23rd Eötvös Competition Problems 1916

1.  a, b are positive reals. Show that 1/x + 1/(x-a) + 1/(x+b) = 0 has two real roots one in [a/3, 2a/3] and the other in [-2b/3, -b/3].

2.  ABC is a triangle. The bisector of ∠C meets AB at D. Show that CD² < CA·CB.

3.  The set {1, 2, 3, 4, 5} is divided into two parts. Show that one part must contain two numbers and their difference. 

Solutions

Problem 1

a, b are positive reals. Show that 1/x + 1/(x-a) + 1/(x+b) = 0 has two real roots one in [a/3, 2a/3] and the other in [-2b/3, -b/3].

Solution

Multiplying up the equation is 3x² + 2(b-a)x - ab = 0 with roots (a-b ±√(a²+ab+b²))/3. Note that these are both real since a and b are both positive. Consider first the larger root k. We have a²+ab+b² < a²+2ab+b², so k < (a-b+a+b)/3 = 2a/3. Also a²+ab+b² > b², so k > (a-b+b)/3 = a/3. Hence k ∈ (a/3, 2a/3).

Let h be the smaller root. Then h > (a-b-(a+b))/3 = -2b/3 and since a²+ab+b² > a², h < (a-b-a)/3 = -b/3, so h ∈ (-2b/3, -b/3). 

Problem 2

ABC is a triangle. The bisector of ∠C meets AB at D. Show that CD² < CA·CB.

Solution

Extend CD to meet the circumcircle again at C'. Then ∠CAB = ∠CC'B, so triangles CAD and CC'B are similar. Hence CA/CD = CC'/CB. So CA·CB = CD·CC' > CD². 

Problem 3

The set {1, 2, 3, 4, 5} is divided into two parts. Show that one part must contain two numbers and their difference.

Solution

Suppose not. Let the two parts be A, B. wlog 2 ∈ A. Hence 1 ∈ B (or 2-1=1 in A) and 4 ∈ B (or 4-2=2 in A). Hence 5 ∈ A (or 5-1=4 in B) and 3 ∈ A (or 4-1=3) in B. But now 5-3=2 in A. Contradiction.