Austria Junior Regional — Original Mock Paper 1

Idea. To bound a, we need to use the fact that real numbers b and c exist. Their squared difference cannot be negative. The two given equations let us write that difference entirely in terms of a.

First square the sum:

a² + b² + c² = (a + b + c)² − 2(ab + bc + ca) = 144 − 72 = 72.

Thus b + c = 12 − a and b² + c² = 72 − a². The identity (b − c)² = 2(b² + c²) − (b + c)² gives

(b − c)² = 2(72 − a²) − (12 − a)² = 3a(8 − a).

The left side is nonnegative. If a were greater than 8, the factor a would be positive and 8 − a would be negative, so the right side would be negative. This is impossible. Therefore a ≤ 8.

Attainment and all equality cases. If a = 8, the identity forces (b − c)² = 0, so b = c. Their sum is 12 − 8 = 4; hence b = c = 2. Conversely, (8, 2, 2) consists of nonnegative numbers, has sum 12 and has pair-product sum 16 + 4 + 16 = 36. It really attains the bound.

The largest value is 8, and the only ordered triple attaining it is (8, 2, 2). The variable being maximised is specifically a, so rearrangements with 8 in another position are not equality cases for this question.

Idea. The statement concerns a trapezoid, but the diagonals divide it into triangles. In each triangle, joining two side midpoints gives a parallel segment of half the third side’s length.

Locate P and the section. Let P′ be the midpoint of AC. In triangle ADC, E and P′ are side midpoints, so EP′ ∥ DC and EP′ = CD/2. In triangle ABC, P′ and F are side midpoints, so P′F ∥ AB and P′F = AB/2.

Since AB ∥ CD, both lines EP′ and P′F are the line through P′ parallel to the bases. Therefore E, P′ and F are collinear. Convexity places E and F on opposite sides of diagonal AC, so P′ lies between E and F. Thus P′ is the stated intersection P. In particular, P is the midpoint of AC, EP = CD/2 and PF = AB/2.

Locate Q and determine the order. Let Q′ be the midpoint of BD. In triangle BCD, F and Q′ are side midpoints, so FQ′ ∥ CD and FQ′ = CD/2. This is again the line EF, so Q′ = Q. Convexity places Q inside segment EF. Because

PF = AB/2 > CD/2 = QF,

Q is closer to F than P is. Therefore the order is E, P, Q, F. We may now subtract lengths on that segment:

EP = CD/2,   PQ = PF − QF = (AB − CD)/2,   QF = CD/2.

Prove both directions. If AB = 2CD, then PQ = (2CD − CD)/2 = CD/2, so all three lengths are equal. Conversely, if all three lengths are equal, EP = PQ gives CD/2 = (AB − CD)/2. Multiplication by 2 yields CD = AB − CD, hence AB = 2CD.

This establishes the order and the required equivalence. No angle condition or symmetry of the trapezoid was used.

First obtain a bound and attain it. Split the 2n numbers into the n disjoint pairs

{1, 2n}, {2, 2n − 1}, …, {n, n + 1}.

The two members of every pair sum to 2n + 1, so an admissible set contains at most one member of each pair. It therefore has at most n elements. The set {1, 2, …, n} is admissible: two distinct members have sum at most n + (n − 1) = 2n − 1 when n ≥ 2, while for n = 1 there is no pair of distinct members. Thus the maximum size is n.

Describe every set attaining the bound. Such a set must choose exactly one member from each pair. Number the pairs by i = 1, 2, …, n. In pair i, call i the lower choice L and 2n + 1 − i the higher choice H.

For any i from 1 to n − 1, choosing H from pair i and L from pair i + 1 would choose the two numbers 2n + 1 − i and i + 1. They are distinct, and their sum is 2n + 2. Therefore the pattern HL in adjacent pairs is forbidden.

A word in L and H with no adjacent HL must consist of some L choices followed by H choices. Indeed, if an H had an L somewhere after it, the first subsequent L would be immediately preceded by H, producing the forbidden pattern. Hence there is a unique k with 0 ≤ k ≤ n for which the choices are k copies of L followed by n − k copies of H.

The corresponding set is

S_k = {1, 2, …, k} ∪ {n + 1, n + 2, …, 2n − k}.

Here the first block is empty when k = 0, and the second is empty when k = n. Together the blocks always contain k + (n − k) = n elements.

Check that every listed set works. Two distinct numbers in its lower block have sum at most 2n − 1. Two distinct numbers in its higher block have sum at least (n + 1) + (n + 2) = 2n + 3. A number from each block has sum at most k + (2n − k) = 2n. Thus none of these three types can give either forbidden sum. Empty blocks simply contribute no pair of that type.

Every largest admissible set has been listed, and different k give different sets because k is exactly the number of elements at most n. There are n + 1 choices of k. The answers are therefore maximum size n and exactly n + 1 sets of that size.

Idea. The square condition concerns three factors. Coprimality prevents a prime factor in one factor from being completed to an even exponent by another factor. We will first prove that each factor is itself a square.

Separate the prime factors. We are given gcd(a, b) = 1. Any common divisor of a and a + b also divides their difference b, so gcd(a, a + b) = 1. Similarly gcd(b, a + b) = 1. Thus a, b and a + b are pairwise coprime.

In the prime factorisation of a perfect square every exponent is even. A prime dividing one of these three pairwise coprime factors divides neither of the others, so its exponent in the product comes entirely from that one factor. Every exponent in each factor is therefore even. Consequently, for positive integers u, v and w,

a = u²,   b = v²,   a + b = w²,   u² + v² = w².

Use parity to get a lower bound. The integers u and v are coprime: a common divisor greater than 1 would also divide both a and b. They cannot both be even. They cannot both be odd either, because an odd square leaves remainder 1 on division by 4; their sum would leave remainder 2, whereas a square leaves only 0 or 1.

Thus one of u and v is even and the other is odd. Their squared sum is odd, so w is odd. The even one is at least 2. The odd one cannot be 1: if it were, writing t for the other positive integer would give w² − t² = 1, or

(w − t)(w + t) = 1.

Here w > t, so both factors are positive integers. They would both have to equal 1, forcing t = 0, a contradiction. The odd one is therefore at least 3. It follows that w² ≥ 2² + 3² = 13. As w is an odd positive integer, w ≥ 5. Therefore a + b = w² ≥ 25.

Check attainment and classify equality. If a + b = 25, then w = 5. The positive even member of {u, v} is smaller than 5, so it is 2 or 4. If it is 2, the other square would be 25 − 4 = 21, which is not a square because 4² < 21 < 5². If it is 4, the other square is 25 − 16 = 9, so the other member is 3.

Hence the only ordered pairs are (a, b) = (9, 16) and (16, 9). Both are coprime, and their product condition is verified by 9 × 16 × 25 = (3 × 4 × 5)² = 60². The minimum is therefore 25, attained exactly by the two stated pairs.