46th Eötvös Competition Problems 1942

1.  Show that no triangle has two sides each shorter than its corresponding altitude (from the opposite vertex).

2.  a, b, c, d are integers. For all integers m, n we can find integers h, k such that ah + bk = m and ch + dk = n. Show that |ad - bc| = 1.

3.  ABC is an equilateral triangle with area 1. A' is the point on the side BC such that BA' = 2·A'C. Define B' and C' similarly. Show that the lines AA', BB' and CC' enclose a triangle with area 1/7.

Solutions

Problem 1

Show that no triangle has two sides each shorter than its corresponding altitude (from the opposite vertex).

Solution

Suppose AD > BC and BE > AC. Then AD > BC ≥ BE > AC ≥ AD. Contradiction. Similarly in the other cases.

Problem 2

a, b, c, d are integers. For all integers m, n we can find integers h, k such that ah + bk = m and ch + dk = n. Show that |ad - bc| = 1.

Solution

In particular we can take h, k so that ch + dk = 1, so c and d have no common factor. Now take m = 1, n = 0. Then ch + dk = 0, so c divides dk and hence k. Suppose k = cr, then h = -dr. So -adr + bcr = 1. Hence ad - bc = ±1. 

Problem 3

ABC is an equilateral triangle with area 1. A' is the point on the side BC such that BA' = 2·A'C. Define B' and C' similarly. Show that the lines AA', BB' and CC' enclose a triangle with area 1/7.

Solution

Take the enclosed triangle to be DEF as shown. Let the line AA' meet the line through C parallel to BB' at X.

By symmetry, DEF is equilateral and AD = BE = CF. So ∠CFX = ∠CXF = 60^o, so CFX is also equilateral. So CX = CF = BE. Hence CXBE is a parallelogram. So CF is parallel to BX and hence CA'F and BA'X are similar. So BX = 2 CF. But BX = CE (parallelogram), so F is the midpoint of CE. Similarly, D is the midpoint of AF, and E is the midpoint of BD. Now area BEC = 2 area BEF = 2 area DEF. Similarly, area AFC = 2 area DEF and area BDA = 2 area DEF. But area ABC = area BEC + area AFC + area BDA + area DEF. Hence result.