Download Equilateral triangle ABC has side length 400 cm and a

(1)
Equilateral triangle ABC has side length 400 cm and a perimeter equal to
100 times the perimeter of equilateral triangle DEF . How long is each side of triangle
DEF ?
(2)
In the figure, point A is the center of the circle, the measure of angle RAS
is 74 degrees, and the measure of angle RT B is 28 degrees. What is the measure of minor
arc BR, in degrees?
R
S
B
(3)
(4)
A
T
A right cylinder with a base radius of 3 units is inscribed in a sphere of
radius 5 units. What is the total volume, in cubic units, of the space inside the sphere and
outside the cylinder? Express your answer as a common fraction in terms of π.
√
In parallelogram ABCD, AB = 16 cm, DA = 3 2 cm, and sides AB and
DA form a 45-degree interior angle. In isosceles trapezoid W XY Z with W X 6= Y Z,
segment W X is the longer parallel side and has length 16 cm, and two interior angles each
have a measure of 45 degrees. Trapezoid W XY Z has the same area as parallelogram
ABCD. What is the length in centimeters of segment Y Z?
(5)
An octagon only has sides of length 1 unit and x units. In a similar octagon,
the corresponding sides have length x units and 9 units, respectively. What is the value of
x?
(6)
A smaller rectangular box has a length and width of 10 cm each and a
height of 1 cm. A larger box is twice the length, three times the width, and 10 times the
height of the smaller box. What is the greatest number of the smaller boxes that can fit
inside one of the larger boxes?
(7)
A regular hexagon is inscribed in a circle and another regular hexagon is
circumscribed about the same circle. What is the ratio of the area of the larger hexagon to
the area of the smaller hexagon? Express your answer as a common fraction.
(8)
The point at (a, b) on a Cartesian plane is reflected over the y -axis to the
point at (j, k). If a + j = 0 and b + k = 0, what is the value of b?
(9)
Triangle ABC with vertices A(−2, 0), B(1, 4) and C(−3, 2) is reflected over
the y -axis to form triangle A′ B ′C ′ . What is the length of a segment drawn from C to C ′ ?
(10)
What is the midpoint of the segment connecting (3, 4) and (3.8, 5.6) in the
Cartesian plane? Express the coordinates as decimals to the nearest tenth.
(11)
The second hand of a clock is 4 inches long. To the nearest inch, how far
does the tip of the second hand travel in 45 seconds?
(12)
The shape of Yellowstone National Park is nearly a rectangle 55 miles wide
and 64 miles long. What is the area of a rectangle with these dimensions?
(13)
A circle with a radius of 2 units has its center at (0, 0). A circle with a
radius of 7 units has its center at (15, 0). A line tangent to both circles intersects the
x -axis at (x , 0) to the right of the origin. What is the value of x ? Express your answer as a
common fraction.
(14)
The coordinates of A and B on a number line are −7 and 2, respectively.
What is the length of segment AB?
(15)
To be able to walk to the center C of a circular fountain, a repair crew
places a 16-foot plank from A to B and then a 10-foot plank from D to C, where D is the
midpoint of AB . What is the area of the circular base of the fountain? Express your
answer in terms of π.
A
D
B
C
Side View
Top View
(16)
Point A and line m are in the same plane, but A is not on m. How many
lines containing A are parallel to m ?
(17)
A parallelogram has three of its vertices at (−1, 0), (2, 4) and (2, −4).
What is the positive difference between the greatest possible perimeter and the least
possible perimeter of the parallelogram?
(18)
In a trapezoid ABCD with AB parallel to CD, the diagonals AC and BD
intersect at E. If the area of triangle ABE is 50 square units, and the area of triangle ADE
is 20 square units, what is the area of trapezoid ABCD?
(19)
A 4 × 4 × 4 inch cube originally built from 1 × 1 × 1 inch cubes is cut into
exactly 29 cubes with integer edge lengths and no material left over. How many of the 29
cubes are 2 × 2 × 2 inch cubes?
(20)
When the point (4, 1) is rotated 90 degrees counterclockwise about the
point (1, 0), what are the coordinates of the point at which it lands?
(21)
What is the total volume in cubic feet of three boxes if each box is a cube
with edge length 4 feet?
(22)
How many non-congruent triangles are there with sides of integer length
having at least one side of length five units and having no side longer than five units?
(23)
What is the area in square units of the convex quadrilateral with vertices
(−1, 0), (0, 1), (2, 0) and (0, −3)?
(24)
What is the area, in square units, of triangle ABC?
B (0, 6)
A (−4, 3)
C (2, −2)
(25)
The consecutive angles of a particular trapezoid form an arithmetic
sequence. If the largest angle measures 120◦ , what is the measure of the smallest angle?
(26)
An isosceles right triangle is removed from each corner of a square piece of
paper, as shown, to create a rectangle. If AB = 12 units, what is the combined area of the
four removed triangles, in square units?
A
B
(27)
In this quilt pattern, points E, F , G and H are midpoints of the sides of
square ABCD, and square EF GH is divided into nine congruent unit squares, as shown.
What percent of the total area of square ABCD does the total shaded area represent?
Express your answer to the nearest whole percent.
A
E
H
D
(28)
B
F
G
C
Trapezoid ABCD has vertices A(−1, 0), B(0, 4), C(m, 4) and D(k, 0), with
m > 0 and k > 0. The line y = −x + 4 is perpendicular to the line containing side CD, and
the area of trapezoid ABCD is 34 square units. What is the value of k?
What is the greatest number of interior right angles a convex octagon can
(29)
have?
(30)
The scale of a certain map is 45 inch = 16 miles. A square park is
represented on this map by a square with side length 58 inch. What is the actual area of this
park in square miles? Express your answer as a decimal to the nearest hundredth.
Copyright MATHCOUNTS Inc. All rights reserved
Answer Sheet
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Answer
4 cm
81 degrees
284π/3 cubic units
8 centimeters
3
60 boxes
4/3
0
6 units
(3.4, 4.8)
19 inches
3520 square miles
10/3
9 units
164π square feet
1 line
6 units
98 sq units
5 cubes
(0, 3)
192 cubic feet
9 triangles
6 square units
19 sq units
60
72 sq units
22 percent
6
3 right angles
156.25 sq miles
Problem ID
1CB
DD212
A15C
DDD
44A
10B
4C551
134B
C31
15A
0011
3A21
4C4C
23A
1342
4031
24CB
BA422
32A1
AC01
0233
25D
0D33
BA322
C03
B5212
43D5
1A02
5A2C
1501
Copyright MATHCOUNTS Inc. All rights reserved
Solutions
(1) 4 cm
ID: [1CB]
No solution is available at this time.
(2) 81 degrees
ID: [DD212]
Let C be the point where line segment AT intersects the circle. The measure of ∠RT B
half the difference of the two arcs it cuts off:
d − m SC
c
m RB
m∠RT B =
.
2
c = 74◦ , m SC
c = 180◦ − 74◦ − m RB.
d Substituting this expression for m SC
c as
Since m RS
well as 28◦ for m∠RT B, we get
d − (180◦ − 74◦ − m RB)
d
m RB
.
28◦ =
2
d = 81 degrees.
Solve to find m RB
R
S
B
A
C
T
(3) 284π/3 cubic units
ID: [A15C]
To begin, we need to visualize the cylinder inscribed in the sphere. We can draw the
cylinder as shown:
6
10
A diagonal drawn in the cylinder will have length 10, which is the diameter of the sphere.
We can see that a 6-8-10 right triangle is formed by the height of the cylinder, the
diameter of the sphere, and the diameter of the base of the cylinder. Now that we know
the height of the cylinder, we have everything we need to compute the desired volume:
4
4
500π
Vspher e = πr 3 = · π · 53 =
3
3
3
2
2
Vcy linder = πr · h = π · 3 · 8 = 72π
The volume inside the sphere and outside the cylinder is the difference of the above values:
Vspher e − Vcy linder =
500π − 216π
284π
500π
− 72π =
=
3
3
3
(4) 8 centimeters
ID: [DDD]
Below are the two shapes implied by the problem:
B
C
16
Z
√
3 2
A
D
W
M
Y
16
X
We have drawn the dotted lines BD and altitude ZM. Notice that the area of △ABD is
the same as the area of △BCD,
√ and the area of △ABD is given by
AB · AD · sin(45)/2 = 16 · 3 2 · sin(45)/2 = 24. So the area of the parallelogram is
24 · 2 = 48.
Let ZY = x . Since ZY XW is an isosceles trapezoid, W M = ZM = 16−x
2 . The area of
x+16 16−x
ZY +W X
· ZM = 2 · 2 . We are told that this area equals 48, so
the trapezoid is then
2
x+16 16−x
we have the equation 2 · 2 = 48. Simplifying, we get
x + 16 16 − x
·
= 48
2
2
(x + 16)(16 − x ) = 192
−x 2 + 256 = 192
x =8
So the length of ZY is 8 centimeters.
(5) 3
ID: [44A]
No solution is available at this time.
(6) 60 boxes
ID: [10B]
Doubling the length allows us to fit 2 times as many boxes. Tripling the width lets us fit 3
times as many boxes. Similarly, the increase in height lets us fit 10 times as many boxes.
Overall, we can fit 2 × 3 × 10 = 60 smaller boxes.
(7) 4/3
ID: [4C551]
Form a triangle whose first vertex is the center of the circle and whose other two vertices
are the midpoint and one of the endpoints of a side of the larger hexagon, as shown in the
diagram. Since each interior angle of a regular hexagon is 120 degrees, this triangle is a
30-60-90 right triangle. Let r be the radius of the circle.
√ The length of the longer leg of
the triangle
is r , so the length of the shorter leg is r / 3 and the length of the hypotenuse
√
is 2r 3. Since for the smaller hexagon the length of the segment
connecting a vertex to
√
the center is r , the dimensions of the larger hexagon are 2/ 3 times larger than the
dimensions
√ 2 of the smaller hexagon. Therefore, the area of the larger triangle is
(2/ 3) = 4/3 times greater than the area of the smaller triangle.
(8) 0
ID: [134B]
If the point (a, b) is reflected over the y -axis, it will land on the point (−a, b). Thus,
j = −a and k = b. We were given that a + j = 0, and a + (−a) = 0 so this is satisfied.
From b + k = 0, we find
b + (b) = 0 ⇒ 2b = 0
b=0
(9) 6 units
ID: [C31]
Reflecting a point over the y -axis negates the x -coefficient. So if C is (−3, 2), C ′ will be
(3, 2). The segment is a horizontal line of length 3 + 3 = 6 .
(10) (3.4, 4.8)
ID: [15A]
No solution is available at this time.
(11) 19 inches
ID: [0011]
No solution is available at this time.
(12) 3520 square miles
ID: [3A21]
The area is equal to 55 · 64 = 55 · 60 + 55 · 4 = 3300 + 220 = 3520 square miles.
(13) 10/3
ID: [4C4C]
To begin, we can draw a diagram as shown:
7
2
By drawing in radii to the tangent line, we have formed two right triangles, one with
hypotenuse x and the other with hypotenuse 15 − x . Notice that the angles at the x axis
are vertical angles and are also congruent. So, these two triangles are similar, and we can
set up a ratio:
2
x
=
15 − x
7
7x = 30 − 2x
9x = 30
10
x=
3
(14) 9 units
ID: [23A]
No solution is available at this time.
(15) 164π square feet
ID: [1342]
Since triangle ABC is isosceles (both AC and BC are radii), CD is perpendicular to AB.
We can use the Pythagorean Theorem to find the radius: (16/2)2 + 102 = R2 , so
R2 = 164. The area is πR2 = 164π square feet .
(16) 1 line
ID: [4031]
No solution is available at this time.
(17) 6 units
ID: [24CB]
9
D3
8
7
6
5
B
4
3
2
1
A
−2
D1
0
−1
0
−1
1
2
3
4
5
6
−2
−3
−4
C
−5
−6
−7
−8
D2
−9
The three given points are labeled A, B, and C. The three possible values of the fourth
point in the parallelogram are labelled D1 , D2 , and D3 , with D1 being the opposite point of
A, D2 the opposite point of B, and D3 the opposite point of C. The parallelogram AD3 BC
has the same perimeter as the parallelogram ABCD2 by symmetry, so we disregard point
D3 .
We will find the perimeter of ABCD2 . To calculate where point D2 is, we notice that
AD2 must be parallel to the vertical segment BC, so the x value of point D2 must be −1.
In addition, the length AD2 must be equal to the length BC, which is 8. Thus, the y value
of point D2 must be −8. So point D2 is at (−1, −8). The vertical segments of
parallelogram ABCD2 have length 8. To find the length of the diagonal segments AB and
CD2 , we
p use the distance formula between points A and B:
AB = (−1 − 2)2 + (0 − 4)2 = 5. Thus, the perimeter of this parallelogram is
8 + 8 + 5 + 5 = 26.
We will find the perimeter of ABD1 C. To calculate where point D1 is, we note that
since figure ABC is symmetric about the x -axis, D1 must lie on the x -axis, so its y value is
0. We also know that the diagonals in a parallelogram bisect each other, so in order for
diagonal AD1 to bisect BC (which crosses the x -axis at x = 2), the x value of D1 must be
5. So point D1 is at (5, 0). In finding the perimeter, we note that all the sides are equal in
length. Since we already found side AB to have length 5, the entire perimeter is 5 · 4 = 20.
Thus, the positive difference between the greatest perimeter and the smallest is
26 − 20 = 6 units.
(18) 98 sq units
ID: [BA422]
A
B
E
D
C
We will use [XY Z] to denote the area of triangle XY Z. Since triangles ADC and BCD
share a base and have the same altitude length to that base, they have the same area.
Since [BCD] = [ADC], we have [BCE] + [CDE] = [ADE] + [CDE], so
[BCE] = [ADE] = 20.
To find the area of triangle [CDE], we note that triangles CDE and ABE are similar,
and the ratio of their sides is DE/BE. Triangles ADE and ABE share an altitude, so
DE/BE = [ADE]/[ABE] = 20/50 = 2/5. Since the ratio of the areas of two similar
triangles is the square of the ratio of their sides, [CDE]/[ABE] = (DE/BE)2 = 4/25, and
[CDE] = (4/25)[ABE] = (4/25)(50) = 8. Thus, the area of trapezoid ABCD is
[ABE] + [ADE] + [BCE] + [CDE] = 50 + 20 + 20 + 8 = 98 .
(19) 5 cubes
ID: [32A1]
No solution is available at this time.
(20) (0, 3)
ID: [AC01]
No solution is available at this time.
(21) 192 cubic feet
ID: [0233]
Each box has volume 43 = 64 cubic feet. Thus, three boxes have volume 64 · 3 = 192
cubic feet.
(22) 9 triangles
ID: [25D]
At least one side must be exactly 5 units long. Let a and b be the other two sides, and
without loss of generality, assume that a ≤ b. If a = 1, then b must be 5. If a = 2, then b
can be 4 or 5. If a = 3, then b can be 3, 4, or 5. If a = 4, then b can be 4 or 5. If a = 5,
then b must be 5. In total, there are 9 non-congruent triangles that meet the conditions.
(23) 6 square units
ID: [0D33]
2
B
1
A
C
0
−2
0
−1
O
1
2
3
−1
−2
−3
D
−4
The quadrilateral implied by the problem is shown above, with vertices labeled. O is the
origin. We will find the desired area by breaking the quadrilateral into two triangles ABC
3
AC·DO
9
= 3·1
= 3·3
and CDA. The area of ABC is AC·BO
2
2 = 2 . The area of CDA is
2
2 = 2.
The total area is 3/2 + 9/2 = 6 square units.
(24) 19 sq units
ID: [BA322]
Define points D E and F as shown. The area of rectangle CDEF is the sum of the areas
of the four triangles BEA, BF C, CDA, and ABC. The areas of the first three triangles
may be found directly using the area formula 12 (base)(height). The area of triangle ABC is
the area of the rectangle minus the areas of the three other triangles:
8 · 6 − 12 · 4 · 3 − 21 · 6 · 5 − 12 · 2 · 8 = 19 .
E
B (0, 6)
F
A (−4, 3)
D
C (2, −2)
(25) 60
ID: [C03]
Let the angles be a, a + d, a + 2d, and a + 3d, from smallest to largest. Note that the
sum of the measures of the smallest and largest angles is equal to the sum of the measures
of the second smallest and second largest angles. This means that the sum of the
measures of the smallest and largest angles is equal to half of the total degrees in the
trapezoid, or 180◦ . Since the largest angle measures 120◦ , the smallest must measure
180◦ − 120◦ = 60◦ .
(26) 72 sq units
ID: [B5212]
Each of the sides of the square is divided into two segments by a vertex of the rectangle.
Call the lengths of these two segments r and s. Also, let C be the foot of the perpendicular
dropped from A to the side containing the point B. Since AC = r + s and BC = |r − s|,
(r + s)2 + (r − s)2 = 122 ,
from the Pythagorean theorem. This simplifies to 2r 2 + 2s 2 = 144, since the terms 2r s
and −2r s sum to 0. The combined area of the four removed triangles is
1 2
1 2
1 2
1 2
2
2
2
2
2 r + 2 s + 2 r + 2 s = r + s . From the equation 2r + 2s = 144, this area is
144/2 = 72 square units.
A
r
s
C
B
(27) 22 percent
ID: [43D5]
The shaded region is 4/9 of the area of square EF GH. Because E, F, G, and H are
midpoints of the sides of square ABCD, the area of EF GH is 1/2 of the area of ABCD. It
follows that the shaded region is (4/9)(1/2) = 2/9 of the area of ABCD. 2/9 = .2, so our
answer is 22 .
(28) 6
ID: [1A02]
The line y = −x + 4 has slope −1. Since line CD is perpendicular to this, it has slope
1
− −1
= 1. Thus,
4−0
= 1 ⇒ m − k = 4.
m−k
Trapezoid ABCD has height 4 and bases BC = m and AD = k + 1. The area of the
trapezoid is thus 12 (4)(m + k + 1). Setting this equal to 34 gives
1
(4)(m + k + 1) = 34 ⇒ m + k = 16.
2
Subtracting m − k = 4 from m + k = 16, we get 2k = 12, or k = 6 .
(29) 3 right angles
ID: [5A2C]
The sum of the interior angles in an octagon will be (8 − 2) · 180 = 1080◦ . We know that
the angles which are not right must have measure less than 180 if the polygon is to be
convex. So, let n equal the number of right angles in the octagon. The average measure of
the remaining angles must be less than 180◦ , which is equivalent to:
1080 − 90n
< 180
8−n
We can simplify this inequality:
1080 − 90n < 1440 − 180n
90n < 360
n<4
So, the greatest possible number of right angles will be 3 .
(30) 156.25 sq miles
ID: [1501]
No solution is available at this time.
Copyright MATHCOUNTS Inc. All rights reserved