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