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Previous
Puzzles of the Month + Solutions


MayJune
2009, Puzzle nr 121 
Back to PuzzleoftheMonth
page  Home 
Puzzle
# 121 Difficulty
level: ,
general math knowledge.
Circle
vs Square
The
diameter a of the large semicircle below is 10 cm long.
Knowing that one of the vertices of the square meets the circumference of the
small circle at point P (see diagram above), try to
guess the area of the blue square WITHOUT using trigonometry!
Keywords:
sangaku, tangent property, bisector theorem.
Related
puzzles:
 Perpendicular
or not?
 Ratio
of similar triangles.
 Area
incognita
Il diametro a del semicerchio
qui sopra vale 10 cm; sapendo che un vertice del
quadrato (punto P) appartiene
anche alla circonferenza piccola, qual è l'area
del quadrato? È vietato utilizzare la trigonometria
per risolvere questo problema!
Parole
chiave: sangaku.
Suggerisci un'altra
soluzione Chiudi
 La
surface X
Le diamètre a du
demicercle cidessus vaut 10 cm; sachant qu'un
des sommets du carré (point P)
appartient aussi à la circonférence
du petit cercle, peuxtu nous dire quelle est l'aire
du carré bleu? Il est interdit d'utiliser
la trigonométrie pour résoudre ce
problème!
Mots
clés: sangaku.
Propose une
autre solution Fermer

Source
of the puzzle:
BrainTrainer,
issue #23. © G. Sarcone. You
cannot reproduce any part of this page without prior written
permission. 


There
are 2 possible solutions to this 'sangaku' problem:
SOLUTION
1
A) Build
a large square with base QR (see
illustration opposite) and prolong the diagonal d to
the vertex S, as illustrated.
According
to the drawing:
q = a/2  R
d = QP =
a/2 + x
x = d  a/2
QR = RS = a
QR ⊥ ON
QP ⊥ OP
∠NQO = ∠OQP
B) According
to the ‘angle
bisector theorem’: if a ray or segment
bisects an angle of a triangle, then it divides
the two segments on either side proportionally.
In other terms, given in the triangle QRS the
bisectrix QM, then:
MS / MR = QS / QR that
is
(am)/m = a2/a
We obtain, by resolving the equation:
a/(1 + 2)
= m
m = 10/2.414 =
4.14 cm
C) The
right triangles QNO and QRM are
similar, then:
(a/2 + x) / R = d / R = a / m =
10 / 4.14
d = R(2.42)
D) Thank
to the ‘tangent
property’, we find that: a/2
+ x = d
Substituting x with d  a/2 in
the following Pythagorean
equation:
(a/2  R)^{2} = x^{2} +
R^{2} = (d  a/2)^{2} +
R^{2}
We can also substitute d with R(2.42)
(a/2  R)^{2} = (d  a/2)^{2} +
R^{2} = [R(2.42)  a/2]^{2} +
R^{2}
and we will obtain:
(5  R)^{2} = [R(2.42)  5]^{2} +
R^{2}
25  10R + R^{2} = (5.86R^{2}  24.2R
+ 25) + R^{2}
R = 14.2/5.86 = 2.423
E) The
area of the blue square is therefore:
d^{2}/2
= [R(2.42)]^{2}/2 = [2.423 x 2.42]^{2}/2
= 17.2 cm^{2}
SOLUTION
2
F) According
to the drawing:
q = a/2  R ; d = QP =
a/2 + x ; x = d  a/2
QP,
the
diagonal d of
the
blue
square,
is
tangent
to point P.
OP is perpendicular to QP,
then:
the angle QPS = the angle OPS = 45
degrees.
G) The
triangle OMP is isosceles,
then the height PS of
the blue square equals: ON + PM =
R + R/2
= R(1 + 2)/2
= R(1.71)
and the diagonal QP equals:
R(1 + 2)
= R(2.42)
H) Thank
to the ‘tangent
property’ (as argument D further
above), we find that:
a/2 + x = d (the diagonal QP)
Substituting x with d  a/2 in
the following Pythagorean
equation:
(a/2  R)^{2} = x^{2} +
R^{2} = (d  a/2)^{2} +
R^{2}
We can also substitute d with R(2.42)
(a/2  R)^{2} = (d  a/2)^{2} +
R^{2} = [R(2.42)  a/2]^{2} +
R^{2}
and we will obtain:
(5  R)^{2} = [R(2.42)  5]^{2} +
R^{2}
25  10R + R^{2} = (5.86R^{2}  24.2R
+ 25) + R^{2}
R = 14.2/5.86 = 2.423
I) The
area of the blue square is therefore:
PS^{2} =
R(1.71)^{2} = (2.423 x 1.71)^{2} =
17.2 cm^{2}
The
Winners of the Puzzle of the Month are:
Nheo Le, USA
Mauro Zoffoli, Italy
Hasan Kurt, the Netherlands
Amaresh G. S., India
Congratulations!

© 2006 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.

Reactions
from our visitors
 Posted:
Fri, 28 May 2010 18:12:51 by Philippe
Chevanne
Dear
Gianni,
I
just discovered your web site from a friend. Great
site, and congratulations!
Lurking a few problems there, I got at problem #76
(mayJune 2009)
I think the solution given is false, as it is not
stated that
the diagonal QP should
be tangent to the small yellow circle.
This unstated condition is necessary to solve the
puzzle, and is
used in the given solutions.
Discarding this condition, the area of the square
can take ANY value.
For instance see attached picture.
To construct the figure from *any* point P,
that is here any
point P on a 45° diagonal line, is an "Appolonius
problem" in the
case Point(P)  Line(diameter)  Circle(large
circle).
That is to construct a circle (yellow) touching the
three objects
considered as three "circles":
P = circle with null radius, Line = circle
with infinite radius, and given (half)circle.
This can be solved in several classical ways.
(for instance at my own web site
http://mathafou.free.fr/pbg_en/jsp136e.html for
english version)
The result is that for any point P on the
45° line, there is a
corresponding square of area = 2QP^{2} =
any value, and a yellow
circle through P and tangent to diameter
and large half circle.
(In fact two such circles, the other being "on
the left" to P)
The condition "QP tangent to yellow
circle" is equivallent to
"yellow circle minimal", restricting the yellow circle to be
centered in the right quarter circle, otherwise the
minimum is 0,
with P=Q.
So that the missing condition "when yellow circle
is minimal"
should be added to the statement.
Best Regards.

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
Angle
bisector and proportions
If a segment bisects an angle of a triangle then it
divides the two segments on either side proportionally:
CA / CP = BA / PB and
CA x PB = CP x BA
Angle
bisector theorem on Wikipedia.


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