MathDB

1993 IMO Shortlist

Part of IMO Shortlist

Subcontests

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Which IMO Shortlist Problem formulation do you prefer?

Let c1,…,cn∈Rc_1, \ldots, c_n \in \mathbb{R} with n≥2n \geq 2 such that 0≤∑i=1nci≤n. 0 \leq \sum^n_{i=1} c_i \leq n. Show that we can find integers k1,…,knk_1, \ldots, k_n such that ∑i=1nki=0 \sum^n_{i=1} k_i = 0 and 1−n≤ci+n⋅ki≤n 1-n \leq c_i + n \cdot k_i \leq n for every i=1,…,n.i = 1, \ldots, n. [hide="Another formulation:"] Let x1,…,xn,x_1, \ldots, x_n, with n≥2n \geq 2 be real numbers such that ∣x1+…+xn∣≤n. |x_1 + \ldots + x_n| \leq n. Show that there exist integers k1,…,knk_1, \ldots, k_n such that ∣k1+…+kn∣=0. |k_1 + \ldots + k_n| = 0. and ∣xi+2⋅n⋅ki∣≤2⋅n−1 |x_i + 2 \cdot n \cdot k_i| \leq 2 \cdot n -1 for every i=1,…,n.i = 1, \ldots, n. In order to prove this, denote ci=1+xi2c_i = \frac{1+x_i}{2} for i=1,…,n,i = 1, \ldots, n, etc.

Highly recommended by the Problem Committee

The vertices D,E,FD,E,F of an equilateral triangle lie on the sides BC,CA,ABBC,CA,AB respectively of a triangle ABC.ABC. If a,b,ca,b,c are the respective lengths of these sides, and SS the area of ABC,ABC, prove that DE≥2⋅2⋅Sa2+b2+c2+4⋅3⋅S. DE \geq \frac{2 \cdot \sqrt{2} \cdot S}{\sqrt{a^2 + b^2 + c^2 + 4 \cdot \sqrt{3} \cdot S}}.
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American Shortlist Inequality

Prove that ab+2c+3d+bc+2d+3a+cd+2a+3b+da+2b+3c≥23 \frac{a}{b+2c+3d} +\frac{b}{c+2d+3a} +\frac{c}{d+2a+3b}+ \frac{d}{a+2b+3c} \geq \frac{2}{3} for all positive real numbers a,b,c,da,b,c,d.

Orthic triangle inequality

Let triangle ABCABC be such that its circumradius is R=1.R = 1. Let rr be the inradius of ABCABC and let pp be the inradius of the orthic triangle A′B′C′A'B'C' of triangle ABC.ABC. Prove that p≤1−13⋅(1+r)2. p \leq 1 - \frac{1}{3 \cdot (1+r)^2}. [hide="Similar Problem posted by Pascual2005"]
Let ABCABC be a triangle with circumradius RR and inradius rr. If pp is the inradius of the orthic triangle of triangle ABCABC, show that pR≤1−(1+rR)23\frac{p}{R} \leq 1 - \frac{\left(1+\frac{r}{R}\right)^2}{3}.
Note. The orthic triangle of triangle ABCABC is defined as the triangle whose vertices are the feet of the altitudes of triangle ABCABC.
SOLUTION 1 by mecrazywong:
p=2Rcos⁡Acos⁡Bcos⁡C,1+rR=1+4sin⁡A/2sin⁡B/2sin⁡C/2=cos⁡A+cos⁡B+cos⁡Cp=2R\cos A\cos B\cos C,1+\frac{r}{R}=1+4\sin A/2\sin B/2\sin C/2=\cos A+\cos B+\cos C. Thus, the ineqaulity is equivalent to 6cos⁡Acos⁡Bcos⁡C+(cos⁡A+cos⁡B+cos⁡C)2≤36\cos A\cos B\cos C+(\cos A+\cos B+\cos C)^2\le3. But this is easy since cos⁡A+cos⁡B+cos⁡C≤3/2,cos⁡Acos⁡Bcos⁡C≤1/8\cos A+\cos B+\cos C\le3/2,\cos A\cos B\cos C\le1/8.
SOLUTION 2 by Virgil Nicula:
I note the inradius r′r' of a orthic triangle.
Must prove the inequality r′R≤1−13(1+rR)2.\frac{r'}{R}\le 1-\frac 13\left( 1+\frac rR\right)^2.
From the wellknown relations r′=2Rcos⁡Acos⁡Bcos⁡Cr'=2R\cos A\cos B\cos C
and cos⁡Acos⁡Bcos⁡C≤18\cos A\cos B\cos C\le \frac 18 results r′R≤14.\frac{r'}{R}\le \frac 14.
But 14≤1−13(1+rR)2⟺13(1+rR)2≤34⟺\frac 14\le 1-\frac 13\left( 1+\frac rR\right)^2\Longleftrightarrow \frac 13\left( 1+\frac rR\right)^2\le \frac 34\Longleftrightarrow
(1+rR)2≤(32)2⟺1+rR≤32⟺rR≤12⟺2r≤R\left(1+\frac rR\right)^2\le \left(\frac 32\right)^2\Longleftrightarrow 1+\frac rR\le \frac 32\Longleftrightarrow \frac rR\le \frac 12\Longleftrightarrow 2r\le R (true).
Therefore, r′R≤14≤1−13(1+rR)2⟹r′R≤1−13(1+rR)2.\frac{r'}{R}\le \frac 14\le 1-\frac 13\left( 1+\frac rR\right)^2\Longrightarrow \frac{r'}{R}\le 1-\frac 13\left( 1+\frac rR\right)^2.
SOLUTION 3 by darij grinberg:
I know this is not quite an ML reference, but the problem was discussed in Hyacinthos messages #6951, #6978, #6981, #6982, #6985, #6986 (particularly the last message).

representation of the number a in the base b

Let a,b,na,b,n be positive integers, b>1b > 1 and bn−1∣a.b^n-1\mid a. Show that the representation of the number aa in the base bb contains at least nn digits different from zero.
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Pairs of integers for ad^2 + 2bde + ce^2= n

a>0a > 0 and bb, cc are integers such that acac – b2b^2 is a square-free positive integer P. [hide="For example"] P could be 3∗53*5, but not 32∗53^2*5. Let f(n)f(n) be the number of pairs of integers d,ed, e such that ad2+2bde+ce2=nad^2 + 2bde + ce^2= n. Show thatf(n)f(n) is finite and that f(n)=f(Pkn)f(n) = f(P^{k}n) for every positive integer kk.
Original Statement:
Let a,b,ca,b,c be given integers a>0,a > 0, ac−b2=P=P1⋯Pnac-b^2 = P = P_1 \cdots P_n where P1⋯PnP_1 \cdots P_n are (distinct) prime numbers. Let M(n)M(n) denote the number of pairs of integers (x,y)(x,y) for which ax2+2bxy+cy2=n. ax^2 + 2bxy + cy^2 = n. Prove that M(n)M(n) is finite and M(n)=M(Pk⋅n)M(n) = M(P_k \cdot n) for every integer k≥0.k \geq 0. Note that the "nn" in PNP_N and the "nn" in M(n)M(n) do not have to be the same.

Poland goes Combinatorics

Let SnS_n be the number of sequences (a1,a2,…,an),(a_1, a_2, \ldots, a_n), where ai∈{0,1},a_i \in \{0,1\}, in which no six consecutive blocks are equal. Prove that Sn→∞S_n \rightarrow \infty when n→∞.n \rightarrow \infty.

Composited function and relatively prime positive integers

Let SS be the set of all pairs (m,n)(m,n) of relatively prime positive integers m,nm,n with nn even and m<n.m < n. For s=(m,n)∈Ss = (m,n) \in S write n=2k⋅non = 2^k \cdot n_o where k,n0k, n_0 are positive integers with n0n_0 odd and define f(s)=(n0,m+n−n0). f(s) = (n_0, m + n - n_0). Prove that ff is a function from SS to SS and that for each s=(m,n)∈S,s = (m,n) \in S, there exists a positive integer t≤m+n+14t \leq \frac{m+n+1}{4} such that ft(s)=s, f^t(s) = s, where ft(s)=(f∘f∘⋯∘f)⏟t times(s). f^t(s) = \underbrace{ (f \circ f \circ \cdots \circ f) }_{t \text{ times}}(s). If m+nm+n is a prime number which does not divide 2k−12^k - 1 for k=1,2,…,m+n−2,k = 1,2, \ldots, m+n-2, prove that the smallest value tt which satisfies the above conditions is [m+n+14]\left [\frac{m+n+1}{4} \right ] where [x]\left[ x \right] denotes the greatest integer ≤x.\leq x.
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India proudly presents an ISL sequence

Define a sequence ⟨f(n)⟩n=1∞\langle f(n)\rangle^{\infty}_{n=1} of positive integers by f(1)=1f(1) = 1 and f(n)={f(n−1)−n if f(n−1)>n;f(n−1)+n if f(n−1)≤n,f(n) = \begin{cases} f(n-1) - n & \text{ if } f(n-1) > n;\\ f(n-1) + n & \text{ if } f(n-1) \leq n, \end{cases} for n≥2.n \geq 2. Let S={n∈N  ∣  f(n)=1993}.S = \{n \in \mathbb{N} \;\mid\; f(n) = 1993\}.
(i) Prove that SS is an infinite set. (ii) Find the least positive integer in S.S. (iii) If all the elements of SS are written in ascending order as n1<n2<n3<…, n_1 < n_2 < n_3 < \ldots , show that lim⁡i→∞ni+1ni=3. \lim_{i\rightarrow\infty} \frac{n_{i+1}}{n_i} = 3.

Partition of the positive rationals into disjoint subsets

a) Show that the set Q+ \mathbb{Q}^{ + } of all positive rationals can be partitioned into three disjoint subsets. A,B,C A,B,C satisfying the following conditions: BA=B;&B2=C;&BC=A; BA = B; \& B^2 = C; \& BC = A; where HK HK stands for the set {hk:h∈H,k∈K} \{hk: h \in H, k \in K\} for any two subsets H,K H, K of Q+ \mathbb{Q}^{ + } and H2 H^2 stands for HH. HH. b) Show that all positive rational cubes are in A A for such a partition of Q+. \mathbb{Q}^{ + }. c) Find such a partition Q+=A∪B∪C \mathbb{Q}^{ + } = A \cup B \cup C with the property that for no positive integer n≤34, n \leq 34, both n n and n+1 n + 1 are in A, A, that is, min{n∈N:n∈A,n+1∈A}>34. \text{min} \{n \in \mathbb{N}: n \in A, n + 1 \in A \} > 34.

Incenter I is the mid point of DE [mixtilinear incircle]

Let ABCABC be a triangle, and II its incenter. Consider a circle which lies inside the circumcircle of triangle ABCABC and touches it, and which also touches the sides CACA and BCBC of triangle ABCABC at the points DD and EE, respectively. Show that the point II is the midpoint of the segment DEDE.