🍾 The Clumsy Sommelier — Champagne Fountain Puzzle

Ages: 11–14  ·  Duration: 90 minutes  ·  Topics: Combinatorics, Recurrence Relations, Geometric Series, Symmetry
Source: FSJM — 38th Competition, Challenge 14 (2024)

Part 0 — Warm-Up: Cascading Cups (10 min)

The Big Idea

"Imagine stacking paper cups in a triangle: 1 cup on top, 2 cups below, 3 cups on the bottom row. You pour water into the top cup. When it overflows, the excess spills equally into the 2 cups directly below. What happens?"

🗣️ "Turn to your neighbour — how much water do you need to pour into the top cup to fill all 6 cups?"

Wait — let's start even simpler!

One-Dimensional Warm-Up

Stack 3 cups in a single column: cup A on top, cup B in the middle, cup C at the bottom. Each cup holds exactly 1 litre. When a cup overflows, all the excess flows down into the next cup.

✋ "Notice: in a single column, there's no waste until you overshoot. You need exactly 3 litres. But what happens when cups branch — when one cup feeds multiple cups below?"

The Triangular Mini-Tower

Now let's try 3 rows: 1 cup on top, 2 in the middle, 3 on the bottom. When a cup overflows, excess splits equally between the 2 cups directly below it.

    [A]           Row 1: 1 cup
   [B][C]         Row 2: 2 cups
  [D][E][F]       Row 3: 3 cups

✍️ "Work independently for 2 minutes: if we pour $C$ litres into cup A, how much does each bottom cup receive?"

Tracing the flow:

• Cup A receives $C$, keeps 1, overflows $C - 1$. Splits equally: cups B and C each get $\frac{C-1}{2}$.
• Cup B receives $\frac{C-1}{2}$, keeps 1, overflows $\frac{C-3}{2}$. Splits to D and E: each gets $\frac{C-3}{4}$.
• Cup C is identical to B (by symmetry). It sends $\frac{C-3}{4}$ to E and F.

So the bottom row receives:

For all cups to be full, the corners must get at least 1:

At $C = 7$: corners get exactly 1, centre E gets 2 (overflows 1, which is wasted).

Key Rule

Part 1 — The Simplified Version: Mini Square Pyramids (≈ 15–20 min)

Setting the Scene

Now let's switch from triangles to squares. A sommelier stacks champagne flutes in a square pyramid:

• Top level: $1 \times 1 = 1$ flute
• Level 2: $2 \times 2 = 4$ flutes
• Level 3: $3 \times 3 = 9$ flutes
• and so on…

Each flute holds exactly 1 unit of champagne. When a flute overflows, the excess splits equally among the 4 flutes directly below it. On the bottom level, any overflow is wasted.

Question: For a 3-storey square pyramid ($1 + 4 + 9 = 14$ flutes), how much champagne is wasted by the time all flutes are full?

Worked Example: 2-Storey Pyramid

Let's start even smaller — a 2-storey pyramid: 1 flute on top, $2 \times 2 = 4$ flutes below (5 total).

Flute T overflows → excess splits 4 ways to A, B, C, D. By symmetry, each bottom flute gets $\frac{C-1}{4}$.

For all to be full: $\frac{C-1}{4} \ge 1 \implies C \ge 5$.

At $C = 5$: each bottom flute gets exactly 1. Waste = 0! 🎉

Worked Example: 3-Storey Pyramid

Now the interesting case: $1 + 4 + 9 = 14$ flutes.

Level 1 (top): 1 flute. Receives $C$. Overflow: $C - 1$.

Level 2: $2 \times 2 = 4$ flutes. By symmetry, each gets $\frac{C-1}{4}$. Overflow from each: $\frac{C-1}{4} - 1 = \frac{C-5}{4}$.

Level 3: $3 \times 3 = 9$ flutes. Each level-2 flute at position $(i, j)$ sends overflow to four level-3 positions: $(i, j)$, $(i+1, j)$, $(i, j+1)$, $(i+1, j+1)$. So each bottom flute receives $\frac{1}{4}$ of the overflow from every level-2 flute sitting above it.

Let's count how many sources feed each level-3 position:

Each source provides $\frac{1}{4}$ of its overflow $\frac{C-5}{4}$, which is $\frac{C-5}{16}$ per contribution.

The corners receive the least. Setting $\frac{C-5}{16} = 1$ gives $C = 21$.

Verification: At $C = 21$, bottom-level flutes receive:

• 4 corners × 1 unit → no waste
• 4 edges × 2 units → waste of 1 each → $4 \times 1 = 4$
• 1 centre × 4 units → waste of 3 → $1 \times 3 = 3$

The Key Insight

Part 2 — The Real Challenge ⭐ (≈ 20–25 min)

The Problem

FSJM Competition — Challenge 14: Clumsy Sommelier

A sommelier prepares a champagne fountain. He places 25 flutes in a square, then places 16 flutes in a square on top, then 9, then 4, then 1 to obtain a 5-storey pyramid.

He then pours champagne into the top flute until all are full. When one flute overflows, the excess flows evenly into the four flutes below. But for the flutes at the very bottom that overflow, the excess is wasted.

How much champagne will have been wasted by the time all the flutes are filled?
Give the answer in number of flutes (whole number or irreducible fraction).

Step 1: Count the Flutes

Total flutes in the pyramid:

✋ "Does anyone recall the formula for the sum of the first $n$ squares?"

Check: $\frac{5 \times 6 \times 11}{6} = 55$. ✓

Step 2: Identify the Bottleneck

🗣️ "From Part 1, which position on the bottom layer fills last? Turn to your neighbour and discuss!"

The corner of each level is the bottleneck. Let's trace the corner chain — the sequence of corner flutes from top to bottom.

The corner at level $k$ receives overflow from only one flute above: the corner at level $k - 1$. The overflow splits 4 ways, and the corner below gets exactly $\frac{1}{4}$ of it.

If $T_k$ = total champagne flowing into the corner of level $k$:

Step 3: Work Backwards from the Bottom

🧩 "Challenge: try this before I show the answer!"

For ALL flutes to be full, the bottom corner must receive at least 1 unit: $T_5 \ge 1$.

The absolute minimum is $T_5 = 1$ (just barely full). Working backwards:

So we must pour $C = 341$ flutes' worth of champagne into the top!

Step 4: Compute the Waste

Every flute in the pyramid holds exactly 1 unit. There are 55 flutes total. All champagne that isn't held by a flute is wasted:

Verification: Does the Conservation Check Out?

Let's verify by adding up the waste from each bottom-layer position. At $C = 341$, the 6 distinct position types on the $5 \times 5$ bottom layer:

Total waste:

Dead-End: Why Tracking Every Flute Is Overwhelming

☕ Break (5 min)

Part 3 — Theory & Formalisation: Recurrence Relations (≈ 10–15 min)

The Theorem: Linear Recurrences with Constant Coefficients

We discovered that the "corner chain" satisfies:

or equivalently, reading bottom-to-top:

This is a linear recurrence relation — a rule that tells you each term from the previous one. It appears everywhere in mathematics and computer science: population growth, compound interest, fractal dimensions, algorithm analysis.

Let's solve it to find a closed-form formula (a direct expression without recursion).

Solving the Recurrence

We want to solve $f(n) = 4\,f(n-1) + 1$ with $f(1) = 1$.

Define $g(n) = f(n) + \frac{1}{3}$. Then:

So $g(n)$ is a geometric sequence with ratio 4!

Therefore:

Concrete Examples

The Waste Formula

For an $n$-storey pyramid, the total number of flutes is $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.

The total champagne poured is $C(n) = \frac{4^n - 1}{3}$.

The waste is:

✋ "Notice how the waste grows explosively! For just 6 storeys, you waste 1274 flutes' worth — 14 times the capacity of the pyramid! The exponential $4^n$ dominates the polynomial $n^3/3$."

Mini-Exercise (pair work)

Solution 1 — Layer-by-Layer Simulation 🥉 Bronze

Prerequisites: Arithmetic, tables

Setup

The most direct approach: compute how much champagne every single flute receives, layer by layer. Each flute either sits in a corner (1 source above), on an edge (2 sources), or in the interior (4 sources).

Computation

We pour $C$ into the top and track everything:

Level 1: 1 flute. Receives $C$. Overflow: $C - 1$.

Level 2: 4 flutes, all identical by symmetry.

• Each receives: $\frac{C - 1}{4}$
• Each overflows: $\frac{C-1}{4} - 1 = \frac{C - 5}{4}$

Level 3: 9 flutes in 3 types.

Level 4: 16 flutes in 3 types.

Level 5 (bottom): 25 flutes in 6 types.

Setting the minimum (corners) to 1: $\frac{C-85}{256} = 1 \implies C = 341$.

Result

Verification

Sum of all waste from bottom layer:

Solution 2 — The Corner Chain 🥈 Silver

Prerequisites: Basic fractions

Key Observation

Forget about 55 flutes — we only need to track one: the corner.

At each level, the corner flute sits at position $(1,1)$. It receives overflow from only the corner of the level above. This creates a chain: the corner of level 1 feeds the corner of level 2, which feeds the corner of level 3, and so on. Each step divides the overflow by 4.

Derivation

Let $T_k$ = champagne flowing into the corner of level $k$.

For all flutes to be full, we need $T_5 \ge 1$ at a minimum. Setting $T_5 = 1$ and working backwards:

Why Is the Corner the Last to Fill?

Every other position on a given level receives from more sources:

• Edge positions receive from 2 flutes above → get $2 \times \frac{1}{4} = \frac{1}{2}$ share
• Interior positions receive from 4 flutes above → get $4 \times \frac{1}{4} = 1$ full share

Meanwhile the corner gets from 1 flute → only $\frac{1}{4}$ share. So corners always receive the least, and the bottom corner fills last.

Result

Solution 3 — The Closed-Form Formula 🥇 Gold

Prerequisites: Geometric series, exponents

Setup

From Solution 2, the minimum pour for an $n$-storey pyramid satisfies:

Derivation

Recognise the pattern:

This is a geometric series with first term 1 and common ratio 4:

For $n = 5$:

Result

Verification

The formula $f(n) = \frac{4^n-1}{3}$ satisfies the recurrence:

Solution 4 — Generalising to Split Factor $s$ 🎓 Diamond

Prerequisites: Geometric series, dimensional reasoning

Setup

What if the overflow split weren't 4-way? In general, with split factor $s$, the corner-chain recurrence is:

The solution is:

Full Table

Result

Interactive Simulation 🎮

Watch champagne cascade through the pyramid! Click Pour to start, or adjust the slider to control how much champagne is poured.

🍾 Open 3D Isometric View (Three.js)

Summary — The Answer

Why Multiple Solutions?

Extensions & Challenge Problems 🧩

🥉 Bronze

1. Tiny pyramid. Compute the waste for a 4-storey square pyramid by hand.
2. Double check. In our 5-storey pyramid at $C = 341$, exactly how much champagne does the centre flute $(3,3)$ on the bottom layer receive? Express as a fraction.
3. Triangular tower. A 4-storey triangular champagne tower has layers of 1, 3, 6, 10 glasses. Each overflow splits 3 ways. What's the waste?

🥈 Silver

4. Efficiency. For an $n$-storey square pyramid, what fraction of champagne is wasted? Show that as $n \to \infty$, this fraction approaches 100%.
5. Equal filling. Is there any pyramid shape (any number of storeys) where the bottom layer can be filled with zero waste? If so, find all such configurations.
6. Non-square layers. Suppose the sommelier uses triangular layers ($1, 3, 6, 10, \ldots$ flutes) with 3-way overflow splitting. How much waste for 5 storeys?

🥇 Gold

7. Exact distribution. For the 5-storey square pyramid, compute the exact champagne received by every position type on every layer. Verify that the total outflow from each layer equals the total inflow minus the layer's capacity.
8. Heat map. Write a program to compute and display a "heat map" of champagne distribution across the bottom layer for $n = 10$. What pattern do you see?
9. Inverted problem. If the sommelier has exactly 500 flutes' worth of champagne, what is the tallest square pyramid he can completely fill?

🎓 Diamond (optional)

10. Continuous limit. As $n \to \infty$ with appropriate scaling, does the champagne distribution across the bottom layer converge to a known probability distribution?
11. Hexagonal pyramid. Design a champagne tower with hexagonal layers where each flute overflows into 6 neighbours. Derive the waste formula.
12. Optimal geometry. Among all regular polygon pyramids, which shape minimises waste for a fixed total capacity? Prove your answer.

Answer Key (for instructors)