MO Schulrunde Klasse 9 Mock Paper 3 with Solutions

MO Schulrunde Klasse 9 Mock Paper 3 · IMOolympiad.com · Original practice

6 written-solution problems · Take-home practice: choose any 4 of 6 problems

For school year 9. The official shared school-round sheet for years 9–10 offers six problems; the local organiser decides which problems and working arrangements apply. Our practice uses the choose-four option.

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

Prove that there are no integers a,b satisfying $a^2-b^2=2$ .

Hint 1

List the possible square remainders modulo 4.

Hint 2

They are only 0 and 1.

Worked solution 1

If an integer is even, its square is 0 modulo 4; if odd, its square is 1 modulo 4. The difference of two such remainders is 0,1 or -1, corresponding to 0,1,3 modulo 4. It is never 2. Thus the proposed equation is impossible, including for negative a or b.

Conclusion: No integer solutions.

Review the idea: Remainders

Question 2

Real numbers x,y,z satisfy $x+y+z=3$ and $x^2+y^2+z^2=3$ . Determine all possible triples.

Hint 1

Compare each variable with the common mean 1.

Hint 2

Expand $(x-1)^2+(y-1)^2+(z-1)^2$ .

Worked solution 2

Consider $(x-1)^2+(y-1)^2+(z-1)^2$ . Expanding, this sum equals $x^2+y^2+z^2-2(x+y+z)+3=3-6+3=0$ . Each square is nonnegative, so each must be zero. Hence x=y=z=1. This triple satisfies both given equations, so it is the unique answer.

Conclusion: Only (1,1,1).

Review the idea: Sum of squares

Question 3

At a gathering of ten people, friendship is mutual and no person is counted as their own friend. Prove that two people have the same number of friends among the other nine.

Hint 1

The possible counts appear to be 0 through 9.

Hint 2

Can counts 0 and 9 both occur?

Worked solution 3

A person can have 0,1,…,9 friends. However, a person with nine friends is friends with everyone, so no person can then have zero friends. Thus at least one of the two extreme counts 0 and 9 is absent. There are at most nine available counts for ten people. By the pigeonhole principle, two people have equal counts.

Conclusion: Two people necessarily share a friendship count.

Review the idea: Pigeonhole principle

Question 4

In an acute triangle ABC, the perpendiculars from B to AC and from C to AB meet those sides at D and E, respectively, and meet each other at H. Prove that B,C,D,E are concyclic and that $\angle BHC=180^\circ-\angle BAC$ .

Hint 1

Both angles BDC and BEC are right angles.

Hint 2

In triangle BHC, express the angles at B and C using the original triangle angles.

Worked solution 4

Since BD is perpendicular to AC, angle BDC is 90 degrees. Similarly angle BEC is 90 degrees. Both D and E therefore lie on the circle with diameter BC, proving concyclicity. Because the original triangle is acute, H lies inside it. We have $\angle HBC=90^\circ-\angle C$ and $\angle HCB=90^\circ-\angle B$ . The angle sum in BHC gives $\angle BHC=180^\circ-(90^\circ-C)-(90^\circ-B)=B+C=180^\circ-A$ .

Conclusion: The circle has diameter BC; angle BHC is 180°−A.

Review the idea: Circles and power of a point · Angles

Question 5

Let a,b be positive coprime integers. Prove that $\gcd(a+b,a^2-ab+b^2)$ is either 1 or 3, and determine when it equals 3.

Hint 1

Reduce the quadratic modulo a+b.

Hint 2

It becomes $3a^2$ , and a is coprime to a+b.

Worked solution 5

Put s=a+b and t=a²-ab+b². Modulo s we have b≡−a, so t≡3a². Let d=gcd(s,t). Then d divides 3a² and is coprime to a, because d divides s and gcd(a,s)=1. In prime factorisation, none of the prime factors of d can come from a². All their powers must therefore come from the factor 3, so d divides 3. Hence d is 1 or 3. If 3 divides s, then t≡3a²≡0 modulo 3, giving d=3. If 3 does not divide s, d cannot be 3, so d=1. This gives the exact criterion.

Conclusion: The gcd is 3 exactly when 3 divides a+b; otherwise it is 1.

Review the idea: Greatest common divisor · Remainders

Question 6

Cards 1,2,…,n are initially in that order in a row. A move reverses any three consecutive cards. Determine for which positive integers n the complete reverse order n,…,2,1 can be reached. When n<3 no move is available.

Hint 1

A three-card reversal exchanges the first and third positions and fixes the middle one.

Hint 2

Thus it swaps neighbouring positions within one parity class of positions.

Worked solution 6

Every move preserves the parity of the position occupied by each card, because it only exchanges positions i and i+2. In the target, card k occupies position n+1-k. This has the same parity as k exactly when n is odd. Thus even n is impossible. If n is odd, every card’s target has its original position parity. A reversal starting at i exchanges neighbouring members of the list of odd positions or of the list of even positions, without changing other cards. To arrange one parity list, locate the card required in its first position and move it left within that list one place at a time by these exchanges. Keep it fixed, then do the same for the required second card among the remaining positions, and continue. Every exchange corresponds to an allowed three-card reversal and leaves the other parity list unchanged. This arranges both lists in the desired order, so we reach the target. For n=1 it is already reached. Exactly odd n work.

Conclusion: Exactly the positive odd integers n.

Review the idea: Permutations and arrangements · Parity

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.

Choose another paper · Check your German selection route

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