27th Eötvös Competition Problems 1923

1.  The circles OAB, OBC, OCA have equal radius r. Show that the circle ABC also has radius r.

2.  Let x be a real number and put y = (x + 1)/2. Put a_n = 1 + x + x² + ... + xⁿ, and b_n = 1 + y + y² + ... + yⁿ. Show that ∑₀ⁿ a_m (n+1)C(m+1) = 2ⁿ b_n, where aCb is the binomial coefficient a!/( b! (a-b)! ).

3.  Show that an infinite arithmetic progression of unequal terms cannot consist entirely of primes. 

Solutions

Problem 1

The circles OAB, OBC, OCA have equal radius r. Show that the circle ABC also has radius r.

Solution

Let P, Q, R be the centers of the circles OBC, OCA, OAB. Then OQAR, ORBP, OPCQ are rhombi side r. Hence AR and CP are equal and parallel to OQ and hence to each other. So ARPC is a parallelogram. Hence AC = PR. Similarly, AB = PQ and BC = QR, so the triangles ABC and PQR are congruent. Hence the circumradius of ABC is the same as the circuradius of PQR. But O lies on a circle center P radius r, so OP = r. Similarly OQ = OR = r, so O is the circumcenter of PQR, and r is the circumradius of PQR. 

Problem 2

Let x be a real number and put y = (x + 1)/2. Put a_n = 1 + x + x² + ... + xⁿ, and b_n = 1 + y + y² + ... + yⁿ. Show that ∑₀ⁿ a_m (n+1)C(m+1) = 2ⁿ b_n, where aCb is the binomial coefficient a!/( b! (a-b)! ).

Solution

The coefficient of x^k on the lhs is n+1Ck+1 + n+1Ck+2 + n+1Ck+3 + ... + n+1Cn+1.The coefficient on the rhs is 2^n-kkCk + 2^n-k-1k+1Ck + 2^n-k-2k+2Ck + ... + nCk. If n-k=0 these are both 1. We use induction on n-k. Suppose the result is true for 0, 1, ... , n-k-1. We have n+1Ck+1 + n+1Ck+2 + n+1Ck+3 + ... + n+1Cn+1 = (nCk + nCk+1) + (nCk+1 + nCk+2) + ... + (nCn-1 + nCn) + n+1Cn+1 = nCk + 2(nCk+1 + nCk+2 + ... + nCn) = nCk + 2(2^n-k-1kCk + 2^n-k-2k+1Ck + 2^n-k-3k+2Ck + ... + n-1Ck) = 2^n-kkCk + 2^n-k-1k+1Ck + 2^n-k-2k+2Ck + ... + 2 n-1Ck + nCk, which is the required result for n-k. 

Problem 3

Show that an infinite arithmetic progression of unequal terms cannot consist entirely of primes.

Solution

Let the difference be d. Take any term b > 1. Then b terms later we get b + bd = b(d+1) which is not a prime.