Geometry

Tessellations

Which shapes can cover a floor with copies of themselves, leaving no gaps and no overlaps, and what decides it?

You need

Angle sums of polygonsGeometry

Where you are 3 in Polygons

Geometry11

Two-dimensional shapes15

Polygons3

Naming and classifying polygons Angle sums of polygons

Tessellations

Why this exists

A tiled floor, a brick wall, the cells of a honeycomb and a chessboard all cover a flat surface with copies of one shape, and each shape seems to fit against itself without any effort. Cut a few regular pentagons out of card and try the same thing, and it fails at once: three of them around a point leave a thin gap, and a fourth will not go in.

So which shapes can do it? The question sounds like one about shapes, but it is mostly a question about angles. Wherever tiles meet at a point, their corners have to fill the space around that point exactly, and the angle sum of a polygon says what each tile’s corners measure. That one requirement decides which regular polygons tile, shows that every triangle and every quadrilateral tiles, and rules out mixtures of shapes that look as if they ought to work.

Definition

A tessellation, or tiling, of the plane is a collection of shapes, called tiles, that cover the whole plane with no gaps and no overlaps: every point of the plane lies in at least one tile, and two tiles share at most points on their boundaries. In this lesson every tile is a polygon. A shape tiles the plane when there is a tessellation in which every tile is a copy of it. A copy may be slid, turned or flipped over, but not stretched, so all the tiles are congruent.

A vertex of a tiling is a point that is a corner of at least one tile. A tiling is edge-to-edge when no corner of a tile lies partway along a side of another tile; then two tiles that touch along a side share the whole of it. The square grid on the left of Figure 1 is edge-to-edge. The brick wall on the right is not, because two bricks have their corners in the middle of the long side of the brick above them.

Figure 1Left, edge-to-edge. Right, not edge-to-edge; the two corners at the marked point fill only a straight angle.

A tiling is regular when it is edge-to-edge and every tile is the same regular polygon. It is semiregular when it is edge-to-edge, uses regular polygons of two or more kinds, and the polygons are arranged the same way around every vertex.

To describe the arrangement at a vertex, go once round it and list the number of sides of each polygon you meet, separated by dots. The square grid has four squares at every vertex, written 4.4.4.4. The tiling in Example 1 has a square and two octagons at every vertex, written 4.8.8. The order matters: 3.3.4.3.4 and 3.3.3.4.4 use the same five polygons placed differently.

The angles of regular polygons are worked out in regular polygons and their angles. This lesson uses the two forms of the result: each interior angle of a regular n-gon is

\frac{(n-2)\cdot 180^\circ}{n} = 180^\circ - \frac{360^\circ}{n}

which is 60^\circ for the equilateral triangle, 90^\circ for the square, 108^\circ for the regular pentagon, 120^\circ for the regular hexagon and 135^\circ for the regular octagon.

Intuition

Look at any vertex of the three tilings in Figure 2. The tiles around it close up completely: no sliver is left over and nothing is folded over anything else. So their corners at that point together make one full turn, 360^\circ. With triangles that is 6 \times 60^\circ, with squares 4 \times 90^\circ, with hexagons 3 \times 120^\circ.

Figure 2The three regular tilings. At every vertex the corners fill a full turn.

Now try regular pentagons, each with corners of 108^\circ. Three of them around a point use 3 \times 108^\circ = 324^\circ and leave a gap of 36^\circ, too narrow for another pentagon. Four would need 432^\circ, more than a full turn, so the fourth would overlap the first.

Figure 3Three regular pentagons leave a gap of 36^\circ, and a fourth will not fit.

There is one other way tiles can meet, the one the brick wall shows. When corners land partway along the side of another tile, that side is a straight line through the point, and it takes up half the turn on its own. The corners there only have to fill the other half, 180^\circ.

Everything in this lesson comes from these two pictures: the corners at a point fill 360^\circ, or they fill 180^\circ against a straight side.

Results

The vertex rule. At a vertex of a tiling where every tile that touches the vertex has a corner there, the angles of those corners add to 360^\circ. At a vertex that lies partway along a side of a tile, the angles of the corners there add to 180^\circ.

The regular tilings. A regular n-gon tiles the plane if and only if n = 3, 4 or 6. So there are exactly three regular tilings, 3.3.3.3.3.3, 4.4.4.4 and 6.6.6.

n	polygon	interior angle	tiles at each vertex
3	equilateral triangle	60^\circ	6
4	square	90^\circ	4
6	regular hexagon	120^\circ	3

Every triangle tiles the plane, and every quadrilateral tiles the plane, convex or not. In the tilings built below, each angle of the triangle appears twice around every vertex, and each angle of the quadrilateral appears once.

The semiregular tilings. Exactly eight arrangements of polygons around a vertex give semiregular tilings. This is stated here, not proved:

\begin{aligned} &3.12.12, \quad 4.6.12, \quad 4.8.8, \quad 3.6.3.6, \\ &3.4.6.4, \quad 3.3.3.4.4, \quad 3.3.4.3.4, \quad 3.3.3.3.6 \end{aligned}

Each one passes the vertex rule. Showing that no others exist takes a longer case-by-case argument that this lesson does not give, because, as Example 5 shows, passing the vertex rule is not enough.

Why it is true

The vertex rule

Take a point P of the tiling that is a corner of at least one tile, and look at a small circle around P. Each tile that touches P covers part of the space around it: a tile with a corner at P covers a wedge equal to its interior angle there, and a tile with P partway along a side covers a half-turn, 180^\circ. No tile can have P inside it, since then that tile alone would fill all the space around P and leave no room for the tile whose corner is there.

Because there are no gaps, the wedges cover the whole turn around P, and because there are no overlaps, no part is covered twice. So by angle addition the wedges add to a full turn, 360^\circ. At most one tile can have P partway along its side: two would use 180^\circ + 180^\circ and leave nothing for a corner. With no such tile the corners add to 360^\circ; with one, they add to 360^\circ - 180^\circ = 180^\circ.

Only three regular polygons tile

Suppose regular n-gons tile the plane, and look at any corner of any tile. By the vertex rule, a whole number k of corners, each of \displaystyle \frac{(n-2)\cdot 180^\circ}{n}, adds to 360^\circ or to 180^\circ.

In the first case,

k = \frac{360n}{(n-2)\cdot 180} = \frac{2n}{n-2} = 2 + \frac{4}{n-2}

which is a whole number only when n - 2 divides 4, that is n - 2 = 1, 2 or 4. So n = 3, 4 or 6, with k = 6, 4 or 3.

In the second case,

k = \frac{n}{n-2} = 1 + \frac{2}{n-2}

which is a whole number only when n - 2 = 1 or 2, giving n = 3 with k = 3 or n = 4 with k = 2. These are rows of triangles or of squares slid along each other, and they bring in no new polygon.

For any other n, a corner of a tile can be neither kind of vertex, so regular n-gons cannot tile. Figure 2 shows that the three that pass really do tile.

Every triangle tiles

Let the triangle be ABC with angles a, b and c. Turn a copy half a turn about the midpoint M of \overline{BC}. The half-turn swaps B and C and sends A to a point D with M the midpoint of \overline{AD}. In ABDC the diagonals bisect each other, so it is a parallelogram (properties of a parallelogram).

Parallelograms tile by sliding. Opposite sides are equal and parallel, so a copy slid along by the vector from A to B fits exactly against the first, and repeating this makes an endless strip whose two edges are straight lines. Copies of the strip slid by the vector from A to C stack against each other with nothing between them. Cutting every parallelogram along the same diagonal gives a tiling by copies of ABC, as in Figure 5.

Six triangles meet at each vertex, two with each angle, and

2(a + b + c) = 2 \cdot 180^\circ = 360^\circ

by the triangle angle sum.

Every quadrilateral tiles

Let the quadrilateral be ABCD with angles a, b, c and d, which add to 360^\circ by the angle sum. Place four copies around A:

the tile itself, with corner a at A;
its half-turn about the midpoint of \overline{AB}, which sends B to A, so it brings corner b to A and fits exactly along \overline{AB};
its half-turn about the midpoint of \overline{AD}, which brings corner d to A and fits exactly along \overline{AD};
the tile slid by the vector from C to A, which brings corner c to A.

The four corners add to a + b + c + d = 360^\circ, so they close up around A exactly. The slid copy’s sides at A are copies of \overline{CB} and \overline{CD}, and they lie exactly along the sides that the two turned copies have there, which are also copies of \overline{CB} and \overline{CD}. So neighbors meet side to side. The same four-tile pattern appears at every vertex, and the whole arrangement repeats when slid along either diagonal, which is how Figure 6 is built. The argument never used convexity, so it works for a concave quadrilateral too.

Worked examples

Example 1

Squares and octagons

Can regular octagons tile the plane on their own? If not, which regular polygon fills the space two of them leave at a vertex?

Solution

Each angle of a regular octagon is

\frac{(8-2)\cdot 180^\circ}{8} = 135^\circ

At a vertex where only corners meet, the angles must add to 360^\circ. Two octagons give 2 \times 135^\circ = 270^\circ, which leaves a gap. Three give 405^\circ, more than a full turn, so they would overlap. No whole number of octagons works, since \displaystyle 360 \div 135 = \frac{8}{3}. The other kind of vertex fails too: against a straight side the corners must add to 180^\circ, and one octagon gives 135^\circ while two give 270^\circ. So regular octagons cannot tile on their own.

Two octagons leave 360^\circ - 270^\circ = 90^\circ, exactly the angle of a square. Figure 4 shows that this is more than a fit at one point: octagons and squares alternate across the whole plane, with a square and two octagons at every vertex. That is the semiregular tiling 4.8.8.

Figure 4The tiling 4.8.8. Two octagons and a square at every vertex.

Check. 90^\circ + 135^\circ + 135^\circ = 360^\circ.

Example 2

The third polygon at a vertex

A square, a regular hexagon and one more regular polygon meet corner to corner at a vertex of an edge-to-edge tiling, and nothing else is there. How many sides does the third polygon have?

Solution

The square’s angle is 90^\circ and the hexagon’s is 120^\circ, together 210^\circ. By the vertex rule the third angle is 360^\circ - 210^\circ = 150^\circ. A regular n-gon’s angle is \displaystyle \frac{(n-2)\cdot 180^\circ}{n}, so

\frac{(n-2)\cdot 180}{n} = 150

Multiplying both sides by n, which is allowed because n is not zero, and then collecting the n terms:

\begin{aligned} 180n - 360 &= 150n \\ 30n &= 360 \\ n &= 12 \end{aligned}

The third polygon is a regular dodecagon, and the vertex is 4.6.12, one of the eight semiregular arrangements.

Check. \displaystyle \frac{(12-2)\cdot 180^\circ}{12} = 150^\circ, and 90^\circ + 120^\circ + 150^\circ = 360^\circ.

Example 3

Any triangle tiles

A triangle has angles 50^\circ, 60^\circ and 70^\circ. Show that copies of it tile the plane, and find the angles that meet at each vertex of the tiling.

Solution

Call the angles a = 50^\circ, b = 60^\circ and c = 70^\circ. Turning a copy half a turn about the midpoint of one side makes a parallelogram, and parallelograms slide into strips that stack, as in “Every triangle tiles” above. The result is Figure 5.

Figure 5Copies of one triangle. Each of its angles appears twice around every vertex.

At the marked vertex six triangles meet. Going round, the corners are a, c, b, a, c, b: each angle of the triangle twice. Their sum is

2(50^\circ + 60^\circ + 70^\circ) = 360^\circ

so they close up exactly, as the vertex rule requires. The tiling shows one more thing. The edges line up into three families of straight lines, and on one side of any line through the vertex lie three corners, one a, one b and one c. They fill a straight angle, which is the triangle angle sum, 50^\circ + 60^\circ + 70^\circ = 180^\circ, seen in the tiling.

Example 4

A quadrilateral tile

A quadrilateral ABCD has m\angle A = 70^\circ, m\angle B = 85^\circ and m\angle C = 125^\circ. Find m\angle D, and show that the corners meeting at each vertex of the tiling in Figure 6 fill a full turn.

Solution

The angles of a quadrilateral add to 360^\circ, so

\begin{aligned} m\angle D &= 360^\circ - 70^\circ - 85^\circ - 125^\circ \\ &= 80^\circ \end{aligned}

Copies tile by the half-turns described in “Every quadrilateral tiles”: turn the tile half a turn about the midpoint of a side and the copy fits along that side, with the side’s two ends swapped.

Figure 6Copies of one quadrilateral. Each of its four angles appears once around every vertex.

At the marked vertex four tiles meet, and each brings a different corner: a from the tile itself, b and d from the two turned copies, c from the slid copy. Their sum is

70^\circ + 85^\circ + 125^\circ + 80^\circ = 360^\circ

The same four corners meet at every vertex, so the tiling never needs anything but the angle sum of the quadrilateral to close up.

Example 5

A fit at one point that cannot spread

Two regular pentagons and a regular decagon fit exactly around a point. Show this, and then show that regular pentagons and regular decagons together cannot tile the plane.

Solution

The pentagon’s angle is 108^\circ and the decagon’s is \displaystyle \frac{(10-2)\cdot 180^\circ}{10} = 144^\circ, and 108^\circ + 108^\circ + 144^\circ = 360^\circ. So the corners fit around one point.

Now suppose some tiling used only these two polygons. At a vertex with p pentagon corners and q decagon corners and nothing else, 108p + 144q = 360. Dividing by 36,

3p + 4q = 10

With q = 0, 3p = 10 has no whole-number answer; with q = 1, p = 2; with q = 2, 3p = 2 has none; and q \geq 3 is already too big. So every such vertex has exactly two pentagons and one decagon, the type 5.5.10. The vertex against a straight side would need 108p + 144q = 180, that is 3p + 4q = 5, and no whole numbers solve that. So every vertex is of type 5.5.10, and the tiling is edge-to-edge.

Take any pentagon P. At each of its corners, the two other tiles there are the tiles across the two sides of P that meet at that corner, and one of them must be a pentagon and the other a decagon. So, going round P, the tiles across its sides must alternate: pentagon, decagon, pentagon, decagon, and so on. But P has five sides. If the tile across side 1 is a pentagon, then sides 2, 3, 4 and 5 have a decagon, a pentagon, a decagon and a pentagon across them. Sides 5 and 1 meet at a corner of P, and that corner would hold three pentagons, which is not 5.5.10. Starting with a decagon fails the same way. So no such tiling exists, even though the corners fit perfectly around a single point.

Example 6

Half a hexagon

A regular hexagon is cut in half by the segment joining the midpoints of two opposite sides. Show that copies of one half tile the plane, and check the angles at every vertex.

Solution

The half is a pentagon. Three of its corners are corners of the hexagon, each 120^\circ. The cut is a line of symmetry of the hexagon, so reflecting in it carries each side it crosses onto itself. The two angles the cut makes with that side are therefore equal, and since they add to a straight angle, each is 90^\circ. The pentagon’s angles are 90^\circ, 120^\circ, 120^\circ, 120^\circ and 90^\circ. As a check on the angle sum,

\begin{aligned} &90^\circ + 90^\circ + 120^\circ + 120^\circ + 120^\circ \\ &= 540^\circ = (5 - 2) \cdot 180^\circ \end{aligned}

The two halves of one hexagon are copies of each other: a half-turn about the hexagon’s center swaps them. So start from the tiling by regular hexagons and cut every hexagon the same way, as in Figure 7. Every piece is a copy of the half, and the pieces cover the plane because the hexagons did.

Figure 7Halves of hexagons. Every hexagon is cut the same way, and the cuts join into straight lines.

There are two kinds of vertex. At a corner of the original hexagons, three corners of 120^\circ meet, and 3 \times 120^\circ = 360^\circ. At the end of a cut, the cut of the hexagon above meets the same point, because the line through the centers of two hexagons that share a side crosses that side at its midpoint at right angles. So the cuts join into one straight line, which crosses the shared side, and four right angles meet there: 4 \times 90^\circ = 360^\circ.

So a pentagon can tile the plane even though the regular pentagon cannot. What decides it is the angles, not the number of sides.

Boundaries

The vertex rule is needed but not enough. Every tiling passes it at every vertex, but corners that fit around one point need not spread into a tiling. Example 5 and Exercise 21 give two combinations, 5.5.10 and 3.7.42, that fit at a point and still fail.

Not every polygon with the right number of sides tiles. Every triangle and every quadrilateral does. Some pentagons tile, as in Example 6, and the regular one does not. Some hexagons tile, the regular one among them, and some do not. It is a known theorem, stated here without its longer proof, that no convex polygon with seven or more sides tiles the plane.

“Semiregular” asks for the same arrangement at every vertex. Drop that and there are infinitely many edge-to-edge tilings by regular polygons. For example, rows of squares and rows of equilateral triangles of the same side can be stacked in any order, giving vertices of types 4.4.4.4, 3.3.3.4.4 and 3.3.3.3.3.3 mixed however you like.

Only the flat plane. The vertex rule uses the fact that the angles around a point in the plane make 360^\circ. On a curved surface such as a ball, the corners at a point can add to less than 360^\circ and still close up, which is why a pattern that cannot lie flat can still cover a ball (Exercise 23).

A patch is not a tiling. A tessellation covers the whole, endless plane. Laying a few tiles without a gap shows nothing about whether the pattern can go on for ever, which is why each tiling here is described by a rule that repeats: a slide, a half-turn, or a cut applied to every tile.

Where this goes next

A tiling that repeats is a pattern with symmetry. The parallelogram tiling is carried onto itself by translations, the quadrilateral tiling by half-turns about the midpoints of its sides as well, and the regular tilings by rotations about their vertices and centers and by reflections in lines, which is line symmetry and rotational symmetry applied to a whole plane rather than one shape. Sorting patterns by which of these moves carry them onto themselves is how wallpaper and fabric designs are classified, and changing a tile’s sides in matched ways, a bump on one side and the same dent on the side it slides onto, turns these tilings into the interlocking picture tilings of animals and birds.

The parallelogram that two triangles make is also how areas are found: a triangle is half a parallelogram, which is the step behind the area of a parallelogram and of a triangle. Two of the regular tilings fit inside each other: join the center of every hexagon to its corners and the hexagon tiling becomes the triangle tiling. Cutting a regular polygon into triangles from its center in the same way leads to the area of a regular polygon.

Counting corners, as in Exercises 16 to 18, is the start of a longer road. Doing the same count on a polyhedron, faces against edges against corners, leads to a relation between the three numbers that holds for every convex polyhedron, and the shortfall below 360^\circ at each corner, the 12^\circ in Exercise 23, is what makes a surface close up into a solid.

Exercises

Fitting regular polygons around a point

In Exercises 1 to 4, the regular polygons listed meet corner to corner at a point. Do their angles add to exactly 360^\circ? If not, say how far off they are.

Three equilateral triangles and two squares.
Two squares and two regular hexagons.
Two regular octagons and one equilateral triangle.
Four equilateral triangles and one regular hexagon.

The missing polygon

In Exercises 5 to 9, regular polygons meet corner to corner at a point and fill it exactly: the ones listed, and one more regular polygon (two copies of one in Exercise 9). Find how many sides the missing polygon has, or show that none exists.

An equilateral triangle, a regular heptagon (seven sides) and one more.
An equilateral triangle, a regular octagon and one more.
A square, a regular pentagon and one more.
Two equilateral triangles, a square and one more.
A regular dodecagon and two copies of one more regular polygon.

Which tiles fit

Use the vertex rule to decide whether regular nonagons (nine sides) can tile the plane.
In the triangle tiling of Figure 2, one row of triangles can be slid along so that the corners in that row land partway along the sides in the next row. How many triangle corners meet at such a point, and what do their angles add to?
A triangle has angles 35^\circ and 65^\circ. Find its third angle, and list the six angles that meet at each vertex when copies tile as in Figure 5.
A quadrilateral has angles 58^\circ, 104^\circ and 77^\circ. Find its fourth angle, and the sum of the corners that meet at each vertex when copies tile as in Figure 6.
A parallelogram has an angle of 70^\circ. Copies are slid into strips and the strips are stacked. Which four angles meet at each vertex?
In the brick wall of Figure 1, how many corners meet at a point partway along the long side of a brick, and what do their angles add to?

Counting in a tiling

In the tiling by regular hexagons, each vertex belongs to three hexagons and each side to two. Counting this way, how many vertices and how many sides are there for each hexagon?
In the 4.8.8 tiling every vertex has one square and two octagons. Count corners to show that there is one square for every octagon.
In the 3.6.3.6 tiling every vertex has two equilateral triangles and two regular hexagons. How many triangles are there for each hexagon?

Reasoning

Explain why regular pentagons cannot tile the plane, even when a corner may sit partway along another pentagon’s side.
Explain why no regular polygon with seven or more sides tiles the plane.
Exercise 5 shows that an equilateral triangle, a regular heptagon and a regular 42-gon fit exactly around a point. Show that there is no edge-to-edge tiling in which every vertex is of type 3.7.42.
Explain why every tiling by regular hexagons is edge-to-edge, while a tiling by squares need not be.
A common pattern for a soccer is sewn from regular pentagons and regular hexagons, with one pentagon and two hexagons at every corner. Add the angles of the three flat polygons at a corner, and explain what the answer says about laying the pattern out flat.

Answers

Yes. 3 \times 60^\circ + 2 \times 90^\circ = 180^\circ + 180^\circ = 360^\circ. This is the vertex of the semiregular tilings 3.3.3.4.4 and 3.3.4.3.4.
No. 2 \times 90^\circ + 2 \times 120^\circ = 420^\circ, which is 60^\circ too much, so the tiles would overlap.
No. 2 \times 135^\circ + 60^\circ = 330^\circ, which leaves a gap of 30^\circ.
Yes. 4 \times 60^\circ + 120^\circ = 360^\circ. This is the vertex of 3.3.3.3.6.
n = 42. The heptagon’s angle is \displaystyle \frac{5 \cdot 180^\circ}{7} = \frac{900^\circ}{7}, so the third angle is \displaystyle 360^\circ - 60^\circ - \frac{900^\circ}{7} = \frac{1200^\circ}{7}. Setting \displaystyle 180 - \frac{360}{n} = \frac{1200}{7} gives \displaystyle \frac{360}{n} = \frac{60}{7}.
n = 24. The third angle is 360^\circ - 60^\circ - 135^\circ = 165^\circ, so \displaystyle \frac{360}{n} = 180 - 165 = 15.
n = 20. The third angle is 360^\circ - 90^\circ - 108^\circ = 162^\circ, so \displaystyle \frac{360}{n} = 18.
n = 12. The last angle is 360^\circ - 2 \times 60^\circ - 90^\circ = 150^\circ, so \displaystyle \frac{360}{n} = 30.
None exists. Each copy would need \displaystyle \frac{360^\circ - 150^\circ}{2} = 105^\circ, so \displaystyle \frac{360}{n} = 75 and \displaystyle n = \frac{24}{5}, which is not a whole number.
No. Each angle is \displaystyle \frac{7 \cdot 180^\circ}{9} = 140^\circ. Corners alone would need \displaystyle 360 \div 140 = \frac{18}{7} of them, and corners against a straight side would need \displaystyle 180 \div 140 = \frac{9}{7}; neither is a whole number.
Three corners, and 3 \times 60^\circ = 180^\circ, a straight angle. With the side they sit on, which takes the other half-turn, they fill the space around the point.
The third angle is 180^\circ - 35^\circ - 65^\circ = 80^\circ. The six angles are 35^\circ, 35^\circ, 65^\circ, 65^\circ, 80^\circ and 80^\circ, which add to 360^\circ.
The fourth angle is 360^\circ - 58^\circ - 104^\circ - 77^\circ = 121^\circ. One corner of each kind meets at every vertex, and 58^\circ + 104^\circ + 77^\circ + 121^\circ = 360^\circ.
70^\circ, 110^\circ, 70^\circ, 110^\circ. Neighboring angles of a parallelogram add to 180^\circ, so the other angle is 110^\circ, and each of the four corners of the parallelogram appears once at every vertex.
Two corners, and 90^\circ + 90^\circ = 180^\circ. The brick’s long side fills the other half-turn.
Two vertices and three sides. Each hexagon has 6 corners, each shared by 3 hexagons, so \displaystyle \frac{6}{3} = 2; it has 6 sides, each shared by 2, so \displaystyle \frac{6}{2} = 3.
In a large patch with S squares, O octagons and V vertices: each vertex holds one square corner and each square has four corners, so V = 4S. Each vertex holds two octagon corners and each octagon has eight, so 2V = 8O. Then 8S = 8O, so S = O. The tiles at the edge of the patch upset the count slightly, by less and less as the patch grows.
Two triangles for each hexagon. Counting triangle corners, 2V = 3T; counting hexagon corners, 2V = 6H. So 3T = 6H, that is T = 2H.
Each angle of a regular pentagon is 108^\circ. Corners alone would need k \times 108 = 360, so \displaystyle k = \frac{10}{3}; corners against a straight side would need k \times 108 = 180, so \displaystyle k = \frac{5}{3}. Neither is a whole number, so no corner of any pentagon can be a vertex of either kind, and the vertex rule fails wherever a corner is.
For n \geq 7 the angle \displaystyle 180^\circ - \frac{360^\circ}{n} is more than 120^\circ and less than 180^\circ. Against a straight side, one such corner is less than 180^\circ and two are more. With corners alone, two are less than 360^\circ and three are more than 360^\circ. So no whole number of corners fits either way.
Suppose every vertex is 3.7.42, and take a triangle T. At each corner of T the other two tiles are a heptagon and a 42-gon, and they are the tiles across the two sides of T that meet there. So across the sides of T heptagons and 42-gons must alternate. With three sides that is impossible: two of the sides would have the same kind across them, and those two sides meet at a corner of T, which would then hold two heptagons or two 42-gons.
A hexagon corner partway along another hexagon’s side would need corners adding to 180^\circ there, but one corner gives 120^\circ and two give 240^\circ. So hexagon corners only ever meet corners, which is edge-to-edge. Squares can do it, since 2 \times 90^\circ = 180^\circ: slide one row of squares half a side along, as in the brick wall.
108^\circ + 120^\circ + 120^\circ = 348^\circ, which is 12^\circ short of a full turn. Laid flat, every corner would leave a gap of 12^\circ, so the pattern is not a tiling of the plane. On the ball the surface bends at each corner, and that bending closes the gap, which is what lets the pattern wrap round into a closed shape.