7th Junior Balkan Mathematical Olympiad Problems 2003

1.  Let A = 44...4 (2n digits) and B = 88...8 (n digits). Show that A + 2B + 4 is a square.

2.  A₁, A₂, ... , A_n are points in the plane, so that if we take the points in any order B₁, B₂, ... , B_n, then the broken line B₁B₂...B_n does not intersect itself. What is the largest possible value of n?

3.  ABC is a triangle. D is the midpoint of the arc BC not containing A. Similarly E, F. DE meets BC at G and AC at H. M is the midpoint of GH. DF meets BC at I and AB at J, and N is the midpoint of IJ. Find the angles of DMN in terms of the angles of ABC. AD meets EF at P. Show that the circumcenter of DMN lies on the circumcircle of PMN.

4.  Show that (1+x²)/(1+y+z²) + (1+y²)/(1+z+x²) + (1+z²)/(1+x+y²) ≥ 2 for reals x, y, z > -1. 

Solutions

Problem 1

Let A = 44...4 (2n digits) and B = 88...8 (n digits). Show that A + 2B + 4 is a square.

Solution

We show that the square root is D = 6...68 (with n-1 6s). We have D = 6(10ⁿ - 1)/9 + 2 = (2/3)(10ⁿ + 2), so D² = (4/9)(10²ⁿ + 4·10ⁿ + 4) = (4/9)(10²ⁿ - 1) + 2(8/9)(10ⁿ - 1) + 4 = A + 2B + 4. 

Problem 2

A₁, A₂, ... , A_n are points in the plane, so that if we take the points in any order B₁, B₂, ... , B_n, then the broken line B₁B₂...B_n does not intersect itself. What is the largest possible value of n?

Answer

4

Solution

We cannot have three points collinear, because if B lies on the segment AC, then the broken line ...ACB... intersects itself. If we take any four points, then their convex hull must be a triangle. This works for 4 points. Suppose we have 5 points A, B, C, D, E. Then we can take D to be inside the triangle ABC.

But there is now nowhere to put E. If it is in 1, then ABDE is self-intersecting. Similarly, in 2, ACDE is self-intersecting, in 3 BCDE is self-intersecting, in 4 ADCE is self intersecting and so on.

Thanks to Suat Namli

Problem 3

ABC is a triangle. D is the midpoint of the arc BC not containing A. Similarly E, F. DE meets BC at G and AC at H. M is the midpoint of GH. DF meets BC at I and AB at J, and N is the midpoint of IJ. Find the angles of DMN in terms of the angles of ABC. AD meets EF at P. Show that the circumcenter of DMN lies on the circumcircle of PMN.

Answer

D = B/2 + C/2, M = A/2 + B/2, N = C/2 + A/2

Solution

∠BFD = A/2, ∠FBC = B + ∠FBA = B + C/2, so ∠BIJ = 180^o - A/2 - (B + C/2) = A/2 + C/2. Similarly, ∠BJI = A/2 + C/2. So BIJ is isosceles. Hence BN is the angle bisector of B. Similarly, CM is the angle bisector of C. So they meet on AD (the angle bisector of A) at the incenter I. Then ∠DNI = ∠DMI = 90^o, so DI is the diameter of the circumcircle of DMN. Hence ∠DNM = ∠DIM = 90^o - ∠IDM = 90^o - B/2 = A/2 + C/2. Similarly, ∠DMN = A/2 + B/2. Hence ∠MDN = B/2 + C/2.

Let O be the midpoint of DI, so that O is the circumcenter of DMN and we have to show that it lies on the circumcircle of PMN. I is the orthocenter of DEF and it is a well-known result that the midpoints of the segments joining the orthocenter to the vertices lie on the circumcircle of the orthic triangle (known as the nine-point circle, because it also contains the midpoints of the sides of DEF).

Thanks to Cristian Ilac

Problem 3

Show that (1+x²)/(1+y+z²) + (1+y²)/(1+z+x²) + (1+z²)/(1+x+y²) ≥ 2 for reals x, y, z > -1.

Solution

Thanks to Michael Bolton

If x < 0, then replacing x by -x, does not change the first two terms and reduces the third term, so we can assume x, y, z ≥ 0.

Since (x-1)2 ≥ 0 with equality iff x = 1, 3(x² + 1) ≥ 2(x² + x + 1) with equality iff x = 1.

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6th Junior Balkan Mathematical Olympiad Problems 2002

1.  The triangle ABC has CA = CB. P is a point on the circumcircle between A and B (and on the opposite side of the line AB to C). D is the foot of the perpendicular from C to PB. Show that PA + PB = 2·PD.

2.  The circles center O₁ and O₂ meet at A and B with the centers on opposite sides of AB. The lines BO₁ and BO₂ meet their respective circles again at B₁ and B₂. M is the midpoint of B₁B₂. M₁, M₂ are points on the circles center O₁ and O₂ respectively such that angle AO₁M₁ = angle AO₂M₂, and B₁ lies on the minor arc AM₁ and B lies on the minor arc AM₂. Show that angle MM₁B = angle MM₂B.

3.  Find all positive integers which have exactly 16 positive divisors 1 = d₁ < d₂ < ... < d₁₆ such that the divisor d_k, where k = d₅, equals (d₂ + d₄) d₆.

4.  Show that 2/( b(a+b) ) + 2/( c(b+c) ) + 2/( a(c+a) ) ≥ 27/(a+b+c)² for positive reals a, b, c. 

Solutions

Problem 1

The triangle ABC has CA = CB. P is a point on the circumcircle between A and B (and on the opposite side of the line AB to C). D is the foot of the perpendicular from C to PB. Show that PA + PB = 2·PD.

Solution

Take Q on the ray PB so that PD = DQ. Then CA = CB, CP = CQ, ∠CAP = ∠CBQ, so triangles CAP and CBQ are congruent, so AP = BQ. Hence 2·PD = PQ = PB + PA.

Thanks to Cristian Ilac

Problem 3

Find all positive integers which have exactly 16 positive divisors 1 = d₁ < d₂ < ... < d₁₆ such that the divisor d_k, where k = d₅, equals (d₂ + d₄) d₆.

Answer

2002

Solution

If d₂ > 2, then all divisors are odd. But (d₂ + d₄) is a divisor and must be even. Contradiction. So d₂ = 2.

If d₃ = 3, then 6 divides N (where we write the number as N), so either d₄ = 6, or d₅ = 6 or d₆ = 6. If d₄ = 6, then (d₂ + d₄) = 8 is a divisor, so 4 is a divisor, so d₄ = 4. Contradiction. If d₅ = 6, then we are given that d₆ = (d₂ + d₄) d₆ > d₆. Contradiction. If d₆ = 6, then d₄ = 4, d₅ = 5. But we are given that d₅ = (d₂ + d₄) d₆ > d₅. Contradiction. So d₃ > 3. So 3 is not a factor of N.

We are told that (d₂ + d₄) d₆ is a factor, so (d₂ + d₄) = d₄ + 2 must be a factor. If d₄ + 1 is also a factor, then we have three consecutive factors. One of them must be a multiple of 3. But we have just shown that 3 does not divide N. So d₄ + 1 is not a factor. Hence d₅ = d₄ + 2. Neither d₄ nor d₅ are multiples of 3. So d₅ must be 1 greater than a multiple of 3.

We have d₅ = k and that d_k is a factor. So in particular k ≤ 16 (since there are only 16 factors). Thus d₅ = 1, 4, 7, 10, 13 or 16. Obviously d₅ ≥ 5, so 1 and 4 are too small.

If d₅ = 7, then since 3 and 6 are not factors we must have d₃ = 4, d₄ = 5. So 10 and 14 are factors. So d₇ ≤ 10. But we are given that d₇ = (d₂ + d₄) d₆ > (2 + 5) 7 = 49. Contradiction.

If d₅ = 10, then d₄ = 8. So 4 and 5 are also factors, but only d₃ is available. Contradiction.

If d₅ = 16, then d₄ = 14. So 4, 7 and 8 are also factors, but only d₃ is available. Contradiction.

The only remaining possibility is d₅ = 13. That implies d₄ = 11. So d₃ is the only factor between 4 and 10. So it cannot be 5, 6, 8, 9 or 10. So it must be 4 or 7.

Suppose it is 4. N has prime factors 2, 11, 13. If it has another prime factor p, then it would have 16 factors, leaving aside multiples of 4 (because each of 2, 11, 13 and p could be in or out). Contradiction. So 2, 11, 13 are its only prime factors. If 11² or 13² divides N, then N has at least 3·3·2 = 18 factors (there are 3 possibilities for 2, 3 for one of 11/13 and 2 for the other). Contradiction. So 2³must divide N. Contradiction (we are supposed to have exhausted the factors under 10).

So the only remaining possibility is d₃ = 7. That gives N = 2002 and it is easy to check that d₁₃ = 182 = (d₂ + d₄) d₆.

Comment. This all seems rather complicated (although it did not take me long). Does anyone have a simpler solution?

Problem 4

Show that 2/( b(a+b) ) + 2/( c(b+c) ) + 2/( a(c+a) ) ≥ 27/(a+b+c)² for positive reals a, b, c.

Solution

Thanks to Michael Lipnowski

By AM/GM 1/( b(a+b) ) + 1/( c(b+c) ) + 1/( a(c+a) ) ≥ 3/(XY), where X = (abc)^1/3, Y = ((a+b)(b+c)(c+a))^1/3. Using AM/GM again gives X ≤ (a+b+c)/3 and Y ≤ 2(a+b+c)/3, so 3/XY ≥ (27/2) 1/(a+b+c)².

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5th Junior Balkan Mathematical Olympiad Problems 2001

1.  Find all positive integers a, b, c such that a³ + b³ + c³ = 2001.

2.  ABC is a triangle with ∠C = 90^o and CA ≠ CB. CH is an altitude and CL is an angle bisector. Show that for X ≠ C on the line CL, we have ∠XAC ≠ ∠XBC. Show that for Y ≠ C on the line CH we have ∠YAC ≠ ∠YBC.

3.  ABC is an equilateral triangle. D, E are points on the sides AB, AC respectively. The angle bisector of ∠ADE meets AE at F, and the angle bisector of ∠AED meets AD at G. Show that area DEF + area DEG ≤ area ABC. When do we have equality?

4.  N is a convex polygon with 1415 vertices and perimeter 2001. Prove that we can find three vertices of N which form a triangle of area < 1.

Solutions

Problem 1

Find all positive integers a, b, c such that a³ + b³ + c³ = 2001.

Answer

10,10,1

Solution

Recollect that the cubes are 1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64, 5³ = 125, 6³ = 216, 7³ = 343, 8³ = 512, 9³ = 729, 10³ = 1000, 11³ = 1331, 12³ = 1728, 13³ = 2197.

8³+8³+8³ = 1536 < 2001 and 13³ > 2001, so the largest of a, b, c must be 9, 10, 11 or 12. Assume a ≥ b ≥ c. If a = 9, then b³ + c³ = 1272, but 8³ + 8³ = 1024, so b = 9, leaving c³ = 543, which fails.

If a = 10, then b³ + c³ = 1001, so b = 10, giving the solution 10,10,1 or b = 9, giving c³ = 272, which fails or b = 8 giving c³ = 489 which fails.

If a = 11, then b³ + c³ = 670, so b = 8 giving c³ = 158, which fails, or b = 7, giving c³ = 327, which fails.

If a = 12, then b³ + c³ = 273, so b must be 6, giving c³ = 57, which fails.

Thanks to Suat Namli

Problem 2

ABC is a triangle with ∠C = 90^o and CA ≠ CB. CH is an altitude and CL is an angle bisector. Show that for X ≠ C on the line CL, we have ∠XAC ≠ ∠XBC. Show that for Y ≠ C on the line CH we have ∠YAC ≠ ∠YBC.

Solution

If ∠XAC = ∠XBC, then ∠AXC= ∠BXC, so triangles AXC and BXC are congruent, so AC = BC. Contradiction.

Suppose ∠YAC = ∠YBC. Let YA meet BC at D, and YB meet AC at E. Then by Ceva, (EC/EA)(HA/HB)(DB/DC) = 1. But CDA and CEB are similar, so CD/CE = CA/CB. Hence (CB/CA)(HA/HB)(DB/EA) = 1.

But HA/HB = (HA/HC)(HC/HB) = (CA/CB)² (triangles ACB, CHB, AHC all similar), so CA/CB = EA/DB, so ED is parallel to AB, so ∠CDE = ∠B. But AEDB is cyclic, so ∠CDE = ∠A. So CA = CB. Contradiction.

Thanks to Cristian Ilac

Problem 3

ABC is an equilateral triangle. D, E are points on the sides AB, AC respectively. The angle bisector of ∠ADE meets AE at F, and the angle bisector of ∠AED meets AD at G. Show that area DEF + area DEG ≤ area ABC. When do we have equality?

Answer

D = B, E = C

Solution

We have area DEF/area DEA = EF/EA = DE/(DE+AD). Similarly area DEG/area DEA = DG/DA = DE/(DE+AE). Thus area DEF + area DEG = DE area DEA (1/(DE+AD) + 1/(DE+AE)). We have area DEA = ½ AD·AE sin A, and area ABC = ½ AB·AC sin A, so (area DEF + area DEG)/area ABC = (DE·AD·AE/AB²)(1/(DE+AD) + 1/(DE+AE)).

Obviously DE ≤ AB, so it is sufficient to prove that AD·AE(1/(DE+AD) + 1/(DE+AE)) ≤ DE (*).

The cosine rule applied to ADE gives DE² = AD² + AE² - AD·AE. Hence also DE² - AD·AE = (AD - AE)² ≥ 0. Thus we have DE(AD - AE)² + DE²(AD + AE) ≥ AD·AE(AD + AE). So AD·AE(AD + AE + 2DE) ≤ DE(AD² + AE² + DE(AD + AE)) = DE(DE + AD)(DE + AE), which is (*).

Thanks to Cristian Ilac

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4th Junior Balkan Mathematical Olympiad Problems 2000

1.  x and y are positive reals such that x³ + y³ + (x + y)³ + 30xy = 2000. Show that x + y = 10.

2.  Find all positive integers n such that n³ + 3³ is a perfect square.

3.  ABC is a triangle. E, F are points on the side BC such that the semicircle diameter EF touches AB at Q and AC at P. Show that the intersection of EP and FQ lies on the altitude from A.

4.  n girls and 2n boys played a tennis tournament. Every player played every other player. The boys won 7/5 times as many matches as the girls (and there were no draws). Find n.

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3rd Junior Balkan Mathematical Olympiad Problems 1999

1.  a, b, c are distinct reals and there are reals x, y such that a³ + ax + y = 0, b³ + bx + y = 0 and c³ + cx + y = 0. Show that a + b + c = 0.

2.  Let a_n = 2³ⁿ + 3⁶ⁿ⁺² + 5⁶ⁿ⁺². Find gcd(a₀, a₁, a₂, ... , a₁₉₉₉).

3.  A square has side 20. S is a set of 1999 points inside the square and the 4 vertices. Show that we can find three points in S which form a triangle with area ≤ 1/10.

4.  The triangle ABC has AB = AC. D is a point on the side BC. BB' is a diameter of the circumcircle of ABD, and CC' is a diameter of the circumcircle of ACD. M is the midpoint of B'C'. Show that the area of BCM is independent of the position of D. 

Solutions

Problem 1

a, b, c are distinct reals and there are reals x, y such that a³ + ax + y = 0, b³ + bx + y = 0 and c³ + cx + y = 0. Show that a + b + c = 0.

Solution

Subtracting the first two equations and dividing by (a - b) we get (a² + ab + b²) + x = 0. Similarly we get (b² + bc + c²) + x = 0. Subtracting we get (a - c)(a + b + c) = 0. Hence a + b + c = 0.

or more elegantly, thanks to Suat Namli

a,b,c are the roots of the polynomial p(z) = z³ + xz + y, so their sum is the coefficient of z2, which is 0. 

Problem 2

Let a_n = 2³ⁿ + 3⁶ⁿ⁺² + 5⁶ⁿ⁺². Find gcd(a₀, a₁, a₂, ... , a₁₉₉₉).

Answer

7

Solution

a₀ = 35, so we need only consider whether 5 and 7 must be divisors. Consider first 5. Consider a₁. 2³ⁿ = 8 = 3 mod 5. 3⁶ⁿ⁺² = 3⁸ = 1 mod 5, and 5⁶ⁿ⁺² = 0 mod 5, so a₁ = 4 mod 5, and is not divisible by 5.

Now consider 7. We have 2³ⁿ = 1ⁿ = 1 mod 7. Since 3⁶ = 5⁶ = 1 mod 7 (Fermat's little theorem), we have 3⁶ⁿ⁺² = 2 mod 7, and 5⁶ⁿ⁺² = 4 mod 7, so a_n = 0 mod 7.

Thanks to Suat Namli

Problem 4

The triangle ABC has AB = AC. D is a point on the side BC. BB' is a diameter of the circumcircle of ABD, and CC' is a diameter of the circumcircle of ACD. M is the midpoint of B'C'. Show that the area of BCM is independent of the position of D.

Solution

We show that M must lie on the line through A parallel to BC.

Let O be the center of the circumcircle of ABD and O' the center of the other circumcircle. Let X be the midpoint of AD. Take coordinates with origin the midpoint of BC, x-axis along BC and y-axis through A. It is sufficient to show that the y-coordinates of O and O' sum to that of A, or equivalently that their mean is the same as the y-coordinate of X. But it is easy to see that the x-coordinate of X less that of O equals the x-coordinate of O' less that of X. Hence the y-coordinate differences must also be equal.

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2nd Junior Balkan Mathematical Olympiad Problems 1998

1.  The number 11...122...25 has 1997 1s and 1998 2s. Show that it is a perfect square.

2.  The convex pentagon ABCDE has AB = AE = CD = 1 and angle ABC = angle DEA = 90^o and BC + DE = 1. Find its area.

3.  Find all positive integers m, n such that mⁿ = n^m-n.

4.  Do there exist 16 three digit numbers, using only three different digits in all, so that the numbers are all different mod 16? 

Solutions

Problem 1

The number 11...122...25 has 1997 1s and 1998 2s. Show that it is a perfect square.

Solution

Same as Sweden 1981/1. 

Problem 3

Find all positive integers m, n such that mⁿ = n^m-n.

Answer

(m,n) = (9,3) or (8,2).

Solution

Thanks to José Nelson Ramírez

Obviously m ≥ n. Hence m-n ≥ n. So m ≥ 2n. Put d = gcd(m,n), and m = dM, n = dN. Then M ≥ 2N. We have (dM)^dN = dN^dM-dN, so M^dN = N^dM-dNd^dM-2dN. So if N > 1, then its prime factors divide M. Contradiction. So N = 1. Hence M^d = d^dM-2d, so M = d^M-2.

If M = 3, then d = 3, giving the solution m = 9, n = 3. If M = 4, then d = 2 giving the solution m = 8, n = 2. If M > 4, then d>1. But dM-2 ≥ 2M-2 > M. Contradiction. So there are no solutions for M > 4.

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1st Junior Balkan Mathematical Olympiad Problems 1997

1.  Show that given any 9 points inside a square side 1 we can always find three which form a triangle with area < 1/8.

2.  Given reals x, y with (x² + y²)/(x² - y²) + (x² - y²)/(x² + y²) = k, find (x⁸ + y⁸)/(x⁸ - y⁸) + (x⁸ - y⁸)/(x⁸ + y⁸) in terms of k.

3.  I is the incenter of ABC. N, M are the midpoints of sides AB, CA. The lines BI, CI meet MN at K, L respectively. Prove that AI + BI + CI > BC + KL.

4.  A triangle has circumradius R and sides a, b, c with R(b+c) = a √(bc). Find its angles.

5.  n₁, n₂, ... , n₁₉₉₈ are positive integers such that n₁² + n₂² + ... + n₁₉₉₇² = n₁₉₉₈². Show that at least two of the numbers are even. 

Solutions

Problem 2

Given reals x, y with (x² + y²)/(x² - y²) + (x² - y²)/(x² + y²) = k, find (x⁸ + y⁸)/(x⁸ - y⁸) + (x⁸ - y⁸)/(x⁸ + y⁸) in terms of k.

Solution

k/2 = (x⁴+y⁴)/(x⁴-y⁴), so k/2 + 2/k = 2(x⁸+y⁸)/(x⁸-y⁸). Hence desired expresson = (k²+4)/4k + 4k/(k²+4) = (k⁴+24k²+16)/(4k³+14k).

Thanks to Cristian Ilac

Problem 4

A triangle has circumradius R and sides a, b, c with R(b+c) = a √(bc). Find its angles.

Answer

A = 90^o, B = 45^o, C = 45^o.

Solution

We have a = 2R sin A, so b+c = 2 sin A √(bc). But by AM/GM (b+c)/2 ≥ √(bc) with equality iff b = c. So we must have b = c and A = 90^o.

Thanks to Suat Namli

Problem 5

n₁, n₂, ... , n₁₉₉₈ are positive integers such that n₁² + n₂² + ... + n₁₉₉₇² = n₁₉₉₈². Show that at least two of the numbers are even.

Solution

If two of n₁, n₂, ... , n₁₉₉₇ are even, then there is nothing to prove. If just one of n₁, n₂, ... , n₁₉₉₇ is even, then an even number of them are odd, so n₁₉₉₈ is also even. So it remains to show that n₁, n₂, ... , n₁₉₉₇ cannot all be odd.

Odd squares are all = 1 mod 8 and 1997 = 5 mod 8. So if all of n₁, n₂, ... , n₁₉₉₇ are odd, then n₁₉₉₈² = 5 mod 8. Contradiction.

Thanks to Suat Namli

 

 

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