Breaking a Clock Face Into Three Pieces With Equal Sums
A round clock face carries the numbers $1$ through $12$ arranged in the usual order around its rim. It falls and breaks into exactly three pieces. When you pick up the pieces, you notice that the numbers on each piece add up to the same total. The break was clean and the pieces are not oddly shaped: the clock was cut along two straight lines that do not cross inside the face, so each piece is the region between the rim and a chord, or between two chords.
Which numbers are on each of the three pieces?
Hints
- The numbers $1$ through $12$ add up to $78$, so each piece must sum to $78 / 3 = 26$.
- Two straight cuts that do not cross inside the face produce two outer pieces, each holding a run of consecutive numbers around the rim, and one middle piece that holds the numbers on both remaining sides. Remember that $12$ and $1$ are adjacent on the rim.
- Try the outer pieces $\{11, 12, 1, 2\}$ and $\{5, 6, 7, 8\}$; both sum to $26$, and the middle band $\{9, 10, 3, 4\}$ also sums to $26$.
Worked Solution
How to Think About It: First use the arithmetic to pin down the per-piece total. Then translate "not oddly shaped" into a structural constraint: pieces cut by two straight non-crossing lines. That means the outer pieces are runs of consecutive numbers on the circular rim (with $12$ wrapping to $1$), and the middle piece takes everything else.
Quick Estimate: $1 + 2 + \cdots + 12 = 78$, so each piece sums to $26$. Numbers opposite each other across a horizontal axis pair up as $12 + 1$, $11 + 2$, $10 + 3$, $9 + 4$, $8 + 5$, $7 + 6$, all equal to $13$. Two such pairs make $26$, so horizontal cuts are natural.
Formal Solution:
*Step 1 -- The target.* $\sum_{k=1}^{12} k = \frac{12 \times 13}{2} = 78$, and $78 / 3 = 26$, so every piece must sum to $26$.
*Step 2 -- Structure of the pieces.* Two straight cuts that do not cross inside the face give two "cap" pieces, each containing a set of consecutive numbers along the rim, and one "band" piece between the chords containing the numbers on both remaining sides of the face. So we need two disjoint runs of consecutive clock numbers (cyclically, so $12$ is followed by $1$) each summing to $26$, with the leftover numbers also summing to $26$.
*Step 3 -- Find a run summing to 26 that contains 12.* Runs of consecutive numbers around the top: $12 + 1 = 13$, $11 + 12 + 1 = 24$, $11 + 12 + 1 + 2 = 26$. So the top cap $\{11, 12, 1, 2\}$ works.
*Step 4 -- Find a second run summing to 26 among the rest.* On the bottom, $5 + 6 + 7 + 8 = 26$. That is the bottom cap.
*Step 5 -- The band.* The remaining numbers are $\{3, 4\}$ on the right side and $\{9, 10\}$ on the left side, and $3 + 4 + 9 + 10 = 26$. Geometrically, this is the horizontal band between the two chords, so the three pieces are consistent with two parallel straight cuts.
*Step 6 -- Uniqueness.* An exhaustive search over all pairs of disjoint cyclic runs with the remainder forming the third piece finds this as the only solution with all three sums equal to $26$. The trap is forgetting that $12$ and $1$ are neighbours: without wrapping around, no run summing to $26$ contains $12$, and the puzzle looks impossible.
Answer: Each piece sums to $26$: the top piece $\{11, 12, 1, 2\}$, the bottom piece $\{5, 6, 7, 8\}$, and the middle band $\{9, 10, 3, 4\}$.
Intuition
Summation fixes the target immediately: 78 split three ways is 26 per piece. The geometry then does the rest. Two non-crossing chords cut a disc into two caps, each holding a consecutive run of numbers, and one middle band that collects the remaining numbers from both sides. Symmetry suggests cutting horizontally through the middle: the top cap $11, 12, 1, 2$ and the bottom cap $5, 6, 7, 8$ each sum to 26 by pairing numbers that add to 13, and the band picks up the rest. The habit of first computing the forced total, then searching under a physical constraint, is exactly how you sanity-check any allocation problem.