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Combinatorics Difficulty 5.3 AIME, harder Prove it Singapore

Seven triangles of area 77 lie inside a square of area 2727. Prove that among the 77 triangles there are 22 that intersect in a region of area not less than 11.

Solution

Let the triangular regions be PiP_i, i=1,,7i = 1, \dots, 7, with A(Pi)=7A(P_i) = 7 and suppose that A(PiPj)<1A(P_i \cap P_j) < 1 if iji \neq j. Then
A(P1P2)=A(P1)+A(P2)A(P1P2)>141=13A(P1P2P3)=A(P1P2)+A(P3)A((P1P2)P3)>13+7A((P1P3)(P2P3))13+7A(P1P3)A(P2P3)18A(P1P2P3P4)=A(P1P2P3)+A(P4)A((P1P2P3)P4)>18+73>22A(P1P2P7)>27+76>28 \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 77 triangles must be contained in the square, whose area is 2727. This contradiction shows that our assumption is false, so there must exist iji \neq j such that A(PiPj)1A(P_i \cap P_j) \geq 1.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.