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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 4.9 AIME Prove it Soviet Union

Problem:

The point $P$ lies inside the triangle $ABC$ . A line is drawn through $P$ parallel to each side of the triangle. The lines divide $AB$ into three parts length $c$ , $c'$ , $c''$ (in that order), and $BC$ into three parts length $a$ , $a'$ , $a''$ (in that order), and $CA$ into three parts length $b$ , $b'$ , $b''$ (in that order). Show that $abc = a'b'c' = a''b''c''$ .

Solution

Solution:

The three small triangles are similar, so $a/a'' = c'/c = b''/b'$ and $a/a' = c'/c'' = b''/b$ . Hence $(a/a'')(b/b'') = (c'/c)(c''/c') = c''/c$ , so $abc = a''b''c''$ . Similarly, $(a/a')(c/c') = (b''/b)(b''/b'') = b/b$ , so $abc = a'b'c'.$

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