| The following sketches
are recomposed line drawings
of various SanGaku
that can be studied at lengh
on the authors'page:
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From a 1803 Sangaku
found in Gumma Prefecture. The base of an isosceles
triangle sits on a diameter of the large circle.
This diameter also bisects the circle on the left,
which is inscribed so that it just touches the inside
of the container circle and one vertex of the triangle.The
top circle is inscribed so that it touches the outsides
of both the left circle and the triangle, as well
as the inside of the container circle. A line segment
connects the center of the top circle and the intersection
point between the left circle and the triangle.
Show
that this line segment is perpendicular
to the drawn diameter of the
container circle. (T.Rothman)
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1788 Sangaku,
Tokyo Prefecture. Also called the Pappus chain. It
asks for the radius of the nth largest inner circle
in terms of r, the radius of the container circle.
Note that the two large vertical circles are identical,
each with radius r/2.
The
original solution to this problem
deploys the Japanese equivalent
of the Descartes circle theorem.(H.
Boas)
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The figure is symmetric. AD, BE
and CF are the diameters of the circle O, and AOB
= AOF = 60°. O1(a) and O4(b) touch each other externally,
touch BE and CF, and O1 touches O internally. O7(c
) touches O8(d ) and O9(d ) externally . O7 , O8
and O9 touch AF and touch the circle O internally.
Find d in terms
of b. (H.Kotera)
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If CB0A is an isosceles triangle
with CB0 = CA, and B is a point
lying on the line B0A, then
one of the circles touching
the circumcircle of BCB0 internally
and also touching AB and AC
is twice the size of the incircle
of ABC. (H. Okumura)
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1874 tablet from
Gumma Prefecture. A large circle lies within a square.
Four smaller orange circles, each with a different
radius, touch the large circle as well as the adjacent
sides of the square.
What
is the relation between the radii of the four small
circles and the length of the side of the square?
(The
problem can be solved by applying
the Casey theorem, which describes
the relation between four circles
that are tangent to a fifth
circle or to a straight line)
(T. Rothman)
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If ABC is a right triangle
with right angle at A, then
the circle touching the circumcircle
of ABC internally and also touching
AB and AC is twice the size
of the incircle of ABC. (H.
Okumura)
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In a square PQRS, two circles touch
SP and the incircle of the square where one
touches PQ and the other touches RS. Let A be the
point of tangency of QR and the incircle and let the
tangents of the two small circles through A intersect
the segment SP at B and C.
Given the inradius of the square,
find the inradius of the circle in the triangle
ABC
The answer
is that the medium circle is
also half the size of the largest
circle. (H.Okumura)
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Two congruent
regular pentagons with a side
in common are inscribed in
a large circle. A circle of
radius r touches the large
circle and the sides DE and
EF of the pentagons, and the
incircle of the triangle ABH
has radius t.
Show that r =
2t. (J. Rigby)
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Two equilateral triangles are inscribed into
a square. Their side lines cut
the square into a quadrilateral and a few triangles.
Find a relationship
between the radii of the two
incircles. (H. Kotera)
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The Egyptian Triangle
PART I
The following three sangaku require
to determine the relative radii of the circles.
It can be solved
by a direct application of the
Pythagorean theorem. (A.Bogomolny)
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The
Egyptian Triangle
PART
II (A.Bogomolny)
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The
Egyptian Triangle
PART
III (I.
Rubin)
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This
configuration was solved by
L. Bankoff and C. W. Trigg. The
famous 3:4:5 triangle also referred
to as the Rope-stretchers triangle
and sometimes as the Egyptian triangle,
both because of the belief that
this simplest of the Pythagorean
triangles was used by the ancient
rope-stretchers in construction
and, in particular, in the construction
of the Great Pyramids. Sometimes,
however, the term Egyptian triangle
is preserved for the one related
to the dimensions of the pyramid
of Cheops and the golden ratio.
(A.
Bogomolny)
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For convenience,
assume the side of the square
is 24 so that DF = FC = 12. Also,
for simplicity, a circle with
center X say, will be denoted
(X). AH is tangent to (F) in
G. Thus also FG = 12. The radius
FG extends to meet BC in J and
is perpendicular to AH.
HC = HG as tangents to (F) from H. For the same reason AG
= AD = 24. FH bisects CFG and
FA bisects its supplementary
angle DFG.
From here, AFH = 90o and FG is the altitude to the hypotenuse
of right ΔAFH. We, therefore, have FG2 = AG·GH, or 122 =
24·GH. Wherefrom GH = 6 and so HC = 6. (A.
Bogomolny)
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Then HB = 18, AH = 30, AK = 15, KG = 9, EK = 9,
FK = 15, QK = 3, DJ = 16, FI = 8, EI = 16, AI = 20
(by the Pythagorean theorem), HJ = 10 (ΔFGK is similar
to ΔJGH), BJ = 8, GJ = 8, FJ = 20, CJ = 16, IQ =
4.
We see that triangles FCJ, HGJ,
FGK, AKE, AEI, DFI, and ABH are
all 3:4:5 triangles. (A.
Bogomolny)
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Let T be the intersection of AI extended with (A).
Then AT = 24 and, since AI = 20, IT = 4 = IQ. Which
says that the circle (I) of radius 4 is tangent to
(A), (B), (F), and (E). If we extend AT further and
let R on AT be on the vertical tangent SR to (I),
then, for one, ΔIRS is similar to AEI and, hence,
is also 3:4:5. Since IS = IQ = 4, IR = 20/3, SR =
16/3, RT = 8/3. If U is the intersection of SR and
CD, then RU = FI - SR = 8 - 16/3 = 8/3 = RT.
In addition, if V is on BC and
VC = RU, then VH = 10/3, RV =
8 and, by the Pythagorean theorem,
HR = 26/3, so that again RX =
8/3. It follows that the circle
(R) with radius 8/3 is tangent
to (A), (I), (H) and CD. As an
extra, it is also tangent to (W)
with radius 3/2. HVR is 5:12:13.
The right triangle with vertical
and horizontal legs and hypotenuse
RW is 7:24:25. (A.
Bogomolny)
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Three small
congruent circles, two of which
lie in the curvilinear triangles
made by the two external tangents
and the medium circle. The
other lies in one of the curvilinear
triangles made by the two larger
circles and one of the external
tangents. The problem states
that the ratio of the size
of the two larger circles is
2:1. If there are two externally
touching circles with different
radii with external common
tangents and if there are 4n
congruent small circles, n
of which lie in the curvilinear
triangle made by the two external
tangents and the medium circle,
and 3n of which lie in one
of the curvilinear triangles
made by the two larger circles
and one of the external tangents,
then the ratio of the size
of the two larger circles is
4:1 for any natural number
n. (H.
Okumura)
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A triangle ABC
is inscribed in a circle O(R) . In ABC , circles
O2(r ), O4(r1 ( ), O5(2 ( ) are tangent one after
another, as shown. The circles O1(t1), O3(t3)touch
the circle O(R) and touch chords AC and BC .
Find r1 , r2, r3, in terms of
t1, t2 and r. (H Kotera)
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Sangaku from the Miyagai
prefecture. circa 1912. 6 congruent right triangles
fan out along the sides of a regular pentagon of
side a.
Find the length
of the hypotenuse t of these
triangles in terms of a. (J.Rigby)
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Fractal image
based on the fact that the
ratio of the size of the two
inscribed circles in the problem
is 2:1. This provides an example
of recursive computer programming.
(H. Okumura)
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