A circle is drawn so that no part of it lies outside a regular hexagon. If such a circle does not touch all six sides of the hexagon, what is the maximum number of sides that it could touch?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
After fiddling around for a few minutes, we can see that the answer should be 2 sides. While we can convince ourselves that the answer is 2 , how can we justify this mathematically? This turns out to be tricky.

Consider hexagon $A B C D E F$ and a circle contained entirely inside it.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-218.jpg?height=323&width=371&top_left_y=2251&top_left_x=931)

Clearly, it is possible for the circle to touch 1 or 2 sides only.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-219.jpg?height=336&width=371&top_left_y=255&top_left_x=931)

Next, we make a few notes:

- If the circle touches all 6 sides, then its radius is half the distance between opposite sides (eg. $A B$ and $D E$ ).
- If the circle actually touches one pair of opposite sides, then its radius is half the distance between opposite sides, so in order to be contained entirely inside the hexagon, it must touch all 6 sides.
- If a circle inside the hexagon touches at least 4 sides, then it must touch at least one pair of opposite sides, so must touch all 6 sides. So if the circle touches fewer than six sides, it must touch 1, 2 or 3 sides.

Is it possible for the circle to touch 3 sides and not all 6 ?

If so, then we need to make sure that the circle does not touch opposite sides.

There are two ways in which this can be done - if the circle touches three consecutive sides (for example, $A B, B C, C D$ ) or three sides no two of which are consecutive (for example, $A B, C D$, $E F$ ).

In order to complete our justification, we need to examine these two cases and use the fact that if a circle is tangent to two lines, then its centre must lie on the angle bisector of the angle formed at the point of intersection of these two lines.

Case 1: Circle touches $A B, B C, C D$

Since the circle is tangent to $A B$ and $B C$, then its centre must lie on the angle bisector of $\angle A B C$, which is diagonal $B E$ of the hexagon.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-219.jpg?height=321&width=356&top_left_y=1902&top_left_x=947)

Since the circle is tangent to $B C$ and $C D$, then its centre must lie on the angle bisector of $\angle B C D$, which is diagonal $C F$ of the hexagon.

Since the centre lies on $B E$ and on $C F$, then the centre of the circle lies at the centre of the hexagon, and so the circle must touch all 6 sides of the hexagon.

Case 2: Circle touches $A B, C D, E F$

Since the circle is tangent to $A B$ and $C D$, then its centre must lie on the angle bisector of the angle formed by extending $A B$ and $D C$ to their point of intersection. This angle bisector will be, by symmetry, the perpendicular bisector of $B C$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-220.jpg?height=314&width=371&top_left_y=412&top_left_x=931)

Since the circle is tangent to $C D$ and $E F$, then its centre must lie on the angle bisector of the angle formed by extending $C D$ and $E F$ to their point of intersection. This angle bisector will be, by symmetry, the perpendicular bisector of $D E$.

Since the centre lies on the perpendicular bisectors of $B C$ and $D E$, then the centre of the circle lies at the centre of the hexagon, and so the circle must touch all 6 sides of the hexagon.

Therefore, if the circle touches at least 3 sides of the hexagon, then it touches all 6 sides. Thus, the maximum number of sides it can touch without touching all 6 sides is 2 sides.

(This problem is an example of one where the answer can be obtained relatively quickly, but the justification is quite difficult. We have included this justification for completeness.)

ANswer: (B)

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