Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2008 Grade 8 Geometry
Problem
Solution
Call the diameter of the larges circle AB. 1) A garden is laid out as a set of four half circles, see diagram. The ratio of the diameters of the three smaller circles is 2 to 1 to 3. What is the ratio of the distance for walking from A to B along the outer edge of the single large circle compared to the distance for walking from A to B along the outer edges of the three smaller circles?
  1. The distance around the larger half circle is:
    AB/2
  2. distances around each oof the other half circles, starting with the smallest, is:
    • Smallest: AB/12
    • Next larger: AB/6
    • Largest: AB/4
  3. The ratio is
          AB/2      =
    AB/12 + AB/6 + AB/4

  4. AB/2      =
    AB (1/12 + 1/6 + 1/4)
  5. 1⁄2 ÷ 1⁄2 = 1
2) The folding pattern below folds to a pyramid with base a square with side length 2 units and isosceles triangles for sides with height 3. What is the height of the resulting pyramid?
Express your answer to the nearest hundredths.
  1. The height of the pyramid is the height of a right triangle whose hypotenuse is 3 units and whose base is 1 unit (half of the base width).
  2. Using the pythagorean theorem:

    h = √ 3 2 - 1 2  = √8 = 2.83 units.

Problem
Solution
3) In the big 3 x 3 square, what is the ratio of the shaded four triangles to the unshaded octagon in the middle of the square? The octagon is not regular but is a truncated square. All line segments are intended to be straight.

Please note: The unshaded figure inside the 3x3 square has cutoff corners making it an octagon (they are small cutoffs). Also, all the small squares are of the same size.
  1. You have to look close, but what looks like a white square in the middle of the 3x3 larger square actually has litle sides that cut off the corners, making it an octagon.
  2. Consider the squares in the lower right corner. The shaded triangle that is inside that 2x2 square is 1⁄4 of the area of the 2x2 square.
  3. At this point we must assume the small squares of that 2x2 square are the same size as the squares in the larger 3x3 square.
  4. Therefore the 4 shaded triangles are 2x2=4 squares in area
  5. The larger 3x3 square is 3x3=9 squares in area
  6. Therefore the unshaded octagon is 9 - 4 squares in area = 5 squares.
  7. The ratio of shaded areas to the unshaded areas is 4/5
4) The line given by the equation y = 3x - 3 is translated 3 units up and then reflected across the line y = 0. What is the equation of the new line?
  1. To translate an equation up 3 units (in y) is to add 3 units to y: y = 3x
  2. To reflect an equation across the y axis (y=0) is to negate the x value: y = -3x
5) A circular cone can be created by joining the two straight edges of the shown section of a circle with central angle 135 degrees and radius 5 centimeter. What is the diameter of the created cone? (The diameter of a cone is the diameter of the circular opening where it is the widest.)
  1. The length of the circle section is
    10 x (135/360) = 11.78 cm.
  2. When you make this into a cone, this length becomes the circumference of the base, so the diameter is:
    11.78 cm / = 3.75 cm