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Leonid Chindelevitch
August 24, 2008
1
Monday, July 21, 2008
1.
Problem:
Let
ABC
be an equilateral triangle inscribed in a circle
ω
, and
let
P
be a point on
ω
, with
A
the vertex farthest away from
P
. Show that
d
(
A,P
) =
d
(
B,P
) +
d
(
C,P
).
Solution:
This follows immediately from a more general result known as
Ptolemy’s theorem
,
http://en.wikipedia.org/wiki/Ptolemaios’
theorem
,
which says that the product of the diagonals of a cyclic quadrilateral equals
the sum of the products of its opposite sides. In other words, if
PQRS
is
a cyclic quadrilateral, then
PR
×
QS
=
PQ
×
RS
+
PS
×
QR.
Using
Q
=
B,R
=
A,S
=
C
, with
AB
=
AC
=
BC
, gives
PA
×
BC
=
PB
×
AC
+
PC
×
BA
=
⇒
PA
=
PB
+
PC.
2.
Problem:
Show that the squares of side lengths 1
/
2
,
1
/
3
,...,
1
/n
can be
ﬁt inside a unit square for every
n
∈
N
,n >
1.
Solution:
Group the squares of sides 1
/
2
,
1
/
3 together, those of sides
1
/
4
,
1
/
5
,
1
/
6
,
1
/
7 together, and so on. In general, group the squares of
sides 1
/
2
m
,
1
/
(2
m
+ 1)
,...,
1
/
(2
m
+1

1) together, for
m
∈
N
. Let us call
this the
n
th group. Allocate a slot of width 1 and height 1
/
2
m
to the
m
th group, for
m
∈
N
. Clearly, the slots will always ﬁt into the square
because the total width is always 1 and the total height is at most
∞
X
m
=1
1
2
m
= 1
.
To prove that the squares of the
m
th group ﬁt into the allocated slot,
note that he maximum height of an element in the
m
th group is 1
/
2
m
,
and
1
2
m
+
1
2
m
+ 1
+
···
+
1
2
m
+1

1
<
2
m
1
2
m
= 1
,
so placing the squares one next to the other inside a slot will never ﬁll the
slot completely.
1
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This note was uploaded on 12/10/2011 for the course MATH 348 taught by Professor Karigiannis during the Summer '06 term at McGill.
 Summer '06
 Karigiannis
 Math, Geometry

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