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Previous
Puzzles of the Month + Solutions


SeptemberOctober
2008, Puzzle nr 119 
Back to PuzzleoftheMonth
page  Home 
Puzzle
# 119 Difficulty
level: ,
general math knowledge.
Perpendicular
or not?
The
large circle (O_{1}, B)
above circumscribes 2 smaller circles (with centers O_{2} and O_{3},
respectively) and an isosceles triangle, whose base stands on the diameter of
the large circle. As shown in the drawing, each geometric shape touches the 3
other ones. Points A, O_{1}, O_{2} and B are
aligned. Demonstrate that the segment O_{3}A is
perpendicular to the diameter AB.
Keywords:
incircles, isosceles, perpendicular, secanttangent.
Related
puzzles:
 Ratio of
similar triangles.
 Soccer
ball problem.
 Ad
angolo retto?
Il grande cerchio (O_{1}, B)
circoscrive 2 cerchi minori (con i rispettivi centri O_{2} e O_{3})
e un triangolo isoscele, la cui base si confonde
col diametro del grande cerchio. Ogni figura geometrica
tocca ognuna delle altre tre. I punti A, O_{1}, O_{2} e B sono
allineati. Dimostrare che il segmento O_{3}A è perpendicolare
al diametro AB.
Parole
chiave: cerchio inscritto, isoscelo, perpendicolare.
Suggerisci un'altra
soluzione Chiudi
 A
angle droit?
Le grand cercle (O_{1}, B)
circonscrit 2 autres cercles plus petits (ayant
pour centres O_{2} et O_{3},
respectivement) ainsi qu’un triangle isocèle,
dont la base se trouve sur le diamètre du
grand cercle. Chaque figure géométrique
touche les 3 autres. Les points A, O_{1}, O_{2} et B sont
alignés. Démontrer que le segment O_{3}A est
bien perpendiculaire au diamètre AB.
Mots
clés: cercle inscrit, isocèle,
perpendiculaire.
Propose une
autre solution Fermer

Source
of the puzzle:
Sangaku tablet of Gunma prefecture (群馬県; Gunmaken),
1803. ©G. Sarcone. 


Some
interesting considerations
• In fig. 1, we can see that triangles A'BC' and ABC are
similar.
• In fig. 2, we can see that AC' and AE' are tangent
to the circle with center O_{3}. The segment O_{3}A represents
then the symmetry axis along which the right triangle AO_{3}D is
reflected on AE'.
What we have to prove
If we prove that angle γ (gamma,
see fig. 2) of the right triangle AO_{3}D is complementary to
angle α (alpha) of the isosceles
triangle AC'A', then we prove that points E and E' are
identical, and consecutively the segment O_{3}A is
perpendicular to the diameter AB.
a)
It is given:
radius O_{1}B = R (see
fig. 3)
radius O_{2}B = r
radius O_{3}D = p
segment O_{3}A = t
β = 90°  α
b)
According to the drawing:
A’C’ = AC’
MA = MA' = R  r
h ⊥ AA'
p ⊥ AC’
O_{1}O_{3} = R  p
O_{1}A = 2r  R
O_{1}M = O_{2}B = r
O_{2}O_{3} = r + p
c)
According to the 'SecantTangent' theorem (see
further below):
t^{2} = p(p +
2r)
d)
According to the Pythagorean theorem:
t^{2} + O_{1}A^{2} = O_{1}O_{3}^{2 } →
t^{2} + (2r  R)^{2} =
(R  p)^{2}
e)
Solving the equations in c) and d)
together, we obtain:
t = 2r[2R(R  r)]
/ (R + r)
and
p = 2r(R  r)
/ (R + r)
f)
According to the Pythagorean theorem:
1) h = (O_{1}C'^{2}  O_{1}M^{2})
= (O_{1}B^{2}  O_{1}M^{2})
= (R^{2}  r^{2})
and
2) AC’^{2} = h^{2} + MA^{2} =
(R^{2}  r^{2})
+ (R  r)^{2} →
AC’ = [2R(R + r)]
g)
By solving the following equations:
1) (p / t)^{2} = p^{2 }/ t^{2} =
[2r(R  r)]^{2}/ 4r^{2}[2R(R  r)]
=
= [2r(R  r)]^{2} / 4rR[2r(R  r)]
= (R  r) / 2R
and
2)
(MA / AC')^{2} = MA^{2} / AC'^{2} =
(R  r)^{2} / 2R(R + r)
= (R  r) / 2R
We proved that p / t = MA / AC',
then the triangles AO_{3}D and AC'M are
similar, hence angle DÂO_{3 }=
angle β = 90°  α.
In conclusion, angle γ (or DÂO_{3})
being complementary to
angle α, the segment O_{3}A is
perpendicular to the diameter AB.
(You
will find here a
more geometric way to solve this puzzle)
The
Winners of the Puzzle of the Month are:
No winners... Yes, the puzzle was
a little bit harder.
Sorry!

© 2005 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
Secanttangent
theorem: fig. 1) c^{2} = a(a + b) and
fig. 2) c^{2} = a(a +
2r)
The
angle, formed between a tangent line and a chord,
is congruent to the inscribed angle on the other
side of the chord and subtended by the chord (angle α' =
angle α, see fig. 3). Then, with similar
triangles:
c / (a + b) = a / c


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