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Combinatorics Difficulty 8.1 Shortlist Prove it Hong Kong

At a party there are 12341234 participants, and each of them has shaken hands with exactly 137137 other participants. It is known that no three participants have shaken hands with each other. Furthermore, for any two participants AA and BB who have not shaken hands with each other, there must be exactly kk other participants who have shaken hands with both AA and BB, where kk is a fixed constant. Find the value of kk.

Solution

Answer: 1717

Consider any participant xx. He has shaken hands with 137137 other participants, say y1,y2,,y137y_1, y_2, \dots, y_{137} (collectively known as Group Y participants). Also, there are 12341137=10961234 - 1 - 137 = 1096 participants who have not shaken hands with xx; let's call them z1,z2,,z1096z_1, z_2, \dots, z_{1096} (collectively known as Group Z participants).

We count the number of times a Group Y participant has shaken hand with a Group Z participant. This can be done in two ways:

* Since no two participants in Group Y can shake hands (otherwise together with xx there will be three participants shaking hands with each other), every Group Y participant must have shaken hands with exactly 136136 participants in Group Z. The total number of such handshakes is thus equal to 137×136137 \times 136.

* On the other hand, every participant in Group Z (say zz) must shake hands with exactly kk participants in Group Y, because xx and zz have not shaken hands implies there are exactly kk participants (who must be in Group Y) who have shaken hands with both xx and zz. It follows that the total number of such handshakes is also equal to 1096k1096k.

From the above two ways of counting we obtain the equality 1096k=137×1361096k = 137 \times 136, which gives k=17k = 17.

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