41st Eötvös Competition Problems 1937

1.  a₁, a₂, ... , a_n is any finite sequence of positive integers. Show that a₁! a₂! ... a_n! < (S + 1)! where S = a₁ + a₂ + ... + a_n.

2.  P, Q, R are three points in space. The circle C_P passes through Q and R, the circle C_Q passes through R and P, and the circle C_R passes through P and Q. The tangents to C_Q and C_R at P coincide. Similarly, the tangents to C_R and C_P at Q coincide, and the tangents to C_P and C_Q at R coincide. Show that the circles are either coplanar or lie on the surface of the same sphere.

3.  A₁, A₂, ... , A_n are points in the plane, no three collinear. The distinct points P and Q in the plane do not coincide with any of the A_i and are such that PA₁ + ... + PA_n = QA₁ + ... + QA_n. Show that there is a point R in the plane such that RA₁ + ... + RA_n < PA₁ + ... + PA_n.

Solutions

Problem 1

a₁, a₂, ... , a_n is any finite sequence of positive integers. Show that a₁! a₂! ... a_n! < (S + 1)! where S = a₁ + a₂ + ... + a_n.

Solution

We use induction on n to prove the stronger result that the lhs < S!. For n = 1 there is nothing to prove. For n = 2, we have (a₁+a₂)!/(a₁!a₂!)= (a₁+1)(a₁+2)...(a₁+a₂)/(1·2 ... ·a₂) > 1. Now given the result for 2 and n we have (a₁ + a₂ + ... + a_n+1)! > (a₁ + a₂ + ... + a_n)! a_n+1! > a₁! a₂! ... a_n+1!, which is the result for n+1.

Problem 2

P, Q, R are three points in space. The circle C_P passes through Q and R, the circle C_Q passes through R and P, and the circle C_R passes through P and Q. The tangents to C_Q and C_R at P coincide. Similarly, the tangents to C_R and C_P at Q coincide, and the tangents to C_P and C_Q at R coincide. Show that the circles are either coplanar or lie on the surface of the same sphere.

Solution

Let L be the common tangent to C_P and C_Q at R. Let π be the plane normal to L at R. Then the center O_P of C_P lies in π and so does the line L_P normal to the plane of the circle C_P at O_P. Similarly, we take O_Q to be the center of C_Q, and the line L_Q normal to the plane of the circle C_Q at O_Q. L_Q also lies in the plane π, so if L_P and L_Q do not meet then they must be parallel, so C_P and C_Q are coplanar. On the other hand if they meet at X, then X is equidistant from every point of C_P and also from every point of C_Q, so C_P and C_Q must lie on the sphere center X radius XR.

Similarly, C_Q and C_R are either coplanar or both lie on the surface of a sphere. If C_P and C_Q are coplanar, and C_Q and C_R are coplanar, then all three circles are coplanar. If C_P and C_Q lie on the surface of a sphere S and C_Q and C_R lie on the surface of a sphere S' ≠ S or lie in a plane S', then S ∩ S' must be C_Q. But C_P ∩ C_R = Q lies in S and S' and hence on C_Q. So P, Q, R are not distinct. Contradiction. Similarly if C_P and C_Q are coplanar and C_Q and C_R lie on the surface of a sphere. So the only remaining possibility is that all three circles lie on the same sphere.

Problem 3

A₁, A₂, ... , A_n are points in the plane, no three collinear. The points distinct P and Q in the plane do not coincide with any of the A_i and are such that PA₁ + ... + PA_n = QA₁ + ... + QA_n. Show that there is a point R in the plane such that RA₁ + ... + RA_n < PA₁ + ... + PA_n.

Solution

Take R to be the midpoint of PQ. If A lies inside the segment PQ, then RA < (PA + QA)/2. If A lies elsewhere on the line PQ, then RA = (PA + QA)/2. Now suppose A does not lie on the line PQ, then rotate A about R through 180^o to A', so that APA'Q is a parallelogram. Then 2 RA = AA' > AP + PA' = PA + QA, so RA > (PA + QA)/2. Since no three points are collinear, at least one A_i must lie off the line PQ, and for that point we have RA_i < (PA_i + QA_i)/2. For the others we certainly have RA_i ≤ (PA_i + QA_i)/2. So summing ∑ RA_i < &sum PA_i.