Geometry - Level 4

(1) Angle A is the complement of angle B and the supplement of angle C. The

measure of angle C is 106 degrees. How many degrees are in the measure of angle B?

(2) The area of the semicircle in Figure A is half the area of the circle in Figure

B. The area of a square inscribed in the semicircle, as shown, is what fraction of the area

of a square inscribed in the circle? Express your answer as a common fraction.

Figure A Figure B

(3) Let �ABC be an equilateral triangle. How many squares in the same plane

as �ABC share two vertices with the triangle?

(4) Find the distance between the points (1; 1) and (4; 7). Express your answer

in simplest radical form.

(5) A rectangular �eld is half as wide as it is long, and it is completely enclosed

by 54 meters of fencing. What is the number of square meters in the area of the �eld?

(6) In the �gure, point A is the center of the circle, the measure of angle RAS

is 74 degrees, and the measure of angle RTB is 28 degrees. What is the measure of minor

arc BR, in degrees?

TAB

R

S

(7) The Crunchy Peanut Butter Company plans to increase the height of its jars

by 40%. By what percent will the radius of the jar need to be decreased to keep the

volume the same? Express your answer as a decimal to the nearest tenth.

(8) A circular sheet of paper of radius 6 inches is cut into 3 equal sectors, and

each sector is formed into a cone with no overlap. What is the height in inches of the

cone?

(9) The length of one leg of a right triangle is 9 meters. The lengths of the

other two sides are consecutive integer numbers of meters. What is the number of meters

in the perimeter of the triangle?

(10) The giant earthmover used for open-air coal mining has rubber circular tires

11.5 feet in diameter. How many revolutions does each tire make during a six-mile trip?

Express your answer to the nearest whole number.

(11) On the number line shown, AE = 6, AB = 123, BC = 11

4, and DE = 1 1

12.

What is CD?

A B C D E

(12) What is the maximum number of square centimeters in the area of a triangle

whose perimeter is 30 cm? Express your answer in simplest radical form.

(13) How many square inches are in the area of a square inscribed in a circle of

radius 6 inches?

(14) The graph shows six labeled points. How many distinct circles of radius 2

units are in the coordinate plane and pass through exactly two of the labeled points on this

graph?

(−1, 1)

(−1,−1)

(1, 1)

(1,−1)

(3, 1)

(3,−1)

(15) Convex hexagon ABCDEF has exactly two distinct side lengths. Side AB

measures 5 units, and side BC measures 6 units. The perimeter of hexagon ABCDEF is

34 units. How many sides of hexagon ABCDEF have measure 6 units?

(16) If each edge of a cube is increased by 50%, what is the percent increase in

the surface area of the cube?

(17) Two rectangles have integer dimensions, and both have a perimeter of 144

cm. What is the greatest possible di�erence between the areas of two such rectangles?

(18) Richard has a 9-inch by 13.5-inch sheet of cardboard. He wants to create a

closed cube with the largest possible volume. There are no restrictions on the way he can

cut the cardboard or tape it back together, and no cardboard is lost when he makes a cut.

What is the greatest volume possible? Express your answer in cubic inches as a decimal to

the nearest thousandth.

(19) From each vertex of a regular octahedron of edge length 6 units, a

square-based pyramid of height 1 unit will be cut away. How many vertices will the new

solid have?

(20) A rectangle is called \cool" if the number of square units in its area is equal

to twice the number of units in its perimeter. A cool rectangle also must have integer side

lengths. What is the sum of all the di�erent possible areas of cool rectangles?

(21) For what value of a is there a right triangle with sides a+1, 6a, and 6a+1?

(22) Each triangle in this �gure is an isosceles right triangle. The length of BC is

2 units. What is the number of units in the perimeter of quadrilateral ABCD? Express

your answer in simplest radical form.

A B

C

D

(23) The card (the �rst �gure) will be placed on a table with unit cubes stacked

on top of it to create a solid such that the number in each square indicates the number of

cubes to be stacked in the column above the square (the second �gure shows two of the

eight stacks already in place). Including the 8 square units of the bottom face of the solid,

what is the surface area of the entire solid formed once all the cubes are in place?

1 3

5 2

2 3

3 4

(24) A cube of side 3 inches has a cube of side 1 inch cut from each corner. A

cube of side 2 inches is the inserted in each corner. What is the number of square inches in

the surface area of the resulting solid?

(25) Right triangle ABC has one leg of length 6 cm, one leg of length 8 cm and

a right angle at A. A square has one side on the hypotenuse of triangle ABC and a vertex

on each of the two legs of triangle ABC. What is the length of one side of the square, in

cm? Express your answer as a common fraction.

A

B C

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Answer Sheet

Number Answer Problem ID

1 16 degrees 2ABC

2 2/5 A4D5

3 9 squares 42A4

4 3p5 units DBA4

5 162 square meters 53B4

6 81 degrees DD212

7 15.5 percent 25A4

8 4p2 520D1

9 90 meters 5DBC

10 877 revolutions D1C1

11 2 32BB

12 25p3 square centimeters 2BA4

13 72 square inches B3B4

14 22 circles A102

15 4 sides 325

16 125 percent 2AA4

17 1225 square centimeters 4DA4

18 91.125 cubic inches 25C

19 24 vertices 4302

20 236 sq units DDB1

21 10 A2501

22 4 +p2 units 0102

23 68 square units 135

24 198 in2 055D1

25 120/37 cm 15D5

Copyright MATHCOUNTS Inc. All rights reserved

Solutions

(1) 16 degrees ID: [2ABC]

If angle C measures 106 degrees, then angle A measures 180� 106 = 74 degrees.

Therefore, angle B measures 90� 74 = 16 degrees.

(2) 2/5 ID: [A4D5]

Let s be the side length of the square in Figure A.

Because the area of the semicircle in Figure A is half the area of the circle in Figure B,

these two �gures have the same radius, r . In �gure A, if we draw a radius of the semicircle

to a vertex of the inscribed square, we obtain a right triangle whose sides are s=2, s, and r .

The Pythagorean Theorem tells us that r 2 = s2 + s2=4. After some manipulation, we see

that

s =2p5r:

In Figure B, we see that the diameter of the circle makes up a diagonal of the square.

Because the diagonal has length 2r , it follows that the side length of the square is

2r=p2 = r

p2.

To calculate the ratio of the areas, we square the ratio of the sides:(2rp5

rp2

)2

=

(2p10

)2

=4

10=

2

5:

(3) 9 squares ID: [42A4]

No square shares more than two vertices with the equilateral triangle, so we can �nd the

number of squares having two of their vertices at two given points and triple the result.

Given 2 points, 3 squares may be drawn having these points as vertices (see �gure).

Therefore, 9 squares share two vertices with the equilateral triangle.

(4) 3p5 units ID: [DBA4]

The horizontal separation between (1; 1) and (4; 7) is 4� 1 = 3 units. The vertical

separation between the points is 7� 1 = 6 units. Therefore, the segment whose endpoints

are (1,1) and (4,7) is the hypotenuse of a right triangle whose legs measure 3 units and 6

units. By the Pythagorean, the length of this segment isp32 + 62 = 3

p12 + 22 = 3

p5 .

(5) 162 square meters ID: [53B4]

Let the rectangle's witdth be w , then its length is 2w . So its perimeter is

2(w + 2w) = 6w = 54. Thus w = 9, and the rectangle's area is 9(2 � 9) = 162 square

meters.

(6) 81 degrees ID: [DD212]

Let C be the point where line segment AT intersects the circle. The measure of \RTB

half the di�erence of the two arcs it cuts o�:

m\RTB =mRB �mSC

2:

Since mRS = 74�, mSC = 180� � 74� �mRB. Substituting this expression for mSC as

well as 28� for m\RTB, we get

28� =mRB � (180� � 74� �mRB)

2:

Solve to �nd mRB = 81 degrees.

TAB

R

S

C

(7) 15.5 percent ID: [25A4]

Recall that increasing a quantity by p% is equivalent to multiplying it by 1 + p%, and

decreasing a quantity by p% is equivalent to multiplying it by 1� p%. Let r and h be the

original radius and height of the jar of peanut butter. Let k be the factor by which the

radius is decreased, so that the new radius and height of the jar of peanut butter are kr

and 1:4h. The original volume is �r 2h, and the new volume is �(kr)2(1:4h). Setting these

expressions equal we �nd k =√1=1:4; which as a decimal is approximately 0.8451.

Multiplying by 0.8451 is equivalent to a 1� 0:8451 = 15:49% decrease, which to the

nearest tenth of a percent is 15:5% .

(8) 4p2 ID: [520D1]

A circular sheet of paper with radius 6 inches has a circumference of 2�r = 2�(6) = 12�

inches. Since this sheet of paper is cut into 3 equal sectors, each sector must have an arc

length of 1=3 that of the original circle, so each sector has an arc length of 4� inches.

Because of the way that a cone is formed, the original radius of the circle now becomes the

slant height of the cone and the arc length becomes the perimeter of the cone's base. If we

let the radius of the cone base be R, then we know that the circumference 4� must equal

2�R, so 2�R = 4�. Dividing both sides of this equation by 2�, we get that the radius of

the cone base is 2 inches.

We then notice that the radius of the base, the altitude of the cone, and the slant height

of the cone form a right triangle where the slant height is the hypotenuse. Since we know

both the radius and the slant height, we can use the Pythagorean Theorem to �nd the

height h of the cone. We have that 22 + h2 = 62, so h2 = 32. Taking the square root of

both sides, we have that h =p32 = 4

p2 inches.

(9) 90 meters ID: [5DBC]

By the Pythagorean theorem, we have

92 + x2 = (x + 1)2 =)81 + x2 = x2 + 2x + 1 =)

2x = 80 =)x = 40;

where x is the shorter missing side. It follows that the sides of the triangle are 9, 40, and

41 meters, and the perimeter of the triangle is 9 + 40 + 41 = 90 meters.

Note: for any odd integer n, the two integers closest to n2=2 along with n form a

Pythagorean triple.

(10) 877 revolutions ID: [D1C1]

The circumference of the wheels is 11:5� feet. Thus the wheels make one full revolution

every 11:5� feet. On the other hand, the total trip is 6 � 5280 = 31680 feet. Thus the

total number of revolutions made is 3168011:5�

� 876:8. Rounding to the nearest whole number,

our answer is 877 revolutions.

(11) 2 ID: [32BB]

Since AB = 123and BC = 11

4, we have

AC = AB + BC = 12

3+ 1

1

4=

5

3+

5

4=

20

12+

15

12=

35

12:

We have AC + CD +DE = AE = 6, so

CD = AE � AC �DE = 6� 35

12� 13

12= 6� 48

12= 2 :

(12) 25p3 square centimeters ID: [2BA4]

Intuitively, the triangle should be equilateral. The height of an equilateral triangle whose

sides measure (30 cm)=3 = 10 cm is 5p3 cm, since the altitude of an equilateral triangle

splits the triangle into two 30-60-90 triangles (see �gure). So the area of the triangle is12(10 cm)(5

p3 cm) = 25

p3 square centimeters.

5

5√

3

Let's prove the upper bound 25p3 cm2 for the area of a triangle whose perimeter is 30

cm. Recall Heron's formula: a triangle with sides of length a, b, and c has area

Area =√s(s � a)(s � b)(s � c);

where the semiperimeter s is (a + b + c)=2. Recall also that the geometric mean of a set

of positive numbers is less than or equal to the arithmetic mean of the numbers. Therefore,

3

√(s � a)(s � b)(s � c) � (s � a) + (s � b) + (s � c)

3

=3s � (a + b + c)

3

=3s � 2s

3

=s

3:

Since the cubing function is monotonically increasing, we may cube both sides to �nd

(s � a)(s � b)(s � c) �(s3

)3:

Going back to Heron's formula, we have

Area =√s(s � a)(s � b)(s � c) �

√s(s3

)3=

s2p3

9:

Substituting s = 15 cm gives the result.

(13) 72 square inches ID: [B3B4]

The diagonal of this square has the same length as the diameter of the circle, of 12 inches.

Hence the area of the square is 122

2= 72 square inches.

(14) 22 circles ID: [A102]

If two points are less than four units away, then there are exactly two circles of radius 2

containing both of them. If they are more than four units away, then no circles of radius 2

contain both of them. If the points are exactly four units apart, then there is one circle of

radius 2 that contains both of them.

We would also like to know when a circle might contain three of these points. Because

of this array, any three of these points (which are non-collinear) must form a right angle. If

three points on a circle form a right angle, then two of them (the endpoints of the right

angle) must form a diameter. This would imply that those two points are four units apart.

It follows that if two points are fewer than four units apart, then none of the circles that

contain both of them contain any other points on this grid.

We now label the points for clarity:

(−1, 1)

(−1,−1)

(1, 1)

(1,−1)

(3, 1)

(3,−1)

A

D

B

E

C

F

The only pairs of points which are not fewer than four units apart are the following: AF

CD AC DF

The other(6

2

)� 4 = 11 pairs must be fewer than four units apart, producing 22 circles.

The two pairs that are exactly four units apart, AC and DF, lie on a circle that also

contains another one of the six points (E and B, respectively), so neither of these pairs

contribute any circles.

Therefore, the total number of circles is 22 .

(15) 4 sides ID: [325]

If all sides measured 6 units, the total perimeter would be 36. But this is 2 too many units,

so we must change two of these sides to 5 units. Thus 4 sides measure 6 units.

(16) 125 percent ID: [2AA4]

Increasing the edge length of the cube by 50% is equivalent multiplying the edge length by32. So if the original edge length is e, then the new edge length is 3

2e. The original surface

area is 6e2, and the new surface area is 6(32e)2

= 6(94e2)= 9

4(6e2). The surface area has

been multiplied by 94, which is equivalent to a 9

4� 1 = 5

4= 125% increase.

(17) 1225 square centimeters ID: [4DA4]

Let the dimensions of the rectangle be l and w . We are given 2l + 2w = 144, which

implies l + w = 72. Solving for w , we have w = 72� l . The area of the rectangle is

lw = l(72� l). As a function of l , this expression is a parabola whose zeros are at l = 0

and l = 72 (see graph). The y -coordinate of a point on the parabola is maximized when

the x-coordinate is chosen as close to the x-coordinate of the vertex as possible. The

x-coordinate of the vertex is halfway between the zeros at x = (0 + 72)=2 = 36, so the

maximum area is (36)(36) = 1296 square units. Similarly, to minimize the area we choose

the length to be as far from 36 as possible. The resulting dimensions are 1 unit and 71

units, so the minimum area is 71 square units. The di�erence between 1296 square units

and 71 square units is 1225 square units.

Area

l

(18) 91.125 cubic inches ID: [25C]

The cube will have maximal volume if all of the cardboard is used. Richard has

9 � 13:5 = 121:5 square inches of cardboard available. Now suppose that he makes a cube

of side length l . Then its surface area will be 6l2 = 121:5. Solving, l = 4:5. Thus the

maximal volume of the cube is l3 = 91:125 cubic inches.

(19) 24 vertices ID: [4302]

The octahedron currently has 6 vertices. Each removal of a square-based pyramid removes

one vertex and adds 4 more; this occurs 6 times. Hence we end up with 4 �6 = 24 vertices.

(20) 236 sq units ID: [DDB1]

Let the side lengths of the rectangle be a and b: It follows that ab = 4(a + b): Expanding

and moving all the terms to the left hand side, ab � 4a � 4b = 0: Adding 16 to both sides

allows us to factor:

a(b � 4)� 4(b � 4) = (a � 4)(b � 4) = 16:

From this point, the pairs (a; b) that provide di�erent areas are (5; 20); (6; 12); and

(8; 8); and the sum of the possible areas is 236 :

(21) 10 ID: [A2501]

Since 6aAB, as in the diagram. From the given information, we know that AC = 8,

BC = 10, and DC = 8� x . We can �nd that AG = AB � AC=BC = 4:8. Also, because of

all the parallel and perpendicular lines, all of the triangles in the diagram are similar. We also

know that 4DFC � 4AGC since the corresponding sides of these triangles are parallel.

From similar triangles AED and ABC, we �nd that s=10 = x=8. From similar triangles

DFC and AGC, we have s=4:8 = (8� x)=8. Summing these two equations, we have

s

10+

s

4:8=

x

8+

8� x

814:8s

48= 1:

Solving for s, we �nd that s =120

37.

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