Transcription of 2021 Euclid Contest - University of Waterloo
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2021 Euclid Contest - University of Waterloo
The CENTRE for EDUCATION. in MATHEMATICS and COMPUTING. 2021 euclid contest Wednesday, April 7, 2021. (in North America and South America). Thursday, April 8, 2021. (outside of North America and South America). Solutions 2021 University of Waterloo 2021 euclid contest Solutions Page 2. 1. (a) Since (a 1) + (2a 3) = 14, then 3a = 18 and so a = 6. (b) Since (c2 c) + (2c 3) = 9, then c2 + c 3 = 9 and so c2 + c 12 = 0. Factoring, we obtain (c + 4)(c 3) = 0 and so c = 3 or c = 4. (c) Solution 1. Manipulating algebraically, we obtain the following equivalent equations: 1 3. 2. + 2 = 10. x 2x 2 + 3 = 20x2 (multiplying through by 2x2 , given that x 6= 0). 5 = 20x2. 1. x2 =. 4. 1. and so x = . 2. Solution 2.
2021 Euclid Contest Solutions Page 4 and so x= 4 and x= 2. This means that Eand F, in some order, have coordinates (4;0) and ( 2;0). Therefore, 4DEF has base EF of length 4 ( 2) = 6 and height 16 (vertical distance
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