Sunday Times Teaser 3286 – Water Stop

by BRG on September 14, 2025

by Howard Williams

Published Sunday September 14 2025 (link)

Chuck and his brother live on a long straight road that runs from west to east through their ranches. Chuck’s ranch is 13 km west of his brother’s. The two brothers often go from one ranch to the other on horseback, but go via a nearby river so that their horses may be watered. The shortest route via the river consists of two straight sections, one being 11 km longer than the other. The point at which they reach the river is 6 km north of the road.

What is the total length of the route between the ranches that goes via the river?

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BRG permalink

This one requires the solution of two simultaneous equations, one linear and one quadratic. This is straightforward in Mathematica and once the equations are produced there is no Python left to code. Here is my Mathematica code:


# (* r is answer, x is the distance east on the line between *)
# (* the ranches to the perpendicular to the water           *)
#
# e1 = (r - 11)^2 - 4 (x^2 + 6^2)^2
#
# e2 = (r + 11)^2 - 4 ((13 - x)^2 + 6^2)
#
# s1 = Solve[e1 - e2 == 0]
#
# Solve[e1 == 0 /. s1]

# (* r is answer, x is the distance east on the line between *)

# (* the ranches to the perpendicular to the water *)

# e1 = (r - 11)^2 - 4 (x^2 + 6^2)^2

# e2 = (r + 11)^2 - 4 ((13 - x)^2 + 6^2)

# s1 = Solve[e1 - e2 == 0]

# Solve[e1 == 0 /. s1]

Frits permalink

The answer doesn’t seem to make much sense.


# https://github.com/enigmatic-code/py-minizinc
from minizinc import MiniZinc

north, more, road = 6, 11, 13

def get_model(tp):
  return f"""
  % A Solution in MiniZinc
  include "globals.mzn";

  % variables 
  var {tp}: x2;  % twice the distance from Chuck to the river
  var float: y;  % other side of a Pythagorean triangle besides x and north
  var {tp}: d;  % distance travelled

  % constraints
  constraint x2 * x2 == 4 * (y * y + {north} * {north});
  constraint (x2 + 2 * {more}) * (x2 + 2 * {more}) == 
              4 * (({road} - y) * ({road} - y) + {north} * {north});
  constraint d == x2 + {more};
  constraint d > {road};

  solve satisfy;
  """

# first try to see if there is an integer solution
model = get_model("1..50")
# load the model
p = MiniZinc(model, solver="minizinc --solver scip --all", verbose=0)

ans = ""
# and solve it
for s in p.solve():
  print(f"answer: {(ans := s['d'])} km")

if ans == "":  
  # try to see if there is an non-integer solution
  model2 = get_model("float")
  # load the model
  p2 = MiniZinc(model2, solver="minizinc --solver scip --all", verbose=0)
  for s in p2.solve():
    print(f"answer: {s['d']} km")

# https://github.com/enigmatic-code/py-minizinc

from minizinc import MiniZinc

north, more, road = 6, 11, 13

def get_model(tp):

return f"""

% A Solution in MiniZinc

include "globals.mzn";

% variables

var {tp}: x2; % twice the distance from Chuck to the river

var float: y; % other side of a Pythagorean triangle besides x and north

var {tp}: d; % distance travelled

% constraints

constraint x2 * x2 == 4 * (y * y + {north} * {north});

constraint (x2 + 2 * {more}) * (x2 + 2 * {more}) ==

4 * (({road} - y) * ({road} - y) + {north} * {north});

constraint d == x2 + {more};

constraint d > {road};

solve satisfy;

"""

# first try to see if there is an integer solution

model = get_model("1..50")

# load the model

p = MiniZinc(model, solver="minizinc --solver scip --all", verbose=0)

ans = ""

# and solve it

for s in p.solve():

print(f"answer: {(ans := s['d'])} km")

if ans == "":

# try to see if there is an non-integer solution

model2 = get_model("float")

# load the model

p2 = MiniZinc(model2, solver="minizinc --solver scip --all", verbose=0)

for s in p2.solve():

print(f"answer: {s['d']} km")

Frits permalink

Most MiniZinc solvers (other than SCIP) can’t solve this model
For some reason get_model(“int”) doesn’t seem to work.

Reply

GeoffR permalink

A Solution in Python using ChatGPT to generate the code.


# ST3286_ChatGPT_Solution
 
import sympy as sp
 
x = sp.symbols('x', real=True)
d1 = sp.sqrt(x**2 + 36)
d2 = sp.sqrt((x-13)**2 + 36)
 
solutions = sp.solve(sp.Abs(d1 - d2) - 11, x)
print("River crossing x-values:", solutions)
 
for sol in solutions:
    d1_val = float(d1.subs(x, sol))
    d2_val = float(d2.subs(x, sol))
    print(f"x = {sol}, d1 = {d1_val:.2f}, \
          d2 = {d2_val:.2f}, total = {d1_val+d2_val:.2f}")

# ST3286_ChatGPT_Solution

import sympy as sp

x = sp.symbols('x', real=True)

d1 = sp.sqrt(x**2 + 36)

d2 = sp.sqrt((x-13)**2 + 36)

solutions = sp.solve(sp.Abs(d1 - d2) - 11, x)

print("River crossing x-values:", solutions)

for sol in solutions:

d1_val = float(d1.subs(x, sol))

d2_val = float(d2.subs(x, sol))

print(f"x = {sol}, d1 = {d1_val:.2f}, \

d2 = {d2_val:.2f}, total = {d1_val+d2_val:.2f}")

Frits permalink

perplexity.ai for this Teaser gives an incorrect answer and generates Python code that causes a “ValueError”.

Reply
- GeoffR permalink
  
  @Frits:
  If you ask it to do a manual solution, it gave the answer.for me.
  
  Reply

GeoffR permalink

After ChatGPT produced the SymPy solution, I asked AI to come up with a faster solution. It produced a numerical binary search solution which fulfilled the brief.


def f(x):
  # Difference in path lengths
  return math.sqrt(x * x + 36) - math.sqrt((x - 13)**2 + 36)

def find_root(lo, hi, target, tol=1e-9):
  # Binary search for roots of f(x)
  while hi - lo > tol:
    mid = (lo + hi) / 2
    if f(mid) > target:
      hi = mid
    else:
      lo = mid
  return (lo + hi) / 2

# Find x for f(x) = -11, then f(x) = 11
x1 = find_root(-100, 0, -11)  
x2 = find_root(14, 100, 11)    

for x in (x1, x2):
  d1 = math.sqrt(x * x + 36)
  d2 = math.sqrt((x - 13)**2 + 36)
  total = d1 + d2  
  print(f"Total length of route = {total:.3f} km.")

def f(x):

# Difference in path lengths

return math.sqrt(x * x + 36) - math.sqrt((x - 13)**2 + 36)

def find_root(lo, hi, target, tol=1e-9):

# Binary search for roots of f(x)

while hi - lo > tol:

mid = (lo + hi) / 2

if f(mid) > target:

hi = mid

else:

lo = mid

return (lo + hi) / 2

# Find x for f(x) = -11, then f(x) = 11

x1 = find_root(-100, 0, -11)

x2 = find_root(14, 100, 11)

for x in (x1, x2):

d1 = math.sqrt(x * x + 36)

d2 = math.sqrt((x - 13)**2 + 36)

total = d1 + d2

print(f"Total length of route = {total:.3f} km.")

Sunday Times Teaser 3286 – Water Stop

by Howard Williams

Published Sunday September 14 2025 (link)

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