31st Eötvös Competition Problems 1927

1.  a, b, c, d are each relatively prime to n = ad - bc, and r and s are integers. Show ar + bs is a multiple of n iff cr + ds is a multiple of n.

2.  Find the sum of all four digit numbers (written in base 10) which contain only the digits 1 - 5 and contain no digit twice.

3.  r is the inradius of the triangle ABC and r' is the exradius for the circle touching AB. Show that 4r r' ≤ c², where c is the length of the side AB. 

Solutions

Problem 1

a, b, c, d are each relatively prime to n = ad - bc, and r and s are integers. Show ar + bs is a multiple of n iff cr + ds is a multiple of n.

Solution

d(ar+bs) - b(cr+ds) = (ad-bc)r = nr, so if (ar+bs) is a multiple of n, then so is b(cr+ds). But b is relatively prime to n, so (cr+ds) is a multiple of n. Similarly, if (cr+ds) is a multiple of n, then so is d(ar+bs) and hence (ar+bs). 

Problem 2

Find the sum of all four digit numbers (written in base 10) which contain only the digits 1 - 5 and contain no digit twice.

Solution

Denote the digits as d₃d₂d₁d₀. If we fix d₀ = k, then there are 4·3·2 ways of choosing the other digits, so there are 24 such numbers. The sum of their d₀ digits is 24k. Thus the sum of the d₀ digits for all the numbers is 24(1+2+3+4+5) = 24·15 = 360. A similar argument works for the other digits, so the sum of the numbers is 360(1+10+100+1000) = 399960. 

Problem 3

r is the inradius of the triangle ABC and r' is the exradius for the circle touching AB. Show that 4r r' ≤ c², where c is the length of the side AB.

Solution

Let the incenter be I and the relevant excenter be E. Then ∠IAE = ∠IBE = 90^o, so AIBE is cyclic. Let the center of the circle through A,I,B,E be C. Then r = IP, r' = EQ. But IP·EQ ≤ XZ·ZY, where XY is the diameter perpendicular to AB. We have XZ·ZY = AZ·ZB = c²/4.