# Two puzzles about encompassing convex sets

1. Prove that any convex set of area $1$ is contained in a rectangle of area $2$.

2. Prove that any convex set of area $1$ is contained in a triangle of area $4$.

Notes: (1) is from "The Art of Mathematics: Coffee Time in Memphis" by Bollobás, and is very satisfying to solve in my opinion. (2) is "original", though I imagine someone has come up with these before me. It has a similarly nice solution I think. Note: I edited out a third question to address Daniel Schepler's point below.

• Seems like this would fit better on math.stackexchange.com ? Mar 23 '18 at 21:26
• In my opinion, (1) and (2) are definitely puzzles -- I know the solutions and they can be solved without any advanced mathematics and have elegant, simple solutions. However, problem (3) is definitely better on math stack exchange since it probably doesn't have an elegant solution. I can edit it out. Mar 23 '18 at 22:45
• For the benefit of the many math-puzzle fans on PSE who don't want to comb through the thousands of homework qs on MSE to find math puzzles, it's good that Tyler posted it here! @DanielSchepler Mar 23 '18 at 23:40

Here is a somewhat sketch proof for (1) that may be missing some less obvious details:

There exist two points along the boundary of the convex figure. The distance between them is L. This means that all other points along the curve has distance less than $L$. Place the rectangle so that it covers these points and make the rectangle $\frac{2}{L}$ wide.

Some graphical example:

So...

...assume for contradiction that there exists a figure which cannot be contained in the rectangle. This mean that there exists two points with distance $d > \frac{2}{L}$ along the perpendicular axis. Two reach these points, yielding the smallest area (to maximize the distance, we need to minimize the area), to keep the convex property, these must be straight lines. Now, the area of this figure is $A = d \times \frac{2}{L}$. Hence $d > \frac{2}{L}$, so $A > 1$. But $A = 1$, so we have a contradiction.

For (2) even sketchier...

...but basically the same argument as above. Pick three points along the boundary of the figure and draw a triangle, so that the area of the triangle is maximized, let the area by $A$. Then draw another larger equilateral triangle which has size $4 \times A$ in the other direction

Some more badly-drawn graphics:

To...

...maximize the outer triangle, we need to maximize the inner one, which means that the figure must be a triangle. Assume that the outer triangle has area larger than $4$. This means that $A>1$, which is a contradiction as well.

• Nice answer! For problem (2) your answer is the same as mine, and I think for problem (1) as well. My only question would be: how are you choosing the blue points (at distance $L$) so that you do it like the second diagram in the following link, and not like in the first diagram in the link: link to image Mar 24 '18 at 6:26
• Don't you just pick two points of the figure at maximal distance, and then place the rectangle parallel to the line between those two points? Those two points will be the only ones touching the short sides of the rectangle (else there would be a larger maximal distance). Mar 24 '18 at 6:46
• @TylerSeacrest Right, as Jaap Scherphuis said. So, in the first example in your image the rectangle is not correctly placed. Mar 24 '18 at 9:40
• @JaapScherphuis what if no maximum exists (say, for an open disk)? I guess this can be fixed by using the closure of the original set, but some care is needed here. Mar 24 '18 at 10:43
• Thanks for clearing that up -- I think I just misinterpreted "This means that all other points along the curve has distance less than L" the first time. Mar 24 '18 at 19:10

For part 1, see Carl Löndahl's answer.

Here is a trivial proof for part 2.

Start with the rectangle of area 2 from part 1. Extend it to a right-angled triangle of area 4 like this:
The legs of the triangle are twice the length of the sides of the rectangle.

• Wow -- good catch. Way easier than the intended answer =). Mar 24 '18 at 6:26

Since I usually prefer no-math solutions for geometry problems, here's one for question 1.

First, clamp the convex thingy between two verticals. The verticals are on springs, and will tightly press against whatever is between them.

Similarly, clamp the thingy between horizontal lines.

Then, rotate the convex thingy until the distance between the verticals is as big as it gets, or in other words, the longest diameter is horizontal. (This is important, since for this particular proof, we want to have the thing touching both the verticals at points that are on the same horizontal line.)

Now, the lines form a rectangle around the thingy. If the thingy was a triangle, it will look like this:

I have added a dotted vertical that splits the rectangle in two smaller rectangles. Each of these is exactly halved into parts inside and outside the triangle, so the area of the rectangle is exactly twice that of the triangle.

If the convex thingy has a more complex shape, it's all the better for us:

In the image, I have split the thingy along the maximal diameter, which is horizontal, because that's how we aligned the thingy.

Also, we know that there will be a point, or possibly a line segment, touching each side of the rectangle, because of how we clamped the lines. I chose a point on each horizontal that touches the thingy, and connected those points to the ends of the maximal diameter. This gave us a triangle on both sides of the maximal diameter.

As seen above, such a triangle's area is half of the encircling rectangle, so the combined area of the triangles is half of the outside rectangle.

Because the thingy is convex, we know from the definition of convex (a straight line segment connecting any two points of the thingy is entirely within the thingy), that all the triangle sides, and therefore the triangles themselves, are completely inside the thingy.

The convex thingy therefore completely contains within itself a quadrangle (or a triangle, if a side coincides with the longest diameter) that is half of the area of the outside rectangle. Thus, the convex thingy takes up at least half of the rectangle, and if the thingy's area is exactly 1, the rectangle cannot possibly have an area larger than 2.

For question 2, we can copy the rectangle that is not larger than 2, split the copy along a diagonal, and assemble a triangle out of the pieces, like @Jaap has already done.

• I like the verticals-on-springs visual. That makes it really easy to understand. Mar 24 '18 at 19:09