416
DOCS.
405,
406 DECEMBER
1917
In the
hope
that
I have
thus
eliminated
I
am
with best
regards
to
you
and
your
sister,[10]
who
visited
me
last
summer
at
my
sister’s
home in
Lucerne, yours truly,
A.
Einstein.
406. To Hans Albert Einstein
[Berlin,] 9
December
1917
Dear
Albert,
Your
last little letter
me
very
much. The
money
for
the
insurance,
which
I
immediately
instructed
to have
sent,
has
arrived,
I
hope.
In
your
letter
the
postscript
was
the
most
amusing, though;
it
was
not
from
you
at all
but
from
the
censor.
He
provided
the
entirely
correct
solution
to
the
problem
I
posed
for
you[1]
in
place
of
yours
and
written in
pencil
under
your
letter: It
goes like
this:[2]
given
Z
A,
height ha
and
(a
+
b
+
c),
you
noticed
correctly
that
it involves
the
construction of
the
AAB'C',
but
then
you
ran
out
of
steam.
You
re-
alized
that the
Z
at
the
tip is
equal
to
a+
B+y/2
or
equal
to
a/2
+
90°.
Therefore,
in the
triangle
AB'C',
the
base
line,
the Z
at
the
tip,
and
the
height
are given.
The A
can
be
constructed
from
it,
though.
For
there
are
two
geometrical
locations for
the
tip A
1)
the
parallel
to
the
base line
at
dis-
tance
ha
2)
the
circle
describing
(a
+
b
+
c)
with
the
peripheral
angle
a/2
+ 90°.
When
you
consider
that the
peripheral
angle
is
equal
to
half the
central
angle,
you easily
find the
center
by
the method
indicated
in the
figure.
The
apothems
of
AB' and AC' intersect
the
points
B
and C.
In
this
way
two
triangles
result,
which
you
should also
prove are congruent
to
each other.
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