Seven buoys A, B, C, D, E, F and G are released in a lake at regular intervals in a manner to make a

Q: Why don't the buoys just form a square that floats down the river?

If the buoys float, their relative positions at any single instant would actually still form a square! However, the problem asks for the pattern of their drop locations (as if they are anchored). Because the boat drifts with the river while traveling between drops, the drop locations are sheared into a triangle.

Q: What does 'in the same manner' mean here?

It means the boat's velocity relative to the water is identical to its velocity in the lake. The boat's engine and steering are doing the exact same thing, but now the entire body of water is moving.

Q: How did we know the time intervals are equal?

The problem explicitly states they are released at 'regular intervals'. Since the boat moves at a constant speed relative to the water, equal time intervals correspond to equal distances along the path.

Q: Why is the shift of buoy C's drop location twice that of buoy B?

The shift is caused by the river carrying the boat while it travels. Buoy C is dropped at $t = 2\Delta t$, while buoy B is dropped at $t = \Delta t$. The boat has been drifting with the river for twice as long when it drops C compared to when it drops B.

Q: Could the boat have moved with the same velocity relative to the ground instead?

If it did, the drop locations would form a perfect square on the ground. But the problem states 'G falls on A', which is impossible for a square pattern since A and G are at different corners. This confirms the boat's velocity was constant relative to the water.

Question

Seven buoys A, B, C, D, E, F and G are released in a lake at regular intervals in a manner to make a square pattern as shown in the figure. The buoys A, C, E and G are on the vertices and the buoys B, D and F are at the midpoints of the sides of the square. If the buoys were released in a uniformly flowing river in the same manner, the buoy G falls on A. What pattern would they make in the river?

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are the captain of a boat tasked with dropping heavy, anchored buoys to mark a specific path. In a perfectly still lake, you drive your boat at a constant speed along three sides of a square, dropping buoys A through G at perfectly equal time intervals, $\Delta t$.

Because your speed is constant, the time it takes to reach each drop point is directly proportional to the distance traveled. If we set up a coordinate system where buoy A is dropped at $(0, a)$, then buoy C is at the origin $(0, 0)$, E is at $(a, 0)$, and G is at $(a, a)$. The midpoints B, D, and F fall exactly halfway between these vertices. The drop times are $t_A = 0$, $t_B = \Delta t$, $t_C = 2\Delta t$, all the way up to $t_G = 6\Delta t$.

### The River's Moving Treadmill

Now, let's move the entire operation to a flowing river. The problem states the boat moves 'in the same manner'. This means the boat's engine and steering are doing the exact same thing relative to the water. However, the river itself is a moving treadmill flowing with some constant velocity $\vec{u}$.

Are the buoys floating away, or are they anchored? If they float, their relative positions at any given moment would actually remain a perfect square! Why? Because every buoy would drift by the exact same velocity $\vec{u}$. But the problem gives us a massive clue: **'buoy G falls on A'**. If they were just a drifting square, G and A would never touch. This tells us we are looking at the pattern of their **drop locations**—imagine them as heavy anchored markers.

Because the boat is being carried by the river while it travels, the actual drop location of any buoy $i$ relative to the ground is its original lake position plus the river's drift over time:

$$\vec{R}_i = \vec{r}_{	ext{lake}}(t_i) + \vec{u} t_i$$

### The Master Equation

The magic happens with the condition that buoy G falls exactly on A. This means the final drop location of G perfectly coincides with the initial drop location of A. Let's equate their position vectors:

$$\vec{R}_G = \vec{R}_A$$

Substituting our drift equation, we get:

$$\vec{r}_G + \vec{u} t_G = \vec{r}_A + \vec{u} t_A$$

We know G is at $(a, a)$ and was dropped at $t = 6\Delta t$. A is at $(0, a)$ and was dropped at $t = 0$. Plugging these in:

$$(a, a) + \vec{u}(6\Delta t) = (0, a) + \vec{u}(0)$$

Solving for the drift per time interval, we find:

$$6\vec{u}\Delta t = (-a, 0) \implies \vec{u}\Delta t = \left(-\frac{a}{6}, 0ight)$$

This tells us that in every time interval $\Delta t$, the river pushes the boat to the left by exactly one-sixth of the square's side length!

### Final Calculation and The Geometric Shear

Now, let's apply this calculated drift to the other buoys to reveal the final geometric shape. Let's look at the bottom row buoys: C, D, and E.

Buoy C is dropped at $t = 2\Delta t$, so it shifts left by two units of drift:
$$\vec{R}_C = (0, 0) + 2\left(-\frac{a}{6}, 0ight) = \left(-\frac{a}{3}, 0ight)$$

Buoy D is dropped at $t = 3\Delta t$:
$$\vec{R}_D = \left(\frac{a}{2}, 0ight) + 3\left(-\frac{a}{6}, 0ight) = (0, 0)$$

Buoy E is dropped at $t = 4\Delta t$:
$$\vec{R}_E = (a, 0) + 4\left(-\frac{a}{6}, 0ight) = \left(\frac{a}{3}, 0ight)$$

Notice their new y-coordinates are all zero! They lie perfectly on a straight horizontal line, with D exactly in the middle.

What about the final shape? A and G are merged at the top apex $(0, a)$. C, D, and E form the flat horizontal base. If you calculate the positions of B and F, you will find they sit perfectly on the slanted straight edges connecting the top apex to the base. The original square has been sheared by the river's flow into a perfect triangle! This beautifully matches option D.

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Visualized Solution

$I ni t ia l S e t u p in t h e L ak e$

$P os i t i o n s in t h e L ak e$

$M o t i o n in t h e R i v er$

$T h e K ey C o n d i t i o n$

$C a l c u l a t in g R i v er D r i f t$

$N e w P os i t i o n s o f B u oy s$

$F ina l P a tt er n$

The Sigma Insight: Relative Velocity

Analyzing the Setup

The River's Moving Treadmill

The Master Equation

Final Calculation and The Geometric Shear

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup in the Lake

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Positions in the Lake

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Motion in the River

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Key Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating River Drift

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } New Positions of Buoys

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Pattern