Making Your Own Sense

Reflections on maths, learning and maths learning support, by David K Butler

When perimeter is equal to area

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davidkeithbutler

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Some triangles have the same perimeter (in cm) as area (in cm²). For example, the 6-8-10 right-angled triangle:

A diagram of a right-angled triangle with the two short sides labelled 6 cm and 8 cm and the hypotenuse labelled 10 cm. Below it are two calculations. Perimeter = 6 + 8 + 10 = 24 cm Area = half times 8 times 6 = 24 cm squared

What properties of the triangle might signal that this sort of alignment of perimeter and area is possible?

One answer I discovered to this question is completely surprising and delightful to me:

If any of the side lengths is 4 cm or less, then the perimiter (in cm) can’t be equal to the area (in cm²).

It seems remarkable that knowing about just one side of the triangle could rule out a property that ostensibly requires all three sides to confirm, but it’s definitely true, because I proved it. Indeed, the proof itself is one of the reasons I love this fact so much.

You may want to attempt to prove it yourself before I show you my proof…

Ok, here’s my proof.

Consider a triangle with side lengths aa, bb and cc in cm. It’s possible to arrange the sides in order of size, so choose the labels so that a≤b≤ca \leq b \leq c.

First we work with the perimeter.

Perimeter=a+b+c>b+c≥b+b=2b\begin{aligned} \text{Perimeter} &= a + b + c \\ &\gt b + c \\ &\geq b + b \\ &= 2b\end{aligned}

That is, the perimieter is strictly greater than 2b2b.

Now we work with the area. The area of a triangle is half the base times the height. Choose the side with length bb as the base and let the height be hh.

The side with length aa goes from the end of the side with length bb to the apex of the triangle. If it goes straight upwards, then the height is equal to aa, but otherwise, the height will necessarily be less than aa. Either way h≤ah \leq a.

Three triangles, all with the same length base, all with base labelled b. The side to the left of the base is labelled a and is oriented differently in each triangle. In the one triangle, it heads diagonally up and to the left. In another, it goes vertically. In the third, it goes diagonally up and to the right. The vertical height is drawn with a dotted line in the two side triangles and labelled with h in all three triangles.

So, when working with the area,

Area=12bh≤12ba\begin{aligned} \text{Area} &= \frac12 bh \\ &\leq \frac12 b a \\ \end{aligned}

Suppose that at least one of the sides is 4 units or less. Then the shortest side must be 4 cm or less. That is a≤4a \leq 4. And so

Area=12bh≤12ba≤12b×4≤2b\begin{aligned} \text{Area} &= \frac12 bh \\ &\leq \frac12 b a \\ &\leq \frac12 b \times 4\\ &\leq 2b \end{aligned}

That is, the area is less than or equal to 2b2b.

Combining these two facts gives

Area≤2b<Perimeter\text{Area} \leq 2b < \text{Perimeter}

And so the perimeter is strictly greater than the area. In particular, they can’t be equal.

End of proof! Yay!

There are so many things about this proof that I love. I love that it uses so much inequality reasoning, which I have a particular fondness for. I love that you decide whether the area and perimeter are equal based on comparing them both to a third thing rather than to each other directly. This feels like such a ninja move. I love that the thing you compare them to is just the middle-length side. Indeed, the longest side doesn’t really feature much in the argument, which is surprising. I also love that the 4 appears more-or-less to cancel out in just the right way with the half in the area formula. A different number wouldn’t have worked, so 4 is the perfect one.

I hope you like my little fact and its proof as much as I do.

Comments

5 responses to “When perimeter is equal to area”

  1. Susan Jones Avatar
    Susan Jones

    I figured that out at some point (one of those things I do when I need to give a student time ;P ) but hadn’t done the actual proofy stuff 😉 THanks for putting it into symbols!

    Reply
    1. davidkeithbutler Avatar
      davidkeithbutler

      I’m glad it was useful to have it written down. 🙂

      Reply
  2. Bryn Avatar
    Bryn

    I went with A = 0.5 abSinC
    So the area must be between 0 and 0.5ab.

    The perimeter is a+b+c and similar to you I said the minimum would be a + 2b.

    Extreme case 1: The area is big and the perimeter is small.
    Set the maximum area to the minimum perimeter:
    0.5ab = a + 2b
    ab = 2a + 4b
    a=2a/b +4
    So the smallest a is 4.

    Extreme case 2: The area is small and the perimeter is big.
    Min area is some 0.5abSinC (Assuming we’re ruling out the 0,0,0 triangle).
    Set minimum area to max perimeter (3c).
    0.5abSinC= 3c
    But a=b=c so
    0.5c^2sin60=3c
    csin60=6
    c = 6/sin60
    c=4root 3

    But as this is above 4, 4 is still the minimum for a.

    Reply
    1. davidkeithbutler Avatar
      davidkeithbutler

      Thanks for engaging! Interesting approach.

      Reply
  3. Playful Math 184: Carnival of Living Math – Denise Gaskins' Let's Play Math

    […] Butler plays with a geometric puzzle: When perimeter is equal to area. “One answer I discovered to this question is completely surprising and delightful to me. You […]

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