A Multiplicative Identity of Areas in a Triangle
Draw, through point R in the interior of ΔABC, lines JQ, KM, and NP parallel to sides BC, AB, and AC, respectively. Let (Γ) denotes the area of shape Γ. Then (JMR)(PQR)(KNR) = (JNR)(KQR)(MPR). |Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander Bogomolny Draw, through point R in the interior of ΔABC, lines JQ, KM, and NP parallel to sides BC, AB, and AC, respectively. Let (Γ) denotes the area of shape Γ. Then (JMR)(PQR)(KNR) = (JNR)(KQR)(MPR).
ProofObserve that quadrilaterals JMPQ, KMJN, and KNPQ supply the premises for the corollary of Bui Quang Tuan's lemma. Thus we may conclude that
(MPR)² = (JMR)(PQR), Multiply all three: (MPR)²(JNR)²(KQR)² = (JMR)²(PQR)²(KNR)². Since all the areas involved are positive, taking square root we obtain the required identity: (JMR)(PQR)(KNR) = (JNR)(KQR)(MPR). At the site of Vanni Gorni the problem is solved a little differently. It appears that it was first published in 1838 in the Belgian magazine Correspondance Mathématique et Physique edited by A. Quetelet (1796-1874). The same issue was expanded in 1859 in the French Nouvelles Annales de Mathématiques edited by Gerono Terquem. Consider ΔCKM. For definiteness sake, assume MR/KM = λ. Then (CJRN)/(CMK) = 1 - λ² - (1 - λ)² = 2λ(1 - λ), implying (CJRN)/(CMK) = 2·√(JMR)/(CMK) · √(KNR)/(CMK), or (CJRN)² = 4(JMR)(KNR). Similarly, in triangles AJK and BNP,
(AMRP)² = 4(PQR)(JMR), and Multiplying all three and taking the square root, we obtain (AMRP)(BKRQ)(CJRN) = 8(JMR)(PQR)(KNR) - the same identity as before but now in terms of parallelograms. Acknowledgement: I am grateful to a former correspondent for bringing Vanni Gorni's site to my attention. |Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander Bogomolny |
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