Eight Fair Slices
Eight equal slices that look nothing alike, one 7-inch mark. How big is the triangle?
The July/August 2026 issue of Scientific American ran a lovely little puzzle: a square cut into eight rectangles of equal area, one width labeled, find the side of the square. It got us wondering — what happens when you play the same game with triangles? The answer turns out to be richer than the original, with a hidden numerical jewel and, lurking one step further out, one of the strangest impossibility theorems in geometry.
Picture a bakery’s triangular sheet cake — a perfect equilateral triangle — that has to feed eight people, each of whom will be watching the knife very closely. The baker makes a single long cut from the top corner down to the base, and then fans straight cuts from each of the two bottom corners out to points along that first cut. Eight triangular pieces. They look wildly unequal — three long thin slivers on the left, five stubbier ones on the right — and yet every piece contains exactly the same amount of cake.
The figure shows the cuts. All eight triangles have equal area, and one marked stretch of the long cut — between two of the fan points — has length 7.
How long is each side of the big triangle?
Everything you need is in the picture. The equal areas are not decoration — they are the machine that determines every length in the figure, the same way the equal areas pinned down the Scientific American square. There are exactly two ideas standing between you and the answer, plus one finishing move that Pythagoras would recognize. No trigonometry required, though the number 7 is doing something quietly miraculous that the solution will confess to.
And once you’ve cracked it, a stranger question is waiting: we cut a triangle into eight equal triangles — but could the baker have done this with a square cake and, say, seven triangular pieces? The interactive supplement has a machine that fair-slices any regular polygon and will show you which counts are possible — and which are forbidden by two of the most surprising theorems in the subject.
Explore this puzzle visually with an interactive diagram — drag sliders, watch the geometry update in real time, and build intuition before you solve.
Everything flows from one humble fact: triangles that share an apex, with their bases lying along one straight line, have areas in proportion to their bases (their heights are identical).
First use it on the two big triangles flanking the long cut. You know their area ratio without measuring anything - just count how many of the eight equal slices sit on each side. That tells you exactly where the long cut meets the base.
Then use the very same fact again, this time on the slices within one fan: they share an apex too, and their bases all lie along the long cut. Equal areas means the marked 7 is one of several identical stretches - so you know the whole cut's length.
To finish, drop the altitude from the top corner (in an equilateral triangle it lands on the base's midpoint) and let Pythagoras relate the cut's length to the side. No trigonometry needed.
Each side of the triangle is 24. Here is the chase, one idea at a time.
The one tool you need
Two triangles that share the same apex, with their bases lying along the same straight line, have areas in proportion to their bases — because they have identical heights (the distance from that shared apex down to the line). Same height, so area is just “half the base times the same thing.” That single fact, used twice, unlocks the whole figure.
Step 1 — count the slices to find the base split
The long cut runs from the top corner C to a point M on the base. To its left sit 3 of the equal slices; to its right sit 5. So the two big triangles flanking the cut have areas in the ratio 3 : 5. But those two big triangles share the apex C, and their bases AM and MB lie along the same line — the base of the cake. By our tool, AM : MB = 3 : 5. The point M sits exactly 3/8 of the way along the bottom. The count of pieces wasn’t trivia; it was the first measurement.
Step 2 — the fans slice the long cut evenly
Now look at the three left-hand slices. They all share the apex A, and their bases are stretches of the long cut CM — again, bases on one common line. Equal areas therefore force equal bases: the left fan cuts CM into three identical parts. (The right fan, sharing apex B, cuts the same segment into five identical parts.) The marked stretch is one of the left fan’s thirds, so the whole cut is CM = 3 × 7 = 21.
Step 3 — Pythagoras finishes it
Call the side of the triangle s, and drop the altitude from the top corner C. In an equilateral triangle it lands on the exact midpoint of the base, at s/2 from either end — and M sits at 3s/8. So the altitude’s foot and M are only s/2 − 3s/8 = s/8 apart. The altitude itself has the familiar squared length of 3s²/4. Now the long cut CM is the hypotenuse of a right triangle whose legs are s/8 and the altitude:
CM² = (s/8)² + 3s²/4 = s²/64 + 48s²/64 = 49s²/64, so CM = 7s/8.
A ratio of exactly seven-eighths — from a figure built out of 3s and 5s. Setting 7s/8 = 21 gives s = 24, and the solved figure fills in:
Quick check: a side-24 equilateral triangle has area 144√3 ≈ 249.4, so each slice should hold 18√3 ≈ 31.2. The middle-left slice has base 7 on the cut and its apex A sits 36√3/7 away from the cut’s line — half of 7 times that is 18√3. It all closes.
The confession about the 7
That clean “CM = 7s/8” was not luck. Look at the two triangles flanking the long cut: ACM has sides 24, 9, 21 — three times (8, 3, 7) — and CMB has sides 21, 15, 24 — three times (7, 5, 8). These are the two smallest integer triangles containing a perfect 60° angle, the 60° cousins of the Pythagorean 3-4-5: in a 60° triangle the rule is a² + b² − ab = c², and 3² + 8² − 24 = 49 = 5² + 8² − 40. Glue those two famous triangles together along their shared side of 7 and they reassemble into an equilateral triangle of side 8 — our figure is exactly that gluing, scaled by three. Even better: among all ways to split the base a : b, the cut’s length is s√(a²+ab+b²)/(a+b), and 3 : 5 is the very smallest split that makes that square root come out whole (the next is 7 : 8). Eight slices, split 3 and 5, is the minimal configuration where everything turns integer. The puzzle was rigged by number theory.
Now try a square cake — and meet a wall
Our fan trick cuts a triangle into any number of equal-area triangles: fan from one corner into 2, 3, 7, 100 pieces — totally free. So surely a square is just as cooperative? Cut it corner to corner and fan each half, and you get any even number easily. But try to cut a square into three equal-area triangles. Or five. Or seven. You will fail, and so will everyone else, forever: Monsky’s theorem (1970) says a square cannot be divided into an odd number of equal-area triangles — and, astonishingly, the only known proof runs through 2-adic number theory, a way of measuring numbers by divisibility instead of size. No purely geometric proof has ever been found. For pentagons and beyond the wall closes in further: Kasimatis’s theorem (1989) says a regular polygon with five or more sides can be fair-sliced into triangles only when the number of pieces is a multiple of the number of sides. A pentagon accepts 5, 10, 15 — and nothing else.
So the full landscape, from freewheeling to rigid: triangles allow every count, squares allow exactly the even counts, and every regular polygon from the pentagon up allows only multiples of its own side count. The interactive’s fair-slice machine lets you sweep both dials and watch the forbidden numbers appear — and its second panel shows how, whenever slicing is possible, a single labeled length rebuilds the entire polygon, with all the trigonometry cancelling before your eyes.
Educators: Download this puzzle for use in class here: Eight Fair Slides Worksheet & Solution
One email, one puzzle, no noise — with a hint ladder and a full worked solution.