MO Schulrunde Klasse 9 Mock Paper 5 with Solutions

MO Schulrunde Klasse 9 Mock Paper 5 · 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

Positive coprime integers a,b satisfy $a+b\mid a^2+b^2$ . Determine all such pairs.

Hint 1

Reduce b modulo a+b.

Hint 2

Use that a is coprime to a+b.

Worked solution 1

Let s=a+b. Since b is congruent to -a modulo s, the condition gives $s\mid2a^2$ . Also $\gcd(a,s)=\gcd(a,b)=1$ , so $\gcd(a^2,s)=1$ . By Euclid’s lemma, a factor coprime to s may be cancelled in divisibility, yielding s|2. Positivity gives s at least 2, hence s=2 and a=b=1. This pair is coprime and indeed 2 divides 2, so it is the only answer.

Conclusion: Only (a,b)=(1,1).

Review the idea: Greatest common divisor

Question 2

Positive real numbers x,y satisfy x+y=1. Prove $1/x+1/y\ge4$ , and determine every equality case.

Hint 1

Combine the reciprocals.

Hint 2

Use $(x-y)^2\ge0$ to bound xy.

Worked solution 2

We have $1/x+1/y=(x+y)/(xy)=1/(xy)$ . From $(x-y)^2=(x+y)^2-4xy\ge0$ , we obtain xy at most 1/4. Since xy is positive, taking reciprocals reverses this inequality and gives 1/(xy) at least 4. Equality holds precisely when x=y, hence when both equal 1/2; that pair works.

Conclusion: Minimum 4, exactly at x=y=1/2.

Review the idea: Sum of squares · Inequality rules and signs

Question 3

Every pair among six labelled vertices is joined by an edge coloured red or blue. Prove that there are three vertices whose three connecting edges all have the same colour.

Hint 1

At one vertex, at least three of its five incident edges share a colour.

Hint 2

Inspect the three edges among their other endpoints.

Worked solution 3

Choose a vertex V. At least three of its five edges have the same colour; call it red, with other endpoints A,B,C. If any edge among A,B,C is red, it and the two red edges to V form a red triangle. If none is red, all three edges AB,BC,CA are blue, forming a blue triangle. These possibilities exhaust the edge colours, so a monochromatic triangle is unavoidable.

Conclusion: A monochromatic triangle always exists.

Review the idea: Pigeonhole principle

Question 4

In triangle ABC, the internal angle bisector AD meets side BC at D. Suppose $AB=AD=DC$ . Determine all three angles of ABC.

Hint 1

Use the two isosceles triangles ABD and ADC.

Hint 2

Let angle CAD be t, and compare the two angles at D.

Worked solution 4

Put $\angle CAD=\angle DAB=t$ . Since AD=DC, triangle ADC has equal angles at A and C, so angle C=t and angle ADC=180°−2t. As B,D,C are collinear, angle ADB=2t. Since AB=AD, triangle ABD has equal angles ABD and ADB, hence angle B=2t. The angles of ABC are therefore 2t,2t,t, whose sum is 180 degrees. Thus t=36 degrees, giving A=72°, B=72°, C=36°. These angles are attainable: construct such a triangle and its internal bisector; the equal angles just used in reverse show AB=AD=DC.

Conclusion: A=72°, B=72°, C=36°.

Review the idea: Angles · Parallel lines and angle bisectors

Question 5

The sequence is given by $u_1=2$ and $u_{n+1}=2-1/u_n$ for positive integers n. Prove that all terms are defined and that $u_n=(n+1)/n$ .

Hint 1

Calculate u2 and u3 to guess the pattern.

Hint 2

Use induction, checking that each denominator stays nonzero.

Worked solution 5

The formula gives u1=2, matching the initial value. Suppose $u_n=(n+1)/n$ , which is positive and therefore nonzero. Then the next term is defined and $u_{n+1}=2-n/(n+1)=(n+2)/(n+1)$ . This is the claimed formula with n+1 in place of n and is again positive. Induction proves simultaneously the formula and that no division by zero occurs.

Conclusion: $u_n$ =(n+1)/n for every n≥1.

Review the idea: Mathematical induction the first principle · Nonlinear recurrences and substitutions

Question 6

A row of n coins, where $n\ge2$ , initially shows only heads. A move turns over two adjacent coins. Prove that a prescribed final pattern is reachable if and only if it has an even number of tails.

Hint 1

Each move preserves the parity of the number of tails.

Hint 2

To prove sufficiency, settle the coins from left to right, using the last coin to absorb the parity condition.

Worked solution 6

A move changes the number of tails by −2, 0 or 2, so every reachable pattern has an even number. Conversely, suppose the target has an even number. For i=1,…,n−1, compare coin i with its target. If it is wrong, turn over coins i and i+1; otherwise do nothing. Later moves do not touch the settled positions to the left. At the end the first n−1 coins are correct. The current pattern still has even tail count, as does the target, so the last coin cannot differ: one difference would change parity. Hence the whole target is reached.

Conclusion: Exactly the patterns with an even number of tails are reachable.

Review the idea: Parity · Proof methods

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.