IMC_2012_web_solutions

# So the sum of the angles is 180 6 180 3 2 1 note

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So the sum of the angles is 0 0 180 6 180 ) 3 2 1 ( . Note: There is more than one way to see that the sum of the angles of a pentagon is 0 540 . Here is one method. Join the vertices of the pentagon to some point, say P , inside the pentagon. This creates 5 triangles whose angles sum to 0 180 5 .The sum of the angles in these triangles is the sum of the angles in a pentagon plus the sum of the angles at P , which is 0 0 180 2 360 . So the sum of the angles in the pentagon is 0 0 0 180 3 180 2 180 5 . Extension Problems 3.1 What is the sum of the angles in a septagon? 3.2 What is the sum of the angles in a polygon with n vertices? 3.3 Does your method in 3.2 apply to a polygon shaped as the one shown where you cannot join all the vertices by straight lines to a point inside the polygon? If not, how could you modify your method to cover this case? P

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4 4. All four digits of two 2-digit numbers are different. What is the largest possible sum of two such numbers? A 169 B 174 C 183 D 190 E 197 Solution: C To get the largest possible sum we need to take 9 and 8 as the tens digits, and 7 and 6 as the units digits. For example, 3 8 1 6 8 7 9 Extension Problem 4.1 All nine digits of three 3-digit numbers are different. What is the largest possible sum of three such numbers? 5. How many minutes will elapse between 20:12 today and 21:02 tomorrow? A 50 B 770 C 1250 D 1490 E 2450 Solution: D From 20:12 today until 20.12 tomorrow is 24 hours, that is 1440 60 24 minutes. There are 50 minutes from 20:12 tomorrow to 21:02 tomorrow. This gives a total of 1490 50 1440 minutes. 6. Triangle QRS is isosceles and right-angled. Beatrice reflects the P-shape in the side QR to get an image. She reflects the first image in the side QS to get a second image. Finally, she reflects the second image in the side RS to get a third image. What does the third image look like? A B C D E Solution: A The effect of the successive reflections is shown in the diagram. S Q R Q R S 1st reflection – in QR 2nd reflection – in SQ 3rd reflection – in RS
5 7. The prime numbers p and q are the smallest primes that differ by 6. What is the sum of p and q ? A 12 B 14 C 16 D 20 E 28 Solution: C Suppose q p . Then . 6 p q The prime numbers are 2, 3, 5, 7, …. . With 2 p , 8 q , which is not prime. Similarly if 3 p , 9 q , which is also not prime. However, when 5 p , 11 q , which is prime. So, 5 p , 11 q gives the smallest primes that differ by 6. Then q p 16 11 5 . 8. Seb has been challenged to place the numbers 1 to 9 inclusive in the nine regions formed by the Olympic rings so that there is exactly one number in each region and the sum of the numbers in each ring is 11. The diagram shows part of his solution. What number goes in the region marked * ? A 6 B 4 C 3 D 2 E 1 Solution: A We let u , v , w , x , y and z be the numbers in the regions shown. Since the sum of the numbers in each ring is 11, we have, from the leftmost ring, that 11 9 u and so 2 u . Then, from the next ring, 11 5 2 v and so 4 v . From the rightmost ring, 11 8 z and so 3 z .

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