Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2011 Grade 7 Geometry
Problem
Solution
1) In the diagram below, EFLM is a square. B is the midpoint of segment AC and I is the midpoint of segment JH. All interior angles are right angles. Segments AB and JI both have length 8 units and segments BE, EF, and FI all have the same length. The total perimeter is 128 units. What is the length of KL?
  1. There are 3 different segment lengths:
    1. The square side length, ML: 6 of these
    2. The shorter side, AB = 8: 6 of these
    3. The KL length: 2 of these
  2. The equation for the perimeter is
    6x8 +6ML + 2KL = 128, so
    6ML + 2KL = 80
    3ML + KL = 40
    KL = 40 - 3ML
  3. Now let's try some values for ML:
    • ML =10: KL = 40 - 30 = 10, which cannot be since it is less than ML
    • ML =11: KL = 40 - 33 = 7,
    • ML = 12 KL = 40 - 3x12 = 40 -36 = 4
    • ML = 13 KL = 40 - 39 = 1
    • ML = 14 KL = 40 - 3x14 = -2, which cannot be
  4. So, ML must be 12 or 13 to make KL a valid value.
    If ML is 12, then KL is 4.
    If ML is 13, then KL is 1.
    These are both valid solutions (given the problem statement), but the more reasonable one (looking at the diagram) is
    KL = 4 and ML is 12.

Problem
Solution
2) Jerry wants to recreate the drawing at right using a larger circle. A is the center of the circle and segment TC is congruent to segment TB. What is the fewest number of angles he needs to measure and still be able to compute the measure of all the angles in the drawing?
 "Congruent", in this case, means the segments TC and TB are the same length.
  1. If you know angle TAC then you also know angle TAB. They are the same angle because they form isosceles triangles.
  2. Once you know TAC, you can know both ATC and ACT because they form an isosceles triangle with angle TAC.
  3. You also know angles ATB and ABT because they are the same as angles ATC and ACT.
  4. The only other angle at the center of the circle, A, is angle CAB which can be computed because you know angles TAC and TAB, which, when added to CAB equal 360 degrees.
  5. We're getting close!
  6. Once you know angle CAB, you can know both angles ACB and ABC because they, too, make an isosceles triangle with angle CAB. That is All the angles of the figure so, the answer is
    1 measurement.
    All you need to know is angle TAC or TAB and all the other angles can be derived from them.
3) Here is a quick sketch that a student made of a diagram she saw in a book. The student was in a hurry and didn't have time to use a protractor while making the sketch but the student was sure the line segments were straight, the two right triangles are similar, and the labeled angle in the smaller triangle on the left was 32 degrees. What was the measurement of the angle that the student saw in the book, in degrees, of the obtuse angle just to the right of the large triangle?
  1. Angle A is 180 - 32 = 148 degrees because they are the 2 angles of a right triangle.
  2. Angle B is the same measure as angle A because they are vertical with each other.
  3. Angle C is 32 degrees because it, along with B, makes a similar triangle with the one to the left.
  4. Angle D is the same measure as angle C because they, too, are similar. Angle D = 32 degrees.
  5. Finally, angle ? and angle D add to 180 degrees, so it is
    180 - 32 = 148 degrees

Problem
Solution
4) The figure represents a net that you cut out along the solid outside lines and then fold along the interior dotted lines. The length of the side of the square is 5 units. The isosceles triangles have height h. What is the smallest number h must be greater than to guarantee the net can be folded to make a pyramid?
  1. The figure to the right is a side view of the net after it has been folded into a pyramid.
  2. From this figure, if you imagine changing the value of h, the pyramid gets either steeper (if you increase h) or flatter (if you decrease h).
  3. If you decrease the value of h below half the base width then the sides do not close to form a pyramid, therefore the smallest number h must be greater than is 2.5 units
5) Two isosceles triangles are inscribed in a circle so that they share a common base, see example in diagram. Triangle HLJ contains the center of the circle, P, in its interior. Of the angles labeled 1, 2, 3, and 4 how many must be acute?
  1. Imagine point H moving such that HJ goes through the center, P
  2. With this geometry, all of the angles of triangles HLJ and HKJ must be acute.
  3. Moving H to the left, such that triangle HLJ contains P also requires that all the angles of HLJ remain acute.
  4. The only possible non-acute (obtuse) angle is JKH which, as H moves closer to it, is greater than 90 degrees and gets larger, approaching 180 degrees.
  5. Therefore, 3 of the angles must be acute
    These are angles 2,3,and 4.