Solutions by H Reuvers


International Mathematical Olympiad 2014

1) Let a0 , a1 , a2 , ... be a strictly increasing sequence of positive integers.
Prove there exists a unique positive integer n such that (a0 + a1 + ... + an )/n is greater than an but at most as great as an+1 .

Solution:

For each positive integer k, let tk := (ak - a0 ) + (ak - a1 ) + (ak - ak-1 ).
We have to prove there exists a unique n such that tn is smaller than an , but tn+1 is greater than an+1 .
Indeed, t1 is smaller than a1 , but, since (tk+1 - tk ) = (k+1)*(ak+1 - ak ), the sequence {tk } increases (much) faster than the sequence {ak } , so there is a unique n where {tk } goes past {ak }.

3) Let ABCD be a convex triangle with right angles at B and D. Let H on BD be such that AH is perpendicular to BD.
Let S and T be points on AB and AD, resp, such that H lies within triangle SCT and the angles CHS and CHT are 90 degrees greater than the angles CSB and CTD, resp.
Prove that the circumscribed circle of triangle STH is tangent to BD.

Partial solution:

The situation depicted in the picture amounts to a system of 25 linear equations in 24 unknown real numbers :
4 equations say the angles ABC, ADC, AHB, AHD are right, 4 others say the angles ASB, ATD, BHD are straight, 17 others say the sum of the angles in any of the 17 triangles is 180 degrees.

We want to prove that these 25 equations together with the premises e+d-u = 90 = b+c-v imply j-a = w-f , which is equivalent to the existence of M on AH such that MH = MS = MT.

I solved the system of 27 equations by expressing 18 of the variables in the other 6 (p,a,m,k,e,x), but I still need an extra equation which must yield m = e+x-90.

4) The points P and Q are lying on the side BC of the acute angled triangle ABC so that the angles PAB and BCA are equal, and likewise the angles CAQ and ABC.
The points M and N are lying on the straight lines AP and AQ, respectively, so that P is the center of segment AM and Q the center of segment AN.
Prove that the intersection point of the straight lines BM and CN is lying on the circumscribed circle of the triangle ABC.

Solution:


International Mathematical Olympiad 2011

1) For any set A of four distinct positive integers a1,a2,a3,a4 let sA be the sum of the elements of A and nA the number of pairs {ai,aj} of its elements such that ai + aj divides sA. Determine the sets A such that nA is maximal.

Solution:

Without loss of generality we assume a1 < a2 < a3 < a4.

nA is maximal iff ((a1+a2 divides a3+a4) and (a1+a3 divides a2+a4) and (a1+a4 divides a2+a3)) and (a2+a3 divides a1+a4)), so iff ((a1+a2 divides a3+a4) and (a1+a3 divides a2+a4) and (a1+a4 = a2+a3)).

So the elements of A are (in increasing order) a,b,c,b+c-a, where for some positive integers m,n we must have 2c+b-a = (1+m)(a+b) and 2b+c-a = (1+n)(a+c) and m is greater than n.

When we procede to express b and c in a, we get, somewhere in the middle of the calculations: (4-mn)c = (mn+4m+4)a.
So mn can only be 2 or 3, so ((m=2 and n=1) or ((m=3 and n=1)).
After completing the calculations we get A={a,5a,7a,11a} or A={a,11a,19a,29a}.

4) Let n be a natural number. We have a balance and n weights with masses 2i, i=0,..,n-1. We want to lay all weights successively on the balance, either at the left side or at the right side, in such a way that the right side is never heavier than the left side. In how many ways can we do this?

Solution:

The total number of laying all weights on the balance is n!*2n. We have to determine the fraction of layings that go wrong.
Since 2k is greater than SUM(0≤i≤k-1: 2i) for all k, the laying goes wrong iff we place a weight at the right side that is heavier than all weights laid previously.
I wrote a program to approximate the fraction f of layings that go wrong for n=1,..,100.
Among the output: for n=1 we have f=0.5, for n=2 we have f=0.375, for n=5 we have f=0.247.., for n=10 we have f= 0.177..., for n=30 we have f=0.10.., for n=80 we have f=0.06.., for n=100 we have f=0.05...

The program:

program weegschaal;
var m,n,k,max,keus:0..100; aantal,aantalgoed:real;
gehad:array[1..100] of boolean; fout:boolean;
begin
readln(n);
aantal:=0; aantalgoed:=0;
while true do
  begin
    aantal:=aantal+1;
    for m:=1 to n do gehad[m]:=false;
    k:=0; max:=0; fout:=false;
    repeat
      k:=k+1;
      repeat keus:=random(n)+1 until (not gehad[keus]);
      gehad[keus]:=true;
      if random>0.5 then if keus>max then fout:=true;
      if keus>max then max:=keus
    until (fout or (k=n));
    if not fout then aantalgoed:=aantalgoed+1;
    writeln(aantalgoed/aantal:13:10)
  end
end.

Studying the output, I found the solution as follows:
The probability that with placing the k-th weight the laying goes wrong is 1/(2k). (This is the probability that the k-th weight is the maximum up to now and we place it to the right.)
So the fraction of layings that do not go wrong is PRODUCT(1≤k≤n: 1-1/(2k)).
So the number of correct layings is (2n-1)(2n-3)(2n-5)...5.3.1.


International Mathematical Olympiad 2001 (Allegedly, no solution has been published up to now.)

5) Let ABC be a triangle and let the angle at A be 60 degrees. Let X belong to BC such that AX bisects the angle at A. Let Y belong to AC such that BY bisects the angle at B. Suppose AX+BX = AY+BY. Determine the possible values in degrees of the angle β at B.

Solution:

The possible values are β=60 and β=30 only. This can be proved as follows:

Give coordinates A(0,0), B(c,0), C(1/2,√3/2) = C(c-ρcos(β),ρsin(β)), X(τ(√3/2,1/2)) = X(c-σcos(β),σsin(β)), Y(ζ(1/2,√3/2) = Y(c-ωcos(β/2),ωsin(β/2)).
From AX+BX = AY+BY we have σ+τ = ζ+ω.

Now we can express c,ρ,σ,τ,ω,ζ in β. Subsequently, using σ+τ = ζ+ω we get:

(2sin(β)+1)(sin(β/2)+√3cos(β/2)) = (2sin(β/2)+√3)(cos(β)+√3sin(β)).

Let p:=sin(β/2). We use the wellknown goniometrical formulas and take squares to eliminate roots. Then we get a polynomial equation in p of degree six.
Since p=0 and p=1/2 (corresponding to β=60) clearly are solutions, we can divide by p and 2p-1. Thereafter we have:

(64-32√3)p4 + (-16+16√3)p3 + (-80+48√3)p2 + (20-12√3)p + (24-14√3) = 0.

After a numerical exploration up to 10 decimal places we guess this equation has no other solution than the one corresponding to β=30 and (hence) p=(√6-√2)/4.
For this value of p the left hand side is exactly zero.
Analysis of the derivatives shows that there are no other zeros of the left hand side between 0 and 1.


International Mathematical Olympiad 2013

4) Let ΔABC be an acute-angled triangle with orthocenter H. Let W be a point distinct from B and C on the segment BC.
Let ω1 be the circumscribed circle of ΔBWH and let X be the point on ω1 such that WX is a diameter of ω1.
Analogously, let ω2 be the circumscribed circle of ΔCWH and let Y be the point on ω2 such that WY is a diameter of ω2.
Prove that X,Y,H are collinear.

Solution:

Give coordinates H(0,0), A(-a,-1), B(b,-1), C(0,ab-1) with a,b positive and ab greater than 1. We have W(b-tb,tab-1) for some positive t smaller than 1.

Let M be the center of ω1. We find M((ab2-2b+tb+ta2b-a)/(2ab-2),(ta2b2-2ab+tb2+1-b2)/(2ab-2)) and X((ta2b+tab2-a-b)/(ab-1),(tb2+tab-ab-b2)/(ab-1)).

Let N be the center of ω2. We find N((a+b-tb-ta2b)/2,(ab-1)/2) and Y(a-ta2b,ab-tab).

We see X,Y,H are lying on the line y = mx with m = (b(1-t)/(1-tab)).


International Mathematical Olympiad 1997 (generalized problem)

5) For natural numbers s and t find natural numbers a and b such that abs = bat.

Solution:

Of course, a=b=1 yields a solution, and if s=t then a=b. Henceforth we only observe non-trivial solutions.
Then from unique prime factorization we deduce there must be natural numbers n, a1, ... , an, b1, ... , bn, and primes p1, ... , pn, such that
a = p1a1 ... pnan, b = p1b1 ... pnbn
and aibs = biat for all i ∈ {1,..,n}.
So there must exist natural numbers f and g with gcd(f,g)=1 such that fai = gbi, and natural numbers ci := bi/f = ai/g.
For c := p1c1 ... pncn we have a = cg, b = cf and g/f = ai/bi = at/bs = cgt/cfs, so fcgt = gcfs.
Now for given s and t we readily find the solutions.

For instance, if s=2 and t=1 we have fcg = gc2f. Then we consider three cases:
If g < 2f then f/g = c2f-g, g=1, f = c2f-1. If c=1 we get trivial solutions, else f ≥ 22f-1, yielding a contradiction.
If g = 2f then from fcg = gc2f we find g=f, yielding a contradiction.
If g > 2f then g/f = cg-2f, f=1, g = cg-2, so either g=3, c=3, b=3, a=27, or g=4, c=2, b=2, a=16.

Another example: s=3 and t=2, so fc2g = gc3f.
As above we find one nontrivial case with g = c2g-3 and f=1, g=2, c=2, b=2, a=4.



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