Friendly Kiepert's Perspectors
|What if applet does not run?|
Consider a configuration of two squares ACBcBa and BCAcAb with a common vertex C. Bottema's theorem asserts that the midpoint M of the segment AbBa is independent of C. The point M also serves as the apex of two right-angled (at M) isosceles triangles: AMB and AcMBc. In the proof of Bottema's theorem we have established an identity
|(1)||(Ab + Ba)/2 = (a + b)/2 + (b - a)·i/2,|
a and b are complex number representations of vertices A and B, respectively.
We have the following generalization due to Floor van Lamoen.
Fix a real number t and construct four points:
Define Ml and Mu as the midpoints of XlYl and XuYu. Then the three points C, Ml and Mu are collinear.
The proof generalizes that of Bottema's theorem. Following the latter it is using complex numbers. a and b have been already introduced as complex numbers corresponding to points A and B. Let g correspond to C, and ml, mu, xl, xu, yl, yu correspond to Ml, Mu, Xl, Xu, Yl, and Yu.
(tl) and (tu) could be rewritten as
To prove that Ml, Mu, and C are collinear, we only have to show that the ratio
K = (ml - g) / (mu - g)
is real. However from (t'l)
|(2l)||ml - g = (a + b)/2 + i·t·(b - a)/2 - g.|
On the other hand, from (t'u)
|(2u)||mu - g||= g - i·(b - a)/2 - g/t + (a + b)/2t - g|
|= 1/t·((a + b)/2 + i·t·(b - a)/2 - g)|
It is thus clear that K = t, a real number.
(The applet suggests a different proof based on the observations that XlYl||XuYu and angles MlBA and XlCB are equal. This proof elucidates the meaning of t:
It follows that the Kiepert perspectors K(f) and
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