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Proof without words — page 2

Arguments that are complete once they have been looked at properly. Not illustrations of proofs — the proofs themselves.
the cutter's measure, segment by segmentthe chooser's measure, segment by segment52530201553010552030cut at 4/9the left piecethe right piecethe cutterthe chooser5050130/3≈ 43.33170/3≈ 56.67the cutter's piecethe chooser's piecethe cutter's running total crosses 50 inside segment 3, 2/3 of the way through it, so the cut is at 4/9both pieces are worth exactly 50 to the cutter; the chooser takes the right one at 170/3 ≈ 56.67 andgains 20/3 ≈ 6.67 over half Applied

One cuts and the other chooses

The oldest rule in fair division promises each of two people at least half the cake by their own measure, and it keeps that promise exactly. It does not promise what the word "fair" is usually asked to carry, and the gap opens the moment the two measures disagree across the cut.

 person 1 cuts three pieces of equal value to person 1piece 1piece 2piece 3person 2 trims the largest of them down to a tie with the secondthe trimming: 140/3 ≈ 46.67 to person 2person 3 chooses, then person 2, then person 1person 1person 3person 2person 3 cuts the trimming in three; person 2 chooses first, then person 1person 1person 2person 3the trimming, drawn at full widtheach person's value of each share — the diagonal is their ownperson 1's shareperson 2's shareperson 3's shareperson 1 valuesperson 2 valuesperson 3 values140/3≈ 46.6740/3≈ 13.33402040403540/3≈ 13.33155/3≈ 51.67own sharethe trimmingperson 1 cuts three pieces worth 100/3 each; person 2 trims 140/3 ≈ 46.67 off the largest, andperson 3 chooses firstthe verdict is the whole 3×3 matrix, not its diagonal: all nine comparisons hold, 1 of them as anexact tie Applied

Three people and a trimmed piece

For two people, one cut and one choice deliver a division nobody would swap out of. For three, the same promise costs a trimming, a residue and a choosing order contrived so that an advantage once given cannot be taken back — and the verdict is not three numbers but a whole three-by-three matrix.

 primal: max 3x₁ + 4x₂dual: min 12y₁ + 10y₂3x₁ + 2x₂ ≤ 12x₁ + 4x₂ ≤ 103y₁ + y₂ ≥ 32y₁ + 4y₂ ≥ 401234501234x₁x₂78/512100(14/5, 9/5)00.511.522.5301234y₁y₂78/53024(4/5, 3/5)05101520253078/51210078/53024they meet at 78/5primal vertex values, from belowdual vertex values, from aboveweak duality checked on all 12 pairs — each of the 4 feasible primal vertices against each of the 3 feasible dual vertices — andcᵀx ≤ bᵀy held every timestrong duality is the equality of the two optima: max 3x₁ + 4x₂ = 78/5 = min 12y₁ + 10y₂, one number reached from below andfrom abovethe dual region is unbounded upward and the drawing cuts it at the box; the enumeration ignores the cut Applied

Two numbers that have to meet

Every linear program has a shadow — a second program built from the same numbers read the other way, whose minimum can never fall below the first's maximum. That much is a one-line calculation; the theorem is that the two numbers are always exactly equal.

 the row chooser mixes00.20.40.60.81-4-20246weight on row Apayoff to the row choosercol Xcol Ycol Z8/15 → 19/15the most the row chooser can guarantee: 19/15the column chooser mixes00.20.40.60.81-4-20246weight on col Xpayoff to the row chooserrow Arow B7/15 → 19/15the least the column chooser can concede: 19/15both sides name 19/15 = 1.267the value is 19/15 = 1.267, reached by the row chooser mixing 8/15 on A and by the column chooser mixing 7/15 on Xthe lower envelope of 3 lines peaks where two cross; the upper envelope of 2 bottoms out at that heightchecked against every pure reply on both sides, and over 61 mixtures on one side and 91 on the other Applied

The value from both sides

Two choosers move at the same instant, and each asks the cautious question — how much can be guaranteed, whatever the other does. With pure choices the two answers are usually different numbers; allow a probability and they are forced to be the same one.

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