36th Eötvös Competition Problems 1932

1.  Show that if m is a multiple of aⁿ, then (a + 1)^m -1 is a multiple of aⁿ⁺¹.

2.  ABC is a triangle with AB and AC unequal. AM, AX, AD are the median, angle bisector and altitude. Show that X always lies between D and M, and that if the triangle is acute-angled, then angle MAX < angle DAX. 

3.  An acute-angled triangle has angles A < B < C. Show that sin 2A > sin 2B > sin 2C. 

Solutions

Problem 1

Show that if m is a multiple of aⁿ, then (a + 1)^m -1 is a multiple of aⁿ⁺¹.

Solution

We use induction on n. For n = 0 we have to show that (a+1)^m - 1 is a multiple of a. That is obvious if we expand by the binomial theorem. So suppose it is true for n. Put m = am', where m' is divisible by aⁿ. For convenience we also put A = a+1. Then we have A^m - 1 = (A^m' - 1)(A^m'(a-1) + A^m'(a-2) + ... + A^m' + 1). The first bracket is divisible by aⁿ⁺¹ by induction. The second bracket has a terms, so adding a and subtracting a 1s we can write it as a + (A^m'(a-1) - 1) + (A^m'(a-2) - 1) + ... + (A^m' - 1). Now each term is divisible by a and hence the sum is divisible by a. So A^m - 1 is divisible by aⁿ⁺² as required.

Problem 2

ABC is a triangle with AB and AC unequal. AM, AX, AD are the median, angle bisector and altitude. Show that X always lies between D and M, and that if the triangle is acute-angled, then angle MAX < angle DAX.

Solution

wlog ∠B > ∠C. So ∠B + ∠B + ∠A > ∠C + ∠B + ∠A = 180^o. Hence ∠B + ∠A/2 > 90^o. Hence ∠AXB = 180^o - ∠B - ∠A/2 < 90^o. So D must lie inside the segment BX. Since ∠B > ∠C, we have AC > AB. But CX/BX = AC/AB, so CX > BX. Hence M lies inside the segment CX. So X lies between D and M.

Let O be the circumcenter and N the midpoint of the arc BC. Then A, X, N are collinear. Since the triangle is acute-angled, O must lie on the opposite side of BC to N, so ∠MAN < ∠OAN = ∠DAN.

Problem 3

An acute-angled triangle has angles A < B < C. Show that sin 2A > sin 2B > sin 2C.

Solution

If 2B ≤ 90^o, then A+B < 2B ≤ 90^o, so C > 90^o. Contradiction. Also 2C < 180^o. Hence 2C > 2B > 90^o. But sin 2x is decreasing over this range, so sin 2B > sin 2C.

If 2A ≥ 90^o, then we have sin 2A > sin 2B for the same reason. So we need to consider the case 2A < 90^o < 2B. Since C < 90^o, we have A + B > 90^o and hence 2B > 180^o - 2A. Hence sin 2B < sin(180^o - 2A) = sin 2A.