Geometry One-a-Days, #2
Name:
All work must be shown on these problems, which are done on separate sheets of
paper, with answers in boxes. The levels of difficulty:
Easy, worth 1 point: 2, 5, 6, 9, 12, 13, 15, 19, 20, 24, 25
Medium, worth 2 points: 1, 3, 11, 14, 16, 17, 18, 23, 26
Hard, worth 3 points: 4, 7, 8, 10, 21, 22
Points
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
27
29
31
33
35
Grade
60
62
64
66
68
70
72
74
76
78
80
82
84
86
88
90
92
94
96
98
100
Problems:
1
The Pythagorean Club of a local high school designed a membership pin made up
of circles, a square, and an equilateral triangle. If the diameter of the larger circle is
three inches, what is the length of a side of the equilateral triangle?
2
Determine the measure (in degrees) of an angle whose complement is four-ninths its supplement.
3
If RS is parallel to AB and if the lengths are as labeled in the figure, find x.
4
A pentagon made up of equilateral triangle ABC with side length 2 on top of square BCDE is inside a circle
passing through points A, D, and E. Find the radius of the circle.
5
A square piece of paper is folded in half vertically. If the resulting figure has a perimeter of 12 cm, what was
the area of the original square?
6
What is the length of the side of a square if its area and perimeter are equal?
7
If GRAW is a square of side a and ΔGMR is equilateral, what is the area of ΔRCA?
8
9
In the figure, ABCD is a square and E, F, G, and H are midpoints of the sides. By
what number must the area of square A´B´C´D´ be multiplied to obtain the area of
square ABCD?
The sides of a triangle are 15, 20, 25. Find the length of the shortest altitude.
10
A square is divided into three pieces of equal area by using two parallel cuts,
as shown. The distance between the parallel lines is 10 inches. What is the
area of the square?
11
A solid rectangular block is formed by gluing together n congruent 1-cm cubes face to face. When the block
is viewed so that three of its faces are visible, exactly 231 of the 1-cm cubes cannot be seen. Find the smallest
possible value of n.
12
Determine the area of the rectangle of smallest area containing the parallelogram given, assuming that
A⁢B=6, B⁢C=4, and m∠A⁢B⁢C=60°
13
Two opposite sides of a square are increased by 10 cm. The other two opposite sides of the square are
decreased by 10 cm. Does either the area or the perimeter stay the same?
14
On hypotenuse A⁢B of right triangle A⁢B⁢C, D is the point for which C⁢B=B⁢D. If m∠A⁢B⁢C=40°, find
the measure of ∠A⁢C⁢D.
15
In the diagram, the tangents to the two circles intersect at 90 degree angles, as shown. If the radius of the
smaller circle is 2 and if the radius of the larger circle is 5, what is the distance between the centers of the two
circles?
16
The figure is composed of twenty-five unit squares. What is the area of the unshaded region?
17
If a wheel completes exactly 5 revolutions while rolling 80π feet, what is the diameter of the wheel?
18
What is the area of an equilateral triangle inscribed in a circle whose circumference is 6π?
19
A man walks 3 miles east, then 3 miles north, and then 2 miles northeast. How far is he from his starting
point?
20
Triangle PQR is equilateral with QR = 30 units. A is the foot of the perpendicular from Q to PR¯ and B is the
midpoint of QA¯. What is the length of PB?
21
A rectangle is called a silver rectangle if, when you remove two squares of side length equal to the height of
the rectangle, the remaining rectangle is similar to the original rectangle. For instance, B⁢C⁢F⁢E is similar to
A⁢B⁢C⁢D in the figure.
Find the silver ratio, which is the ratio of width to height, of a silver rectangle.
22
In a rectangular room, 30'×12'×12', a spider is on the middle of an end wall, one foot from the ceiling. A fly is
on the end wall of the far side, in the middle and one foot above the floor, too frightened to move. What is the
shortest distance that the spider must travel to get to the fly? (Hint: the distance is less than 42'.)
23
A Koch snowflake begins with an equilateral triangle with a side of length 1 (stage 0). Trisect each side and
attach a smaller equilateral triangle to the center of each side (stage 1). Repeat for each triangle to obtain
stage 2. Find the perimeter and area of the stage-2 snowflake.
24
Of the following triangles given by the lengths of their sides, which one has the greatest area?
(a) 5, 12, 12
(b) 5, 12, 13
(c) 5, 12, 14
(d) 5, 12, 15
(e) 5, 12, 16
25
Find a Pythagorean triple where each integer is odd.
26
In the right triangle shown, MB + MA = BC + AC. If BC = 8 and AC = 10, compute MB.
Solutions:
1
(36)/4. Let x represent the length of one side of the square. Then 2x is the length of the diameter of the large circle,
and 2x = 3 implies that x = 3/2 = (AB). Therefore,
12(AB)=12(32)=(AC)=322.
Since (AC) is equal to 2/3 (AD),
(AD) = 942.
An equilateral triangle whose altitude has measure 9/(42) has side lengths of (36)/4.
2
18. If x equals the measure of the angle in degrees, then the stated relationship can be represented by the equation 90
− x = (4/9)(180 − x), yielding x = 18.
3
x = 4. Since RS is parallel to AB, triangles ABC and RSC are similar. Therefore,
x+2x−1=3x−22x−3,
so 2x2 + x – 6 = 3x2 – 5x + 2, which implies that x2 – 6x + 8 = 0. Therefore, (x – 4)(x – 2) = 0, so x is either 2 or 4.
However, x must be greater than 2 for RS and AB to be parallel, so x = 4.
4
2.
Consider the equilateral triangle EDF with F inside the square. Since the measure of angle ABE is 150 degrees, it
follows that ABEF is a parallelogram with each side of equal length (i.e., a rhombus). Since F is equidistant from A,
D, and E, it is the center of the circle, and the radius is 2.
5
16
Let the side of the square piece of paper be s. Then
s+s+s2+s2=12,3s=12,s=4;
and the area of the original square was 4 × 4, or 16, square centimeters.
6
4. By letting s be the side of such a square, we have
s2 = 4s,
s2 − 4s = 0,
s(s − 4) = 0,
s = 0, 4.
Clearly, 4 is the solution that we seek.
7
a2(3−1)4
Construct CL¯⊥RA¯, let LA = x, and let RL = a – x. Since ΔALC is a 45-45-90 triangle, CL = x. Also, since ΔRCL is
a 30-60-90 triangle, we have RL = x3, so x3 = a – x and
x=a3+1.
We then conclude that the area of ΔRCA is
12a(a3+1)=a22(3+1)
=a2(3−1)4.
8
5. Let the square have side 2a. Let x=DD′ and y=C′F.
Then, since △GDD′∼△FDC,
2a5a=xa,x=2a5;
and since △GDD′≅△FCC′,
2a5a=a2−y2a,y=a5.
9
12. First notice that this triangle is a right triangle, so two of the altitudes are the legs, whose lengths are 15 and 20.
The third altitude, whose length is x, is the one drawn to the hypotenuse. The area of the triangle is (1/2)(15)(20) =
150. Using 25 as the base and x as the altitude, we have (1/2)(25)(x) = 150, hence
x=30025=12.
Source: vol. 96, no. 6, Sept. 16, 2003
10
1300 square units. Let x be the length of a side of the square, y the shorter side of the parallelogram, and z the longer
side of the parallelogram, as labeled in the diagram. The area of a parallelogram equals height times base, and the
area of this parallelogram is one-third the area of the square. Thus, xy = (1/3)x² and y = x/3. Now consider the right
triangle with z as its hypotenuse. By the Pythagorean theorem,
z2=x2+(x−y)2=x2+49x2,
so
z=133x.
Finally, reconsider the area of the parallelogram, with height = 10 and base =
z=133x.
Thus,
10·133x=13x2,
and, since x ≠ 0, we find
x=1013,
and the required area is 1300 square units.
11
384. Let the dimensions of the block be p cm × q cm × r cm. The invisible cubes form a rectangular solid whose
dimensions are p – 1, q – 1, and r – 1. Thus (p – 1)⋅ (q – 1)⋅ (r – 1) = 231. There are only five ways to write 231 as a
product of three positive integers:
231 = 3⋅ 7⋅ 11 = 1⋅ 3⋅ 77
= 1⋅ 7⋅ 33 = 1⋅ 11⋅ 21
= 1⋅ 1⋅ 231
The corresponding blocks are 4 × 8 × 12, 2 × 4 × 78, 2 × 8 × 34, 2 × 12 × 22, and 2 × 2 × 232. Their volumes are
384, 624, 544, 528, and 928, respectively. Thus the smallest possible value of n is 384.
12
163. The required rectangle is either Z⁢D⁢Y⁢B or S⁢B⁢T⁢D, as shown in the two figures.
In the figure on the left, Z is the foot of the altitude from B to A⁢D↔, and Y is the foot of the perpendicular from D
to B⁢C↔. The height of rectangle Z⁢D⁢Y⁢B is the length of altitude B⁢Z¯, which is
632=33,
whereas the width of the rectangle is
B⁢Y=B⁢C+C⁢Y
=4+3
=7.
The area of rectangle Z⁢D⁢Y⁢B is therefore 213. Similarly, the area of rectangle S⁢B⁢T⁢D is
(6+2)23=163,
which is therefore the area of the smallest rectangle containing the given parallelogram.
13
The perimeter stays the same; the area decreases by 100 cm2. Suppose that the side length of the original square is a
Then the area of the original square is a2 cm2. The new figure will be a rectangle with side lengths (a + 10) cm and (a
− 10) cm and with an area of (a + 10) • (a − 10) cm2 = (a2 − 100) cm2. Thus, the area decreases by 100 cm2. The
perimeter of the original square is 4a cm, while the perimeter of the rectangle is 2 • (a + 10) + 2 • (a − 10) = 2a + 20
+ 2a − 20 = 4a cm. So the perimeter remains unchanged.
14
20°. Since ΔB⁢C⁢D is isosceles and m∠C⁢B⁢D=40°, we have
m∠B⁢C⁢D=12(180−40)
=70
Because ∠A⁢C⁢D and ∠B⁢C⁢D are complementary,
m∠A⁢C⁢D=90−70=20°.
15
72. The distance from the center of the smaller circle to the point of intersection is 22. The distance from the center
of the larger circle to the point of intersection is 52. Thus, the total distance is 72.
16
20 square units. The area of the figure is 25 square units. The shaded region consists of four triangles, each with base
1, two of whose heights are 2 and two of whose heights are 3. The sum of the area of these triangles is (1/2)(2 + 2 + 3
+ 3), or 5. So the area of the unshaded region is 25 – 5, or 20, square units.
17
16 feet.
The circumference of the wheel, πd, multiplied by 5 must be equal to 80π feet. So 5πd = 80π. Thus, d = 16 feet.
18
2734
We let C denote the center of the circle, and we let X and Y be a vertex and the midpoint of the side opposite the
vertex, respectively. Since the radius of the circle is 6π/2π, we have XC = 3. Also, since the center of the circle is the
circumcenter of the inscribed equilateral triangle, we know that XC is two-thirds of XY, so XY = 4.5. If a is the length
of each side of the triangle, then the Pythagorean theorem gives 4.5² + (a/2)² = a² and the area of the triangle is
.5(4.5)a. Solving for a in the Pythagorean equation, we get a = 27 = 33. Therefore, the triangle's area is
.5(4.5)33=6.753
=2734.
19
2 + 32. As indicated in the first figure, the man is walking along a path from point A to point D, followed by point C
and point B. The distance we are looking for is represented by AB. As shown in the second figure, ΔABE and ΔCBF
are isosceles right triangles. Since BC = 2,
CF = 1/2.
DE = CF, so
DE = 1/2.
Then
AE = 3 + 1/2.
Since AE = BE,
BE = 3 + 1/2
also. Therefore,
AB = 2AE = 2(3 + 1/2) = 2 + 32.
20
1527≈19.84.
Since triangle APQ is a 30-60-90° triangle, AP = 15, and
QA=153
so
BA=1532.
Using the Pythagorean theorem, x2=152+(1532)2→x=1527≈19.84.
21
1+2. Let w and h denote the width and height, respectively, of a silver rectangle, using the figure below as a guide.
Since A⁢B⁢C⁢D~B⁢C⁢F⁢E,
hw=wh−2w,
so h2−2⁢h⁢w=w2, and therefore h2−2⁢h⁢w−w2=0. Dividing both sides of the equation by w2 gives
(hw)2−2hw−1=0,
so h/w is a solution to the equation x2−2x−1=0. By the quadratic formula, h/w is either
2+4+42=1+2
or
2−4+42=1−2.
Since the ratio must be greater than 0, the only possibility for the silver ratio is 1+2.
Extensions
× A bronze rectangle is one such that, if three squares of side length equal to the height of the rectangle are removed,
the remaining rectangle is similar to the original. What is the bronze ratio?
× If you call the gold ratio the first metal ratio and call the silver ratio the second metal ratio, then the bronze ratio is
the third metal ratio. What is the nth metal ratio?
22
40 feet. The walls, floor, and ceiling of the room can be represented as a plane figure:
As shown, the shortest distance between the spider's location and the fly's location is a straight line that is the
hypotenuse of a right triangle with legs of 32 feet and 24 feet. From the Pythagorean theorem, the distance is equal to
(32 ft.)2+(24 ft.)2, or 40 feet.
23
Perimeter = 16/3; area = 10327.
The original equilateral triangle (stage 0) had side s = 1 and perimeter 3s = 3. In the stage 1 snowflake, each side is
replaced with 4 segments, each one-third the length of the original: length = 4(1/3) = 4/3. So the perimeter is 3(4/3).
In the stage 2 snowflake, each of the 4 × 3 new sides of length 1/3 is replaced with 4 segments, each of length onethird of one-third:
4·3·13·43=3·(43)2=163.
The area of the original triangle is
1234.
The three added triangles in stage 1 are each (1/3)2 = 1/9 the area of the original, so we add
(39·34)+34.
Finally, we add the areas of 12 equilateral triangles, each (1/9)2 the area of the original:
(12·(19)2·34)+(39·34)+34=(3·492·34)+(39·34)+34.
We see the pattern emerge. The expression simplifies to
10327.
24
(b) 5, 12, 13.
Note that 52 + 122 = 132. With two sides of length 5 and 12, the area is maximum when there is a right angle between
them.
25
There is none. If each leg is an odd integer, their squares are odd integers and the sum of the squares would be even.
Source: vol. 80, no. 8, Nov. 27, 1987
26
3 1/13. We know that