Maths Olympiad Prep

Library / /10 of 11

, 2013

Combinatorics Difficulty 9.1 IMO level Prove it Saudi Arabia

A Saudi company has two offices. One office is located in Riyadh and the other in Jeddah. To insure the connection between the two offices, the company has designated from each office a number of correspondents so that
(a) each pair of correspondents from the same office share exactly one common correspondent from the other office.
(b) there are at least 10 correspondents from Riyadh.
(c) Zayd, one of the correspondents from Jeddah, is in contact with exactly 8 correspondents from Riyadh.
What is the minimum number of correspondents from Jeddah who are in contact with the correspondent Amr from Riyadh?

Solution

Assume that we have two correspondents, RR from Riyadh and JJ from Jeddah, who are not connected to each other. Let JR\mathcal{J}_R be the set of correspondents from Jeddah connected to RR and RJ\mathcal{R}_J the set of correspondents from Riyadh connected to JJ. Define the function
f:JRRJ f: \mathcal{J}_R \longrightarrow \mathcal{R}_J
in the following way:
For each correspondent JJ' JR\in \mathcal{J}_R, since the correspondents JJ and JJ' are different, they share a unique correspondent in RJ\mathcal{R}_J. Let f(J)f\left(J'\right) be this unique correspondent. So ff is well-defined.
Assume that there exist two correspondents J1,J2JRJ_1, J_2 \in \mathcal{J}_R such that f(J1)=f(J2)=RRJf\left(J_1\right)= f\left(J_2\right)=R' \in \mathcal{R}_J. This means that the correspondent RR' from Riyadh shares with the correspondent RR two correspondents J1,J2J_1, J_2 from Jeddah. Hence J1=J2J_1=J_2 and the map ff is injective.
Consider a correspondent RRJR' \in \mathcal{R}_J and let JJ' be the unique correspondent in JR\mathcal{J}_R that RR' shares with RR. Then RR' is the unique correspondent in RJ\mathcal{R}_J that JJ' shares with JJ. Thus R=f(J)R'=f\left(J'\right) and ff is surjective. This proves that RR and JJ have the same number of correspondents from the other city.
We deduce from this that if the correspondent Amr from Riyadh is not connected to the correspondent Zayd from Jeddah, the correspondent Amr has exactly eight correspondents from Jeddah.
Assume now that the correspondent Amr is connected to the correspondent Zayd from Jeddah. Since there are at least ten correspondents in Riyadh and Zayd is connected only with eight, let R1,R2R_1, R_2 be two correspondents from Riyadh who are not connected to Zayd. Let J1J_1 be the unique correspondent that Amr and R1R_1 share. Let J2J_2 be the unique correspondent that R1R_1 shares with R2R_2. Since R1R_1 is not in contact with Zayd, he has eight correspondents from Jeddah. So there exists at least six correspondents in contact with R1R_1 who are not in contact neither with Amr nor with R2R_2. Let JJ be one of these correspondents.
The correspondent R2R_2 from Riyadh has exactly eight correspondents from Jeddah since he is not connected to the correspondent Zayd. The correspondent JJ from Jeddah has exactly eight correspondents from Riyadh since he is not connected to the correspondent R2R_2 from Riyadh. The correspondent Amr from Riyadh has exactly eight correspondents from Jeddah since he is not connected to the correspondent JJ from Jeddah.
This proves that in all cases the correspondent Amr from Riyadh has exactly eight correspondents from Jeddah.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.