MO Schulrunde Klasse 9 Mock Paper 1 with Solutions

MO Schulrunde Klasse 9 Mock Paper 1 · 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 $n(n+1)(n+2)(n+3)+1$ is a perfect square for every integer n.

Hint 1

Pair the first factor with the last and the middle two together.

Hint 2

Put $t=n^2+3n$ .

Worked solution 1

The outer pair has product $n(n+3)=n^2+3n=t$ , and the inner pair has product $(n+1)(n+2)=t+2$ . Therefore the expression is $t(t+2)+1=(t+1)^2$ . Since t is an integer, this is an integer square, including when n is negative or one of the factors is zero. Explicitly, the square is $(n^2+3n+1)^2$ .

Conclusion: The expression equals the integer square (n²+3n+1)².

Review the idea: Algebraic identities

Question 2

Positive real numbers x,y,z satisfy $x+y+z=1$ . Prove $(x+y)(y+z)(z+x)\le8/27$ , and find every equality case.

Hint 1

The three factors have a fixed sum.

Hint 2

Apply AM–GM to x+y, y+z and z+x.

Worked solution 2

The three positive factors sum to $2(x+y+z)=2$ . The arithmetic–geometric mean inequality therefore gives $(x+y)(y+z)(z+x)\le((2)/3)^3=8/27$ . Equality in this inequality holds exactly when the three factors are equal. From x+y=y+z=z+x we obtain x=y=z, and the required sum then gives x=y=z=1/3. This triple does attain the bound.

Conclusion: Maximum 8/27, exactly at x=y=z=1/3.

Review the idea: Arithmetic geometric and harmonic means

Question 3

Five points lie inside or on a square of side 1. Prove that two of them are at distance at most $1/\sqrt2$ . Points need not be distinct.

Hint 1

Divide the square into four equal smaller squares.

Hint 2

Assign points on dividing lines consistently to one neighbouring small square.

Worked solution 3

Cut the square along its two midlines. Assign a point on the vertical midline to the right half, and a point on the horizontal midline to the upper half. Thus the centre belongs to the upper-right quarter and every point belongs to exactly one of the four boxes. Five points in four boxes force two in the same quarter-square. Their horizontal and vertical separations are each at most 1/2. Pythagoras gives squared distance at most $(1/2)^2+(1/2)^2=1/2$ , hence distance at most $1/\sqrt2$ . Coincident points already satisfy the conclusion.

Conclusion: A pair is forced into one quarter-square, whose diagonal is 1/√2.

Review the idea: Pigeonhole principle · Pythagoras and stewart

Question 4

In a convex trapezoid ABCD, with vertices in boundary order and $AB\parallel CD$ , diagonals AC and BD meet at O. Prove that triangles AOD and BOC have equal areas.

Hint 1

Compare triangles ABD and ABC first.

Hint 2

They share base AB and have equal altitudes because C and D lie on a parallel line.

Worked solution 4

Write [XYZ] for the area of triangle XYZ. Since C and D lie on a line parallel to AB, their perpendicular distances to AB are equal. Hence [ABD]=[ABC]. Convexity puts O on both diagonal segments, so [ABD]=[ABO]+[AOD] and [ABC]=[ABO]+[BOC]. Subtracting the common area [ABO] proves [AOD]=[BOC].

Conclusion: The two areas are equal by subtracting the common triangle ABO.

Review the idea: Triangle area ratios

Question 5

Find all ordered pairs of positive integers $x,y$ satisfying $xy=2x+3y+6.$

Hint 1

Move the two linear terms to the left and complete the product $(x-3)(y-2)$ .

Hint 2

First show $x>3$ and $y>2$ . Then list the positive factor pairs of 12.

Worked solution 5

The mixed product suggests collecting terms into two factors. Rearranging gives $xy-2x-3y=6$ . Adding 6 to both sides completes the product: $(x-3)(y-2)=12.$ We must check the factors are positive before listing them. The original equation gives $(x-3)y=2x+6>0$ . Since y is positive, x>3. Similarly, $x(y-2)=3y+6>0$ gives y>2.

Thus x−3 is a positive divisor d of 12, and y−2=12/d. The choices $d=1,2,3,4,6,12$ give, respectively, $(x,y)=(4,14),(5,8),(6,6),(7,5),(9,4),(15,3).$ Each pair is positive and satisfies the completed-product equation. Expanding that equation back gives the original equation, so every listed pair works and no others are possible.

Conclusion: Exactly $(4,14),(5,8),(6,6),(7,5),(9,4),(15,3)$ .

Review the idea: Factorisation integer solutions · Divisibility

Question 6

Determine all positive integers n for which there is a permutation $(a_1,\ldots,a_n)$ of $1,\ldots,n$ such that every sum $i+a_i$ is odd.

Hint 1

Odd positions must contain even values.

Hint 2

Compare the numbers of odd positions and even values; for even n exchange neighbouring pairs.

Worked solution 6

If n is odd, there are (n+1)/2 odd positions but only (n-1)/2 even values. Since every odd position would need an even value, this is impossible. If n is even, use $(a_1,a_2,a_3,a_4,\ldots)=(2,1,4,3,\ldots,n,n-1)$ . Each position and its value have opposite parity, so every sum is odd. Thus exactly the even positive integers n work.

Conclusion: Exactly the positive even integers n.

Review the idea: Parity · Pigeonhole principle

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.