# A hungry ant on a matchbox

A hungry ant is standing on the top of an empty matchbox cover (without its drawer) and has detected the smell of a drop of honey on the floor of the cover as shown in the figure below. What is the shortest distance for this little creature to crawl and reach the drop and drink it?

• But, you have watermarked someone else's artwork with your own name (you added the ant & the dimensions, but "borrowed" the matchbox). Is it under a Creative Commons license? :-) Nice artwork, though :-)
– Mawg
Commented Oct 22, 2017 at 8:24
• How accurate are the dimensions which you give? Ship matches are given as 2 inches = (5.08 mm) long. Are the real dimensions in fact denominated in inches? Commented Oct 23, 2017 at 16:33

So I drew a picture where the red part is outside the box, the gray part is inside the box, black dot is the ant, yellow dot is the honey drop. The shortest way to crawl for the ant is to take the most direct path, which is the hypothenuse of the right-angle triangle (drawing) called $d$. To find $d$, apply simple geometry : $d = \sqrt{(1.75+1.5+1)^2+(1+4)^2} = 6,56...$

• Hypoteneuse, not tangent, and geometry, not trigonometry, but very good explanation. Commented Oct 21, 2017 at 22:14
• My bad it’s fix for the hypothenuse ^^ but isn’t trigonometry the geometry of triangles basically ? Commented Oct 22, 2017 at 19:24
• @DamienBannerot not quite, trig has to do specifically with angles and side lengths, and their relations. This doesn't really fall under that umbrella.
– user18141
Commented Oct 22, 2017 at 20:17
• @DamienBannerot - Not quite; you're only working at the c^2=a^2+b^2 level, not the angles-and-ratios level. Commented Oct 22, 2017 at 22:32

Depending on what we allow, he could have it after crawling

5.06 cm and falling 1.5 cm. The total CRAWL is then 5.06cm, but the total travel is 6.56cm

My reasoning is:

He's an ant, so he can crawl along the top of the box for 4cm, flip under it and be upside down (because ants can do that), then crawl 1cm towards the back of the box. If he then lets go of the box he will land laterally beside the honey drop at a distance of .75cm, which he then traverses, for a total "crawl distance" of 5.75cm, plus the drop. But this is not optimal.

So:

Instead let us consider that he must traverse 5cm "back and forth" and .75cm "side to side." With the seemingly negligible thickness of the box, this can represent a right triangle of side 5cm and .75 cm, which give a hypotenuse of 5.06cm, which is therefore the direct distance to the point directly above the honey. All the ant must do is make the drop at any point after flipping upside down and before arriving above the honey (which I've also assumed to have negligible radius).

Note that

This total distance is less than the best pure-walking path, already found by Gareth McCaughan [Edit - his answer was reduced to the correct distance, having previously been over 8cm], whose answer may be more in the spirit of the asker's intentions. I will still submit this as another view. Additionally, the 6.56cm solution becomes even better when we consider the request was for "the shortest distance for this little creature to crawl," which here is 5.06cm.

Although I cheesed in the "drop distance" to reduce the "crawl distance," it's interesting to note that this achieved the same final result (of 6.56cm) as the "correct" answers, even though theirs included direct-path triangles along a doubly-unfolded box. They've already laid out their math well, so I'm not going to edit mine, but it's mathematically interesting to see that "walk above it and drop" has the exact same result as "walk the most direct route along a three-dimensional series of walls." Just thought I'd point that out, because I thought it was neat!

• As it is mentioned in the puzzle, the crawling distance is required and no dropping is allowed. If it was a spider we could consider dropping but ants may fall but not dropping intentionally. :-) Commented Oct 20, 2017 at 16:08
• Fair enough. I added a fun note about the result, though. I didn't see that restriction, but I was kind of cheating anyway. =-) Commented Oct 20, 2017 at 16:26
• To tell you the truth, you had a good way of thinking, well done. :-) Commented Oct 20, 2017 at 16:44
• @Seyed "and no dropping is allowed" not at all, no such thing is mentioned in the question. it says he must crawl and reach. so he crawls, then drops, and thus reaches. if it said "he must crawl and thus reach ..." then you'd have a point. this here is the correct answer. Commented Oct 22, 2017 at 16:36
• @WillNess, Read all other comments. Ants won't drop themselves, it is NOT in their nature, don't mix them with spiderman. Commented Oct 22, 2017 at 17:14

The question implies Ant can crawl, but does not imply that that Ant cannot travel by other means. It also does not imply that Ant cannot alter the environment around it.

So:

Suppose that Ant can chew through the box with its mandibles. It can then drop down to the lower level of the box, 1.5 cm, if the thickness of the box material is negligible), which would hardly be enough to hurt a hardy ant. Then, it would crawl $\sqrt{(3.5-1.75-1)^2 + (5-1-1)^2} = \sqrt{(0.75)^2 + (3)^2} \approx 3.0923$ cm, for a total distance traveled of 4.5923 cm.

But we can do better:

Suppose that Ant is strong enough to jump an arbitrary distance (ants are very strong, after all). Once it chews through the box, it ought to be able to jump directly to its destination once it has a foothold in the hole, since it has line of sight. Since Ant is strong enough to jump an arbitrary distance, the effects of gravity become negligible, and the distance can be calculated as a line between the goal and Ant's starting position (since it's approximately the same as in the hole it chewed, exactly the same if Ant is a particle): $\sqrt{(3.5-1.75-1)^2 + (5-1-1)^2 + (1.5)^2} = \sqrt{(0.75)^2 + (3)^2 + (1.5)^2} \approx 3.4369$ cm.

But we're not done yet:

3.4369 cm is approximately the total distance traveled; however, since Ant is directly leaping from the starting position to the end position, it never actually crawls at all. Hence the minimum total distance crawled is 0 cm.

Note that this answer makes even more sense in a zero-gravity environment.

• Putting the "lateral" in lateral-thinking ?
– Rubio
Commented Oct 21, 2017 at 16:40
• "does not imply that that Ant cannot travel by other means" I would like to submit the answer that the ant gets a bee to carry him and crawls a total of 0cm. Commented Oct 26, 2017 at 3:34

Nice problem and solutions!

The surface is effectively a torus, so can be unfolded as a rectangle which can tile the plane. A straight-line path can be drawn from an ant-start point to the honey-point in any rectangle in the tiling. So no cutting is necessary, unlike the case with a cuboid’s surface.

Curiously

the tile is a square for the dimensions given.

I couldn’t help wondering if there isn’t a Pythagorean triple lurking there, but the current dimensions just miss a nice one.

A classic Bryant & May UK matchbox, with glue for the matches supplied by my grandmother’s little factory in Dumbarton, Scotland, and as featured in Agatha Christie, had dimensions (checked on eBay) 6 x 3.5 x 2 cm (longer and deeper than the previous dimensions). If we keep the ant at its current location and shift the honey from (1,1) to:

(1.5,0)

Then the arithmetic uses the triple:

(20,21,29)

To give a distance of exactly:

7.25 cm

If you

unfold the matchbox cover, also splitting into "inside" and "outside" and unfolding those at the exposed edges of the box,

you will find that

the ant and the honey are 1+5-1=7cm apart along the length-of-matchbox axis, and (taking the shorter direction) 1.75+1.5+1=4.25cm apart along the around-the-matchbox axis, for a distance of $\sqrt{5^2+4.25^2}\simeq6.56$cm.

I confess

I was expecting this to turn out to be a nice round number and wonder whether I have slipped up somewhere above...

(Thanks to frodoskywalker for pointing out an error in the comments.)

• I think you'll find the long-axis distance is 5cm... Commented Oct 20, 2017 at 16:07
• D'oh, so it is! Commented Oct 20, 2017 at 16:18

$$\sqrt{41}$$.

Because

If you cut the box in half(upper and lower part is cut through the way there will be right and left piece), you can easily see you won't need the right part. At this point you can unfold the left part making it easy to identify the best way to get to the honey(with mirroring the painted side, simulating the unpainted side). This way you will be able to easily identify the best way to go there is a line which goes left on top part, goes to side of the box, switch to unpainted side, and continue the way on unpainted side on the lower part. There will be a right triangle where a = 5, b = 4, c = $$\sqrt{41}$$, because of Pythagoras. If you can post an edit with the image i am describing i would appreciate it, you should only mirror the 5 x 5 part to the edge which is on the leftmost part on the original picture(where the ant will switch sides).

Can the ant drop in?

The honey is only offset by .75 (1.75 - 1). Drop in perpendicular to the honey. Then crawl 1 cm to the honey.

to the drop
strt(.75^2 + 4^2) = 4.07
1 cm back to honey
total 5.07

• No, the ant can't drop, he crawls. This was suggested before. Thanks for your solution. :-) Commented Oct 21, 2017 at 11:37
• @Seyed Maybe you could have been explicit in the question. Commented Oct 21, 2017 at 11:45
• It has been clearly mentioned in the question that you should find the distance that the ant "crawl". By the way, ant will never drop willingly, they my fall but not drop intentionally. Commented Oct 21, 2017 at 12:05
• @Seyed You still could have explicit stated cannot drop. Great question. Nice graphic. Commented Oct 21, 2017 at 12:10
• @Seyed I disagree that the question is clear on this point. You ask for the minimum distance crawled; that in no way implies that other methods of moving are forbidden. Indeed, it suggests that other methods are encouraged, since they help reduce crawling. If I wanted to get downtown with as little walking as possible, you'd recommend I drive or take a bus, not advise me on the shortest all-walking route. Commented Oct 21, 2017 at 23:39