Carnot's Theorem
Lazare Nicolas Margurite Carnot (1753-1823) was a person of many gifts and strong political views. During the Revolution he directed the Army of the North and under Napoleon Bonaparte he served as the Minister of War. In 1794, together with Gaspard Monge he founded the now famous Ecole Polytechnique. His geometric and military publications are considered masterpieces in their respective areas. The nice little theorem that bears his name shows an unexpected relationship between the radii of the circumcircle and incircle of a given triangle. In the formulation and the proof of Carnot's theorem I'll use the standard notations. The theorem read as follows In any triangle ABC, the (algebraic) sum of the distances (suitably signed) from the circumcenter O to the sides, is R + r, the sum of the circumradius and the inradius. OMa + OMb + OMc = R + r. In acute triangles, the circumcenter is always located inside the triangle. In this case, all three segments OMa, OMb, and OMc lie entirely inside the triangle. If one of the angles is obtuse, the circumcenter falls outside the triangle. One of the segments (the one that corresponds to the side opposite the obtuse angle) lies entirely outside, while the other two segments lie only partially outside the triangle. In the above sum, the segments that intersect the interior of the triangle are taken with the sign plus, the remaining side with the sign minus. Segments OMa, OMb, and OMc serve as altitudes of triangles OBC, OAC, and OAB with the bases on the sides a, b, and c. The sign convention guarantees that the areas of these triangles (taken with the proper signs) always add up to the area of ABC.
Proof
I shall consider only the case of an acute triangle. The other case is left as an exercise. For the inradius we always have r(a + b + c) = 2Area(ABC) From the above, we also have Area(OAB) + Area(OBC) + Area(OAC) = Area(ABC). Therefore, (*) r(a + b + c) = aOMa + bOMb + cOMc
A remark is now in order. O being the circumcenter of ABC, in the isosceles triangle AOB, AOB = 2 C. And similarly, BOC = 2 A and AOC = 2 B. Equipped with this knowledge we may consider several triples of similar (right-angled) triangles:
ABHb, ACHc, and BOMa (or, an equal COMa) BAHa, BCHc, and COMb (or, an equal AOMb) CBHb, CAHa, and AOMc (or, an equal BOMc)
From the first triple we derive AHb/c = AHc/b = OMa/R which leads to OMa(b + c) = R(AHb + AHc) Similarly, we get two additional identities OMb(a + c) = R(BHa + BHc) OMc(a + b) = R(CHa + CHb) Summing the three up yields after simplifications OMa(b + c) + OMb(a + c) + OMc(a + b) = R(a + b + c) Adding this to (*) and dividing by (a + b + c) completes the proof.
References
1. R. Honsberger, Mathematical Gems III, MAA, 1985
Copyright 1996-2006 Alexander Bogomolny
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