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Previous
Puzzles of the Month + Solutions


AugustSeptember
2007, Puzzle nr 113 
Back to PuzzleoftheMonth
page  Home 
Puzzle
# 113 Italiano Français
Difficulty
level: ,
basic geometry knowledge.
Indecisive
triangle
If ED = 23 units, and the value of the sides of the green square
ABCD is a multiple of 11, what is the area of the red triangle AFE? Hint:
the length value of EF is not irrational.
Find the very shortest way to solve this puzzle and use only basic geometry,
trigonometry is not allowed!


(the
drawing is only representative, it is not scaled!)
[enlarge] 

You
can solve this puzzle by using only visual thinking
skills and logic... 
• Since
ABCD is a square and n = AD = AB.
• Then, triangles ADF and BDF are of the same area (DF x n/2).
• Triangles EDF and BDF cover the surface of triangle EDB.
• Thus, the area A of
triangle AFE = the area A of
triangle EDB, which is m x n/2 (see visual proofs below). 
We
can also prove that the area A of
triangle AFE equals m x n/2 with
the following equation:
AFE = ABE  ABF = (AB x AE)/2  (AB x AD)/2 =
= (AB/2)(AEAD) = (AB/2)(DE) = m x n/2
= 23n/2
To
get a numerical solution, we have now to find a suitable
value for n (or AB).
Triangles
ABE and EDF are similar. Ratios for similar triangles:
m/n = AE/DE = BE/EF
AE,
DE, and EF all have rational lengths, so BE must
also be rational. Since AE and AB are integers, BE
can be rational only if it is also integer.
The
only right angled triangles with integer side lengths
are called Pythagorean triangles, and their respective
sidelength values, Pythagorean
triples. Therefore, we have to look
for Pythagorean triples of the form: 11y, 11y+23,
z
The
triple 33, 56, 65 is
found easily.
The
value of n (or AB) being 33 units, the area
of triangle AFE is:
23 x 33/2 = 379.5 square units
Amaresh
G. S. has sent us another visual solution
that proves the area A of
triangle AFE equals m x n/2.
In
rectangle ABGE opposite, EB is a diagonal, and line
HI passes through F and is perpendicular to AB.
Area A of
rectangle AIFD = Area A of
rectangle HFCG.
Area A of
triangle AFD is half the area A of
rectangle AIFD.
Area A of
triangle FCG is half the area A of
rectangle HFCG.
Therefore,
area A of
triangle AFD = area A of
triangle FCG.
DFE
+ FCG = half the area A of
rectangle DCGE, i.e. m x n/2.
The
Winners of the Puzzle of the Month are:
Larry Bickford, USA
Amaresh G.S., India
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! 

More
Math Facts behind the puzzle
It
is possible to generate Pythagorean
triples by supplying two different positive
integer values for m and n in the
diagram opposite. You can multiply out these terms
and check that:
(m^{2}– n^{2})^{2} +
(2mn)^{2} = (m^{2} + n^{2})^{2}
Once
one triple has been found, you can generate many
others by just scaling up all the sides by the same
factor.
Some
properties of Pythagorean triangles include:
 Exactly one length value of a leg is divisible by
3.
 Exactly one length value of a leg is divisible by
4.
 Exactly one sidelength value is divisible by 5.
 At most one length value of a leg is a square or
a multiple of a square.
Here
are the first few triples for m and n between
1 and 4

n
= 1 
n
= 2 
n
= 3 
n
= 4 
m
= 2
m = 3
m = 4
m = 5
m = 6
m = 7
m = 8
m = 9
m = 10 
[3,
4, 5]
[8, 6, 10]
[15, 8, 17]
[24, 10, 26]
[35, 12, 37]
[48, 14, 50]
[63, 16, 65]
[80, 18, 82]
[99, 20, 101]

[5,
12, 13]
[12, 16, 20]
[21, 20, 29]
[32, 24, 40]
[45, 28, 53]
[60, 32, 68]
[77, 36, 85]
[96, 40, 104]

[7,
24, 25]
[16, 30, 34]
[27, 36, 45]
[40, 42, 58]
[55, 48, 73]
[72, 54, 90]
[91, 60, 109] 
[9,
40, 41]
[20, 48, 52]
[33, 56, 65]
[48, 64, 80]
[65, 72, 97]
[84, 80, 116] 



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