MO Schulrunde Klasse 9 Mock Paper 2 with Solutions

MO Schulrunde Klasse 9 Mock Paper 2 · 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

For every nonnegative integer n, prove that the fraction $(2n+1)/(3n+1)$ is already in lowest terms.

Hint 1

A common divisor divides the difference of numerator and denominator.

Hint 2

Use n and 2n+1 to force that divisor to divide 1.

Worked solution 1

Let d be a positive common divisor of 2n+1 and 3n+1. It divides their difference n. It then divides $(2n+1)-2n=1$ . Thus d=1, so the numerator and denominator are coprime. The denominator is positive for every stated n; at n=0 the fraction is 1/1 and the same argument applies.

Conclusion: The greatest common divisor is 1 for every n≥0.

Review the idea: Greatest common divisor

Question 2

For positive real numbers a,b, prove $a^3+b^3\ge(a+b)^3/4$ . Determine exactly when equality holds.

Hint 1

Move the right side to the left and factor.

Hint 2

Look for the square $(a-b)^2$ .

Worked solution 2

Multiplying the desired inequality by 4, the difference is $4(a^3+b^3)-(a+b)^3=3(a+b)(a-b)^2$ . This is nonnegative because a+b is positive and a square is nonnegative. It is zero exactly when a=b. Conversely, a=b makes both original sides equal, so the equality condition is complete.

Conclusion: Equality holds exactly when a=b.

Review the idea: Algebraic identities · Sum of squares

Question 3

Every cell of a 3-by-3 board initially contains 0. A move adds 1 to all three entries of one chosen row or one chosen column. Prove that no sequence of moves can leave the top-left entry equal to 1 and all other entries equal to 0.

Hint 1

Express each final entry using its row and column move counts.

Hint 2

For any two rows and columns, opposite-corner sums must agree.

Worked solution 3

Let $r_i$ and $c_j$ count moves on row i and column j. The final entry is $a_{ij}=r_i+c_j$ . Consequently $a_{11}+a_{22}=r_1+c_1+r_2+c_2=a_{12}+a_{21}$ . The proposed final board would make the left side 1 and the right side 0, a contradiction. This argument even rules out the target if subtracting whole rows or columns were allowed.

Conclusion: Impossible: every reachable board has a11+a22=a12+a21.

Review the idea: Proof methods

Question 4

In triangle ABC, D is the midpoint of BC. The line through D parallel to AC meets AB at E, and the line through D parallel to AB meets AC at F. Prove that AEDF is a parallelogram and that its perimeter is $AB+AC$ .

Hint 1

Use similarity to locate E and F on their sides.

Hint 2

Both are midpoints, and opposite sides of AEDF are parallel.

Worked solution 4

Because DE is parallel to AC, triangles BDE and BCA are similar. From BD/BC=1/2 we obtain BE/BA=1/2, so AE=AB/2. Similarly DF parallel to AB gives CF/CA=1/2, so AF=AC/2. In quadrilateral AEDF, AE is parallel to DF and AF is parallel to ED, so it is a parallelogram. Its perimeter is $2(AE+AF)=AB+AC$ .

Conclusion: AEDF is a parallelogram with perimeter AB+AC.

Review the idea: Similar triangles · The midpoint theorem

Question 5

Prove that $n^2+n+1$ is not divisible by 5 for any integer n.

Hint 1

Only the remainder of n modulo 5 matters.

Hint 2

Compute the expression at remainders 0,1,2,3,4.

Worked solution 5

Every integer has one of the remainders 0,1,2,3,4 modulo 5. The corresponding values of n²+n+1 have remainders 1,3,2,3,1. None is 0, so divisibility by 5 is impossible. Negative integers also belong to these same five residue classes, so they require no separate exception.

Conclusion: No integer n produces a multiple of 5.

Review the idea: Remainders

Question 6

Find all real x for which $|x-1|+|x-3|=|2x-4|$ .

Hint 1

View the right side as $|(x-1)+(x-3)|$ .

Hint 2

Equality holds when the two summands have the same sign, allowing zero.

Worked solution 6

For x at most 1, both x-1 and x-3 are nonpositive, so both sides equal 4-2x. For x at least 3, both are nonnegative, so both sides equal 2x-4. If 1<x<3, the left side is 2, while $|2x-4|\lt 2$ . Thus no interior value works. The complete solution is $( -\infty,1]\cup[3,\infty)$ .

Conclusion: x≤1 or x≥3.

Review the idea: Absolute value inequalities

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.