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Cut The Knot!An interactive column using Java appletsby Alex Bogomolny |
Napoleon's Propeller
July 2002
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As the two most recent columns have been devoted to synthetic proofs of a curious result, I've been looking for an example or two of an illuminating analytic proof. I found quite a few. Two such appear below. In the process I made a small, but surprising, discovery that is reflected in the title of the present column. Three altitudes of a triangle meet at a point known as the orthocenter of the triangle. There are many proofs of that result. Here's one that uses complex numbers. Given ΔABC, we may assume its vertices lie on a circle centered at the origin of a Cartesian coordinate system. Let's think of points in the plane as complex numbers. Define Indeed, for AH and BC to be orthogonal, the ratio
where * denotes the conjugate operator. If X denotes the latter expression, The proof admits an elegant shortcut [Hahn, p. 71] that appeals to geometric intuition. Note that
H - A = B + C = 2·(B + C)/2,
which in particular means that H - A is parallel to the line joining the origin - the circumcenter of ΔABC - with the midpoint of BC, a chord in the circumcircle. In other words, AH is perpendicular to BC, and similarly for BH and CH. The proof delivers more than was expected. With just a few sentences, not only we get an explicit expression for the orthocenter, the form of the expression The modified variant is short, powerful and illuminating, at least on a par with several synthetic proofs. Following is an example where the analytic apparatus of complex numbers is used with clarity unmatched by purely geometric proofs I can think of.
Let two triangles be similar and similarly oriented [Wells, p. 20]. Then the midpoints of the segments joining their corresponding vertices form a third triangle similar to the other two.
Two triangles ABC and A'B'C' are similar iff, say,
If l = 1/2 and m = 1/2, this is equivalent to
The assertion is thus immediate as is the generalization for l + m = 1, or in fact, for any l and m not simultaneously 0. (This is a particular case of the Fundamental Theorem of Directly Similar Figures: if the lines connecting the corresponding vertices of two directly similar polygons are devided in equal ratios, then the resulting polygon is directly similar to the given two. Steve Gray has reminded me in a private correspondence that the Fundamental Theorem of Directly Similar Figures, or more specifically (1'), with complex coefficients yields an elegant theorem concerning two triples of similar triangles.) With (1), it is easy to determine when a triangle is equilateral. The condition is 
The latter is equivalent to A2 + B2 + C2 - AB - BC - AC = 0, which is the same as
(A + jB + j2C)·(A + j2B + jC) = 0,
where j is a rotation through 120° (in the positive direction): j2 + j + 1 = 0. Therefore, depending on the orientation of ΔABC, either
Of course, (2) could be used to derive Napoleon's theorem. The proof is exceptionally clear. Another proof with complex numbers, although straightforward, is longer and might appear somewhat obscure. However, it delivers an easily overlooked surprise. Let OC, OA, and OB be the centers of the Napoleon triangles erected on the sides of ΔABC. Then easy computations show that
where d is the rotation through 60° in the positive direction XA equals two thirds of the median in ΔABC (looked at as a complex number) drawn from vertex A, and similarly for XB and XC. 
(Indeed, (B + C)/2 - A = (B + C - 2A)/2 is the complex number "from A to (B+C)/2," a median of ΔABC. But It's clear then that ΔXAXBXC is equal to ΔABC, but has its center at the origin. (Triangles ABC and XAXBXC are not just equal. They are centrally symmetric to each other.) (3), therefore, says something about ΔABC. If ΔXAXBXC is rotated through 60° and its vertices are added pairwise after a cyclic permutation to the vertices of its image, the three complex numbers thus obtained form an equilateral triangle. (To make the picture more compact, we may join the vertices of ΔXAXBXC to those of its rotated image and consider the triangle formed by the midpoints of these segments. The midpoints, too, form an equilateral triangle.) Here's a surprise. The origin, XA and dXA also form an equilateral triangle, and the same holds for the other two triples. The whole picture is exactly that of the Asymmetric Propeller. In the applet below, we may see either two equal triangles at 60° to each other, or the three equilateral triangles formed by the pairs of complex numbers corresponding to the matching vertices of those triangles, or a combination of the above. Napoleon's theorem is equivalent to the Asymmetric Propeller's theorem! How small is the world! Now, both the original Asymmetric Propeller and Napoleon's theorem start with three equilateral triangles and discover the fourth one by construction. It might have been natural to look for a link between the two results. I never saw the link established by synthetic means. Martin Gardner wrote about the Asymmetric Propeller in 1999 in a paper that since has been reprinted in his latest collection Gardner's Workout. In the Postscript to the corresponding chapter, Gardner mentions another problem [see also Honsberger, p. 274-276] described to him by Leon Bankoff. (This is in fact the Finsler-Hadwiger Theorem.) That problem admits a simple synthetic solution and, as a consequence of Neuberg's Theorem, another one using complex numbers. It's also a special case of the Fundamental Theorem of Directly Similar Figures. References
 Napoleon's Theorem
Asymmetric Propeller
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