16th Eötvös Competition Problems 1909

1.  Prove that (n + 1)³ ≠ n³ + (n - 1)³ for any positive integer n.

2.  α is acute. Show that α < (sin α + tan α)/2.

3.  ABC is a triangle. The feet of the altitudes from A, B, C are P, Q, R respectively, and P, Q, R are distinct points. The altitudes meet at O. Show that if ABC is acute, then O is the center of the circle inscribed in the triangle PQR, and that A, B, C are the centers of the other three circles that touch all three sides of PQR (extended if necessary). What happens if ABC is not acute? 

Solutions

Problem 1

Prove that (n + 1)³ ≠ n³ + (n - 1)³ for any positive integer n.

Solution

If we have equality, then n³ = 6n² + 2. But if n ≤ 6, then n³ ≤ 6n² < 6n² + 2. If n ≥ 7, then n³ ≥ 7n² > 6n2 + 2. Contradiction. 

Problem 2

α is acute. Show that α < (sin α + tan α)/2.

Solution

Take A and B on a circle center O radius 1 with ∠AOB = α. Let the tangents at A and B meet at D, and let the tangent at A meet the line OB at C. Then the area of the sector OAB = α/2 < area OADB. Area OAC = ½ tan α and area OAB = ½ sin α. So it is sufficient to show that area OADB < (area OAB + area OAC)/2 or equivalently that area ABD < ½ area ABC. That in turn is equivalent to AD < ½ AC or AD < CD. But AD = BD (equal tangents), and BD < CD, because CD is the hypoteneuse. 

Problem 3

The feet of the altitudes from A, B, C are P, Q, R respectively, and P, Q, R are distinct points. The altitudes meet at O. Show that if ABC is acute, then O is the center of the circle inscribed in the triangle PQR, and that A, B, C are the centers of the other three circles that touch all three sides of PQR (extended if necessary). What happens if ABC is not acute?

Solution

Suppose first that ABC is acute-angled. ∠OPC = ∠OQC = 90^o, so OPCQ is cyclic. Hence ∠OPQ = ∠OCQ = 90^o - ∠A. Similarly, ∠OPR = ∠OBR = 90^o - ∠A. Hence OP bisects ∠QPR. Similarly for the other angles. So O is the incenter of PQR.

Now suppose ∠A is obtuse. Then A is the orthocenter of OBC, which is acute-angled, so A is the incenter of PQR. O the the intersection of the bisector of ∠QPR and the perpendicular to the bisector of ∠PRQ, so it is the excenter of PQR opposite to P.