Harmonic Mean in GeometryThe harmonic mean
It makes a most dramatic appearance in the problem of average speed, but also pops up in various geometric circumstances. Below I collect a few examples.
AGBbe a semicircle and AC=a, BC=b.
CG⊥ABand CH⊥OGthen GH=H(a,b).
In a trapezoid
ABCD,let EFbe the line parallel to the bases ABand CDthrough the point of intersection of the diagonals ACand BD:
CD=aand AB=b,then EF=H(a,b).
It is worth observing that the point of intersection of the diagonals divides
EFinto two equal parts, which leads to another common example:
In a 1877 sangaku from the Hyogo prefecture,
AEand BDare both perpendicular to AB. Pis the intersection of ADand BE,and CP⊥AB.
AE=aand BD=b,then CP=2⋅H(a,b).
The same motif reemerges in the See-Saw Lemma where
AE, BF,and EFare tangent (the latter at X)to the semicircle AXB.
AE=aand BF=b,then XY=H(a,b).
Vladimir Nikolin from Serbia has observed appearance of the harmonic mean in his Rhombus Lemma.
ADbe the bisector of ∠Ain ΔABC, B′D||ACand C′D||AB.Then AB′DC′is a rhombus and its side pis half of the harmonic mean of sides b=ACand b=AB.
Choosing the angle at
Ato be right leads to a special case where the rhombus becomes a square.
Let a square be inscribed into a right triangle as in the diagram
aand bare the legs of the triangle and p the side of the square, 1p=1a+1b.In other words, p=12H(a,b).
As a consequence of the Rhombus Lemma there is a simple way of constructing the harmonic ratio:
ΔABC,with AB=cand AC=b,draw AD- the bisector of ∠A.At Derect a perpendicular to ADand let Ebe its intersection with either ABor AC.Then AE=H(b,c).