Three runners are located at the corners of an equilateral triangle, 100 meter a side. They run to a point inside the triangle and their goal is to do it as fast as possible. If they run at the same speed the natural point is the middle of the triangle. The challenge is to devise the meeting point if their speeds are 3, 4, and 5 meters per second respectively.

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    $\begingroup$ I presume the puzzle is to find the point so that they all arrive there at the same time? $\endgroup$ – xnor Jun 2 at 4:41
  • $\begingroup$ To get the shortest time you can show that they need to arrive at the same time. $\endgroup$ – Moti Jun 2 at 7:36

Now that you've specified that you want a geometric construction...

The basic idea here is that we'll flip the problem. Instead of directly constructing the point inside the triangle, we'll consider the following alternate problem:

Given three concentric circles of radii 3, 4, and 5. construct an equilateral triangle, each of whose vertices lie on a different one of these circles.

If we construct such a triangle $\triangle DEF$, then the center of the circles $O$ would be a point 3, 4, and 5 units from $D$, $E$, and $F$, in some order. This is the point we want, and then we simply need to transfer the point over to the original triangle by mapping the two triangles using a dilation, a rotation, and a translation.

So here's our three circles. Choose an arbitrary point $A$ on the outermost circle.


Now here's the clever idea of the day. Suppose our desired equilateral triangle is $\triangle ABC$, with $A$ on the outermost circle, $B$ on the second circle, and $C$ on the innermost circle. Then the point $C$ is what you get when you rotate $B$ 60 degrees around $A$. It follows that if I rotate the entire second circle around $A$, it will intersect the innermost circle at $C$. But that means I'm done!

We rotate the second circle 60 degrees counter-clockwise about $A$:


Mark the intersection point with the innermost circle:


Then constructing the last point is easy:


The observant reader will note that depending on what direction I choose to rotate the circle, or which intersection point I mark, there could be other equilateral triangles that satisfy the conditions. From some brief thought, I believe there are 4 in total, two of which work because $O$ is inside the triangle, whereas the other two do not because $O$ would lie outside said triangle for those constructions.

Anyways, we have our magic point inside a dummy triangle, so transfer it over and you're done.

  • $\begingroup$ Very nice. I can see that it is "equilateral":) Not the way I had in mind - actually more elegant than mine. I hope you see the value in delaying my "award"! The only thing missing is, how the 60 degrees rotation translates it with a different way the rotating by 70? Why ABC is 60? $\endgroup$ – Moti Jun 10 at 16:06
  • $\begingroup$ It seems that you do not see the difference between a construction and a proof. You need to have a set of statements that result mathematically the conclusion that BAC = 60. When you get have the proof I will accept the solution. $\endgroup$ – Moti Jun 17 at 18:50
  • $\begingroup$ @Moti I'm going to restate the proof I've been giving you: Suppose the equilateral triangle $\triangle ABC$ exists on those three circles satisfying our conditions. Then since $B$ rotated 60 degrees around $A$ will give $C$, rotating the set of possible points for $B$ (the middle circle) 60 degrees around $A$ will necessarily intersect the set of possible points for $C$ (the inner circle) exactly at $C$. Thus $A$ and $C$ are two correct vertices of an equilateral triangle satisfying our conditions, and the proof that $B$ is constructed correctly is trivial from the two green circles. $\endgroup$ – greenturtle3141 Jun 18 at 0:06
  • $\begingroup$ I believe that I understand your claims. You should start your proof by stating that BAC by definition is 60. Than you continue by stating that you rotate all the structure by 60 - showing the rotated triangle ,may be with the two circles demonstrating that the order of creating the intersection is important. $\endgroup$ – Moti Jun 18 at 2:07

I'm not sure what exactly you want but here you go:

Suppose the equilateral triangle is $\triangle ABC$. Suppose further that the 5 m/s runner starts at $A$, the 4 m/s runner starts at $B$, and the 3 m/s runner starts at $C$. Then the point where the runners need to meet is the unique point $P$ satisfying:

  • $AP = \dfrac{500}{\sqrt{25+12\sqrt{3}}}$
  • $BP = \dfrac{400}{\sqrt{25+12\sqrt{3}}}$
  • $CP = \dfrac{300}{\sqrt{25+12\sqrt{3}}}$

We can draw circles centered at $A$, $B$, $C$ with the above radii to verify that they are concurrent:


This also shows that this is a perfectly constructable, well-defined point. Your question does not specify to what degree we must define this point, so I'll leave it like this.

I will now prove that the above distances are the desired.

We are essentially solving for $x$ in the following diagram:


Rotate the entire diagram 60 degrees clockwise around $B$:


Because we rotated 60 degrees, $\angle PBP' = 60^\circ$. But then $\triangle PBP'$ is equilateral, so $PP' = 4x$.


We conclude that $\angle P'PC$ must be right, so $\angle BPC = 90+60 = 150^\circ$. By the Law of Cosines on $\triangle BPC$:

$$100^2 = 16x^2 + 9x^2 + 12\sqrt{3}x^2$$

Solving gives $\boxed{x = \dfrac{100}{\sqrt{25+12\sqrt{3}}}}$, as desired.

  • $\begingroup$ This is a solution but I am looking for a more elegant one without the need of higher level of knowledge in trigonometry - basic Euclidian approach. $\endgroup$ – Moti Jun 7 at 5:33
  • $\begingroup$ But this WAS a very basic Euclidean approach. I don't use any "advanced" trigonometry until the very end with the Law of Cosines. How exactly do you want us to define this point? Multiple correct answers were given, I don't really think you should just expect us to guess the method you're thinking of because simplicity is subjective. $\endgroup$ – greenturtle3141 Jun 7 at 13:02
  • $\begingroup$ You use trig - Law of Cosine. There is a way to prove this with basic diagrams. $\endgroup$ – Moti Jun 7 at 19:21
  • $\begingroup$ @Moti Prove what? $\endgroup$ – greenturtle3141 Jun 7 at 19:49
  • $\begingroup$ To show a way/proof where the ratio can be any - a, b, c. Your approach using trig extension to the general case will be way too complicated. $\endgroup$ – Moti Jun 7 at 19:56

Well, maybe there's a more puzzleriffic solution to this, but here's my go with a compass and a straightedge, because why not. :-)

I'm not going to bother with spoiler tags, since this answer is kind of long, and just seeing a couple of the first images in not going to spoil anything for anyone. Also, this method feels a bit.. forceful, there's probably a neater way to do all this. But anyway.

Because naming things is hard, I chose A as the runner with speed 4, B's speed is 5, and C's speed is 3.

We'll want to find the points which the runners would reach at the same time. We'll do that by plotting such loci (Oo, Latin!) pairwise, and then finding the intersection(s) of those plots.

Let's start by drawing a circle with radius 4 around A, and another with radius 5 around B. As long as those circles intersect, the actual units don't matter. I tried to pick something that gives a halfway readable diagram. Let's name the points where those circles intersect D and E. If the runners A and B would choose either of those points, they would obviously arrive at the same time.

Next, we choose one of those points (E), and consider the angle $\angle$AEB:

enter image description here

We want one more point whose distance is at the desired ratio of "4 from A, 5 from B". We could do that by choosing another length unit, and drawing two more circles at A and B, but instead of that, we can also bisect the angle $\angle$AEB to find point I, and then use the angle bisector theorem to show that point I splits the segment AB at the desired proportions: line EI will bisect the angle $\angle$AEB if and only if the ratio $\frac{\mid\text{AI}\mid}{\mid\text{BI}\mid}$ is equal to to the ratio $\frac{\mid\text{AE}\mid}{\mid\text{BE}\mid}$, which is, by our construction, $\frac{4}{5}$.

enter image description here

Now we have three points, E, I and D, all located so that their distance from A is 4/5 of their distance from B. Thus we can construct (using perpendicular bisectors of EI and ID, not shown) a circle through those points. This is called a Circle of Apollonius, and by its construction, we know that all points on this circle share the property that their distance from A is 4/5 of their distance from B. In other words, all points that runners A and B would reach at the same time lie exactly on the circle $\circ$EID.

enter image description here

Then we can just repeat the procedure for B and C with circle size ratio 5:3, and we'll find the second Apollonian circle through J, O and K:

enter image description here

And now we can see that the two Apollonian circles intersect twice, at P and Q. P is inside the triangle, so that's going to be our spot:

enter image description here

Since we want to be thorough, let's add the third Apollonian circle for A and C (at ratio 4:3) to confirm that it too passes through P (and Q).

This should be trivially true, since P (or for that matter, Q) is on both of the earlier Apollonian circles, meaning that runners A and B would reach it simultaneously, and also that runners B and C would reach it simultaneously. The third Apollonian circle plots all the the points that runners A and C would reach simultaneously, so by the transitivity of "simultaneity", P (and Q) should obviously be on the third circle, too.

enter image description here

How nice, it all worked out as expected! So there we have it, the runners should meet at P.

(If they chose to meet at Q instead, they would all arrive at the same time there, too. However, that would take a lot longer.)

All images plotted with the excellent GeoGebra tool, and screenshotted from there.

  • $\begingroup$ So complicated - will take me some time to digest... $\endgroup$ – Moti Jun 2 at 7:25
  • $\begingroup$ There is way simpler way without the need to know about Circle of Apollonius (is it easy to prove it? - did not try) $\endgroup$ – Moti Jun 2 at 7:34
  • $\begingroup$ @Moti, yeah, that's often the problem with using just a compass and straightedge; it would be a lot easier to find point I by saying "starting at A, go $44.44..$ metres towards B", but it's kind of the whole point of the exercise to show that it can be done without using such superfluous tools as a tape measure :-) $\endgroup$ – Bass Jun 2 at 7:36
  • $\begingroup$ You did not need to do it that way - you can use different circles -) I can email you the elegant solution with basic geometry tools. $\endgroup$ – Moti Jun 2 at 7:38
  • $\begingroup$ @Moti no need, we'll see if someone else can come up with it, and if they don't, you can post it as a self-answer after a couple of days. $\endgroup$ – Bass Jun 2 at 7:39

There is a little dilemma with this question;

since there are two targets A would like to go, to which direction should A move to meet with B and C? and visa versa.?

which forces you assume something like;

They know their own speed and decide where to meet which will be shortest distance to meet for sure since they move straight all the time.


they have to check others' location at any time they move and move towards between two distance which will be some other kind of hard geometry question in my opinion.

So I will go for the first one first,

This is a question to find a point inside a triangle where

the distance from A (5), B (4) and C (3) will be 5x, 4x and 3x respectively.


enter image description here

so it becomes a pure geometry question but the answer is

about 15 seconds.

For the second case the answer will be something like:

enter image description here

but geometrically a better way to show and proof is needed. It takes maybe a little more than first one but the shortest will be for sure the first one.

  • $\begingroup$ It is not a tracking problem - they know the speed and task and can coordinate between them. $\endgroup$ – Moti Jun 7 at 5:37

This is a solution, but it assumes that we have time to do some coordination with the runners before running towards the critical point.

Label the vertices A, B, and C. Get the 4m/s runner to run towards (and beyond) point B from point A, and the 3m/s runner to run away from the 4m/s runner (or away from point A) from point B (it may help to first use a straightedge to mark a line for the runners to follow). Mark the point that they meet. Now do the same thing again, except this time the 3m/s runner should run towards the 4m/s runner. You should now have two points. It should be fairly easy to draw a circle that has those points as it's diameter (you might want to find the midpoint first, then it's easy to mark the circle with a compass).
Now repeat this whole process with the 3m/s runner starting from point B, and the 5x runner from point C, this time, the 3m/s runner always running towards the 5x. The point of intersection of the two circles that are marked that lies inside the equilateral triangle should be the correct point for all the runners to run towards. (each runner starts from the same point that they started from in all of the trial runs)
Here is a diagram, showing how each part is constructed: Created using GeoGebra Classic: https://www.geogebra.org/classic

note that:

this solution also uses the circle of Apollonius, but more directly

Is this to be considered a satisfactory answer?

  • $\begingroup$ No need for Apollonius - basic Euclidian. $\endgroup$ – Moti Jun 7 at 5:36

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