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Puzzles of the Month + Solutions


JuneJuly
2007, Puzzle nr 112 
Back to PuzzleoftheMonth
page  Home 
Puzzle
# 112 Italiano Français
Difficulty
level: ,
basic geometry knowledge.
Achtung,
Minen!
In the late 40’s a soldier named Mark entered accidentally
a square mined field (point A, see drawing below)... Two friends
of him, at the edge of the field, shouted him and asked to continue walking directly
towards the shelter B made from sand bags...
So,
all together from different directions started going
straight ahead to the shelter. The average walking
speed of the three soldiers was habitually 6 km/h,
but Mark while he was crossing the mined field walked
carefully at an average speed of 3 km/h. The two
friends, at the edge of the field, were separated
from each other by approx. 1060 meters.
Nevertheless, all three arrived simultaneously at the shelter 10
minutes later. Can you indicate on the drawing where initially
did Mark’s friends stand and what is the area of the mined field?
To solve this puzzle use only basic geometry, trigonometry is not allowed.


[enlarge] 

This
problem isn't actually very difficult... We know
that the friends of Mark at the edges of the
square field walked directly toward the shelter
B and arrived at the same time; then, their initial
position can be defined by points C and D where
the perimeter of the square intersects the arc
of a circle having center B. But we need just
an extra information to find the exact places
where Mark’s
friends stand... 
• Draw
a line through points A and B (see drawing below).
• Draw a circle having center A, whose radius r meets the intersection
point between the side of the square field and the line AB. The diameter of this
circle (2 x r) represents the distance Mark might have covered inside
the field with a normal average speed (6 km/h = 2 x 3 km/h) for the same amount
of time.
• Draw a large circle having center B and being tangent to the
circle whose center is A. The arc of circle having center B determines
now two points C and D on the perimeter of
the square, which are the initial positions of Mark's friends.
• BC = BD, then triangle CDB is isosceles. 
CB
and DB are the distances that Mark's friends effectively
covered which are in meters: 10/60 x 6 [km/h] = 1
[km] = 1000 meters
Note
that the height h of the triangle CDB, is
also the height of the square field.
In
any triangle it is possible to find the height h,
if the value of its sides a, b and c are
known, using the following formula which is a variant
of Hero's
formula:
h = 2(s(sa)(sb)(sc))/b
where s is the semiperimeter of the
triangle:
s = (a+b+c)/2
Now
adapting the above formula to our triangle CDB, we
obtain:
h = 2(s(sb)(sd)^{2})/d
where s = (2 x 1000 + 1060)/2 = 1530
then,
h = 1/500 x (1530
x (1530  1060) x (1530  1000)^{2}) = 898.88
[m]
Thus,
the Area A of the square field is: 898.88^{2} =
807,980.76 [m^{2}] or approximately 0.808
[km^{2}]
Simple,
isn't it?
The
Winners of the Puzzle of the Month are:
Paris Karagounis, Greece
Thierry Lebordais, France
Congratulations!

© 2003 G.
Sarcone, www.archimedeslab.org
You can reuse content from Archimedes’ Lab
on the ONLY condition that you provide credit to the
authors (© G.
Sarcone and/or M.J.
Waeber) and a link back to our site. You CANNOT
reproduce the content of this page for commercial
purposes.

You're
encouraged to expand and/or improve this article. Send
your comments, feedback or suggestions to Gianni
A. Sarcone. Thanks! 

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