Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2013 Grade 7 Geometry
Problem
Solution
1) Write the letters M, T, and H on Figure 2 so that when the figure is folded into a box, it will spell MATH around the sides of the box. Figure 1 shows the correct placement of the letters.
  1. Well, the first 2 letters are easy:




  2. The problem is where does the "H" go and what is it's orientation? Here's what you do:
  3. Fold the square below the "T" up under the "T"
  4. You should note what square, after this fold, is the only candidate to hold the "H". Clearly it is neither the square above or below the "T". They get folded above and below the T. It is the bottom right one.
  5. If you visualize folding the bottom right one, you should come up with this orientation
2) The figure shows a drawing of a pyramid tower climbing toy located in a park. It is constructed with large congruent wooden cubes solidly nailed together. The volume of one wooden cube is 1000 cu. inch. What is the visible surface area in square inches? This includes the exposed north, south, east and west faces and looking top down.
 The length of the side of one of the cubes is:
    l = 10001⁄3 = 10 inches,
    so, the area of one face is 10x10 = 100 sq. in.
  • Looking down from the top there are 5x5 surfaces exposed
  • The only other exposed surfaces are the sides and there are 4 of each layer's side length in cubes:
    Sides = 4 x (5 + 4 + 3 + 2 + 1) = 15 x 4 = 60 side surfaces
  • Total exposed surfaces = 60 + 25 =
    85 surfaces x 100 sq. in. = 8500 sq. in

Problem
Solution
3) In the diagram of nested squares each next square is inscribed in the previous square with its vertices bisecting the sides of the previous square. What fraction of the area of the largest outside square is shaded?
  1. Each inscribed square is 1⁄2 the area of the square it is inscribed in.
  2. The innermost square is 1⁄2 x 1⁄2 x 1⁄2 = 1⁄8 of the area of the outermost square.
  3. The 2 smaller shaded triangles, together, are 1⁄2 the area of the innermost square =
    1⁄2 x 1⁄8 = 1⁄16 the area of the outermost square
  4. The bottom shaded triangle is 1⁄4 the area of the second inscribed square = 1⁄4 x 1⁄2 x 1⁄2 = 1⁄16 the area of the outer square
  5. Together, the 2 smaller triangles + the bottom triangle =
    1⁄16 + 1⁄16 = 1⁄8 of the area of the outermost square.
4) A dog food company wants a bigger can size and a new look. It considers doubling the diameter or doubling the height. What is the ratio of the volume of the doubled diameter can to the volume of the doubled height can?
(Volume of a cylinder is r2 h )
  1. Volume of the doubled diameter can =
    (2r)2 h
  2. Volume of the doubled height can =
    r2 2h
  3. Their ratio =
    (2r)2 h = 4r2 = 2
    r2 2h       2r2    1
  4. Their ratio is 2/1
5) The triangle with vertices (0,0), (3,0) and (0,4) is rotated by 90 degrees counterclockwise around the origin, the image is rotated by another 90 degrees counterclockwise around the origin, and that image is rotated by another 90 degrees counterclockwise around the origin to produce four triangles. Create a single figure by shading all four triangles. How many points that have integer values for both the x and y coordinate are in the interior of the shaded figure?
The figure to the right contains those 4 triangles. Just carefully count the integer x and y points that are inside the combined figure.

There are 21 of them.