# Problems with seemingly not enough information

Two of my favorite geometry problems are as follows:

Consider two concentric circles with the property that a chord of the larger circle with length 20 is tangent to the inner circle. What is the area of the region between the circles?

Consider a sphere with a circular hole drilled through its center, such that the height of the remaining ring (in the direction along the hole) is $$h$$. What is the volume of the ring?

The fascinating thing about these problems is that they seem to be under-determined: In the first case, it seems that you should need to know at least one of the circles' radii; in the second case it seems that you should need to know the radius of the sphere or the radius of the hole. It turns out that the answer is independent of these unknown quantities, however, so the questions are well posed.

Another cute fact about these problems is that, supposing them to be well posed, they admit very easy computations of their answers, since we can choose the unknown parameters to be whatever we want to facilitate the computation:

• In the first case, choosing the radius of the inner circle to be zero, the chord is a diameter of the outer circle, and the desired area is just the area of the same circle, $$\pi \times 10^2 = 100 \pi$$.
• In the second case, choosing the radius of the hole to be zero, the volume is just that of a full sphere of radius $$h/2$$, i.e. $$\frac{1}{6}\pi h^3$$.

Question: What are other examples of problems which seem to be ill-posed, but are not?

I once thought that these two problems were anomalous, but I've recently discovered there are other examples. (I will post one if no one else does.) The examples need not come from geometry. Please post only one problem per answer, and if (as with the above problems) a computation is simplified by assuming the problem to be well posed, please explain.

• I first saw that sphere problem posed in a physics class, and solved it just as you did. Of course, a hole of radius zero is not a hole at all, but it is a limiting case. Also, at the time, I did not actually prove that the result was constant over all positive radii. I instead rationalized that if it were not, then that would make the question itself just as wrong as my answer.
– Pope
Jul 29 '21 at 7:42
• Related are porisms, configurations in which some coincidence remains as you vary one of the initial conditions. Examples are Steiner's porism and Poncelet's porism. Jul 29 '21 at 11:47

Two mathematicians sit on a bench:

$$-$$ I have two preschooler kids.

$$-$$ How old are they?

$$-$$ The product of their ages equals the number of pigeons around this bench.

$$-$$ That's not enough to understand.

$$-$$ The older is named Jim.

$$-$$ Now I see.

So, how old are they?

• 1 and 4. Though this problems requires that one have some sense of what counts as a pre-schooler. Jul 29 '21 at 15:11
• This appears at Puzzling Stack Exchange. Jul 29 '21 at 16:00
• I may be missing something, but doesn't this problem make the false assumption that if two children 2 years old, then neither is older than the other? Jul 30 '21 at 21:24
• @tupben it does, of course. Jul 30 '21 at 21:51

A sphere of radius $$R$$ is sliced by two parallel planes separated by distance $$d$$. What is the surface area of the sphere between the two planes?

Apparently you would need to know e.g. the distance of the planes to the equator, or something similar, but this turns out not to matter.

Given that the problem is well posed, you can compute the answer as follows: Let the answer be $$f_R(d)$$. Position the two planes symmetrically about the equator of the sphere. Then surface area between the planes and above (or below) the equator is half the total area between the planes, so $$f_R(d)=2f_R(d/2)$$, which implies that $$f_R(d)$$ is linear in $$d$$. Since $$f_R(2R)=4\pi R^2$$ (the surface area of the whole sphere), we must have $$f_R(d)=2\pi R d$$.

From Catriona Agg:

Find the area of the shaded region:

There are many different configurations of quarter circles which satisfy the constraints shown. In particular, you can make the radius of one of the shaded quarter-circles take any value in $$[0,8\sqrt{2}]$$, and there exist radii for the other two circles which make the diagram true. However, miraculously, the shaded area does not depend on the exact configuration. Knowing this, you can easily find the shaded area by:

• Taking the two shaded quarter circles to have equal radii. Then the shaded area is a semicircle of radius 8, so the area is $$\frac{\pi}{2}8^2 = 32 \pi$$.
• Taking one of the two shaded quarter circles to have radius zero. Then the shaded area becomes a single quarter circle of radius $$8\sqrt{2}$$, so its area is $$\frac{\pi}{4}(8\sqrt{2})^2 = 32 \pi$$.

There's the old one that goes something like:

"I'm thinking of a number. I double it, add nine, subtract three, divide by two and then subtract my original number. What number do I get?"

I'm assuming that this crowd can see why this is answerable without knowing the original number.

This one's more of a logic problem than a mathematics problem, but still fits the criteria of sounding like there's not enough information.

Alice, Bob and Claire want to play table tennis. They only have one table and two bats, so they decide that the first two players will be chosen randomly. After that, the winner of every game stays on, and the loser steps out for a game to let the other person in. By the end of their session, Alice has played 17 games, Bob 15 games and Claire 10 games. Who lost the second game?

In the picture below the largest grey circle in the right triangle has radius 4, the smallest radius 1. Find the radius of the other circle (from a Sangaku problem).

Consider a cubic like the one depicted below and a point P. Draw a tangent that will intersect the cubic in a point Q. Then draw the tangent in Q that intersects the cubic in a point S. If the area of the grey region is 1, which is the area of the dotted region?

• I like the first one, but am not so sure it fits the question. I was able to use similar figures to show that the circle radii must be in geometric progression. For the second one, I might take your word for it that the solution is the same for any case. Then I would put together a very simple simple case, with S(0,0), Q(1,0) and the graph of y=x(x-1)².
– Pope
Jul 29 '21 at 9:50