416

DOCS.

405,

406 DECEMBER

1917

In the

hope

that

I have

thus

eliminated

your difficulties,

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

pleased

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

had

provided

the

entirely

correct

solution

to

the

problem

I

had

posed

for

you[1]

in

place

of

yours

and

had

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.