Composing & Solving Maths Problems Chris Harman 27 years on AMC Problems Committee Well and truly retired to Coolum, 10 years ago from USQ harmanchris@gmail.com • Australian Mathematics Competition (AMC) • Turning a concrete context into a smart maths problem • Turning a smart maths problem into a concrete context • Morphing a good idea into another, and another • Live problem solvers - you the audience
AMC J: Years7,8 I: Years9,10 S: Years11,12 30 Questions - Multiple Choice
Warmup Problem AMC 1990 Question: J19, I13 A normal duck has two legs. A lame duck has one leg. A sitting duck has no legs. There are 33 ducks with a total of 32 legs. The total number of normal ducks and lame ducks is twice the number of sitting ducks. How many lame ducks are there?
Results: J19: 18% correct: 12 lame ducks (28% went for 11 lame ducks) I13: 24% correct (28% went for 11 lame ducks
Solution: N normal ducks, L lame ducks, S sitting ducks N + L + S = 33 2N + L = 32 N + L = 2S etc etc L = 12
AMC 1980 Question: J25, I25, S14 A beetle crawls around the outside of a square of side one metre, at all times keeping precisely one metre from the boundary of the square. What is the area enclosed, in square metres, in one complete circuit by the beetle?
Results: J: 4% correct: ( 5+ )m 2 (26% went for 9 m 2 ) I: 5% correct (36% went for 9 m 2 ) S: 18% correct (39% went for 9 m 2 )
Turning a concrete context into a smart maths problem Design for our new family room extension Planned Room Area: 30 m 2 Pool Table: 2.4 m x 1.2 m (Area 2.9 m 2 ) Cue: 1.5 m
Results: Planned room area: 30 m 2 Pool Table area: 2.9 m 2 Pool Table plus Cue: 20.7 m 2 So we changed planned room area to 50 m 2 $$$$!!!!
Next Problem AMC 1988 Question: J24, I20 Four points P,Q, R and S are spaced at one metre intervals, in that order, along a straight line. Find the length of the shortest path from P to S , in metres, keeping at least one metre from Q and R .
Results: J: 5% correct: (1+ ) metres (25% went for 5 metres) I: 9% correct (22% went for 5 m)
Next Problem AMC 1986 Question: I 27 A solid cube has edges of length one metre. The set of points precisely one metre away from the surface of the cube encloses a solid shape. Find the volume, in cubic metres of this shape.
Results: 6% correct: 7+13 /3 cubic m (10% went for 7 cubic m)
Turning a smart maths problem into a concrete context Smart Problem: 0 ≤ a ≤ 20, 0 ≤ b ≤ 20 , and 0 ≤ c ≤ 20 The average of a, b and c is 16 . What is the smallest that any one of a, b or c could be? Context Problem: A meter records voltages between 0 volts and 20 volts. If the average value for three readings on the meter was 16 volts, then what was the smallest possible reading?
Solution: (a + b + c)/3 = 16 a + b + c = 3 × 16 = 48 Suppose a and b are maximum, ie a + b = 20 + 20 =40, Therefore the remaining minimum value of c is 48 – 40 = 8.
Context Problem: Dead straight metal railway line in the desert is 2km between fixed ends. Sun heats up the line an extra 20 centimetres only. If the line simply bows up rather than buckles, estimate how high will it be off the ground at its centre. Referred by: Emeritus Professor Neville de Mestre, Bond University, Over 70s World Surfing Ironman Champion
Solution: Approximate model is an isosceles triangle with base 2000 metres and equal sides of 1000.1 metres. This consists of two right-angled triangles with sides (h, 1000, 1000.1). Pythagoras yields h = √200.01 =14.15 approx. Therefore line bows up about 14 metres in the centre .
Ocean Race Start (or Triathlon Lake start) from a beach - Neville de Mestre Consider the start of an ocean swim on a beach in groups of 50, say. The first turning buoy is 800 metres straight off the beach. If a person swims with the bunched mob they will be restricted in their swimming for, say, the first 200 metres to 80% of their normal swimming rate. Then the field will start to string out and the last 600 metres can be covered at normal swim rate. Let normal swim rate be X metres /minute. Time for first 200m @4X/5 m/min = 250/X minutes. Time for 600m @ X m/min = 600/X minutes Total time to buoy = 850/X minutes To avoid swimming with the mob a swimmer moves to a starting position 50 m away from the mob along the beach. [Diagram here of (50,800, hypotenuse) triangle.] Using Pythagoras’ theorem he/she has to swim √[50² + 800²} = 801.6 at normal swim rate. This takes 801.6/X minutes. Top rate for world class swimmers over a distance is about 100 m/ minute. Therefore the saving in time is (850-801.6)/100) = 0.48 minutes=29 seconds. This puts them almost 50 m ahead of where they would be if they had swum with the mob. ( NOTE: You can play with the % of normal rate and distances to obtain similar results, but it is usually about 5 to 10 % faster to go along the beach and start.)
Real Life Context Problem 1990 J26, I26, S21 On my car, a particular brand of tyre lasts 40 000 kilometres on a front wheel or 60 000 kilometres on a rear wheel. By interchanging the front and rear tyres, what is the greatest distance, in kilometres, I can get?
Solution: Consider just the two left hand tyres. Assume that the tyres are rotated to maximise the distance of wear. The total average wear on the two tyres in 1 km is ½(1/40000+1/60000) = 1/48000 of their capacity. Thus the maximum distance the car can travel is 48 000 km. Results: J: 8% correct. Most popular answer 50 000 km – 27%. I: 8% correct. Most popular answer 50 000 km – 34% S: 9% correct. Most popular answer 50 000 km – 29%
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