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Singapore Mathematical Olympiad (SMO)

Combinatorics Difficulty 5.3 AIME, harder Prove it Singapore

Seven triangles of area $7$ lie inside a square of area $27$ . Prove that among the $7$ triangles there are $2$ that intersect in a region of area not less than $1$ .

Solution

Let the triangular regions be $P_i$ , $i = 1, \dots, 7$ , with $A(P_i) = 7$ and suppose that $A(P_i \cap P_j) < 1$ if $i \neq j$ . Then
$\begin{align*} A(P_1 \cup P_2) &= A(P_1) + A(P_2) - A(P_1 \cap P_2) > 14 - 1 = 13 \\ A(P_1 \cup P_2 \cup P_3) &= A(P_1 \cup P_2) + A(P_3) - A((P_1 \cup P_2) \cap P_3) \\ &> 13 + 7 - A((P_1 \cap P_3) \cup (P_2 \cap P_3)) \\ &\geq 13 + 7 - A(P_1 \cap P_3) - A(P_2 \cap P_3) \geq 18 \\ A(P_1 \cup P_2 \cup P_3 \cup P_4) &= A(P_1 \cup P_2 \cup P_3) + A(P_4) - A((P_1 \cup P_2 \cup P_3) \cap P_4) \\ &> 18 + 7 - 3 > 22 \\ \dots \\ A(P_1 \cup P_2 \cup \dots \cup P_7) > 27 + 7 - 6 > 28 \end{align*}$

But the union of all $7$ triangles must be contained in the square, whose area is $27$ . This contradiction shows that our assumption is false, so there must exist $i \neq j$ such that $A(P_i \cap P_j) \geq 1$ .

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