what all do we know? are the circles true circles?

and like do we have enough information to say a number?

I'm not that mathy

It's a good question, and the simple answer is that we aren't given enough hard information for there to be a unique solution, or in other words, for the puzzle to be a puzzle. But then, this is true of all puzzles.

In order to derive a solvable puzzle you have to make a few assumptions, and I think they are these:

The "rectangle" is indeed a rectangle of sides 8 and x.

The two "circles" are indeed circles.

The line that appears to pass through the point where the two circles touch does indeed pass through that point.

That line is parallel with the side length 8 and perpendicular to the side length x.

It is somewhat surprising that that is enough to generate a genuine puzzle with a single solution, but I believe it is.

what all do we know? are the circles true circles?

and like do we have enough information to say a number?

I'm not that mathy

What we know:

(it's pretty much what you'd guess by looking at the diagram)

The rectangle has vertical and horizontal sides, and is 8 high. The puzzle is to find the width, x.

The two inscribed circles are true circles and touch each other tangentially (they just touch without overlapping). One circle tangentially touches one side and the top of the rectangle, the other circle tangentially touches the other side and the bottom.

A vertical line (parallel to the rectangle, side 8) has one end touching one circle, the other end touching the other circle. The line passes through the point where the circles touch each other but you'll find that last condition isn't necessary because the line always passes through that point if it meets the other conditions. The line is 6 long.

You're not told the size of the two circles, other than that they fit inside the rectangle as shown.

You've got enough information to find the answer, which is a constant.

One way of finding an approximate answer is to draw the diagram accurately and measure x. If you know how, you can use a CAD program to draw it, and get a very accurate approximation - but it's still just an approximation, and the real challenge is to use geometry or mathematics to find the answer.

JoeP's (numerical) answer is a reasonable guess... but

It's wrong

Clue

Because you're not told the size of the circles but (correctly) assume that the answer is constant, then either there's only one (two?) possible size(s) the circles can be to fit in that rectangle, or you can make one circle smaller and the other correspondingly larger without changing the answer. Which of those possibilities is correct? If it's the latter, can you can draw an extreme version where one of the circles shrinks to the smallest possible size, and does that simplify the puzzle?

__________________

Last edited by ceptimus; 09-16-2020 at 09:38 AM.
Reason: I realized that the part I've now struck out was nonsense.

One srs question: could this be solved with GCSE level maths?

It looks like the kind of thing that, when I did GCSE maths tuition (recent years but not this year), students would be horrified by but is actually p. straightforward. I say this not actually having tried to solve it yet.

The solution to these puzzles is almost always drawing the right lines.

Yes ... currently busybusy but I have a mental picture of the lines. And statements going round my head "opposite angles are equal", "the angle at the circumference is half the angle at the centre" ...

O1A and O2A are parallel, being radii at the point the two circles touch.
Angles AO1C1 and AO2C2 are equal, being opposite angles. Let's call this angle because the first angle in any maths problem is always called .
Angle AD1C1 = half angle AO1C1, which is a thing you are supposed to learn.

And now it's just trig ... I think

__________________ Free thought! Please take one!

Last edited by JoeP; 09-15-2020 at 03:38 PM.
Reason: Spoilerific

Looking at it again, I found a solution that doesn't involve sines or cosines, just Pythagoras. If people are interested I'll post in when I'm back at the computer (phone posting right now).

Looking at it again, I found a solution that doesn't involve sines or cosines, just Pythagoras. If people are interested I'll post in when I'm back at the computer (phone posting right now).

I'm fairly confident that the centers of the circles and the point where they touch all lie on the diagonal of the rectangle, but it would take me a couple minutes to prove it.

Question: are the red line segments necessarily parallel? They look that way no matter what scale I draw them at, but I can't think of a reason they would be. Or really I guess my problem is I can't think of a reason they would be perpendicular to the line the centers lie on. AKA parallel to the line that is tangent to both circles, through the point where they meet.

I'm fairly confident that the centers of the circles and the point where they touch all lie on the diagonal of the rectangle, but it would take me a couple minutes to prove it.

Question: are the red line segments necessarily parallel? They look that way no matter what scale I draw them at, but I can't think of a reason they would be. Or really I guess my problem is I can't think of a reason they would be perpendicular to the line the centers lie on. AKA parallel to the line that is tangent to both circles, through the point where they meet.

Well, reaching back many years, I seem to recall that for your first statement to be true, that would have to be a square.
But then the reason I went into tech support was so I wouldn't have to do all the Math.

__________________
We are called to be architects of the future, not its victims.
- Buckminster Fuller

AD = AE = AG = AK (the radius of the small circle)

similarly

BD = BF = BH = BL (the radius of the large circle)

Triangle ADK is an isosceles triangle, with A as its apex. The same goes for triangle BDL, with apex B. From this one can conclude that AC = KL / 2 = 3

Now AG + AC + BH = 8 (the height of the rectangle)

Subtracting AC = 3 gives AG + BH = 5 (the sum of the radii)

Since AD + BD = AG + BH = 5, we have a right triangle ABC with a hypothenuse (AB) 5 and one side (AC) 3, the other side (BC) must be 4.

X = AE + BF + BC = AG + BH + BC = 5 + 4 = 9

Thanks for the puzzle, ceptimus. It was a fun one (and made me feel SMRT)

Question: are the red line segments necessarily parallel? They look that way no matter what scale I draw them at, but I can't think of a reason they would be. Or really I guess my problem is I can't think of a reason they would be perpendicular to the line the centers lie on. AKA parallel to the line that is tangent to both circles, through the point where they meet.

They are parallel. In the figure in my previous post it should be obvious (right?) that the angle ABC is equal to angle BA(all the way to the right), and that these are the half angles of the apices of isosceles triangle DBL and DAK. Therefore, the angle AK makes with the horizontal is equal to the angle BL makes with the horizontal, and thus they are parallel. Does that make sense?

Here's a 3:4:5 triangle drawn in Fusion 360, with a square drawn inside. The funny little symbols are the way you specify things like 'parallel', 'perpendicular', and 'equal length' in Fusion 360. The question is, what's the area of the square?