Moscow Math Olympiad
The Moscow Math Olympiad has a different set of problems for every grade. Students need to write a proof for every problem. These are the 8th grade problems from this year’s Olympiad:
Problem 1. There were 6 seemingly identical balls lying at the vertices of the hexagon ABCDEF: at A — with a mass of 1 gram, at B — with a mass of 2 grams, …, at F — with a mass of 6 grams. A hacker switched two balls that were at opposite vertices of the hexagon. There is a balance scale that allows you to say in which pan the weight of the balls is greater. How can you decide which pair of balls was switched, using the scale just once?
Problem 2. Peter was born in the 19th century, while his brother Paul was born in the 20th. Once the brothers met at a party celebrating both birthdays. Peter said, “My age is equal to the sum of the digits of my birth year.” “Mine too,” replied Paul. By how many years is Paul younger than Peter?
Problem 3. Does there exist a hexagon which can be divided into four congruent triangles by a single line?
Problem 4. Every straight segment of a non-self-intersecting path contains an odd number of sides of cells of a 100 by 100 square grid. Any two consecutive segments are perpendicular to each other. Can the path pass through all the grid vertices inside and on the border of the square?
Problem 5. Denote the midpoints of the non-parallel sides AB and CD of the trapezoid ABCD by M and N respectively. The perpendicular from the point M to the diagonal AC and the perpendicular from the point N to the diagonal BD intersect at the point P. Prove that PA = PD.
Problem 6. Each cell in a square table contains a number. The sum of the two greatest numbers in each row is a, and the sum of the two greatest numbers in each column is b. Prove that a = b.
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udalrich:
Problem 1: There are three possible cases. AD were switched, BE were switched or CF were switched. Since we can only use the scale once, we need to find a grouping of the balls that, depending on which case is the actual one, results in the first group being heavier, the second group being heavier or the two groups being identical.
Place C and D on the left side, and A and F on the right side. If A and D were swapped, the sides are actually CA (3+1=4) and DF (4+6=10) and the right side is heavier. If B and E were swapped, the sides are actually CD (3+4=7) and AF(1+6=7) and the sides are balanced. If C and F were swapped, the sides are actually FD (6+4=10) and AC (1+3=4) and the left side is heavier.
That’s all I have time for now.
2 June 2011, 11:37 amJohnS:
I think I’ve misunderstood problem 3 – if I split a polygon with a single line, I always get two surfaces (Or one if I draw the line through an edge, but that doesn’t help), how am I meant to get four triangles?
2 June 2011, 2:52 pmChrist Schlacta:
@JohnS: It has to be an irregular polygon. I can split a hexagon into 3 congruent triangles, but not into four 🙁
2 June 2011, 7:04 pmkcaze:
Problem 2: Say Peter was born in 18AB, and Paul born in 19CD and say they have an age difference of X years. Then X must be 100 – (10*A + B) + (10*C + D). Since the two ages of the brothers both equal the sum of their digits, we have: 1 + 8 + A + B = 1 + 9 + C + D + X. Substituting for X and simplifying, we get 11(A – C) + 2(B-D) = 101. Since we have the limits of 0 <= A,B,C,D <= 9, A-C must = 9 and so we have A = 9 and C = 0. Then, any choice of B and D such that B = D+1 will satisfy the conditions of the problem. For example, we could have 1892 and 1901 and so their age difference must be 9 years.
Problem 3: You can take create such a hexagon with a right triangle that’s not isosceles. Call the two legs of the triangle A and B and draw a line that has length A+B. That line will be the line that you cut on the hexagon. To form the hexagon, on one side of the line, put two right triangles, the first of which has leg A touching the line and the second of which has leg B touching the line. Then, for the other side of the line, just do the same thing, but with the two triangles flipped so that the first triangle has leg B touching the line and the second triangle has leg A touching the line.
For problem 4, if I’m understanding the question correctly, isn’t it trivial to prove it impossible? Graphs with more than 2 vertices that have odd degrees implies that it’s impossible to trace the graph without backtracking on an edge. But the fact that the question gives a lot more restrictions than necessary (each segment must consist of an odd number of sides of cells) makes me doubt it’s that simple though. I’m probably misunderstanding some part of the question.
3 June 2011, 6:08 amJBL:
More specifically, it would have to be a concave (i.e., not convex) hexagon: for example, it’s easy to cut something Pac-man shaped into 3 pieces using a single cut.
3 June 2011, 9:42 amLJY:
Problem 1. Label all sides A … F in clockwise order and balls have increasing weight. Weigh 2 adjacent balls from RHS ie either A,B or BC.Adjacent balls should weigh only 1 unit different if they have not been swapped. If the scales show one unit different then it is the unweighed Ball on the RHS that has been swapped with its opposite. If the scales show more than one unit different then it is the heavier ball and its pair that have been swapped.
Problem 2: Let Paul’s age = 19ab where a,b have possible values 0,…9, then possible age range for Paul is 1+9+0+0=10 …. to 1+9+9+9 = 28. Similarly Peter’s age ranges from 9 to 27. But peter is older than paul therefore Peter’s age range narrows to 11 —>27
and Paul’s from 9 —>26. Earliest year in the 20th century when Paul could be born is 1900 which would make the year they are celebrating their b’days 1910, when Paul would be 10 (1+9+0+0 = 10 and 1910-1900 = 10) latest year is 1926 which would make peter 27 if born in 1899. By examiming possible combinations, answer is that age difference is 13 years. (not very elegant but best i coudl do)
Peter could be born 1910-27 = 1873 –> 1899 (1910-1899=11) 1+8+7+3 = 19, 1+8+7+4 = 20,
4 June 2011, 2:07 amFelix:
Problem 1: Put ACD on one side, and BF on the other. There are three possibilities:
6 June 2011, 11:47 am– AD have been switched, then ACD = 8 and BF = 8, both sides weigh the same
– BE have been switched, then ACD = 8 and BF = 11, so right side is heavier
– CF have been switched, then ACD = 11 and BF = 5, so left side is lighter.
Since all three give different outcomes, the side of the balance (or if it doesn’t move) tells which balls have been switched.
andy01q:
problem 6:
7 June 2011, 3:27 pmThe square has c rows and c collumns.
The sum of all numbers equals the sum of all numbers horizontally alligned and also equals the sum of all numbers vertically alligned. We will call it d.
d / c = a and d / c = b, therefore a=b
ss:
problem 2:
10 June 2011, 10:43 am9 years
example: 1899 1908 and party in 1926 (ages 27, 18)
aspertyx:
problem 6:
Suppose a > b. Call the largest number of each row “big” and the second largest number of each row “small”. (big + small = a). Each column contains only one of the “big” numbers of each row since each “big” > a/2.
Now find the smallest of the “big” numbers (say B). In that row, look at the “small” (2nd largest number in the row, say S). In the column containing this S, there is a “big” number, say BB. BB + S >= B + S since B is the smallest of the “bigs”. But B + S = a > b. So BB + S > b. Contradiction. A similar argument works for b > a.
20 June 2011, 9:41 amMaths olympiad:
Mathematics Olympiad is the biggest competition in which indigenous group of ‘mathletes’ from respective countries compete for Gold, Silver and Bronze medal. The competition tests the innate problem solving skills.
1 July 2011, 1:00 amMr. Briggs:
Problem 3: Yes – if you count degenerate ones. Two isosceles triangles stuck together, apex on apex, is a hexagon, and if you cut it through the perpendicular bisectors of the non-equal sides, then you get four congruent triangles.
29 July 2011, 11:46 amRoman Chokler:
Problem 3 is easy. If you take any scalene right triangle you can glue four congruent copies of it to form a Z-shaped hexagon. This hexagon will not be degenerate. Q.E.D.
28 April 2013, 2:45 pmPW:
For problem 3, better make them (non-isosceles) right triangles. You’ll
28 August 2014, 10:15 amneed the angles at each end of the cutting line to sum to pi, so they’ll
have to be the same angle twice.