Carnot's Theorem: What Is It?
A Mathematical Droodle


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Explanation

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What we see is a generalization of Wallace's Theorem. The theorem due to Lazare Carnot asserts that the feet of the lines equally inclined to the sides drawn from a point on the circumscribed circle, are collinear.

The proof is based on essentially the same property of the configuration as in the Wallace's Theorem, viz. presence of several cyclic quadrilaterals. Taking two at a time, the quadrilaterals share an inscribed angle which in each of the quadrilaterals equals another angle. From here we obtain pairs of similar triangles and the corresponding side ratios, the product of which is shown to be 1. The Menelaus theorem then implies colinearity of the three points.

Remark

My sincere thanks go to Michel Cabart who detected an error in the original text. Michel has not only suggested a way of amending the error but has also offered a much shorter and a more straightforward ending for the proof.


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In more detail, quadrilaterals PB'AC' and PB'CA' are cyclic. In PB'AC', the angles APC' and AB'C' are equal. In PB'CA', the angles A'B'C and A'PC are also equal.

Further ∠APC = ∠B = ∠A'PC', from which we derive

∠APC + ∠CPC' = ∠A'PC' + ∠CPC'

and subsequently
(1) ∠APC' = ∠A'PC.

Now we may proceed in two ways. Michel Cabart's is pretty short:

∠AB'C' = ∠APC' = ∠A'PC = ∠A'B'C = ∠AB'A'.

This may only happen when A', B', C' are collinear. The original ending used similarity of several triangles and Menelaus' theorem. This is how it went:

By construction, ∠AC'P = ∠CA'P, which, combined with (1), makes triangles AC'P and A'CP similar. Therefore, AC'/CA' = PA/PC.

From the other two pairs of similar triangles, we obtain A'B/B'A = PB/PA and CB'/BC' = PC/PB. Multiplying the three identities leads to

AC'/CA'·A'B/B'A·CB'/BC' = 1,

or

AC'/BC'·BA'/CA'·CB'/AB' = 1,

which is the identity from the Menelaus theorem.

The points A', B' and C' are therefore colinear.

References

  1. F. G.-M., Exercise de Géométrie, Éditions Jacques Gabay, 1991

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