A New Geometrical Proposition
Author(s): Y. Sawayama
Source: The American Mathematical Monthly, Vol. 12, No. 12 (Dec., 1905), pp. 222-224
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/2967716 .
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222
A NEW GEOMETRICAL PROPOSITION.
Instructor in The CentralMilitarv School for Boys, Tokyo,Japan.
By Y. SAWAYAMA,
General enunciation.-Describe eight circles, each of which touches a cir-
cle and any two secants of it; next construct a triangle by joining any three of
the points of intersection of the latter circle and the two secants; then the chords
of contact and the line of centers of two of the eight circles taken appropriately
in pairs are concurrent, the point of
concurrency being equidistant from
the three sides of the triangle.
Particular enunciation. - Call
the circle, ABC, the eight circles a,
a', b, b', c, c', d, d', and the two secants / 'V
AD, BC (Fig. 1). Then
1. The line of centers of the
two circles a and a', and their two
chords of contact pass through the
center of the inscribed circle of the
triangle ABC.
2. The line of centers of the
two circles b and bi, and their two
chords of contact pass through the
center of the circle escribed to the side
BC of the triangle ABC.
3. The line of centers of the
two circles c and c', and their two chords of contact pass through the center of
the circle escribed to the side CA of
w 2a the triangle ABC.
4 4. The line of centers of the
\two circles d and d', and their two
chords of contact pass through the
center of the circle escribed to the
side AB of the triangle ABC.
Demonstration.-Let K, E, and
F be the three points of oontact where
any one of the eight circles touches
the circle ABC and the two secants
BC, AD (Figs. 2 and 3). Then as is
well known, we have two properties:
1. The line KE passes through
the middle point M of the arc BC, so
also the line KF through the middle
_____ _____ ____ _____ _ _ |point N of the arc AD.
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223
2. The two straight lines EF
and MN are parallel.
From the last property we have
Z EFK Z MNK, and as the four
points M, N, A, and K lie on the cir-
cle ABC, ZMNK= Z MAX (Fig. 2),
or =its supplementary angle (Fig. 3). :
.-. ZEFK= ZMAKor=-itssup-
plementary angle. Hence the four
points A, K, F, and I (which is the
intersection of EF and AM) are con-\
cyclic.
L AIK= L AFK(Figs. 2 and A'
3), or=-its supplementary angle (fig- Z
ure omitted). Now the line AF
touches the circle EFK, hence Z AFK
or its supplementary angle=zFEK. .X.LAIK L FEK.
.A.A MKI and A MIE are mutually equiangular, from which we can deduce
that the circle passing through the three points I, E, and K touches the line MI,
hence MI2i-=K.ME. On the other hand, since arcMB==arcMC,we have L M3KB
- MBE (in the figures we omit the line BK).
.'.The circle passing through the three points B, E, and K touches the line
MB, hence we have MK.ME B--MB2, or M12 1MB2 MIM IWB.
This shows that I is equidistant from the three sides of the triangle ABC.
Next, call 0 the center of the circle
BEK, L the intersection of BC and AD, P
the intersection of EF and OL, Q another
intersection of OI and the circle BIC, and
draw the straight line passing through Q
and L (Fig. 4).
Then the square of the tangent drawn
from the center of the circle BIC to the circle
EFK is equal to the rectangle MK.ME, or the
square of the radius of the circle BIC.
This last fact shows that the two cir-
cles EFK and BIC are orthogonal, and the
radius OBEof the circle EFK is equal to the
tangent drawn from the point 0 to the circle
BIC, and that 0E2=OQQ0I.
Again, in the right-angled triangle
OLE, OE2 OL.OP. .. OQ.OI OL. OP.
The four points I, P, L, and Q are concyclic.
Noting the angle IPL is a right angle and the L IQL and L IPL are equal
or supplementary, we have Z IQL=right angle.
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224
.'.The line QL passes through a point 1 which is equidistant from the
three sides of the triangle ABC (1 is, in fact, the other extremity of the
diameter passing through I of the circle BIC). And the line OI is perpendicular
to a fixed line I'L passing through the fixed point Q on it.
Similarly, the center of another circle whose chord of contact passes
through the point I, and the point I are on the straight line which is perpen-
dicular to I'L and passes through the fixed point Q on it. Therefore the cen-
ters of the two circles whose chords of contact pass through I and the point I
are collinear.
Corollary I. In the preceding figures, the point of intersection (other
than I) of the circle BIC and the line EF is one of the points which are equidis-
tant from the three sides of the triangle DBC.
One of the two intersections of the line E? and the circle whose center is
N and whose radius is NA is equidistant from the three sides of the triangle
BAD, and the other from the three sides of the triangle CAD.
Corollary II. In Fig. 1 the points of contact of the eight circles with
either of the two secants are in involution, the center of which is the intersection
L of the two secants, and then its constant is equal to the power of the point L
with respect to the circle ABC.
Demonstration. Take L, the intersection of two secants AD and BC as
the center of inversion, and the power of L with respect to the circle ABC for the
constant of inversion. In the figures 9, and 3, call E', F, and K', the inverses
of the three points E, F, and K, respectively. Then as the two secants EC and
AD, and the circle ABC are their own inverses, the circle 'FK' which is the
inverse of the circle EFK, touches the secants BC and AD and the circle ABC.
Corollary III. When AD and BC are perpendicular, the centers of the
four circles a, b, a', b' are concyclic; so also those of a, b, c, d; those of a', b', c',
d', and those of c, d, c', d'.
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