MathDB

2006 IMO Shortlist

Part of IMO Shortlist

Subcontests

(9)
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Miquel circles and a beautiful similarity

Points A1 A_{1}, B1 B_{1}, C1 C_{1} are chosen on the sides BC BC, CA CA, AB AB of a triangle ABC ABC respectively. The circumcircles of triangles AB1C1 AB_{1}C_{1}, BC1A1 BC_{1}A_{1}, CA1B1 CA_{1}B_{1} intersect the circumcircle of triangle ABC ABC again at points A2 A_{2}, B2 B_{2}, C2 C_{2} respectively (A2≠A,B2≠B,C2≠C A_{2}\neq A, B_{2}\neq B, C_{2}\neq C). Points A3 A_{3}, B3 B_{3}, C3 C_{3} are symmetric to A1 A_{1}, B1 B_{1}, C1 C_{1} with respect to the midpoints of the sides BC BC, CA CA, AB AB respectively. Prove that the triangles A2B2C2 A_{2}B_{2}C_{2} and A3B3C3 A_{3}B_{3}C_{3} are similar.
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three "old" circles and four concurrent lines

Circles w1 w_{1} and w2 w_{2} with centres O1 O_{1} and O2 O_{2} are externally tangent at point D D and internally tangent to a circle w w at points E E and F F respectively. Line t t is the common tangent of w1 w_{1} and w2 w_{2} at D D. Let AB AB be the diameter of w w perpendicular to t t, so that A,E,O1 A, E, O_{1} are on the same side of t t. Prove that lines AO1 AO_{1}, BO2 BO_{2}, EF EF and t t are concurrent.

tiling holey triangles with diamonds

A holey triangle is an upward equilateral triangle of side length nn with nn upward unit triangular holes cut out. A diamond is a 60∘−120∘60^\circ-120^\circ unit rhombus. Prove that a holey triangle TT can be tiled with diamonds if and only if the following condition holds: Every upward equilateral triangle of side length kk in TT contains at most kk holes, for 1≤k≤n1\leq k\leq n.
Proposed by Federico Ardila, Colombia

local champion integers

Let a>b>1 a > b > 1 be relatively prime positive integers. Define the weight of an integer c c, denoted by w(c) w(c) to be the minimal possible value of |x| \plus{} |y| taken over all pairs of integers x x and y y such that ax \plus{} by \equal{} c. An integer c c is called a local champion if w(c)≥w(c±a) w(c) \geq w(c \pm a) and w(c)≥w(c±b) w(c) \geq w(c \pm b).
Find all local champions and determine their number.
Proposed by Zoran Sunic, USA
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common tangents and a superb similar triangle with ABC

In a triangle ABC ABC, let Ma M_{a}, Mb M_{b}, Mc M_{c} be the midpoints of the sides BC BC, CA CA, AB AB, respectively, and Ta T_{a}, Tb T_{b}, Tc T_{c} be the midpoints of the arcs BC BC, CA CA, AB AB of the circumcircle of ABC ABC, not containing the vertices A A, B B, C C, respectively. For i∈{a,b,c} i \in \left\{a, b, c\right\}, let wi w_{i} be the circle with MiTi M_{i}T_{i} as diameter. Let pi p_{i} be the common external common tangent to the circles wj w_{j} and wk w_{k} (for all {i,j,k}={a,b,c} \left\{i, j, k\right\}= \left\{a, b, c\right\}) such that wi w_{i} lies on the opposite side of pi p_{i} than wj w_{j} and wk w_{k} do. Prove that the lines pa p_{a}, pb p_{b}, pc p_{c} form a triangle similar to ABC ABC and find the ratio of similitude.
Proposed by Tomas Jurik, Slovakia

"antipodal" points of a polyhaedron

Consider a convex polyhedron without parallel edges and without an edge parallel to any face other than the two faces adjacent to it. Call a pair of points of the polyhedron antipodal if there exist two parallel planes passing through these points and such that the polyhedron is contained between these planes. Let AA be the number of antipodal pairs of vertices, and let BB be the number of antipodal pairs of midpoint edges. Determine the difference A−BA-B in terms of the numbers of vertices, edges, and faces.
Proposed by Kei Irei, Japan

n divides 2^m+m

For all positive integers nn, show that there exists a positive integer mm such that nn divides 2m+m2^{m} + m.
Proposed by Juhan Aru, Estonia
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n variable inequality

Prove the inequality: \sum_{i < j}{\frac {a_{i}a_{j}}{a_{i} \plus{} a_{j}}}\leq \frac {n}{2(a_{1} \plus{} a_{2} \plus{}\cdots \plus{} a_{n})}\cdot \sum_{i < j}{a_{i}a_{j}} for positive reals a1,a2,…,an a_{1},a_{2},\ldots,a_{n}.
Proposed by Dusan Dukic, Serbia

n x n square and strawberries

A cake has the form of an n n x n n square composed of n2 n^{2} unit squares. Strawberries lie on some of the unit squares so that each row or column contains exactly one strawberry; call this arrangement A\mathcal{A}.
Let B\mathcal{B} be another such arrangement. Suppose that every grid rectangle with one vertex at the top left corner of the cake contains no fewer strawberries of arrangement B\mathcal{B} than of arrangement A\mathcal{A}. Prove that arrangement B\mathcal{B} can be obtained from A \mathcal{A} by performing a number of switches, defined as follows:
A switch consists in selecting a grid rectangle with only two strawberries, situated at its top right corner and bottom left corner, and moving these two strawberries to the other two corners of that rectangle.

France TST 2007

A point DD is chosen on the side ACAC of a triangle ABCABC with ∠C<∠A<90∘\angle C < \angle A < 90^\circ in such a way that BD=BABD=BA. The incircle of ABCABC is tangent to ABAB and ACAC at points KK and LL, respectively. Let JJ be the incenter of triangle BCDBCD. Prove that the line KLKL intersects the line segment AJAJ at its midpoint.
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sequence positive

The sequence of real numbers a0,a1,a2,…a_0,a_1,a_2,\ldots is defined recursively by a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0 \text{for}  n\geq 1.Show that an>0 a_{n} > 0 for all n≥1 n\geq 1.
Proposed by Mariusz Skalba, Poland

x is rational implies y is rational

For x∈(0,1) x \in (0, 1) let y∈(0,1) y \in (0, 1) be the number whose n n-th digit after the decimal point is the 2n 2^{n}-th digit after the decimal point of x x. Show that if x x is rational then so is y y.
Proposed by J.P. Grossman, Canada

four points lie on a circle

Let ABCD ABCD be a trapezoid with parallel sides AB>CD AB > CD. Points K K and L L lie on the line segments AB AB and CD CD, respectively, so that AK/KB=DL/LCAK/KB=DL/LC. Suppose that there are points P P and Q Q on the line segment KL KL satisfying \angle{APB} \equal{} \angle{BCD}\qquad\text{and}\qquad \angle{CQD} \equal{} \angle{ABC}. Prove that the points P P, Q Q, B B and C C are concyclic.
Proposed by Vyacheslev Yasinskiy, Ukraine
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Italian WinterCamps test07 Problem5

A sequence of real numbers a0, a1, a2,… a_{0},\ a_{1},\ a_{2},\dots is defined by the formula a_{i \plus{} 1} \equal{} \left\lfloor a_{i}\right\rfloor\cdot \left\langle a_{i}\right\rangle\qquad\text{for}  i\geq 0; here a0a_0 is an arbitrary real number, ⌊ai⌋\lfloor a_i\rfloor denotes the greatest integer not exceeding aia_i, and ⟨ai⟩=ai−⌊ai⌋\left\langle a_i\right\rangle=a_i-\lfloor a_i\rfloor. Prove that ai=ai+2a_i=a_{i+2} for ii sufficiently large.
Proposed by Harmel Nestra, Estionia

n lamps

We have n≥2 n \geq 2 lamps L1,...,Ln L_{1}, . . . ,L_{n} in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows: if the lamp Li L_{i} and its neighbours (only one neighbour for i \equal{} 1 or i \equal{} n, two neighbours for other i i) are in the same state, then Li L_{i} is switched off; – otherwise, Li L_{i} is switched on. Initially all the lamps are off except the leftmost one which is on. (a) (a) Prove that there are infinitely many integers n n for which all the lamps will eventually be off. (b) (b) Prove that there are infinitely many integers n n for which the lamps will never be all off.
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