# The Simple Geometry Behind Brownie Bake Offs and Equal Areas

## Introduction

Gina the geometry student stayed up too late last night doing her homework while watching *The Great British Bake Off*, so when she finally went to bed her sleepy mind was still full of cupcakes and compasses. This led to a most unusual dream.

Gina found herself the judge of the Great Brownie Bake Off at Imaginary University, a school where students learn lots of geometry but very little arithmetic. Teams of Imaginary U students were tasked with making the biggest brownie they could, and it was up to Gina to determine the winner.

Team Alpha was the first to finish, and they proudly presented their rectangular brownie for judging. Gina pulled out a ruler and measured the brownie: It was 16 inches long and 9 inches wide. Team Beta quickly followed with their square brownie, which measured 12 inches on each side. That’s when the trouble began.

“Our brownie is much longer than yours,” said Team Alpha’s captain. “Ours is clearly bigger, so we are the winners!”

“But the short side of your rectangle is much shorter than the side of our square,” said a representative from Team Beta. “Our square is clearly bigger. We have won!”

Gina found it strange to be arguing about this. “The area of the rectangular brownie is 9 times 16, which is 144 square inches,” she said. “The area of the square brownie is 12 times 12, which is also 144 square inches. The brownies are the same size: It’s a tie.”

Both teams looked puzzled. “I don’t understand what you mean by ‘times,’” said one student, who had never been taught multiplication. “Me neither,” said another. A third said, “I heard about students at Complex College measuring area using numbers once, but what does that even mean?” Imaginary University was a strange place indeed, even as dreams go.

What was Gina to do? How could she convince the teams that their brownies were the same size if they didn’t understand how to measure area and multiply numbers? Luckily, 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 smaller ones: one measuring 9-by-12 and the other 9-by-4. With three quick cuts she turned the 9-by-4 piece into three smaller 3-by-4 pieces. A bit of rearranging resulted in audible oohs and aahs from the crowd: Gina had turned the rectangle into an exact replica of the square.

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 area. Dissections like this have been used in geometry for thousands of years to show that figures are the same size, and there are many remarkable results about dissections and equivalence. Even today mathematicians still use dissection and rearrangement to fully understand when certain shapes are equivalent, leading to some surprising recent results.

You’ve probably seen geometric dissections in math class when developing the area formulas for basic shapes. For example, you might remember that the area of a parallelogram is equal to the length of its base times its height: This is because a parallelogram can be dissected and rearranged into a rectangle.

This dissection shows that the area of the parallelogram is equal to the area of a rectangle with the same base and height, which, as anyone who didn’t attend Imaginary University knows, is the product of those two numbers.

Speaking of Imaginary U, the Great Brownie Bake Off was just heating up. Team Gamma approached with a large triangular brownie. “Here is the winner,” they boldly announced. “Both our sides are much longer than the others.”

Gina measured the sides. “This has the same area too!” she exclaimed. “This is a right triangle, and the legs measure 18 and 16, and so the area is … ” Gina paused for a moment, noticing the baffled looks on everyone’s faces. “Oh, never mind. 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 triangle so that it made a perfect rectangle when nestled into the larger piece.

“That’s exactly our brownie!” cried Team Alpha. Sure enough, the resulting rectangle was 9 by 16: exactly the same size as theirs.

Team Beta had their doubts. “But how does this triangle compare to our square?” their team leader asked.

Gina was ready for that. “We already know the rectangle and the square are the same size, so by transitivity, the triangle and the square are the same size.” Transitivity is one of the most important properties of equality: It says that if *a *= *b* and *b* = *c*, then *a *= *c*. Gina continued, “If the area of the first brownie is equal to the area of the second, and the area of the second brownie is equal to the area of the third, the first and the third brownies must have equal areas too.”

But Gina was having too much fun with dissections to stop there. “Or we could just make a few more cuts.”

First Gina rotated the rectangle that was formerly a triangle. Then she cut it using the exact same pattern she had used on Team Alpha’s rectangle.

Then she showed how this new dissection of Team Gamma’s triangle could be turned into Team Beta’s square, exactly as she had done with Team Alpha’s rectangle.

In this situation we say that the triangle and the square are “scissors congruent”: You can imagine using scissors to cut up one figure into finitely many pieces that can then be rearranged to form the other. In the case of the triangle and the square, the brownies show exactly how this scissors congruence works.

Notice that the pattern works in either direction: It could be used to turn the triangle into the square or the square into the triangle. In other words, 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 triangle, the rectangle and the square shows that scissors congruence is also transitive. Since the triangle is scissors congruent to the rectangle and the rectangle is scissors congruent to the square, the triangle is scissors congruent to the square. The proof is in the patterns: Just overlay them on the intermediate shape, as was done with the rectangle above.

If you cut the triangle into pieces that make the rectangle, then cut up the rectangle into pieces that make the square, the resulting pieces can be used to form any of the three shapes.

The fact that scissors congruence is transitive is at the heart of an amazing result: If two polygons have the same area, then they are scissors congruent. This means that, given any two polygons with the same area, you can always cut one up into a finite number of pieces and rearrange them to make the other.

The proof of this remarkable theorem is also remarkably straightforward. First, slice each polygon into triangles.

Second, turn each triangle into a rectangle, similar to how Gina rearranged the triangular brownie.

Now comes the tricky technical part: Turn each rectangle into a new rectangle that is one unit wide.

To do this, start chopping off pieces from the rectangle that are one unit wide.

If you can chop the rectangle into an integral number of pieces of width 1, you’re done: Just stack them on top of each other. Otherwise, stop chopping when the last piece is between 1 and 2 units wide, and stack the rest on top of each other.

Don’t worry if the rectangle itself is less than 1 unit wide: Just slice it in half and use the two pieces to make a new rectangle that’s twice as long and half as thick. Repeat as necessary until you’ve got a rectangle between 1 and 2 units wide.

Now imagine that this final rectangle has height *h *and width *w*, with 1 < *w* < 2. We’re going to cut up that rectangle and rearrange it into a rectangle with width 1 and height *h *×* w*. To do this, overlay the *h *×* w* rectangle with the desired *hw* × 1 rectangle like this.

Then cut from corner to corner along the dotted line, and cut off the little triangle at the bottom right following the right edge of the *hw* × 1 rectangle.

This cuts the *h *×* w* rectangle into three pieces that can be rearranged into an *hw* × 1 rectangle. (Justifying this final dissection requires some clever arguments involving similar triangles. See the exercises below 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 area of the original polygon was *A*, then the height of this rectangle must be *A*, so every polygon with area *A* is scissors congruent to a rectangle with width 1 and height *A*. That means that if two polygons have area *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 area *A* is scissors congruent to every other polygon with area *A*.

But even this powerful 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 what Team Pi showed up with.

The moment Gina saw that circle coming she woke up from her dream in a cold sweat. She knew that it was impossible to cut up a circle into finitely many pieces and rearrange them to form a square, or a rectangle, or any polygon. In 1964 the mathematicians Lester Dubins, Morris Hirsch and Jack Karush proved that a circle is not scissors congruent to any polygon. Gina’s dream had turned into a geometric nightmare.

But as 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 square, as long as you can use infinitely small, infinitely disconnected, infinitely jagged pieces that couldn’t possibly be produced with a pair of scissors.

As surprising and exciting as Laczkovich’s result was, it only proved that such a decomposition is theoretically possible. It didn’t explain how to construct the pieces, only that they could 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 that are 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 it is another step forward in answering a long line of questions that started when Archimedes first invented, or discovered, $latex \pi$. And it keeps us moving toward inventing, or discovering, new mathematics that previous generations couldn’t dream of.

**Exercises**

1. Explain how we know that in the derivation of the area formula for a parallelogram, the triangle we cut off fits perfectly into the space on the other side of the parallelogram.

2. Explain why any triangle can be dissected into a rectangle.

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

3. Explain why $latex \triangle$ *XYQ* is similar to $latex\triangle$ *ABX*. What does this make the length of *QY*?

4. Explain why $latex \triangle$ *PCX* is congruent to $latex \triangle$ *AZQ*.

## Click for Answer 1:

There are many ways to show that the two triangles are congruent. One way is to note that the distance between parallel lines is constant, so the two right triangles have a pair of congruent legs.

And in a parallelogram, opposite sides are congruent, which makes the two triangles congruent by the hypotenuse-leg triangle congruence theorem. You could also make an argument using the angle-side-angle triangle congruence theorem.

## Click for Answer 2:

One of the great elementary results in triangle geometry is the triangle midsegment theorem: If you connect the midpoints of two sides of a triangle, the resulting line segment is parallel to, and half the length of, the third side.

Because 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, which means their measures sum to 180 degrees. Since $latex\angle$ 1 is congruent to $latex\angle$ 3, that means angles 3 and 2 are also supplementary.

Thus, when you flip the top triangle around and to the right, the congruent sides will match up perfectly, and angles 2 and 3 will form a straight line.

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

## Click for Answer 3:

Since *BXYZ *is a rectangle, both $latex\angle$ *ZBC* and $latex\angle$ *ZYX *are right angles. And since opposite sides of a rectangle are parallel, this makes $latex\angle$ *YQX* congruent to $latex\angle$ *AXB*, as they are alternate interior angles. Thus $latex\triangle$ *XYQ* is similar to $latex\triangle$ *ABX *by angle-angle similarity. In similar triangles sides are in proportion, so $latex \frac{XY}{AB} = \frac{QY}{BX}$. Thus, $latex \frac{h}{hw} = \frac{QY}{w}$, and so *QY *= 1. Notice that, since $latex\angle$ *ADC* is a right angle and $latex \angle$ *DAP* and $latex \angle$ *YQX *are congruent corresponding angles, this makes $latex \triangle$ *DAP* congruent to $latex\triangle$ *YQX*. This proves that you can slide $latex\triangle$ *YQX *into the spot currently occupied by $latex \triangle$ *DAP*, as is needed in the scissors congruence argument.

## Click for Answer 4:

Notice that $latex \angle$ *AZQ *and $latex\angle$ *PCX *are both right angles, and thus congruent. Using properties of parallel lines as in exercise 3, we can also see that $latex \angle$ *AQZ* and $latex \angle$ *PXC* are congruent corresponding angles. Also in exercise 3, we showed that *QY *= 1. This makes *QZ* = *w* − 1, which is exactly what *CX* is equal to. Thus, $latex \triangle$ *PCX* is congruent to $latex \triangle$ *AZQ *by angle-side-angle triangle congruence. This justifies the other part of the argument that an *h* × *w *rectangle is scissors congruent to an *hw* × 1 rectangle.