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NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Geometry Difficulty 6.0 AIME, harder Prove it Romania

Consider a convex pentagon $A_0A_1A_2A_3A_4$ so that the rays ( $A_iA_{i+1}$ and ( $A_{i+3}A_{i+2}$ meet at $B_{i+4}$ , for each $i = 0, 1, 2, 3, 4$ – indices are considered modulo 5. Show that
$\prod_{i=0}^{4} A_i B_{i+3} = \prod_{i=0}^{4} A_i B_{i+2}.$

Solution

Let $C_i$ be the projection of $B_i$ on the line $A_{i+2}A_{i+3}$ , $i = 0, 1, 2, 3, 4$ . Notice that the right-angled triangles (possibly degenerate) $A_iB_{i+3}C_{i+3}$ and $A_iB_{i+2}C_{i+2}$ are similar, for angles $A_i$ are vertical. Hence $\frac{A_iB_{i+3}}{A_iB_{i+2}} = \frac{B_{i+3}C_{i+3}}{B_{i+2}C_{i+2}}$ , implying
$\prod_{i=0}^{4} \frac{A_i B_{i+3}}{A_i B_{i+2}} = \prod_{i=0}^{4} \frac{B_{i+3} C_{i+3}}{B_{i+2} C_{i+2}} = 1,$
as claimed.

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