Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it Belarus

All vertices of triangles $ABC$ and $A_1B_1C_1$ lie on the hyperbola $y = 1/x$ . It is known that $AB \parallel A_1B_1$ and $BC \parallel B_1C_1$ . Prove that $AC_1 \parallel A_1C$ .

Solution

Let the coordinates of the given points be $A(a; 1/a)$ , $B(b; 1/b)$ , $C(c; 1/c)$ , $A_1(a_1; 1/a_1)$ , $B_1(b_1; 1/b_1)$ , $C_1(c_1; 1/c_1)$ . It is easy to calculate the slope of the line $AB$ : $k = -1/(ab)$ . Similarly, the slopes of $A_1B_1$ , $BC$ , $B_1C_1$ are $-1/(a_1b_1)$ , $-1/(bc)$ , $-1/(b_1c_1)$ , respectively. Now, the conditions $AB \parallel A_1B_1$ and $BC \parallel B_1C_1$ are equivalent to the equalities $-1/(ab) = -1/(a_1b_1)$ and $-1/(bc) = -1/(b_1c_1)$ . These equalities imply the equality $-1/(ac_1) = -1/(a_1c)$ which is equivalent to the condition $AC_1 \parallel A_1C$ .

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