MO Regionalrunde Klasse 9 Mock Paper 1 with Solutions

MO Regionalrunde Klasse 9 Mock Paper 1 · IMOolympiad.com · Original practice

4 written-solution problems · 240 minutes for this practice paper

For school year 9. The 240-minute reference is the Lower Saxony organiser’s schedule for years 7–13. Your regional invitation takes precedence.

Original independent practice in English. Not an official paper or predicted selection test. Keep hints and solutions closed during your attempt.

How to review your proof

Check that you have used every condition, justified the main idea, covered all cases and stated the conclusion. A different complete proof can also be correct. This is a self-review checklist; written proofs are not automatically marked.

Question 1

Determine all ordered pairs of positive integers a,b satisfying $a^2+b^2=2ab+a+b$ .

Hint 1

The difference of the two variables is useful.

Hint 2

Put d=a−b; then their sum is d².

Worked solution 1

The equation is equivalent to $(a-b)^2=a+b$ . Set d=a−b, an integer. Then a+b=d², so adding and subtracting gives $a=d(d+1)/2$ and $b=d(d-1)/2$ . These are integers because neighbouring integers have even product. They are both positive exactly when d≥2 or d≤−2; d=−1,0,1 makes at least one zero. Conversely every stated d produces positive integers with difference d and sum d², hence satisfies the original equation. Equivalently, the answers are the pairs of neighbouring positive triangular numbers and their reversals.

Conclusion: a=d(d+1)/2, b=d(d−1)/2 for integer |d|≥2.

Review the idea: Factorisation integer solutions

Question 2

Positive real numbers a,b,c satisfy abc=1. Prove $(a+b+c)(ab+bc+ca)\ge9$ , and determine all equality cases.

Hint 1

Use abc=1 to turn ab,bc,ca into reciprocals.

Hint 2

Pair each ratio with its reciprocal.

Worked solution 2

Since abc=1, we have ab=1/c, bc=1/a and ca=1/b. Expanding the product gives $3+(a/b+b/a)+(b/c+c/b)+(c/a+a/c).$ For positive u,v, $u/v+v/u-2=(u-v)^2/(uv)\ge0$ . Each bracket is therefore at least 2, proving the bound 9. Equality requires a=b=c; the product condition then makes all three equal to 1. That triple attains 9.

Conclusion: Equality exactly at a=b=c=1.

Review the idea: Sum of squares · Algebraic identities

Question 3

Eight labelled points occur in a fixed order around a circle. They are paired by four straight chords, with every point used exactly once. How many pairings have no crossing in the interior of the circle? Prove your count.

Hint 1

Look at the partner of one fixed point.

Hint 2

A chord splits the remaining points into two even-sized sets that can be paired independently.

Worked solution 3

Let $C_n$ count noncrossing pairings of 2n cyclically ordered points, with $C_0$ =1. Fix the first point. Its partner must be the 2k-th point for some k=1,…,n: otherwise one side contains an odd number of unused points, which cannot be paired without crossing that chord. The two sides contain 2(k−1) and 2(n−k) points; their pairings are independent, giving $C_n=\sum_{k=1}^n C_{k-1}C_{n-k}$ . This yields $C_1$ =1, $C_2$ =2, $C_3$ =5 and $C_4=1\cdot5+1\cdot2+2\cdot1+5\cdot1=14$ . Every pairing has one unique partner for the fixed point, so there is no double count.

Example: pairing point 1 with point 4 leaves points 2–3 on one side and points 5–8 on the other. Dashed chords show one possible completion.

Conclusion: 14 noncrossing pairings.

Review the idea: Counting with recurrences

Question 4

In triangle ABC, the medians from B and C lie on perpendicular lines. Prove $AB^2+AC^2=5BC^2$ .

Hint 1

Place the centroid at the origin and the two median lines on coordinate axes.

Hint 2

The coordinates of the three vertices average to the centroid.

Worked solution 4

Let G be the centroid. Choose perpendicular coordinate axes along GB and GC, writing B=(p,0), C=(0,q), with p,q nonzero. Let D be the midpoint of BC, so D=(p/2,q/2). The centroid divides median AD in the ratio AG:GD=2:1, with A and D on opposite sides of G=(0,0). Thus the coordinates of A are −2 times those of D, namely A=(−p,−q). Pythagoras now gives $BC^2=p^2+q^2$ , $AB^2=4p^2+q^2$ and $AC^2=p^2+4q^2$ . Adding the last two expressions gives exactly $5(p^2+q^2)=5BC^2$ .

Conclusion: AB²+AC²=5BC².

Review the idea: Concurrency and collinearity · Pythagoras and stewart

After this paper

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Format reference: official organiser information. Questions and explanations are independent practice material.