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28 / ANTIQUITY TO THE RENAISSANCE

Omar Khayyam

1048–1131 CE·Nishapur, Samarkand & Isfahan

A cube and a beam must hold a set amount. Often no neat length works. Two curves can find the length anyway.

Explore the mathematics ↓
Imagined portrait of Omar Khayyam
Imagined likeness

01 / CUBICS FROM CONICS

Where two curves cross, a cubic is solved.

A cube and a 2 × 2 beam share one length x. Together they must hold a target number of small cubes: x³ + 4x = B.

Khayyam drew a parabola and a circle built from the problem. Where they cross, both of their rules hold at once, and that point’s x solves the cubic.

Target volume:
Graph and blocks use one fixed scale: 1 unit = 1 small cube edge
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THE IDEAFor x³ + 4x = B, the parabola gives x² = 2y and the circle on a diameter of B ÷ 4 gives y² = x(B ÷ 4 − x). At their crossing both hold, and together they give x³ + 4x = B.

The person behind the ideas

More of the mathematics

Different cubics, different curves

Khayyam used a suitable pair of conics for each type of cubic. This lesson explores just one type.

Why not the origin?

Both curves also pass through (0, 0), but x = 0 means no cube at all. Only the other crossing solves the problem.