# Triangular pool and three swimmers [closed]

Given a triangular pool 100, 120, 140 yards and three swimmers which swims at rates of 3.5, 4.0, 4.5 yards per second - place them on the edges of the pool in such way that when they start swimming at same moment towards the other locations, 1 to 2 edge position, 2 to 3 edge position, and 3 to 1 edge position, they will arrive at the same time to the the target edge.

• This appears to be a simple math problem, and not a puzzle. Jun 4 '20 at 8:58

The solution

isn't unique, but since this wasn't all that boring, I chose to answer instead of VTCing.

Let's start by placing the swimmers on dry ground.

The smaller numbers are distances in "black units, to be defined later". With these relative positions, each swimmer will reach the next numbered spot in exactly "one black time unit, to be defined later".

Then, we pick two swimmers (in the example, 1 and 2) and a corner of the pool (the one opposite the longest side, let's call it "A"). Since we know the length of all the pool sides, we also know the exact angle at A (about 78.5 degrees), which we'll call $$\alpha$$.

If we plot out all the points from which we can view swimmers 1 and 2 in an angle of $$\alpha$$, we get

a circle arc. (Or more precisely, two circle arcs on opposite sides of the swimmers. Let's ignore the portion that's on swimmer 3's side.)

Along this curve, we can place our chosen pool corner A in any spot that falls between the extensions of the line segments 3-2 and 3-1.

After picking a spot for A, we can draw the pool shape in two ways. Let's pick the one where the shortest side is clockwise from A, and draw an oversized pool for clearness' sake:

The green numbers are distances in "green units of length, to be defined later". Their ratios correspond to the pool sides.

From here, we can get swimmer 3 to the edge of the pool by scaling the pool appropriately: we can move the long pool side perpendicularly towards swimmer 3, which preserves the pool angles, and therefore, the side proportions.

This gives us the remaining pool corners B and C.

We can then call the side AB 100 yards, solve for the units (can't be bothered though), and we have a solution.

We could get other solutions by

• picking other pool corners than A to go between swimmers 1 and 2
• picking other spots along the blue circle arc for the chosen corner, and
• picking the other "handedness" of the pool shape

so the solutions are many and varied.

There are multiple solutions.
Proof:
The question asks us to inscribe a 7:8:9 triangle inside a 5:6:7 triangle.

Using the Pythagorean Theorem to set up a system of two equations gives us that, setting the 9 as the base gives us a height of $$\frac{8\sqrt5}{3}$$ that appears $$\frac{11}{3}$$ units from a vertex.
To inscribe this triangle inside the 5:6:7 triangle, pick any side and find the slice parallel to that side such that $$\frac{\text{distance betw. slice and side}}{\text{length of slice}}=\frac{8\sqrt5}{27}$$. This slice is the base of the inscribed triangle, and drawing in the altitude gives the third point.
For any base parallel to a side, there are two choices of altitude: therefore, there are multiple solutions.