1967 AMC 12 Problem 38

Question

Given a set

S

consisting of two undefined elements “pib” and “maa,” and the four postulates:

\mathrm{P}_1:

Every pib is a collection of maas.

\mathrm{P}_2:

Any two distinct pibs have one and only one maa in common.

\mathrm{P}_3:

Every maa belongs to two and only two pibs.

\mathrm{P}_4:

There are exactly four pibs.

Consider the three theorems:

\mathrm{T}_1:

There are exactly six maas.

\mathrm{T}_2:

There are exactly three maas in each pib.

\mathrm{T}_3:

For each maa there is exactly one other maa not in the same pib with it.

The theorems which are deducible from the postulates are:

\mathrm{T}_3

only

\mathrm{T}_2

and

\mathrm{T}_3

only

\mathrm{T}_1

and

\mathrm{T}_2

only

\mathrm{T}_1

and

\mathrm{T}_3

only

all

Answer: E

Concepts:basic counting graph theory logical deduction

Difficulty rating: 2030

Small hint:

Label the four pibs

1,

2,

3,

and

4

Big hint:

Each maa corresponds to an unordered pair of pibs

Solution:

Label the four pibs

1,

2,

3,

and

4.

By

\mathrm P_2

and

\mathrm P_3,

every maa is exactly the common maa of one unordered pair of pibs. Thus there are

\binom42=6

maas, proving

\mathrm T_1.

A fixed pib

i

contains the three maas

ij

with

j\ne i,

proving

\mathrm T_2.

For maa

ij,

the unique maa sharing neither pib is the one belonging to the complementary pair of pibs, proving

\mathrm T_3.

All three follow.

Therefore, the correct answer is E.

1967 AMC 12 Problem 38

Answer: E

Difficulty rating: 2030

Small hint:

Big hint:

Solution:

Problem 38 in other years