Competition problem — Combinatorics

Combinatorics Difficulty 4.7 AIME Prove it Netherlands

At a congress all attendees are either a mathematician or a biologist and there is no one that is both. The mathematicians all know each other and each of them knows four of the biologists. The biologists also all know each other and each of them knows nine of the mathematicians. It turns out that every mathematician knows twice as many people as every biologist. (If person $A$ knows person $B$ , then person $B$ also knows person $A$ .)
How many mathematicians are at the congress?

Solution

Let $m$ be the number of mathematicians and $b$ the number of biologists.

Each mathematician knows all other mathematicians ( $m-1$ people) and 4 biologists, so each mathematician knows $m-1+4 = m+3$ people.

Each biologist knows all other biologists ( $b-1$ people) and 9 mathematicians, so each biologist knows $b-1+9 = b+8$ people.

We are told that every mathematician knows twice as many people as every biologist:
$m+3 = 2(b+8)$

Also, the number of mathematician-biologist acquaintances can be counted in two ways:
- Each mathematician knows 4 biologists: $m \times 4$
- Each biologist knows 9 mathematicians: $b \times 9$
So:
$m \times 4 = b \times 9$

Now solve the system:

From the second equation:
$m \times 4 = b \times 9 \implies m = \frac{9}{4}b$

Substitute $m$ into the first equation:
$m+3 = 2(b+8) \\ \frac{9}{4}b + 3 = 2b + 16 \\ \frac{9}{4}b - 2b = 16 - 3 \\ \frac{9}{4}b - \frac{8}{4}b = 13 \\ \frac{1}{4}b = 13 \\ b = 52$

Now, $m = \frac{9}{4}b = \frac{9}{4} \times 52 = 9 \times 13 = 117$

Answer:

There are $117$ mathematicians at the congress.

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