SW-MO-ARML Practice Problems
October 18, 2009
ARML Geometry Problems
For each of the following, 10 points is earned with a correct answer with no assistance
from any coach. With a hint sheet, the correct answer is worth 6 points, and with a
directional hint from a coach, worth 4 points. If the team collectively pulls at least 70
points, there will be a reward.
FCD with m
1. 2000 I-2. BCDE is a square. ABC
If AF = 2000, compute the area of square BCDE.
A 120 and AB = AC.
A
C
B
F
E
D
2. 2001 T-6. Compute the ratio of the volume of a sphere to the volume of the
largest regular octahedron that will fit inside it.
3. 2002 I-3. The legs of isosceles trapezoid ABCD are diameters of tangent circles
O and P. If AB = 90 and CD = 1000, compute the height of the trapezoid. NO
CALCULATOR ALLOWED!
A
O
D
B
P
C
4. 2002 I-2. Starting with the outermost ring, a farmer cuts a circular field in rings
of uniform width. As an example, the diagram shows one and a half trips
around the field. If the width of each ring is 5 feet and if after the farmer has
made eight and a half trips around the field, he has cut half the field, compute
the radius of the field in feet.
5. 2004 I-2. In the unit circle with center O, arcAC arcCE arcEA , chords
AD , CF , and EB are equidistant from O, and the four inscribed circles are
congruent. Compute the area of any one of the four small circles.
A
F
B
O
C
E
D
6. 2005 T-1. ABCD is a convex quadrilateral with AC = BC = 10 and
AB + 1 = AD = CD = 13. Compute the area of ABCD.
7. 2006 T-1. A cylinder of height 10 cm and radius 6 cm is turning at 1 revolution
per minute. A paintbrush 4 cm wide starts at the top as shown and slowly
moves vertically down the cylinder at 1 cm/minute. Compute the time in
minutes it takes for the paintbrush to paint three-fifths of the lateral surface area
of the cylinder.
8. 2006 T-3. In
CBA , CD and AE are medians, FC || AB ,
FEH , CGD, and AGE . The area of FCGE = 7 and the area of EGDH is 11.
Compute the area of CBA .
C
F
E
G
B
H
D
A
9. 2006 I-2. Two 3 by 4 rectangles overlap in such a way that their sides are
perpendicular. If the area and the perimeter of the shaded region are 22 and 20
respectively, compute AB.
B
A
10. 2007 T-3. In rectangle ABCD, M is the midpoint of AB , EC = 10 and EM = 4.
Compute the area of ABCD.
M
A
B
4
E
10
D
C
HINTS for Practice Problems 10-18-09
1. The base angles of an isosceles triangle are congruent. Determine angle
measures in the triangles shown and in triangle ACF. Then use your special
right triangles (30-60-90 and 45-45-90) to get the AC and then BC.
2. View just the top half of the figure – a square pyramid inscribed in a
4
r 3 and for a pyramid:
hemisphere. Then, for volumes: for a sphere: V
3
1
V
Bh .
3
3. The median of the trapezoid goes is OP , which is the length of a diameter in
either circle. Find the length of the median and you have the length of a leg of
the isosceles triangle. Drop an altitude from A or from B and use the
Pythagorean Theorem to get the height. Hint for calculation. View the
arithmetic as the difference of two squares and think about how that would
factor!!!!
4. Let r be the radius of the smaller circle left uncut. Then write an equation to
represent the amount left uncut in the middle (add up two semicircles, one with
radius r and one with radius r + 5) multiplied by 2 which is set equal to the
total area of the large circular field.
5. The triangle formed by the three tangents of the inner circle is an equilateral
triangle – how do we know? Find the length of the radius of this equilateral
triangle in terms of r, the radius of the smaller circles. Draw a line segment
from O to a point of tangency of one of the outer circles with the larger circle.
Then use the information found to determine the length of that segment, which
is the radius of the larger circle. Since the larger circle is a unit circle, you can
now find the radius of the smaller circles.
6. Draw the figure. ADC and ABC are isosceles triangles, so draw the altitude
to the base for each and you will see Pythagorean triples!
7. Think of cutting the lateral surface and laying it out as a rectangle. In one
minute, the paintbrush would sweep out a parallelogram across that rectangle.
It is now 1 cm lower as it continues around for the second pass. In the second
pass, there will be overlap, but there will be a new section covered that extends
the first section 1 cm lower. This would continue for each pass around the
cylinder. Now use the circumference of the cylinder as the height of the
parallelogram and then the height painted can be the base in order to compute
the area of the parallelogram.
HBE , and hence their areas are equal. Then, the area of
8. By ASA, FCE
CBD will be equal to the sum of the areas given for the quadrilaterals. Since
D is a midpoint, the area of CBD is ½ the area of CBA .
9. Look at just the rectangle that is the overlap region. Its area is 2 and its
perimeter is 8, based on the amount left out from the total for the two
rectangles. Let the sides be k and l and write equations for perimeter and area.
Then simplify the equation for the perimeter and then square it. With a good
substitution, you get k 2 l 2 without having to find either length!
10. First, use similar triangles to get the lengths AE and DE. Then, let AD = k and
AM = l. Use the Pythagorean Theorem to set up two equations to find k and l.
ANSWERS for Practice Problems 10-18-09
1.
6,000,000
2.
π or π:1
3.
300
4.
145
5.
π/25
6.
108
7.
3
8.
36
9.
2√3
10. 18√39