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Puzzles of the Month + Solutions



Back to PuzzleoftheMonth
page  Home 
Puzzle
# 130


Rising
sun
All
the circles and semicircle are tangent to each other
and are inscribed in a square. The 3 small circles are
the same size, with radius r. R is
the radius of the red circle. Prove that r =
3R/8
Difficulty
level: ,
general math knowledge.
Category:
Geometry.
Keywords:
Square, circle, semicircle, radius.
Related
puzzles:
 Steam
locomotive,
 A
mathematic shield.
 Sol
levante
I cerchi e il semicerchio sono tangenti
tra di loro e iscritti in un quadrato. I 3 cerchi
minori hanno lo stesso perimetro di raggio r. R è il
raggio del cerchio rosso. Dimostra che r =
3R/8
Parole
chiave: quadrato, cerchi, semicerchio, raggio.
Suggerisci un'altra
soluzione Chiudi
 Soleil
levant
Les cercles et le demicercle sont tangents
entre eux et inscrits dans un carré. Les trois
petits cercles ont le même périmètre
de rayon r. R est
le rayon du cercle rouge. Démontrez que r =
3R/8
Mots
clés: carré, cercle, demicercle,
rayon.
Propose une
autre solution Fermer

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



Point A is
the center of the red circle,
B and E are center points of small
circles,
D is the center point of the semicircle,
DEF is a straight line,
EC is perpendicular to AD.
Define s =
DF, i.e. 1/2 length of the square,
then:
DG
= s – 2r = 2s  2R
so:
s = 2(R  r)
Define x =
CE
Applying
Pythagoras theorem to the triangle CDE:
(1) x^{2} + r^{2} =
(s  r)^{2}
x^{2} + r^{2} =
(2R  3r)^{2}
Applying Pythagoras theorem to the triangle ACE:
(2) x^{2} + (R + s 
3r)^{2} = (R + r)^{2}
x^{2} +
(3R  5r)^{2} = (R + r)^{2}
Subtracting (1) from (2):
(3) x^{2} + (3R 
5r)^{2}  r^{2} =
(R + r)^{2}  (2R 
3r)^{2}
Rearranging (3):
12R^{2} 
44Rr + 32r^{2} = 0
or
4(3R 
8r)(R  r) = 0
Taking
the solution where R > r:
3R =
8r or 3R/8 = r
Q.E.D.

The
5 Winners of the Puzzle of the Month are:
Arthur Vause, U.K.  Marc
Hallemans,
Belgium  Tony
Garcia,
Dominican Republic  David
Almeida,
Portugal  Yongting
Chen,
Canada
Congratulations!



Beyond
the challenge

How
to Mathematically Square the Circle
There isn't any method to "geometrically" square
a circle WITH compasses and triangle set squares (tough
several approximation methods exist).
In the picture,
the area A of
the green circle equals the area A of
the yellow square. The distance πR represents
obviously one 1/2 circle rotation. The diameter of
the semicircle is then:
πR + R or R(π +
1)
Discuss
the problem on our FaceBook
page!

© 2012 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.

Puzzle
of the Month by Gianni
A. Sarcone is licensed under a Creative
Commons AttributionNonCommercialNoDerivs 3.0 Unported
License. 
You're
encouraged to expand and/or improve this article. Send
your comments, feedback or suggestions to Gianni
A. Sarcone. Thanks! 




