Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2008 Grade 6 Geometry

Problem
Solution
1) Toleen's math teacher makes a model of a solid rectangular prism out of white one centimeter cubes to explore surface area and volume. She paints the entire outside surface red. If the dimensions of the prism are 3 cm by 4 cm by 6 cm, how many of the cubes have exactly two sides painted red? 1. The cubes which have 2 sides painted are the edges that are not on the corners.
2. There are 4   3 cm edges, 4   4 cm edges and 4   6 cm edges.
3. Subtract 2 for each edge (for the corners) and then multiply by the above numbers to get the total number of edge cubes that are not corners =
    (3 - 2) x 4 + (4 - 2) x 4 + (6 - 2) x 4 =
    28 cubes that have 2 sides painted.
2) In right triangle AMC (see diagram) triangles ANP and PLC are similar. MNPL is a square. Segment MC has length 6 units and segment LC has length 4 units. What is the area of triangle ANP?
1. Write the lengths of MC and LC on the diagram.
2. Compute the length of a side of the square (ML and NP) = 2 units and write those lengths on the diagram.
3. Since the triangles are similar, the scale factor of the triangles is NP/LC = 2/4 = 1/2
4. Apply this scale factor to NP to get the length of AN =
    (1/2) x 2 = 1 unit
5. Compute the area of ANP using the lengths of NP and AN =
    (1 x 2)/2 = 1 sq. unit
3) The points of an isosceles triangle are A=(2,3), B=(3,1), and C=(1,1). If you rotate the triangle 180 degrees about the origin and then reflect the triangle across the x-axis, what are the new coordinates for A, B, and C? 1. If you rotate the triangle 180 degrees around the origin that puts it in quadrant III upside-down. Perform that and draw in the new points in that quadrant.
2. If you reflect the triangle across the x-axis that puts it into quadrant II, right-side up again. Draw in those points.
3. The new coordinates are
    A: (-2,3)
    B: (-3,1)
    C: (-1,1)
.
See figure to the left. Notice that the positions of B and C have reversed!

Problem
Solution
4) A dart board has a bull's-eye surrounded by 5 rings. The bull's-eye and the second and fourth rings are shaded (see diagram). If the bull's-eye has a radius of 2 cm, and each ring has a width of 2 cm, what is the area of all the shaded regions combined?
Express your answer in terms of .
1. The radius of the outer shaded circle is 10 cm.
2. The radius of the inner edge of that circle is 8 cm.
3. Compute the area of this ring using these two radii
    (keep the answer in terms of ). =
    (102 - 82) = 36
4. The radius of the inner shaded circle is 6 cm.
5. The radius of the inner edge of that circle is 4 cm.
6. Compute the area of this ring using these two radii
    (keep the answer in terms of ). =
    (62 - 42) = 20
7. Now, add the area of the bulls-eye to each of these areas to get the total =
    (36 + 20 + 4) = 60
5) When Andreyev wants to finish building the frame of the front of his club house he notices that things aren't quite right. The right post has a bit of a lean inwards and that messes with the top beam. What is the angle between the left post and the top beam? See diagram.
There are 2 ways to approach this problem:
Method 1:
1. Compute the size of the angle adjacent to the 92.12o angle that is supplementary with it = 87.88o
2. Since the angles of a rectangle all add up to 360o, the angle ? = 92.12o.
Method 2:
You don't have to compute anything! Because 2 of the angles of the club house are 90o, that makes a special relationship between the ? angle and the unlabeled angle that is the same relationship the unlabeled angle has with the 92.12o angle.
They are both supplementary with it!
Therefore, angle ? = 92.12o.