MO Regionalrunde Klasse 9 Mock Paper 4 · 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

Positive coprime integers a,b satisfy a2+b2=c2a^2+b^2=c^2 for an integer c. Prove that exactly one of a,b is divisible by 3, and that the even one of a,b is divisible by 4.

Hint 1

Use squares modulo 3 and modulo 4 first.

Hint 2

Then examine what would happen if the even leg were 2 modulo 4, using squares modulo 16.

Worked solution 1

Modulo 3, a square is 0 or 1. If neither a nor b were divisible by 3, c² would be 2 modulo 3, impossible. They cannot both be divisible by 3 because they are coprime, so exactly one is. They cannot both be odd, since their squared sum would be 2 modulo 4; they cannot both be even by coprimality. Thus exactly one is even. If that even integer were 2 modulo 4, its square would be 4 modulo 16. The other, odd square is 1 or 9 modulo 16. Their sum would be 5 or 13, neither a square residue modulo 16 (0,1,4,9). Contradiction: the even leg is divisible by 4.

Conclusion: Exactly one leg is divisible by 3; the even leg is divisible by 4.

Review the idea: Remainders · Greatest common divisor

Question 2

Real numbers x,y,z satisfy x+y+z=1. Find the minimum of x2+y2+z2+xyx^2+y^2+z^2+xy, and give every triple attaining it.

Hint 1

Use s=x+y and d=x−y.

Hint 2

Complete a square in s; the d² term is nonnegative.

Worked solution 2

Put s=x+y, d=x−y, so z=1−s. Then x2+y2+xy=(3s2+d2)/4x^2+y^2+xy=(3s^2+d^2)/4. Therefore the expression equals 3s2+d24+(1−s)2=74(s−47)2+d24+37.\frac{3s^2+d^2}{4}+(1-s)^2=\frac74\left(s-\frac47\right)^2+\frac{d^2}{4}+\frac37. Its minimum is 3/7, achieved exactly when s=4/7 and d=0. Thus x=y=2/7 and z=3/7. These real values meet the given constraint and attain the minimum.

Conclusion: Minimum 3/7, uniquely at (2/7,2/7,3/7).

Review the idea: Sum of squares

Question 3

A family of subsets of a ten-element set has the property that any two distinct members have at least one common element. Determine its greatest possible size, and give a family attaining the bound.

Hint 1

Pair each subset with its complement.

Hint 2

At most one member of each complementary pair may be used.

Worked solution 3

There are 2¹⁰=1024 subsets, paired into 512 disjoint complementary pairs. Two complementary subsets have empty intersection, so the family can contain at most one from each pair. Hence its size is at most 512. To attain this, fix one element and take every subset containing it. There are 2⁹=512 such subsets, and any two share the fixed element. This proves both the upper bound and attainability.

Conclusion: Maximum 512; all subsets containing one fixed element attain it.

Review the idea: Counting with bijections

Question 4

ABCD is a rectangle and O is its centre. For an arbitrary point P in the plane, prove PA2+PB2+PC2+PD2=4PO2+AC2.PA^2+PB^2+PC^2+PD^2=4PO^2+AC^2. Deduce where the left side is smallest.

Rectangle ABCD with centre O and an arbitrary point P outside the rectangleABCDOP

Hint 1

Choose coordinate axes through the centre parallel to the sides.

Hint 2

The linear terms cancel when the four squared distances are added.

Worked solution 4

Write the vertices as (u,v),(−u,v),(−u,−v),(u,−v), where u,v>0, and write P=(x,y). Adding the four Pythagorean distance formulas cancels all terms linear in x or y and gives 4(x2+y2)+4(u2+v2)4(x^2+y^2)+4(u^2+v^2). Here PO²=x²+y² and AC²=(2u)²+(2v)²=4(u²+v²), proving the identity. Since PO²≥0, the sum is at least AC², with equality exactly when P=O.

Conclusion: Minimum AC², attained uniquely at the rectangle centre.

Review the idea: Pythagoras and stewart · Algebraic identities

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.