A Clock Face in Three Pieces of 26, and No Pie Cut Ever Works

A clock face breaks into three pieces, and the numbers on each piece add up to the same total. Where do the cracks go?

The target is easy. The numerals add to

and $78/3 = 26$, so each piece must total twenty-six. The instinct after that is to cut it like a pie, three wedges of four hours each. Try it: $12+1+2+3 = 18$, $4+5+6+7 = 22$, $8+9+10+11 = 38$. Those three add back to seventy-eight, so it is a genuine split of the dial, and not one of the pieces is twenty-six.

No pie cut ever works, and here is why

Sliding the wedges around does not help, and the reason is not that you got unlucky. Every possible pie cut fails, and the proof is three lines of triangular numbers.

Write $T(x) = x(x+1)/2$ for the running total of the dial up to hour $x$. The first two pieces do not wrap around, so their sums are differences of running totals, and both have to be twenty-six:

The third piece then comes out to twenty-six automatically, since the whole dial is seventy-eight. So the question reduces to whether three of the thirteen numbers

form an arithmetic progression with common difference twenty-six. Check the gaps. The only pair in that list differing by exactly twenty-six is $(10, 36)$, at positions four and eight. So $a = 4$ and $b = 8$ are forced, and then $T(c)$ would have to be $36 + 26 = 62$, which is not in the list. Contradiction. No pie cut of the dial gives twenty-six three times.

Exhaustive enumeration of all two hundred and twenty cuts agrees, and it adds one detail the proof does not: the closest miss is off by exactly one. Cracks after hours two, seven and ten give pieces $\{3,4,5,6,7\} = 25$, $\{8,9,10\} = 27$ and $\{11,12,1,2\} = 26$. The pie instinct is defeated narrowly rather than absurdly, which is why it takes an argument to kill it.

Two wedges and a band

Give up on all three pieces being wedges and the answer appears:

The first two are wedges, sitting opposite each other at the top and bottom of the face. The third is not: it collects the two remaining pieces from opposite sides of the dial, nine and ten on the left, three and four on the right, and reaches twenty-six as $19 + 7$.

Geometrically the picture is tidier than the list of numbers suggests. Two straight cracks through the middle, one running between the one and the two and out the other side, the other between the ten and the nine and out the other side, cut the face into four sectors of four, two, four and two hours. Take the two four-hour sectors as pieces, join the two opposite two-hour sectors, and you have (4).

The band is forced, not chosen

Here is the claim I would have got wrong on a first pass. It is tempting to say the answer is the unique way to split the dial into three equal sums, and that is false one hundred and two times over: there are exactly $102$ ways to partition the twelve numerals into three groups summing to twenty-six, if you allow the groups to be any subsets at all. Most of them are scattered collections nobody would call a piece of anything.

What is unique is the answer subject to the geometry, and the same triangular numbers prove it. Ask which wedges total twenty-six. A wedge either wraps past the twelve or it does not.

If it does not wrap, its sum is $T(b) - T(a) = 26$, and we already know the only such pair is $(10,36)$, giving $\{5,6,7,8\}$. If it does wrap, its sum is $78 - T(c) + T(a) = 26$, so

and the only pair in (3) differing by fifty-two is $(3, 55)$, at positions two and ten, giving $\{11,12\} \cup \{1,2\}$.

I like this because the impossibility result and the uniqueness result are the same computation run twice, once looking for a gap of twenty-six and once for a gap of fifty-two. Fig. 1 contains both answers.

Dials of other sizes

For an $n$-hour dial the total is $n(n+1)/2$, and three equal pieces need that to be divisible by three, which happens exactly when $n \equiv 0$ or $2 \pmod 3$. So a ten-hour dial is hopeless before you start, since $55$ is not a multiple of three.

When the total does divide, the pie question becomes the same search for two gaps of $n(n+1)/6$ among the running totals, and the answer is almost always no. Running it for every dial up to a hundred and twenty hours, exactly two sizes admit a genuine three-wedge cut: nine and ninety. The nine-hour dial splits as

with cracks after hours three, six and eight. So the twelve-hour failure is the rule rather than bad luck, and the puzzle would be flat on a nine-hour clock.

What the puzzle does not claim

Three pieces that all meet at the centre of the dial are joined at a single point, so a genuinely shattered clock face would not hold them together as pieces. Nothing above claims otherwise. The geometric statement being made is that two straight cracks through the middle produce four quarters, and that the answer keeps two of them and joins the other two. Whether the joined pair is one physical fragment is a question about glass rather than about arithmetic.

The sums are also sums of numerals rather than of angles or of anything physical. A quarter of the face carries a quarter of the area no matter which quarter it is, and the whole problem exists because the numerals are not distributed evenly around the dial in the way area is.

Sources and further reading

• The function $T$ and the list in (3) — Triangular number
• What the 102 counts — Partition of a set
• The structure that makes a wedge a wedge — Cyclic order
• The dial itself — Clock face

Everything above was checked before publication by exhaustive enumeration rather than sampling, which matters because the load-bearing claim is an impossibility and no drawn counterexample would establish one. All two hundred and twenty pie cuts were enumerated and none gives twenty-six three times; all one hundred and two equal-sum splits were enumerated and exactly one has two or more wedge pieces. The contiguity test was written independently of the enumeration it feeds, so that wrap-around wedges such as $\{11,12,1,2\}$ are recognised rather than rejected.