43rd Eötvös Competition Problems 1939

1.  Show that (a + a')(c + c') ≥ (b + b')² for all real numbers a, a', b, b', c, c' such that aa' > 0, ac ≥ b², a'c' ≥ b'².

2.  Find the highest power of 2 dividing 2ⁿ!

3.  ABC is acute-angled. A' is a point on the semicircle diameter BC (lying on the opposite side of BC to A). B' and C' are similar. Show how to construct such points so that AB' = AC', BC' = BA' and CA' = CB'.

Solutions

Problem 1

Show that (a + a')(c + c') ≥ (b + b')² for all real numbers a, a', b, b', c, c' such that aa' > 0, ac ≥ b², a'c' ≥ b'².

Solution Using the fact that aa', ac, a'c' are all non-negative we find that if a ≥ 0, then so is a', c, c'. Similarly, if a ≤ 0, then so is a', c, c'. Hence ac' and a'c are non-negative. Their geometric mean is bb'. Hence (ac' + a'c)/2 ≥ bb'. Hence (a + a')(c + c') = ac + a'c' + (ac' + a'c) ≥ b² + 2bb' + b'² = (b + b')². 

Problem 2

Find the highest power of 2 dividing 2ⁿ!

Solution

Let X = {1, 2, 3, ... , 2ⁿ}. There are 2^n-1 multiples of 2 in X, 2^n-2 multiples of 2², 2^n-3 multiples of 2³, ... 1 multiple of 2ⁿ. The multiples of 2 contribute 2^n-1 powers of 2. The multiples of 2² contribute an additional 2^n-2 powers of 2 and so on. So the highest power of 2 divding 2ⁿ! is 2^k, where k = 2^n-1 + 2^n-2 + ... + 1 = 2ⁿ - 1.

Problem 3

ABC is acute-angled. A' is a point on the semicircle diameter BC (lying on the opposite side of BC to A). B' and C' are similar. Show how to construct such points so that AB' = AC', BC' = BA' and CA' = CB'.

Solution

Let points A', B', C' lie on the altitudes as shown. Now CA'D and CBA' are similar, so CA'/CD = CB/CA'. Hence CA'² = CD·CB. Similarly, CB'² = CE·CA. But ∠AEB = ∠ADB = 90^o, so AEDB is cyclic. Hence CD·CB = CE·CA, so CA'² = CB'². So CA' = CB'. Similarly for the other two desired equalities.