MathDB

2010 BAMO

Part of BAMO Contests

Subcontests

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2010 BAMO A max no of subsets with primes sum of a 3-element set

We write {a,b,c}\{a,b,c\} for the set of three different positive integers a,ba, b, and cc. By choosing some or all of the numbers a, b and c, we can form seven nonempty subsets of {a,b,c}\{a,b,c\}. We can then calculate the sum of the elements of each subset. For example, for the set {4,7,42}\{4,7,42\} we will find sums of 4,7,42,11,46,494, 7, 42,11, 46, 49, and 5353 for its seven subsets. Since 7,117, 11, and 5353 are prime, the set {4,7,42}\{4,7,42\} has exactly three subsets whose sums are prime. (Recall that prime numbers are numbers with exactly two different factors, 11 and themselves. In particular, the number 11 is not prime.)
What is the largest possible number of subsets with prime sums that a set of three different positive integers can have? Give an example of a set {a,b,c}\{a,b,c\} that has that number of subsets with prime sums, and explain why no other three-element set could have more.
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2010 BAMO C a+b+c \ge \frac{|a|+|b|+|c|}{3}

Suppose a,b,ca,b,c are real numbers such that a+b≥0,b+c≥0a+b \ge 0, b+c \ge 0, and c+a≥0c+a \ge 0. Prove that a+b+c≥∣a∣+∣b∣+∣c∣3a+b+c \ge \frac{|a|+|b|+|c|}{3} .
(Note: ∣x∣|x| is called the absolute value of xx and is defined as follows. If x≥0x \ge 0 then ∣x∣=x|x|= x, and if x<0x < 0 then ∣x∣=−x|x| = -x. For example, ∣6∣=6,∣0∣=0|6|= 6, |0| = 0 and ∣−6∣=6|-6| = 6.)

An easy graphing problem

All vertices of a polygon PP lie at points with integer coordinates in the plane, and all sides of PP have integer lengths. Prove that the perimeter of PP must be an even number.
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A nice PHP BAMO problem

Place eight rooks on a standard 8×88 \times 8 chessboard so that no two are in the same row or column. With the standard rules of chess, this means that no two rooks are attacking each other. Now paint 2727 of the remaining squares (not currently occupied by rooks) red. Prove that no matter how the rooks are arranged and which set of 2727 squares are painted, it is always possible to move some or all of the rooks so that:
• All the rooks are still on unpainted squares. • The rooks are still not attacking each other (no two are in the same row or same column). • At least one formerly empty square now has a rook on it; that is, the rooks are not on the same 88 squares as before.

A brilliant geo problem

Acute triangle ABCABC has ∠BAC<45∘\angle BAC < 45^\circ. Point DD lies in the interior of triangle ABCABC so that BD=CDBD = CD and ∠BDC=4∠BAC\angle BDC = 4 \angle BAC. Point EE is the reflection of CC across line ABAB, and point FF is the reflection of BB across line ACAC. Prove that lines ADAD and EFEF are perpendicular.
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