannot be zero so we can cancel that factor out

$\frac{r}{r + 0.3} = \frac{30}{31} \Rightarrow$

$31\cdot r = 30\cdot r + 9$

$r = 9$

Hence...

...they walked along a circle that had a radius of 9 meters, with the inner foot directly on the circle. I leave it to the reader to figure out how many laps it was.

The key to the question is "Normal"

Today I went for a normal walk

A "normal" in geometry is not something "common" or "ordinary". A geometrical "normal" is always perpendicular to something else.

Let us say that this something is...

...a line that extends from a fixed point, to the walker.

If the walker then walks along the "normal" of this thing, they will...

...always walk in a circle. The point is the center of the circle. The line in question is the radius of the circle. Walking along the normal of a radius of a circle means you walk along the circle itself.

If you walk this way, one foot will move further than the other.

Now let us do the Maths (sorry, Spoiler and MathML cannot be combined):

Let us assume that the distance between the walker's feet are, well, one foot, 30 cm, or 0.3 meters. Let us assume they went $x$ number of laps around the circle. So the distances walked is $(r + 0) \cdot x \cdot 2\pi = 3000$ for the inner foot, and $(r + 0.3) \cdot x \cdot 2\pi = 3100$ for the outer.

Divide one by the other and we get:

$\frac{(r + 0) \cdot x \cdot 2\pi}{(r + 0.3) \cdot x \cdot 2\pi} = \frac{3000}{3100} \Rightarrow$

$x\cdot 2\pi$ cannot be zero so we can cancel that factor out

$\frac{r}{r + 0.3} = \frac{30}{31} \Rightarrow$

$31\cdot r = 30\cdot r + 9$

$r = 9$

Hence...

...they walked along a circle that had a radius of 9 meters, with the inner foot directly on the circle. I leave it to the reader to figure out how many laps it was.