The largest triangle in a square

Put an equilateral triangle in a square as large as it will go. The obvious arrangement stands it on the square’s bottom edge, and the obvious arrangement is not the answer.

Why tilting helps #

A triangle standing on the square’s edge has side 1 and area √3/4, which is 0.4330. Its height is √3/2 ≈ 0.866, so it does not reach the top of the square: there is a strip of unused paper above it, and the triangle is limited by its width rather than its height.

Tilt it and the width shrinks while the height grows. The best tilt is the one where the two limits meet — where the triangle’s extent across the square is the same in both directions, so neither wall is doing more work than the other.

That angle is fifteen degrees, the resulting side is sec 15° = 1.03528, and the area is exactly 2√3 − 3 = 0.46410. The triangle’s side is longer than the square’s, which is a good sanity check on the intuition: the tilted triangle uses the square’s diagonal direction, and the diagonal is longer than the side.

Solved twice #

The site’s habit is that a claim gets a test it could fail, and an optimisation is easy to get wrong in ways that look right. So the answer is obtained two ways with no code in common.

By support widths. A convex shape fits inside an axis-aligned unit square, at some translation, exactly when its width measured across the x direction and its width across the y direction are both at most one — translation is free and cannot be wrong. So the largest scaled copy at a given rotation is the reciprocal of the larger of those two widths, and the whole problem is a one-dimensional maximisation over the rotation angle. It is bracketed on a grid and refined by golden-section search, and the refinement moves the answer by less than a part in ten thousand million.

By a corner construction. Put one vertex of the triangle in a corner of the sheet and the other two on the far sides, each at distance t from the near corner: the vertices are at (0, 0), (1, t) and (t, 1). Equal sides give

1 + t² = 2(1 − t)²

which rearranges to t² − 4t + 1 = 0, whose root in range is t = 2 − √3. And 2 − √3 is tan 15°.

The two methods agree to nine figures. Neither knows about the other, and neither was given the answer.

The family does not climb #

The figure sweeps the polygon count and the natural expectation is a sequence rising toward the square’s own area as the polygon rounds off. It does not, and the reason is a symmetry mismatch worth having.

The triangle takes 46.4 per cent. The square takes 100. A regular polygon with many sides approaches a circle, and the largest circle in a unit square has area $\pi/4 = 78.54$ per cent, so the far end of the sweep converges there rather than to one.

But the approach is not from below. A regular octagon with four of its sides flat against the square’s four sides has area $2/(1+\sqrt2) = 82.84$ per cent — comfortably above the circle’s figure, because an octagon reaches into the square’s corners and a circle cannot.

So the sequence goes 46.4, 100, then down, and the values at eight and twelve sit above the limit while the values at three, five, six and seven sit below it.

Which is a fact about four-fold symmetry #

The pattern has a cause and it is the container rather than the polygon.

A square has four-fold symmetry. A regular $n$-gon whose $n$ is a multiple of four shares that symmetry, so it can be placed with its own axes on the square’s axes and every one of its extreme directions used at once. An $n$-gon whose $n$ is not a multiple of four cannot, and its best placement is a compromise — which is exactly what the fifteen-degree tilt is.

So the family splits by $n \bmod 4$, with the multiples of four converging on $\pi/4$ from above and everything else from below, and the two branches meeting only in the limit where the polygon stops being able to exploit any symmetry at all.

The triangle is the extreme case of the second branch and that is why it is the interesting one. Three shares no factor with four, so no placement lines any of its axes up with the square’s, and the optimum is a genuine compromise between two directions rather than an alignment. The fifteen degrees is half of the thirty by which sixty and ninety fail to agree.

That also predicts which entries in the sweep are worth solving twice. An $n$ divisible by four has an answer somebody can write down by inspection; the others need the maximisation, and the others are where an error would hide.

The thing the quadratic gives away #

t² − 4t + 1 = 0 is a quadratic, and that is the finding this essay exists for.

Folding’s advantage over straightedge and compass is exactly one degree wide: a compass extracts square roots and a single fold extracts cube roots as well. The advantage is real and it settles two classical problems, and it is completely irrelevant here. The largest equilateral triangle in a square is a quadratic construction, so a compass reaches it, and so does a fold, and neither has anything to boast about.

That is worth stating plainly because an essay on this site could easily have been written the other way — folding reaches the optimum! — and it would have been true and misleading. The optimum is reachable by folding because it is reachable by anything.

How to fold it anyway #

The construction is a pleasant one and it uses two of the seven axioms.

A sixty-degree angle first. Take the bottom edge and the corner at (1, 0). Fold so that the corner at (0, 0) lands on the square’s vertical midline, with the crease passing through (1, 0). That is the fifth axiom: a point onto a line, with the fold through a point.

The corner travels a distance of one to a point half a unit from the pivot, which is a 1 : 2 ratio, which is a thirty-sixty-ninety triangle. So the image of the bottom edge makes sixty degrees with the bottom edge.

Then bisect what is left. The square’s corner is ninety degrees, so thirty remain between the sixty-degree ray and the vertical side. Bisecting two lines is the third axiom, and half of thirty is fifteen.

Two folds, two axioms, and the optimum. A compass does the same job in about the same number of steps.

Why an exact construction is worth having #

A fifteen-degree tilt could be produced with a protractor in a few seconds, so it is worth being clear about what the fold construction is for.

A protractor is accurate to about half a degree. A fold is accurate to how well the folder can align two things, which is a fraction of a millimetre over the length of a crease, and on a hundred-millimetre sheet that is a few hundredths of a degree.

More importantly, the fold construction does not accumulate. Every subsequent crease is aligned against features of the paper rather than against a previously measured angle, so an error at one step does not become the reference for the next. That is exactly the difference between a systematic and a random error, and it is why dividing a strip by folding beats measuring it even when the measurement is easy.

The habit generalises past this problem. Haga’s construction produces exact rational divisions of a square’s side by a single fold; Fujimoto’s method converges on a third with an error that halves at every step. In both cases the paper is doing arithmetic, and the arithmetic is exact for the same reason: the fold is defined by an alignment rather than by a reading.

The sequence, which is where it gets strange #

Ask the same question for every regular polygon and the answers stop behaving.

More sides means a rounder shape, and a rounder shape ought to use a square better, and up to eight it does. The octagon reaches 82.84%.

Then it falls. The nonagon manages 75.13%, the ten-sided 75.32%, the eleven-sided 76.26%, the twelve-sided 80.38% — every one of them worse than the octagon.

The reason is symmetry. The square and the octagon are related — an octagon has the square’s four-fold symmetry and more — so an octagon can be placed in a square with all four walls doing equal work. A nonagon cannot: its nine-fold symmetry shares nothing with the square’s four-fold, so some of its vertices are pressed against the walls while others are nowhere near them.

The tilts show the same thing. Every optimum in the sweep is at a simple fraction of a right angle — 15°, 45°, 9°, 15°, 6.43°, 22.5°, 5°, 9°, 4.09°, 15° — and the fraction is set by the polygon’s own symmetry step rather than by anything about the square. The polygon is turned by half a step, which is the position where its two widths come out equal.

The values are worth reading in two groups. For the odd-sided polygons the tilt is a right angle divided by twice the number of sides: 15° at three, 9° at five, 6.43° at seven, 5° at nine, 4.09° at eleven. For the even-sided ones the divisor depends on whether the count is a multiple of four, because a polygon with 4k sides already has the square’s own quarter-turn symmetry and one with 4k + 2 sides does not.

None of that was put in. The search maximises a scale factor over a rotation angle and reports where it landed, and the landings are at half-steps because that is where a symmetric shape’s two widths agree. It is the sort of regularity that a numerical answer either shows or does not, and this one shows it to the resolution the refinement gives — which is another way of saying the maximisation is finding the right point rather than a point nearby.

One more case is worth drawing, because it is the one that does worst after the triangle.

What “largest” is measured in #

One decision in the question deserves stating because it could have gone another way and the answer would change.

The polygons here are compared by area, and the maximisation is over the largest scale factor that fits. Those are the same thing for a fixed shape — scaling a polygon by s scales its area by s² — so maximising the side and maximising the area pick the same rotation.

They stop agreeing the moment polygons with different numbers of sides are compared. A triangle of side 1.035 and a dodecagon of side 0.268 are not comparable by side length, and the sequence in the census is areas for that reason. Reporting sides instead would produce a monotonically decreasing list that said nothing at all, since a polygon with more sides at a fixed size has a shorter side by construction.

The area is also the quantity a folder cares about, because it is the paper. A shape inscribed in a sheet is a claim on the sheet, and what is left over is what everything else in the design has to come from.

It is worth seeing what a construction looks like when it does need folding’s extra degree.

Which theorem was checked, and how #

Three checks, and the first two are exact.

The square must come out at one. The largest square inside a unit square is that square, area exactly 1, and any implementation that had lost a factor of two or measured a width across the wrong axis would fail this immediately. It comes out at 1 to fifteen figures.

The triangle must come out at 2√3 − 3. This is the case that tests the maximisation itself, because its answer is neither trivial nor at the end of the search range. It agrees to better than a part in a million.

The two methods must agree. The corner construction solves a quadratic and shares no line of code with the width maximisation, and the two sides come out equal to nine figures.

The generator also refuses to draw the census if the sequence turns out to be increasing — if every polygon past the peak used more of the sheet than the peak, the figure would have no finding and would be a decoration.

The alternative to an exact construction is worth naming, since this problem happens not to need it.

What is being asked, and what is not #

The question here is narrow and worth stating precisely, because inscribed-figure problems have many variants and they have different answers.

Regular polygons only. The largest triangle of any shape in a square is half the square, area 0.5, and it is not equilateral. Requiring regularity is what makes the problem interesting.

A square sheet. Whether the sheet should be square is a separate question with its own measurements, and the answer there also turns on the arithmetic of what is being fitted.

One polygon. Packing several is a different problem entirely and is the one origami design actually reduces to.

And the whole essay is about the outline of the sheet. Nothing here says anything about creases: an inscribed triangle is a region of paper, not a folded object, and turning one into the other is a separate construction with its own losses.

What the answer looks like at the other extreme #

A useful way to sanity-check an optimisation is to ask what happens at the ends, and the polygon sequence has two ends worth visiting.

At the low end, the triangle is far and away the worst fit at 46.41%. It uses less than half the sheet, and that is not a defect of the triangle — it is a consequence of a triangle being a poor match for a square by any measure. Three-fold symmetry and four-fold symmetry share only the identity, so no orientation puts the triangle’s vertices anywhere useful relative to the square’s corners.

At the high end the polygons converge on the circle. The largest circle in a unit square uses π/4, which is 78.54%, and every polygon with many sides is approaching that value. The approach is not from below: the octagon at 82.84% and the twelve-sided at 80.38% are both above π/4, because a polygon in tune with the square can use the square’s corners in a way a circle cannot.

So the sequence has a floor at the triangle, a ceiling at the octagon, and a long tail settling toward the circle’s value from both sides. That shape is the signature of the symmetry argument rather than of anything about numbers of sides, and it is what would have to change if the sheet were not a square.

The circle’s own value is the calibration worth keeping. Any inscribed shape that manages more than 78.54% is exploiting the square’s corners; any that manages less is failing to exploit them. Read that way the census is a single question asked thirteen times: does this polygon fit the corners.

The construction question this essay barely touches has a table of its own.

Who solved it, and when #

The largest equilateral triangle in a square is a classical recreational result with no clear single author, of the kind that appears in puzzle collections and geometry columns and gets rediscovered every few decades. Its answer is old, its proof is short, and the date a result becomes famous is not the date it was found — a warning this site has had cause to repeat.

The origami construction is likewise traditional. Folding an equilateral triangle from a square is a standard exercise and appears in teaching material going back to the paper-folding of nineteenth-century geometry classrooms, where constructions of this kind were the entire point of the activity.

What is this repository’s is the double solution and the census: the width maximisation, the corner quadratic, the check that they agree, and the sequence over twelve polygons that turns out not to increase.

Where the ladder goes next #

The biggest one that can also be folded is the same census crossed with the constructibility question, and it is where folding’s extra degree finally earns something. The eleven-sided polygon is a respectable 76.3% of the square and no single fold can build it; whether that matters depends on how many creases may be made at once.