3rd Eötvös Competition Problems 1896

1.  For a positive integer n, let p(n) be the number of prime factors of n. Show that ln n ≥ p(n) ln 2.

2.  Show that if (a, b) satisfies a² - 3ab + 2b² + a - b = a² - 2ab + b² - 5a + 7b = 0, then it also satisfies ab - 12a + 15b = 0.

3.  Given three points P, Q, R in the plane, find points A, B, C such that P is the foot of the perpendicular from A to BC, Q is the foot of the perpendicular from B to CA, and R is the foot of the perpendicular from C to AB. Find the lengths AB, BC, CA in terms of PQ, QR and RP. 

Solutions

Problem 1

For a positive integer n, let p(n) be the number of prime factors of n. Show that ln n ≥ p(n) ln 2.

Solution

n ≥ the product of its prime factors. Each prime factor ≥ 2, so n ≥ 2^p(n). Taking logs gives the result. 

Problem 2

Show that if (a, b) satisfies a² - 3ab + 2b² + a - b = a² - 2ab + b² - 5a + 7b = 0, then it also satisfies ab - 12a + 15b = 0.

Solution

One approach is simply to solve the equations. The trick is to notice that we can factorise the first as (a-b)(a-2b+1) = 0. If a = b, then the second equation gives a = 0, so the solution is (a,b) = (0,0). If a-2b+1 = 0, then the second equation gives (4b²-4b+1) - 2b(2b-1) + b² - 5(2b-1) + 7b = b² - 5b + 6 = 0, so b = 2 or 3, giving solutions (a,b) = (3,2) or (5,3). It is easy to check that all of (0,0), (3,2), (5,3) satisfy ab - 12a + 15b = 0.

Another approach is to find a suitable combination of the two equations. We obviously cannot take m(1) + n(2) with m and n reals, but suppose we take m, n to have the form pa + qb + r. So suppose we take (pa + qb + r)(a²-3ab+2b²+a-b) + (p'a+q'b+r')(a²-2ab+b²-5a+7b). To avoid an a³ term we need p' = -p. To avoid a b³ term we need q' = -2q. Then to avoid an a²b term we need -3p+q+2p-2q = 0, so q = -p. So we try (a-b+r)(a²-3ab+2b²+a-b) + (-a+2b+r')(a²-2ab+b²-5a+7b). After some more algebra we find this works with r=-9, r'=3.

Problem 3

Given three points P, Q, R in the plane, find points A, B, C such that P is the foot of the perpendicular from A to BC, Q is the foot of the perpendicular from B to CA, and R is the foot of the perpendicular from C to AB. Find the lengths AB, BC, CA in terms of PQ, QR and RP.

Solution

Take I to be the incenter of PQR and A, B, C to be the three excenters. Then AP is the internal bisector of ∠QPR and is perpendicular to BC, the external bisector. Similarly for the other points.

Of course, we could also take I, B, C or A, I, C or A, B, I as the three points.

We derive some bookwork formulae which may not be familiar. For a triangle ABC with sides a,b,c as usual, put s = (a+b+c)/2 and take the inradius to be r. Suppose the vertex A is a distance x from the two points of contact of the incircle with the sides AB, AC. Then B and C must be c-x and b-x from their points of contact, so a = (c-x)+(b-x). Hence x = s-a. So tan A/2 = r/(s-a). But the area of the triangle is √(s(s-a)(s-b)(s-c)) by Heron or rs by considering it as made up of ABI, BCI, CAI. Hence r = √((s-a)(s-b)(s-c)/s) and tan A/2 = √((s-b)(s-c)/((s-a)s)). Now using sec2A/2 = 1 + tan2A/2 and noting that s(s-a) + (s-b)(s-c) = bc, we find sec²A/2 = bc/(s(s-a)). Hence sin A/2 = tan A/2 cos A/2 = √( (s-b)(s-c)/bc) (*).

ARPC is cyclic (because ∠APC = ∠ARC = 90^o), so x = ∠APR = ∠ACR = 90^o - ∠A. Similarly, y = 90^o - ∠B, z = 90^o - ∠C. Hence ABC is similar to AQR. So QR/BC = AQ/AB = cos A = sin x. Hence BC = QR/sin x.

We are now home, because we can use the formulae (*) to get sin x, sin y, sin z.