# Weekly Dose 11 - Maths Olympiad Practice

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29 Apr 2016
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### Weekly Dose 11 - Maths Olympiad Practice

• 1. Each of the numbers from 1 to 9 is placed, one per circle, into the pattern shown. The sums along each of the four sides are equal. How many different numbers can be placed in the middle circle to satisfy these conditions? Solution: Because the sums along each side must be equal, therefore sums of the 8 numbers must be divisible by 4. Note that 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 = 4 × 11 + 1. Therefore the numbers which have a remainder of 1 when divided by 4 are the candidates to be placed in the middle. Answer:
• 2. When Anura was 8 years old his father was 31 years old. Now his father is twice as old as Anura is. How old is Anura now? Solution: Now, after 𝑥 year, - Anura’s father’s age is 31 + 𝑥 and Anura’s age is 8 + 𝑥 - Anura’s father’s age is 2 × Anura’s age 31 + 𝑥 = 2 × (8 + 𝑥) = 16 + 2𝑥 𝑥 = 15 Anura’s age = 8 + 15 = ____ Answer: 23
• 3. In rectangle 𝐴𝐵𝐶𝐷, 𝐴𝐵 = 12𝑐𝑚 and 𝐴𝐷 = 5𝑐𝑚. Point P, Q, R and S are all on diagonal AC, so that 𝐴𝑃 = 𝑃𝑄 = 𝑄𝑅 = 𝑅𝑆 = 𝑆𝐶. What is the total area of the shaded region, in cm2? Solution: The area for 𝐴𝐵𝐶𝐷 = 5 × 12 = 60 𝑐𝑚2 The area for △ 𝐴𝐵𝐶 and △ 𝐴𝐷𝐶 = 60 ÷ 2 = 30 𝑐𝑚2 Since 𝐴𝑃 = 𝐶𝑆 = 1 5 𝐴𝐶, therefore the area for △ 𝐴𝑃𝐷 =△ 𝐴𝑃𝐵 =△ 𝐶𝑆𝐷 =△ 𝐶𝑆𝐵 = 1 5 △ 𝐴𝐵𝐶 = 1 5 × 30 = 6 𝑐𝑚2 Total area of shaded region = 6 + 6 + 6 + 6 = ____ Answer: 24 𝑐𝑚2
• 4. Solution: 𝐴𝑃 = 𝐴𝑄 ⇒ ∠𝐴𝑃𝑄 = ∠𝐴𝑄𝑃 = ∝ ° 𝐵𝑄 = 𝐵𝑅 ⇒ ∠𝐵𝑅𝑄 = ∠𝐵𝑄𝑅 = 𝛽° ∠𝑃𝐴𝑄 = 180° − 2 ∝ ° ∠𝑅𝐵𝑄 = 180° − 2𝛽° From △ 𝐴𝐵𝐶, 70° + (180° − 2 ∝ °) + (180° − 2𝛽°) = 180° 2 ∝ ° + 2𝛽° = 250° ∝ ° + 𝛽° = 125° ∠𝑃𝑄𝑅 = 180° − ∝ ° − 𝛽° = 180° − (∝ ° + 𝛽°) = 180° − 125° = 55 ° Answer: 55° In triangle 𝐴𝐵𝐶, 𝐴𝑃 = 𝐴𝑄 and 𝐵𝑄 = 𝐵𝑅. Determine angle 𝑃𝑄𝑅 in degrees.