15.
A
Let Professor Rosseforp's usual journey take minutes at an average speed of
metres/
minute.
Then the distance to work is
metres. On Thursday his speed increased by
10%, i.e. it was
metres/minute.
The time taken was
minutes.
t
v
vt
11
v
/ 10
(
t

x
)
Therefore
.
So
, i.e.
.
11
v
/ 10
× (
t

x
) =
vt
11
(
t

x
) =
10
t
t
=
11
x
16.
A
At points and
,
.
So
, i.e.
, i.e.
or
.
A
C x
=
0
y
2

4
y
=
12
(
y

6
)(
y
+
2
) =
0
y
=
6
y
= 
2
So
is
and is
.
At points and
,
.
So
, i.e.
, i.e.
or
.
So
is
and
is (3, 0).
C
(
0
, 
2
)
A
(
0
,
6
)
B
D y
=
0
x
2
+
x
=
12
(
x

3
)(
x
+
4
) =
0
x
=
3
x
= 
4
D
(
4
,
0
)
B
Therefore the areas of triangles
and
are
and
.
DAB
DBC
1
2
×
7
×
6
=
21
1
2
×
7
×
2
=
7
So
has area 28.
{
It is left to the reader to prove that area
.}
ABCD
ABCD
=
1
2
BD
×
AC
17.
B
In the diagram,
is the centre of the quartercircle arc
;
is the
point where the central square touches arc
;
is the point where
the
central square touches arc
;
is the centre of the circle.
B
AC D
AC F
CE O
As both the circle and arc
have radius 1,
is a square of side 1.
AC
OABC
By Pythagoras' Theorem:
.
So
.
Therefore
. By a similar argument,
.
Now
since
.
So the side of the
square is
OB
2
=
1
2
+
1
2
OB
=
2
OD
=
OB

DB
=
2

1
OF
=
2

1
DF
2
=
OD
2
+
OF
2
=
2
×
OD
2
OD
=
OF
2
×
OD
=
2
(
2

1
)
=
2

2
.
A
B
C
D
E
F
O
18.
A
In the diagram,
is the midpoint of
.
Triangle
is
isosceles since
. Therefore,
bisects
and
is perpendicular to
. The angles at a point total
, so
.
Therefore
.
So
.
D
AC
ABC
AB
=
BC
=
1
2
BD
∠
ABC
BD
AC
360
°
∠
ABC
=
360
° 
2
×
90
° 
2
α
=
180
° 
2
α
∠
ABD
= ∠
CBD
=
90
° 
α
∠
BAD
= ∠
BCD
=
α
1
2
1
2
2
α
x
A
B
C
D
Therefore
.
x
=
AC
=
2
×
AD
=
2
×
AB
cos
α
=
2
×
1
2
cos
α
=
cos
α
19.
E
Note that the number represented by
appears in both the horizontal row and
the vertical column. Note also that 2 + 3 + 4 + 5 + 6 + 7 + 8 = 35. Since the
numbers in the row and those in the column have sum 21, we deduce that
.
x
x
=
2
×
21

35
=
7
We now need two disjoint sets of three numbers chosen from 2, 3, 4, 5, 6, 8
x
so that the numbers in both sets total 14. The only possibilities are {2, 4, 8} and {3, 5, 6}.
We have six choices of which number to put in the top space in the vertical line, then two
for the next space down and one for the bottom space. That leaves three choices for the first
space in the horizontal line, two for the next space and one for the final space. So the total
number of ways is
.
6
×
2
×
1
×
3
×
2
×
1
=
72
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 Spring '13
 MRR
 Math, Prime Numbers, Prime number, Square number

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