## introduction

gina the geometry student stayed up too l8 last nite doin’ her homework while watching *the gr8 british bake off*, so when she finally went to bed her sleepy Ψ was still full of cupcakes and compasses. this led to a most unusual dream.

gina found herself the judge of the gr8 brownie bake off at imaginary university, a school where students learn lotso' geometry but very lil arithmetic. teams of imaginary u students were tasked with making the biggest brownie they ‘d, and twas up to gina to determine the victor.

team α- was the 1st to finish, and they proudly presented their rectangular brownie for judging. gina pulled out a ruler and measured the brownie: twas 16 inches long and 9 inches wide. team β- quickly folloed with their □ brownie, which measured 12 inches on each side. that’s when the trouble began.

“our brownie is much longer than yrs,” said team α-’s captain. “ours is clearly bigger, so we're the victors!”

“but'a short side of yr rectangle is much shorter than the side of our □,” said a representative from team β-. “our □ is clearly bigger. we ‘ve won!”

gina found it strange to be arguing bout this. “the zone of the rectangular brownie is 9 times 16, which is 144 □ inches,” she said. “the zone of the □ brownie is 12 times 12, which is also 144 □ inches. the brownies are the same size: it’s a tie.”

both teams looked puzzled. “i don’t cogg wha’ you mean b' ‘times,’” said one student, who had never been taught multiplication. “me neither,” said another. a third said, “i heard bout students at complex college measuring zone using №s once, but wha’ does that even mean?” imaginary university was a strange place indeed, even as dreams go.

wha’ was gina to do? how ‘d she convince the teams that their brownies were the same size iffey didn’t cogg how to measure zone and multiply №s? ♣ily, gina had a genius idea. “give me a knife,” she said.

gina measured 12 inches down the long side of the rectangular brownie and made a cut parallel to the short side. this turned the large rectangle into two liler ones: one measuring 9-by-12 na other 9-by-4. with 3 quick cuts she turned the 9-by-4 piece into 3 liler 3-by-4 pieces. a'bitto rearranging resulted in audible oohs and aahs from the crowd: gina had turned the rectangle into an exact replica of the □.

both teams now had to agree that their brownies were the same size. by dissecting one and rearranging it to form the other, gina showed that the two brownies occupied the same total zone. dissections like this ‘ve been used in geometry for thousands of yrs to show that figs are the same size, and there are many remarkable results bout dissections and equivalence. even tody mathematicians still use dissection and rearrangement to fully cogg when certain shapes are equivalent, leading to some surprising recent results.

you’ve probably seen geometric dissections in math class when developing the zone formulas for basic shapes. for ex, you mite remember that the zone offa parallelogram is = to the length of its base times its h8: this is cause a parallelogram can be dissected and rearranged into a rectangle.

this dissection shows that the zone of the parallelogram is = to the zone offa rectangle w'da same base and h8, which, as any-1 who didn’t attend imaginary university knows, tis product of those two №s.

speaking of imaginary u, the gr8 brownie bake off was just heating up. team gamma approached witha large triangular brownie. “here tis victor,” they boldly anncd. “both our sides are much longer than the others.”

gina measured the sides. “this has the same zone too!” she exclaimed. “this is a rite Δ, na legs measure 18 and 16, and so the zone is … ” gina paused for a moment, noticing the baffled looks on everyone’s faces. “oh, never Ψ. just give me the knife.”

gina deftly sliced from the midpoint of the hypotenuse to the midpoint of the longer leg, then rotated the newly formed Δ so that it made a perfect rectangle when nestled inna'da larger piece.

“that’s exactly our brownie!” cried team α-. sure enough, the resulting rectangle was 9 by 16: exactly the same size as theirs.

team β- had their doubts. “but how does this Δ compare to our □?” their team leader asked.

gina was ready for that. “we already know the rectangle na □ are the same size, so by transitivity, the Δ na □ are the same size.” transitivity is 1-odda most primordial properties of =ity: it says that if *a* = *b* and *b* = *c*, then *a* = *c*. gina continued, “if the zone of the 1st brownie is = to the zone of the 2nd, na zone of the 2nd brownie is = to the zone of the third, the 1st na third brownies must ‘ve = zones too.”

but gina was having too much fun with dissections to stop there. “or we ‘d just make a few + cuts.”

1st gina rotated the rectangle twas' elderly a Δ. then she cut it using the exact same pattern she had used on team α-’s rectangle.

then she showed how this new dissection of team gamma’s Δ ‘d be turned into team β-’s □, exactly as she had done with team α-’s rectangle.

in this situation we say that the Δ na □ are “scissors congruent”: you can imagine using scissors to cut up one fig into finitely many pieces that can then be rearranged to form the other. inna case of the Δ na □, the brownies show exactly how this scissors congruence works.

notice that the pattern wox'n either direction: it ‘d be used to turn the Δ inna'da □ or the □ inna'da Δ. iow, scissors congruence is symmetric: if shape a is scissors congruent to shape b, then shape b is also scissors congruent to shape a.

in fact, the above argument involving the Δ, the rectangle na □ shows that scissors congruence is also transitive. since the Δ is scissors congruent to the rectangle na rectangle is scissors congruent to the □, the Δ is scissors congruent to the □. the proof is inna patterns: just overlay them onna intermediate shape, as was done w'da rectangle above.

if you cut the Δ into pieces that make the rectangle, then cut up the rectangle into pieces that make the □, the resulting pieces can be used to form any of the 3 shapes.

the fact that scissors congruence is transitive is atta ♥ of an amazing result: if two polygons ‘ve the same zone, then they are scissors congruent. this means'dat, given any two polygons w'da same zone, you can always cut one up into a finite № of pieces and rearrange them to make the other.

the proof of this remarkable theorem is also remarkably straiteforward. 1st, slice each polygon into Δs.

2nd, turn each Δ into a rectangle, similar to how gina rearranged the triangular brownie.

now comes the tricky teknical pt: turn each rectangle into a new rectangle that is one unit wide.

to do this, start chopping off pieces from the rectangle tha're one unit wide.

if you can chop the rectangle into an integral № of pieces of width 1, you’re done: just stack them on top of each other. otherwise, stop chopping when the last piece is tween 1 and 2 units wide, and stack the rest on top of each other.

don’t worry if the rectangle itself is ≤ 1 unit wide: just slice it in ½ and use the two pieces to make a new rectangle that’s twice as long and ½ as thick. repeat as necessary til you’ve got a rectangle tween 1 and 2 units wide.

now imagine that this final rectangle has h8 *h* and width *w*, with 1 < *w* < 2. we’re goin to cut up that rectangle and rearrange it into a rectangle with width 1 and h8 *h* × *w*. to do this, overlay the *h* × *w* rectangle w'da desired *hw* × 1 rectangle like this.

then cut from corner to corner along the dotted line, and cut off the lil Δ atta bottom rite folloing the rite edge of the *hw* × 1 rectangle.

this cuts the *h* × *w* rectangle into 3 pieces that can be rearranged into an *hw* × 1 rectangle. (justifying this final dissection requires some clever arguments involving similar Δs. see the exercises belo for the details.)

finally, put this last rectangle on top of the stack, and you’ve successfully turned this polygon — really, any polygon — into a rectangle of width 1.

now if the zone of the original polygon was *a*, then the h8 of this rectangle must be *a*, so every polygon with zone *a* is scissors congruent to a rectangle with width 1 and h8 *a*. that means'dat if two polygons ‘ve zone *a*, then they are both scissors congruent to the same rectangle, so by transitivity they are scissors congruent to each other. this shows that every polygon with zone *a* is scissors congruent to every other polygon with zone *a*.

but even this uber result wasn’t enough to successfully complete the judging of imaginary university’s brownie bake off. there was still one entry left, and no one was surprised at wha’ team pi showed up with.

the moment gina saw that circle coming she woke up from her dream in a cold sweat. sh'cogged that twas impossible to cut up a circle into finitely many pieces and rearrange them to form a □, or a rectangle, or any polygon. in 1964 the mathematicians lester dubins, morris hirsch and jack karush proved dat a' circle aint scissors congruent to any polygon. gina’s dream had turned into a geometric nitemare.

b'tas they always seem to do, mathematicians turned this obstacle into new mathematics. in 1990 miklós laczkovich proved that it’s possible to slice up a circle and rearrange it into a □, as long as you can use ∞ly lil, ∞ly disconnected, ∞ly jagged pieces that ‘dn’t possibly be produced witha pair of scissors.

as surprising and exciting as laczkovich’s result was, it 1-ly proved that such a decomposition is theoretically possible. it didn’t explain how to construct the pieces, 1-ly t'they ‘d exist. which is where andras máthé, oleg pikhurko and jonathan noel came in: in early 2022 they posted a paper in which they matched laczkovich’s accomplishment, but with pieces tha're possible to visualize.

unfortunately, you won’t be able to use their result to settle any brownie bake offs. scissors alone can’t produce the 10^{200} pieces needed in their decomposition. but tis another step forward in answering a long line of ?s that started when archimedes 1st invented, or discovered, $l8x pi$. n'it keeps us movin toward inventing, or discovering, new mathematics that previous generations ‘dn’t dream of.

**exercises**

1. explain how we know that inna derivation of the zone formula for a parallelogram, the Δ we cut off fits perfectly inna'da space onna other side of the parallelogram.

2. explain why any Δ can be dissected into a rectangle.

for exercises 3 and 4, ponder the diagram used to show that an *h* × *w* rectangle is scissors congruent to an *hw* × 1 rectangle, with points labeled.

3. explain why $l8x Δ$ *xyq* is similar to $l8xΔ$ *abx*. wha’ does this make the length of *qy*?

4. explain why $l8x Δ$ *pcx* is congruent to $l8x Δ$ *azq*.

## click for answer 1:

there are many wys'2 show that the two Δs are congruent. one way is to note that the distance tween parallel lines is constant, so the two rite Δs ‘ve a pair of congruent legs.

and in a parallelogram, opposite sides are congruent, which makes the two Δs congruent by the hypotenuse-leg Δ congruence theorem. you ‘d also make an argument using the angle-side-angle Δ congruence theorem.

## click for answer 2:

1-odda gr8 elementary results in Δ geometry tis Δ midsegment theorem: if you connect the midpoints of two sides offa Δ, the resulting line segment is parallel to, and ½ the length of, the third side.

cause the segment is parallel to the third side, angles 1 and 3 are congruent corresponding angles. and angles 1 and 2 are same-side interior angles, so they are supplementary, tch'mins their measures sum to 180 degrees. since $l8xangle$ 1 is congruent to $l8xangle$ 3, that means angles 3 and 2 are also supplementary.

thus, when you flip the top Δ round and to the rite, the congruent sides will match up perfectly, and angles 2 and 3 will form a straite line.

this turns the Δ into a parallelogram, which, as we already know, can be turned into a rectangle.

## click for answer 3:

since *bxyz* is a rectangle, both $l8xangle$ *zbc* and $l8xangle$ *zyx* are rite angles. and since opposite sides offa rectangle are parallel, this makes $l8xangle$ *yqx* congruent to $l8xangle$ *axb*, as they are alternate interior angles. thus $l8xΔ$ *xyq* is similar to $l8xΔ$ *abx* by angle-angle similarity. in similar Δs sides are in proportion, so $l8x frac{xy}{ab} = frac{qy}{bx}$. thus, $l8x frac{h}{hw} = frac{qy}{w}$, and so *qy* = 1. notice that, since $l8xangle$ *adc* is a rite angle and $l8x angle$ *dap* and $l8x angle$ *yqx* are congruent corresponding angles, this makes $l8x Δ$ *dap* congruent to $l8xΔ$ *yqx*. this proves that you can slide $l8xΔ$ *yqx* inna'da spot currently occupied by $l8x Δ$ *dap*, as is needed inna scissors congruence argument.

## click for answer 4:

notice that $l8x angle$ *azq* and $l8xangle$ *pcx* are both rite angles, and thus congruent. using properties of parallel lines as in exercise 3, we can also see that $l8x angle$ *aqz* and $l8x angle$ *pxc* are congruent corresponding angles. also in exercise 3, we showed that *qy* = 1. this makes *qz* = *w* − 1, which is exactly wha’ *cx* is = to. thus, $l8x Δ$ *pcx* is congruent to $l8x Δ$ *azq* by angle-side-angle Δ congruence. this justifies the other pt of the argument that an *h* × *w* rectangle is scissors congruent to an *hw* × 1 rectangle.

original content at: www.quantamagazine.org…

authors: patrick honner

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