MO Regionalrunde Klasse 9 Mock Paper 5 with Solutions

MO Regionalrunde Klasse 9 Mock Paper 5 · 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

For a positive integer d, let $\varphi(d)$ be the number of integers k with $1\le k\le d$ and gcd(k,d)=1; in particular $\varphi(1)=1$ . Prove $\sum_{d\mid n}\varphi(d)=n$ for every positive integer n.

Hint 1

Partition the integers 1,…,n by their gcd with n.

Hint 2

For gcd(k,n)=g, divide both numbers by g.

Worked solution 1

Every k from 1 through n has a unique g=gcd(k,n), which is a divisor of n. Write k=gu and n=gd. Then 1≤u≤d and gcd(u,d)=1, so exactly φ(d) values of k have gcd n/d with n. Conversely every such u produces exactly one k in that class. As d runs through the positive divisors of n, these classes are disjoint and cover all n integers. Summing their sizes proves the identity, including n=1.

Conclusion: The divisor sum of φ equals n.

Review the idea: Greatest common divisor · Counting with bijections

Question 2

Positive real numbers a,b,c satisfy abc=1. Prove $\frac1{1+a+ab}+\frac1{1+b+bc}+\frac1{1+c+ca}=1.$

Hint 1

Represent a,b,c as successive ratios x/y,y/z,z/x.

Hint 2

The three fractions then acquire the same denominator.

Worked solution 2

Take x=ab, y=b, z=1. Then a=x/y, b=y/z and c=z/x, using abc=1. Let D=xy+yz+zx, which is positive. The first denominator is $1+x/y+x/z=D/(yz)$ , so its reciprocal is yz/D. The other two reciprocals are zx/D and xy/D. Their sum is (yz+zx+xy)/D=1. The explicit choice of x,y,z shows that no allowed triple was omitted.

Conclusion: The sum is identically 1.

Review the idea: Algebraic identities · Exact arithmetic

Question 3

In a tournament each pair of players plays once, with exactly one winner and no draw. Prove that the players can be put in a row so that each player defeated the player immediately to their right.

Hint 1

Use induction on the number of players.

Hint 2

Insert a new player just before the first player in the old row whom the newcomer defeated.

Worked solution 3

The assertion is immediate for one player. Suppose an ordered row of n players already has the required property, and add a new player V. If V defeated nobody in that row, place V at the end; the former last player defeated V. Otherwise let W be the first player in the row defeated by V. Insert V immediately before W. If W was first, no left-hand condition is needed. Otherwise W’s predecessor U was not defeated by V, so U defeated V because there are no draws. The two new neighbouring results are U defeating V and V defeating W; all other neighbouring results remain unchanged. This proves the induction.

Conclusion: Such a row always exists; induction gives a construction.

Review the idea: Mathematical induction the first principle

Question 4

Two distinct circles of the same radius intersect at A and B. A line through A, different from AB and not tangent to either circle, meets the first circle again at C and the second again at D. Suppose C and D lie on opposite sides of A. Prove that BC=BD.

Hint 1

Use triangles ABC and ABD and their common circumradius.

Hint 2

Angles BAC and BAD are supplementary, so their sines are equal.

Worked solution 4

Let the common radius be R. The points A,B,C are distinct and noncollinear, as are A,B,D, by the line restrictions. Since rays AC and AD point in opposite directions, $\angle BAC+\angle BAD=180^\circ$ , so their sines agree. The extended sine rule in the two triangles gives $BC=2R\sin\angle BAC$ and $BD=2R\sin\angle BAD$ . Hence BC=BD. The use of the same R is exactly where the equal-circle hypothesis is needed.

Conclusion: BC=BD.

Review the idea: Trigonometry in geometry · Circles and power of a point

After this paper

Choose one gap in your proof to repair, study the linked idea, and write a complete solution again before the next mock.

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