Evening star., 1935-09-08
To,the Editor of The Star:

Toshe Editor OF The star:
r yNDER the title. "Offers For
mula for TriSection," The
Sunday Star of September l
~ on page A-9 prints an inter-
esting geometrical picture with a de-
scription of the alleged trisection OF an
angle with rule and compass" The
copyrighted article has pathetic in
terest, for the method shown does not
trisect an angle. The problem will
be explained here in the hope that
the explanation may save some valu-
able time of those young mathema-
ticians who may be tempted to at
tempt the impossible.
The sketch contains all the neces-
sary points, but omits the most i'm
portant lines which would show the
failure of the young mathematician
at a glance. In his attempt to trisect
the angle AOB the author first halves
and quarters the angle so that the
central angles GOV and BOV are each
equal to one fourth OF the given angle
AOB. He then constructs four equal
isosceles triangles mot shown in the
picture) LNO, Imp, MPN and MPR,
with the yertex angle of each equal
to one-fourth of the given angle. Then
he draws the line OR connecting the
vertices of the lowest and the highest
triangle and claims thus TO have cut
the angle LON in the ratio of 2 to l,
so that the angle ROT would be one
third of the given angle AOB.
Author Has No Proof.
The author offers no proof of the
correctness of his solution. He has
none. He has done just one thing:
He has divided the vertex angle into
two unequal angles, one OF which is
opposite a side equal to the radius
OIL, while the other is opposite side
exactly twice as long. But that does
not prove that the ratio between the
angles is the same IS the ratio be-
tween the sides. In a right triangle
with angles Of 30. so and 80 degrees
the sides are in a ratio Of 1:l.72:2,
but the angles are in q ratio of 1913.
II the author had really cut the ver-
tex angle in the ratio of al, then by
cutting the line OR with the radius
OIL from point NJ to point @ the tri-
angle OMR would be divided into two
isosceles triangles NOR and MOO
with the base angles Of the two in the
ratio Of I'll. In reality. however. only
the upper triangle NOR is isoseeles,
while the lower triangle has sides of
different sizes: MO is equal to the
radius OIL: MO is twice as long and
90 is larger than MO since the sum
of two sides Of a triangle always is
larger than the third side. Now the
angle ROM equals the sum of the
angles MOO and ONO. and these
two angles would be equal If the ver-
tex angle LRM (or LON) had really
been trisected. But they are not
equal, the angle MOO being smaller
than OMQ, as it is opposite the smaller
side. MO. Hence the angle MOO
is less than one-third and the angle
ORM Tor LORI is more than two
thirds of the vertex angle. And the
angle ROT is not one-third of the
given angle, but is really smaller.
The trigonometricai proof is like
wise simple. Angles do not increase
in direct proportion to the increase
of the opposite sides, but their sines
do. The sine of the larger angle GOR
or MRO) is exactly twice the value
of the sine of the smaller angle CMOR
or LRO). Now that the sine of an
angle "a equals sine A multiplied
by cosine A. that is, it is less than
twice the sine A, since the factor
coSine A is true fraction between
zero and one. Hence the angle NOR
is less than half of LOR and the angle
ROT is less than onerthird of the
given angle AOB. In other words,
the angle AOB has not been trisected
Ancient Problem.
The trisection of an angle was q
problem highly popular with the geo.
metricians of antiquity. The problem
can be solved and has been solved
in various ways. The ancient math-
ematician Nicomedes employed in its
solution the conchoid, q beautiful
curve of the fourth degree which is
still popular in architecture Other
curves may be employed. The SIM
plest solution which is given in our
test books is that of Willebrord Smell
of Leyden.
Let AOB be the given angle. Ex-
tend the line 40; with 40 as radius,
draw the semi-circle ABC. Draw a
line BE so that the part DE outside
tho rirr1a ar1foIc Tho 70 diuc An Tojn
Cd. Since DE, DO and 30 are all
equal, 1t follows that the angles DOE
and DEO are each equal to one-third,
and the angles ODD and BDO are
each equal to two-thirds Of the given
angle AOB. This solution requires
AZ &,xrc .-1,laS and