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ASU 101 All Soviet Union MO 1968 parallel and equal products wanted

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June 20, 2019
geometry

Problem Statement

Given two acute-angled triangles ABCABC and A′B′C′A'B'C' with the points OO and O′O' inside. Three pairs of the perpendiculars are drawn: [OA1][OA_1] to the side [BC][BC], [O′A1′][O'A'_1] to the side [B′C′][B'C'], [OB1][OB_1] to the side [AC][AC], [O′B1′][O'B'_1] to the side [A′C′][A'C'], [OC1][OC_1] to the side [AB][AB], [O′C1′][O'C'_1] to the side [A′B′][A'B']; It is known that [OA1]∥[O′A′],[OB1]∥[O′B′],[OC1]∥[O′C′][OA_1] \parallel [O'A'], [OB_1] \parallel [O'B'], [OC_1] \parallel [O'C'] and ∣OA1∣⋅∣O′A′∣=∣OB1∣⋅∣O′B′∣=∣OC1∣⋅∣O′C′∣|OA_1|\cdot|O'A'| = |OB_1|\cdot |O'B'| = |OC_1|\cdot |O'C'| Prove that [O′A1′]∥[OA],[O′B1′]∥[OB],[O′C1′]∥[OC][O'A'_1] \parallel [OA], [O'B'_1] \parallel[OB], [O'C'_1] \parallel[OC] and ∣O′A1′∣⋅∣OA∣=∣O′B1′∣⋅∣OB∣=∣O′C1′∣⋅∣OC∣|O'A'_1|\cdot|OA| = |O'B'_1|\cdot|OB| = |O'C'_1|\cdot|OC|
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