Download Triangle Problems

Solutions: AMC Prep for ACHS: Triangles
ACHS Math Competition Team
3 Feb 2009
Problem 1
The sides of a triangle have lengths 15, 20, and 25. Find the length
of the shortest altitude.
3rd AMC 10 A
tions
Problem2002
1
The sides of a triangle have lengths 15, 20, and 25. Find the length
this isof atheright
triangle, so two of the altitudes are the leg
shortest altitude.
and 20. The third altitude, whose length is x, is the on
First notice
a right triangle,
two of the altitudes
= 150.areUsing 2
use. The
areathat
of this
theis triangle
is 21so(15)(20)
the legs, whose lengths are 15 and 20. The third altitude, whose
he altitude,
have
length is x,we
is the
one drawn to the hypotenuse. The area of the
1 triangle is 2 (15)(20) = 150. Using
30025 as the base and x as the
altitude,
we
have
(25)x = 150, so x =
= 12.
2
25
1
25
20
x
15
OR
1
300
25x = 150 =⇒ x =
= 12
2
25
Problem 2
Points K , L, M, and N lie in the plane of the square ABCD so that
AKB, BLC, CMD, and DNA are equilateral triangles. If ABCD has
an area of 16, findth the area of KLMN.
Solutions
2003
54
4
AMC 12 A
K
A
4
B
N
4
C
D
M
6
OR
◦
L
Problem 2
Points K , L, M, and N lie in the plane of the square ABCD so that
AKB, BLC, CMD, and DNA are equilateral triangles. If ABCD has
an area of 16, findth the area of KLMN.
Solutions
2003
54
4
AMC 12 A
Quadrilateral KLMN is a square
◦
because it has 90 rotational
symmetry. which implies that
each pair of adjacent sides is
congruent and perpendicular.
K
A
4
B
N
4
C
D
M
6
OR
◦
L
Problem 2
Points K , L, M, and N lie in the plane of the square ABCD so that
AKB, BLC, CMD, and DNA are equilateral triangles. If ABCD has
an area of 16, findth the area of KLMN.
Solutions
2003
54
4
AMC 12 A
K
A
4
B
N
4
C
D
M
6
OR
◦
L
Quadrilateral KLMN is a square
◦
because it has 90 rotational
symmetry. which implies that
each pair of adjacent sides is
congruent and perpendicular.
Since ABCD has sides of length
√
4 and K is 2 3 from side AB, the
length√of the diagonal KM is
4 + 4 3.
Problem 2
Points K , L, M, and N lie in the plane of the square ABCD so that
AKB, BLC, CMD, and DNA are equilateral triangles. If ABCD has
an area of 16, findth the area of KLMN.
Solutions
2003
54
4
AMC 12 A
K
A
4
B
N
4
C
D
M
6
OR
◦
L
Quadrilateral KLMN is a square
◦
because it has 90 rotational
symmetry. which implies that
each pair of adjacent sides is
congruent and perpendicular.
Since ABCD has sides of length
√
4 and K is 2 3 from side AB, the
length√of the diagonal KM is
4 + 4 3. Thus the area is
p 2
p
1
4 + 4 3 = 32 + 16 3
2
Problem 3
◦
Let 4XOY be a right triangle with m∠XOY = 90 . Let M and N be
the midpoints of legs OX and OY , respectively. Given that
XN = 19 and YM = 22, find XY .
Carlos takes
R
=
3R
minutes to mow his.
AProblem
A
3
<
, Beth will finish first.
4R
3R
◦
Let 4XOY be a right triangle with m∠XOY = 90 . Let M and N be
Let OM = a and ON = b. Then
the midpoints of legs OX and OY , respectively. Given that
2
= (2a)
+=b222,and
XN = 19
192 and
YM
find 22
XY2 .= a2 + (2b)2 .
X
X
2a
O
e
lows that
M
a
19
b
Y
N
O
22
2b
Y
5(a2 + b2 ) = 192 + 222 = 845.
MN =
p
√
a2 + b2 = 169 = 13.
△XOY is similar to △M ON and XO = 2 · M O, we have XY = 2 · M N =
X
Carlos takes
R
=
3R
minutes to mow his.
AProblem
A
3
<
, Beth will finish first.
4R
3R
◦
Let 4XOY be a right triangle with m∠XOY = 90 . Let M and N be
Let OM = a and ON = b. Then
the midpoints of legs OX and OY , respectively. Given that
2
= (2a)
+=b222,and
XN = 19
192 and
YM
find 22
XY2 .= a2 + (2b)2 .
X
X
2a
O
e
M
a
19
b
Y
N
22
O
2b
Y
Let OM = a and ON = b. Then
2
2
2
192 = (2a)25(a
+b22+
, b ) = 19 + 22 = 845.
lows that
MN =
p
√
a2 + b2 = 169 = 13.
△XOY is similar to △M ON and XO = 2 · M O, we have XY = 2 · M N =
X
Carlos takes
=
R
3R
minutes to mow his.
AProblem
A
3
<
, Beth will finish first.
4R
3R
◦
Let 4XOY be a right triangle with m∠XOY = 90 . Let M and N be
Let OM = a and ON = b. Then
the midpoints of legs OX and OY , respectively. Given that
2
= (2a)
+=b222,and
XN = 19
192 and
YM
find 22
XY2 .= a2 + (2b)2 .
X
X
2a
O
e
M
a
19
b
Y
N
22
O
2b
Y
Let OM = a and ON = b. Then
) =2 19
222b=)2845.
= a+2 +(
,
192 = (2a)25(a
+b2+
, b 22
2
lows that
MN =
2
2
2
p
√
a2 + b2 = 169 = 13.
△XOY is similar to △M ON and XO = 2 · M O, we have XY = 2 · M N =
X
Carlos takes
=
R
3R
minutes to mow his.
AProblem
A
3
<
, Beth will finish first.
4R
3R
◦
Let 4XOY be a right triangle with m∠XOY = 90 . Let M and N be
Let OM = a and ON = b. Then
the midpoints of legs OX and OY , respectively. Given that
2
= (2a)
+=b222,and
XN = 19
192 and
YM
find 22
XY2 .= a2 + (2b)2 .
X
X
2a
O
e
M
a
19
b
Y
N
22
O
2b
Y
Let OM = a and ON = b. Then
) =2 19
222b=)2845.
= a+2 +(
, =⇒ 5(a2 +b2 ) = 192 +222 = 845
192 = (2a)25(a
+b2+
, b 22
2
lows that
MN =
2
2
2
p
√
a2 + b2 = 169 = 13.
△XOY is similar to △M ON and XO = 2 · M O, we have XY = 2 · M N =
X
Carlos takes
=
R
3R
minutes to mow his.
AProblem
A
3
<
, Beth will finish first.
4R
3R
◦
Let 4XOY be a right triangle with m∠XOY = 90 . Let M and N be
Let OM = a and ON = b. Then
the midpoints of legs OX and OY , respectively. Given that
2
= (2a)
+=b222,and
XN = 19
192 and
YM
find 22
XY2 .= a2 + (2b)2 .
X
X
2a
O
e
M
a
19
b
Y
N
22
O
2b
Y
Let OM = a and ON = b. Then
) =2 19
222b=)2845.
= a+2 +(
, =⇒ 5(a2 +b2 ) = 192 +222 = 845
192 = (2a)25(a
+b2+
, b 22
2
lows that
2
2
2
p
√q
MN = p
a2 + b2 = 169 = 13.
MN =
a2 + b 2 =
845/5 =
p
169 = 13
△XOY is similar to △M ON and XO = 2 · M O, we have XY = 2 · M N =
X
Carlos takes
=
minutes to mow his.
3R
Hence
AProblem
A
3
5(a2 + b2 ) = 192 + 222 = 845.
<
, Beth will finish first.
4R
3R
◦
Let 4XOY be a right triangle
with
m∠XOY = 90 . Let
be
It follows
that
p M and N √
Let OM = a and ON = b. Then
2 + b2 =
MN
= athat
169 = 13.
the midpoints of legs OX and OY , respectively.
Given
R
2
= (2a)
+=b222,
and
22
(2b)2 . to △M ON and XO = 2 · M O, w
XN = 19
192 and
YM
find△XOY
XY2 .= a2is+similar
Since
X
2a
O
e
19
b
Y
N
26.
X
X
M
a
a
M
a
22
O
2b
Y
O
26
13
b
N
b
Y
Let OM = a and ON = b. Then
) =2 19
222b=)2845.
= a+2 +(
, =⇒ 5(a2 +b2 ) = 192 +222 = 845
192 = (2a)25(a
+b2+
, b 22
2
lows that
2
2
2
p
√q
MN = p
a2 + b2 = 169 = 13.
MN =
a2 + b 2 =
845/5 =
p
169 = 13
△XOY is similar to △M ON and XO = 2 · M O, we have XY = 2 · M N =
X
XY = 2MN = 26
and D lie on a line, in that order, with AB = CD and BC = 12.
onProblem
the line, and
4 BE = CE = 10. The perimeter of △AED is
eter of △BEC. Find AB.
E
Points A, B, C, and D lie on a line, in
that order, with AB = CD and BC = 12.
10
10
Point E is not on the line, and
D
A
B
C
12
BE = CE = 10. The perimeter of
4AED is twice that of 4BEC. Find AB.
B) 8
(C) 17/2
(D) 9
(E) 19/2
selects two distinct numbers from the set {1, 2, 3, 4, 5}, and Sergio
ts a number from the set {1, 2, . . . , 10}. The probability that
r is larger than the sum of the two numbers chosen by Tina is
) 9/20
(C) 1/2
(D) 11/20
(E) 24/25
esand
△AED
and
△BEC,
andorder, with AB = CD and BC = 12.
D lie on
a line,
in that
p
onProblem
the line, and
= 10. The perimeter of △AED is
4 BE = CE
EHFind
= AB.
102 − 62 = 8.
eter of △BEC.
Points A, B, C, and D lie on a line, in
E
= x and AE = ED = y. Then 2x+2y+12
= 2(32),
= 26−x.
that order,
with ABso=y CD
and BC = 12.
10
10
Point E is not on the line, and
C (26 −Dx)2
82 +A (x + 6)B2 =12y 2 =
and= CE
x ==9.10. The perimeter of
BE
4AED is twice that of 4BEC. Find AB.
E
B) 8
(C)
17/2
y
10
A
x
(D)
9
y
10
B 6 H 6 C
x
D
Let H
be the midpoint of BC. Then EH
(E)
19/2
is the perpendicular bisector of AD, and
4AED is isosceles.
ten ways
Tina numbers
to select from
a pair
sums
9, 8,
selects
twofor
distinct
theofsetnumbers.
{1, 2, 3, 4,The
5}, and
Sergio
obtained
in
just
one
way,
and
the
sums
7,
6,
and
5
can
each
ts a number from the set {1, 2, . . . , 10}. The probability that
two
ways.
The
probability
for
each
of
Sergio’s
choices
is
1/10.
r is larger than the sum of the two numbers chosen by Tina is
selections in decreasing order, the total probability of Sergio’s
) 9/20
(C) 1/2
(D) 11/20
(E) 24/25
ater
is
9
8
6
4
2
1
2
esand
△AED
and
△BEC,
andorder, with AB = CD and BC = 12.
D lie on
a line,
in that
p
onProblem
the line, and
= 10. The perimeter of △AED is
4 BE = CE
EHFind
= AB.
102 − 62 = 8.
eter of △BEC.
Points A, B, C, and D lie on a line, in
E
= x and AE = ED = y. Then 2x+2y+12
= 2(32),
= 26−x.
that order,
with ABso=y CD
and BC = 12.
10
10
Point E is not on the line, and
C (26 −Dx)2
82 +A (x + 6)B2 =12y 2 =
and= CE
x ==9.10. The perimeter of
BE
4AED is twice that of 4BEC. Find AB.
E
B) 8
(C)
17/2
y
10
A
x
(D)
9
y
10
B 6 H 6 C
x
D
Let H
be the midpoint of BC. Then EH
(E)
19/2
is the perpendicular bisector of AD, and
4AED is isosceles.
Segment EH is the common altitude of isosceles triangles 4AED
and 4BEC, and
ten ways
Tina numbers
to select from
a pair
sums
9, 8,
selects
twofor
distinct
theofsetnumbers.
{1, 2, 3, 4,The
5}, and
Sergio
obtained
in
just
one
way,
and
the
sums
7,
6,
and
5
can
each
ts a number from the set {1, 2, . . . , 10}. The probability that
two
ways.
The
probability
for
each
of
Sergio’s
choices
is
1/10.
r is larger than the sum of the two numbers chosen by Tina is
selections in decreasing order, the total probability of Sergio’s
) 9/20
(C) 1/2
(D) 11/20
(E) 24/25
ater
is
9
8
6
4
2
1
2
esand
△AED
and
△BEC,
andorder, with AB = CD and BC = 12.
D lie on
a line,
in that
p
onProblem
the line, and
= 10. The perimeter of △AED is
4 BE = CE
EHFind
= AB.
102 − 62 = 8.
eter of △BEC.
Points A, B, C, and D lie on a line, in
E
= x and AE = ED = y. Then 2x+2y+12
= 2(32),
= 26−x.
that order,
with ABso=y CD
and BC = 12.
10
10
Point E is not on the line, and
C (26 −Dx)2
82 +A (x + 6)B2 =12y 2 =
and= CE
x ==9.10. The perimeter of
BE
4AED is twice that of 4BEC. Find AB.
E
B) 8
(C)
17/2
y
10
A
x
(D)
9
y
10
B 6 H 6 C
x
D
Let H
be the midpoint of BC. Then EH
(E)
19/2
is the perpendicular bisector of AD, and
4AED is isosceles.
Segment EH is the common
altitude of isosceles triangles 4AED
p
2
and 4BEC, and EH = 10 − 62 = 8.
ten ways
Tina numbers
to select from
a pair
sums
9, 8,
selects
twofor
distinct
theofsetnumbers.
{1, 2, 3, 4,The
5}, and
Sergio
obtained
in
just
one
way,
and
the
sums
7,
6,
and
5
can
each
ts a number from the set {1, 2, . . . , 10}. The probability that
two
ways.
The
probability
for
each
of
Sergio’s
choices
is
1/10.
r is larger than the sum of the two numbers chosen by Tina is
selections in decreasing order, the total probability of Sergio’s
) 9/20
(C) 1/2
(D) 11/20
(E) 24/25
ater
is
9
8
6
4
2
1
2
esand
△AED
and
△BEC,
andorder, with AB = CD and BC = 12.
D lie on
a line,
in that
p
onProblem
the line, and
= 10. The perimeter of △AED is
4 BE = CE
EHFind
= AB.
102 − 62 = 8.
eter of △BEC.
Points A, B, C, and D lie on a line, in
E
= x and AE = ED = y. Then 2x+2y+12
= 2(32),
= 26−x.
that order,
with ABso=y CD
and BC = 12.
10
10
Point E is not on the line, and
C (26 −Dx)2
82 +A (x + 6)B2 =12y 2 =
and= CE
x ==9.10. The perimeter of
BE
4AED is twice that of 4BEC. Find AB.
E
B) 8
(C)
17/2
y
10
A
x
(D)
9
y
10
B 6 H 6 C
x
D
Let H
be the midpoint of BC. Then EH
(E)
19/2
is the perpendicular bisector of AD, and
4AED is isosceles.
Segment EH is the common
altitude of isosceles triangles 4AED
p
2
and 4BEC, and EH = 10 − 62 = 8. Let AB = CD = x and
ten ways
Tina
to
a pair
sums
9, 8,
selects
twofor
theofsetnumbers.
{1, 2, 3, 4,The
5}, and
Sergio
AE
=distinct
ED
= ynumbers
. select from
obtained
in
just
one
way,
and
the
sums
7,
6,
and
5
can
each
ts a number from the set {1, 2, . . . , 10}. The probability that
two
ways.
The
probability
for
each
of
Sergio’s
choices
is
1/10.
r is larger than the sum of the two numbers chosen by Tina is
selections in decreasing order, the total probability of Sergio’s
) 9/20
(C) 1/2
(D) 11/20
(E) 24/25
ater
is
9
8
6
4
2
1
2
esand
△AED
and
△BEC,
andorder, with AB = CD and BC = 12.
D lie on
a line,
in that
p
onProblem
the line, and
= 10. The perimeter of △AED is
4 BE = CE
EHFind
= AB.
102 − 62 = 8.
eter of △BEC.
Points A, B, C, and D lie on a line, in
E
= x and AE = ED = y. Then 2x+2y+12
= 2(32),
= 26−x.
that order,
with ABso=y CD
and BC = 12.
10
10
Point E is not on the line, and
C (26 −Dx)2
82 +A (x + 6)B2 =12y 2 =
and= CE
x ==9.10. The perimeter of
BE
4AED is twice that of 4BEC. Find AB.
E
B) 8
(C)
17/2
y
10
A
x
(D)
9
y
10
B 6 H 6 C
x
D
Let H
be the midpoint of BC. Then EH
(E)
19/2
is the perpendicular bisector of AD, and
4AED is isosceles.
Segment EH is the common
altitude of isosceles triangles 4AED
p
2
and 4BEC, and EH = 10 − 62 = 8. Let AB = CD = x and
ten ways
Tina
to
select2x
a +pair
ofset
numbers.
sums
9, 8,
selects
twofor
from
the+
{1,
3, )4,,The
5}, and
Sergio
AE
=distinct
ED
= ynumbers
. Then
2y
12
= 22,(32
obtained
in
just
one
way,
and
the
sums
7,
6,
and
5
can
each
ts a number from the set {1, 2, . . . , 10}. The probability that
two
ways.
The
probability
for
each
of
Sergio’s
choices
is
1/10.
r is larger than the sum of the two numbers chosen by Tina is
selections in decreasing order, the total probability of Sergio’s
) 9/20
(C) 1/2
(D) 11/20
(E) 24/25
ater
is
9
8
6
4
2
1
2
esand
△AED
and
△BEC,
andorder, with AB = CD and BC = 12.
D lie on
a line,
in that
p
onProblem
the line, and
= 10. The perimeter of △AED is
4 BE = CE
EHFind
= AB.
102 − 62 = 8.
eter of △BEC.
Points A, B, C, and D lie on a line, in
E
= x and AE = ED = y. Then 2x+2y+12
= 2(32),
= 26−x.
that order,
with ABso=y CD
and BC = 12.
10
10
Point E is not on the line, and
C (26 −Dx)2
82 +A (x + 6)B2 =12y 2 =
and= CE
x ==9.10. The perimeter of
BE
4AED is twice that of 4BEC. Find AB.
E
B) 8
(C)
17/2
y
10
A
x
(D)
9
y
10
B 6 H 6 C
x
D
Let H
be the midpoint of BC. Then EH
(E)
19/2
is the perpendicular bisector of AD, and
4AED is isosceles.
Segment EH is the common
altitude of isosceles triangles 4AED
p
2
and 4BEC, and EH = 10 − 62 = 8. Let AB = CD = x and
ten ways
Tina
to
select2x
a +pair
ofset
numbers.
sums
9,x.8,
selects
twofor
from
the+
{1,
3, )4,,The
5},yand
Sergio
AE
=distinct
ED
= ynumbers
. Then
2y
12
= 22,(32
so
= 26
−
obtained
in
just
one
way,
and
the
sums
7,
6,
and
5
can
each
ts a number from the set {1, 2, . . . , 10}. The probability that
two
ways.
The
probability
for
each
of
Sergio’s
choices
is
1/10.
r is larger than the sum of the two numbers chosen by Tina is
selections in decreasing order, the total probability of Sergio’s
) 9/20
(C) 1/2
(D) 11/20
(E) 24/25
ater
is
9
8
6
4
2
1
2
esand
△AED
and
△BEC,
andorder, with AB = CD and BC = 12.
D lie on
a line,
in that
p
onProblem
the line, and
= 10. The perimeter of △AED is
4 BE = CE
EHFind
= AB.
102 − 62 = 8.
eter of △BEC.
Points A, B, C, and D lie on a line, in
E
= x and AE = ED = y. Then 2x+2y+12
= 2(32),
= 26−x.
that order,
with ABso=y CD
and BC = 12.
10
10
Point E is not on the line, and
C (26 −Dx)2
82 +A (x + 6)B2 =12y 2 =
and= CE
x ==9.10. The perimeter of
BE
4AED is twice that of 4BEC. Find AB.
E
B) 8
(C)
17/2
y
10
A
x
(D)
9
y
10
B 6 H 6 C
x
D
Let H
be the midpoint of BC. Then EH
(E)
19/2
is the perpendicular bisector of AD, and
4AED is isosceles.
Segment EH is the common
altitude of isosceles triangles 4AED
p
2
and 4BEC, and EH = 10 − 62 = 8. Let AB = CD = x and
ten ways
Tina
to
select2x
a +pair
ofset
numbers.
sums
9,x.8,Thus,
selects
twofor
from
the+
{1,
3, )4,,The
5},yand
Sergio
AE
=distinct
ED
= ynumbers
. Then
2y
12
= 22,(32
so
= 26
−
obtained
in
just
one
way,
and
the
sums
7,
6,
and
5
can
each
ts a number from the set {1, 2, . . . , 10}. The probability that
2
2 of Sergio’s
2
2
two
ways.
The
probability
for
each
choices
is
1/10.
(x two
+ 6)numbers
= y =chosen
(26 − xby
) Tina is
r is larger than the sum 8of +
the
selections in decreasing order, the total probability of Sergio’s
) 9/20
(C) 1/2
(D) 11/20
(E) 24/25
ater
is
9
8
6
4
2
1
2
esand
△AED
and
△BEC,
andorder, with AB = CD and BC = 12.
D lie on
a line,
in that
p
onProblem
the line, and
= 10. The perimeter of △AED is
4 BE = CE
EHFind
= AB.
102 − 62 = 8.
eter of △BEC.
Points A, B, C, and D lie on a line, in
E
= x and AE = ED = y. Then 2x+2y+12
= 2(32),
= 26−x.
that order,
with ABso=y CD
and BC = 12.
10
10
Point E is not on the line, and
C (26 −Dx)2
82 +A (x + 6)B2 =12y 2 =
and= CE
x ==9.10. The perimeter of
BE
4AED is twice that of 4BEC. Find AB.
E
B) 8
(C)
17/2
y
10
A
x
(D)
9
y
10
B 6 H 6 C
x
D
Let H
be the midpoint of BC. Then EH
(E)
19/2
is the perpendicular bisector of AD, and
4AED is isosceles.
Segment EH is the common
altitude of isosceles triangles 4AED
p
2
and 4BEC, and EH = 10 − 62 = 8. Let AB = CD = x and
ten ways
Tina
to
select2x
a +pair
ofset
numbers.
sums
9,x.8,Thus,
selects
twofor
from
the+
{1,
3, )4,,The
5},yand
Sergio
AE
=distinct
ED
= ynumbers
. Then
2y
12
= 22,(32
so
= 26
−
obtained
in
just
one
way,
and
the
sums
7,
6,
and
5
can
each
ts a number from the set {1, 2, . . . , 10}. The probability that
2
2 of Sergio’s
2
2
two
ways.
The
probability
for
each
choices
is
1/10.
(x two
+ 6)numbers
= y =chosen
(26 − xby
) Tina is
r is larger than the sum 8of +
the
selections in decreasing order, the total probability of Sergio’s
) 9/20
(D) 11/20
(E) 24/25
= 9.1/2
ater
isand x(C)
9
8
6
4
2
1
2
Problem 5
How many non-congruent triangles with perimeter 7 have integer
side lengths?
Problem 5
How many non-congruent triangles with perimeter 7 have integer
side lengths?
First, compute the partitions of 7
Problem 5
How many non-congruent triangles with perimeter 7 have integer
side lengths?
First, compute the partitions of 7 (combinations of positive integers
that add up to 7) containing three elements.
Problem 5
How many non-congruent triangles with perimeter 7 have integer
side lengths?
First, compute the partitions of 7 (combinations of positive integers
that add up to 7) containing three elements. These are: {5, 1, 1},
Problem 5
How many non-congruent triangles with perimeter 7 have integer
side lengths?
First, compute the partitions of 7 (combinations of positive integers
that add up to 7) containing three elements. These are: {5, 1, 1},
{4, 2, 1},
Problem 5
How many non-congruent triangles with perimeter 7 have integer
side lengths?
First, compute the partitions of 7 (combinations of positive integers
that add up to 7) containing three elements. These are: {5, 1, 1},
{4, 2, 1}, {3, 3, 1},
Problem 5
How many non-congruent triangles with perimeter 7 have integer
side lengths?
First, compute the partitions of 7 (combinations of positive integers
that add up to 7) containing three elements. These are: {5, 1, 1},
{4, 2, 1}, {3, 3, 1}, and {3, 2, 2}.
Problem 5
How many non-congruent triangles with perimeter 7 have integer
side lengths?
First, compute the partitions of 7 (combinations of positive integers
that add up to 7) containing three elements. These are: {5, 1, 1},
{4, 2, 1}, {3, 3, 1}, and {3, 2, 2}. Each of these must satisfy the
triangle inequality: The length of any side must be less than the
sum of the other two sides but greater than their difference. Of the
above 4 combinations, 5 is not less than 1 + 1,
Problem 5
How many non-congruent triangles with perimeter 7 have integer
side lengths?
First, compute the partitions of 7 (combinations of positive integers
that add up to 7) containing three elements. These are: {5, 1, 1},
{4, 2, 1}, {3, 3, 1}, and {3, 2, 2}. Each of these must satisfy the
triangle inequality: The length of any side must be less than the
sum of the other two sides but greater than their difference. Of the
above 4 combinations, 5 is not less than 1 + 1, and 4 is not less
than 2 + 1. The remaining 2 combinations constitute valid side
lengths.
segments, each of length 1. Point G is not on line AF
and point J lies on GF . The line segments HC, JE, and
Problem 6
HC/JE.
Points A, B, C, D, E, and F lie, in that G
order, on AF , dividing it into five
segments, each of length 1. Point G is
not on line AF . Point H lies on GD, and
H
J
point J lies on GF . The line segments
F
E
C
D
B
A
HC, JE, and AG are parallel. Find
HC/JE.
(A) 5/4
(B) 4/3
(C) 3/2
(D) 5/3
(E)
21. The mean, median, unique mode, and range of a collecti
all equal to 8. The largest integer that can be an elemen
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15
22. A set of tiles numbered 1 through 100 is modified rep
segments, each of length 1. Point G is not on line AF
and point J lies on GF . The line segments HC, JE, and
Problem 6
HC/JE.
Points A, B, C, D, E, and F lie, in that G
order, on AF , dividing it into five
segments, each of length 1. Point G is
not on line AF . Point H lies on GD, and
H
J
point J lies on GF . The line segments
F
E
C
D
B
A
HC, JE, and AG are parallel. Find
HC/JE.
(A) 5/4
(B) 4/3
(C) 3/2
(D) 5/3
(E)
Since 4AGD is similar to 4CHD, we have HC/1 = AG/3.
21. The mean, median, unique mode, and range of a collecti
all equal to 8. The largest integer that can be an elemen
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15
22. A set of tiles numbered 1 through 100 is modified rep
segments, each of length 1. Point G is not on line AF
and point J lies on GF . The line segments HC, JE, and
Problem 6
HC/JE.
Points A, B, C, D, E, and F lie, in that G
order, on AF , dividing it into five
segments, each of length 1. Point G is
not on line AF . Point H lies on GD, and
H
J
point J lies on GF . The line segments
F
E
C
D
B
A
HC, JE, and AG are parallel. Find
HC/JE.
(A) 5/4
(B) 4/3
(C) 3/2
(D) 5/3
(E)
Since 4AGD is similar to 4CHD, we have HC/1 = AG/3. Also,
4AGF is similar to 4EJF , so JE/1 = AG/5.
21. The mean, median, unique mode, and range of a collecti
all equal to 8. The largest integer that can be an elemen
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15
22. A set of tiles numbered 1 through 100 is modified rep
segments, each of length 1. Point G is not on line AF
and point J lies on GF . The line segments HC, JE, and
Problem 6
HC/JE.
Points A, B, C, D, E, and F lie, in that G
order, on AF , dividing it into five
segments, each of length 1. Point G is
not on line AF . Point H lies on GD, and
H
J
point J lies on GF . The line segments
F
E
C
D
B
A
HC, JE, and AG are parallel. Find
HC/JE.
(A) 5/4
(B) 4/3
(C) 3/2
(D) 5/3
(E)
Since 4AGD is similar to 4CHD, we have HC/1 = AG/3. Also,
4AGF is similar to 4EJF , so JE/1 = AG/5. Therefore,
21. The mean, median, unique mode, and range of a collecti
AG/3
HC
all equal to 8.
= The largest integer that can be an elemen
JE
AG/5
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15
22. A set of tiles numbered 1 through 100 is modified rep
segments, each of length 1. Point G is not on line AF
and point J lies on GF . The line segments HC, JE, and
Problem 6
HC/JE.
Points A, B, C, D, E, and F lie, in that G
order, on AF , dividing it into five
segments, each of length 1. Point G is
not on line AF . Point H lies on GD, and
H
J
point J lies on GF . The line segments
F
E
C
D
B
A
HC, JE, and AG are parallel. Find
HC/JE.
(A) 5/4
(B) 4/3
(C) 3/2
(D) 5/3
(E)
Since 4AGD is similar to 4CHD, we have HC/1 = AG/3. Also,
4AGF is similar to 4EJF , so JE/1 = AG/5. Therefore,
21. The mean, median, unique mode, and range of a collecti
AG/3
HC
5
all equal to 8.
= The largest
= . integer that can be an elemen
JE
3
AG/5
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15
22. A set of tiles numbered 1 through 100 is modified rep