These are my proposed solutions to the Harmonic geometry problems posed here.
this
1) a square inscribed in a right triangle whose legs are A ft and B feet (with one corner of the square at the right angle of the triangle), what is the length of a side of the square?
For HS students, the easy approach is to construct the line from (0,0) to the point (x,x) on the opposite vertex of the square, which rests on the hypotenuse of the triangle. This line has equation y=x, and if we find the intersection of this with the hypotenuse through (0, A) and (B,0) we will have our x-value for the length of the square. The hypotenuse has equation \( y= \frac {-Ax}{B}+ A \) Since at the point of intersection, y= x we can substitute x for y to get \( x= \frac {-Ax}{B}+ A \) and eliminating the denominators we get Ax + Bx = AB, which produces \(x = \frac{AB}{A+B}\) which is 1/2 the harmonic mean of A and B.
2) The three dimensional version of the problem is solved by a similar approach in three-space. Using the four vertices (0,0,0) ; A=(A,0,0); B=(0,B,0); and C= (0,0,C) for our right rectangular tetrahedron we know that the diagonal of the cube to the face ABC will have coordinates (r,r,r) . For lack of confusion we can let the equation of the plane ABC be mx+ny+pz = d. Now each point on the plane must satisfy this equation, so we know that (r,r,r) will solve it and mr+nr+zr = d . Each of the points A, B, and C must also solve the equation, so mA=nB=pC = d as well . solving each of these for m, n, and p we see that \( m = \frac{d}{A}\) ; \(n=\frac{d}{B}\) and \( p= \frac{d}{C}\) . Now when we substitute these back into mr+nr+zr = d we get \( \frac{dr}{A}+ \frac{dr}{B}+\frac{dr}{C} = d \) . Now if we factor out the dr in each numerator we can divide both sides by d to get \( r ( \frac{1}{A}+ \frac{1}{B}+\frac{1}{C}) = 1 \). Now keep the r alone and divide the one by the three fractions we get \( \frac{1}{r}= \frac{1}{A}+ \frac{1}{B}+\frac{1}{C}\) which we recognize as 1/3 of the harmonic mean of the lengths A, B, and C.
3) Given a 5-12-13 Right Triangle, Find a relation between the radius of the inscribed circle and the altitudes. Is this property true of a non-right triangle?
Using the hopefully familiar formula that A= r s (where s is the semi-perimeter) gives 30 = r (15) making the solution of r=2 rather easy. But what of the harmonic mean? If we find the three altitudes we see that they are 5, 12, and 60/13. The harmonic mean of these turns out to be 6, and (as we expect from previous problems) the radius turns out to be 1/3 the Harmonic Mean of the altitudes.
And in ANY triangle:
The radius of the incircle is ⅓ of the Harmonic Mean of the altitudes.
and given by twice the area divided by the perimeter.
4) For the student of the conics, In the ellipse \( 9x^2+4y^2=36 \), find the length of the semi-latus rectum.
Using a=3 and b=2 for the lengths of the semi-major and semi-minor axes lengths we have a focal length of \( \sqrt{5} \)
and so the coordinate of B must be \( ( \sqrt{5} , x)\} where x is the length of the semi-latus rectum. Using these in the original equation \( 9x^2+4(5)=36 \) gives us x=4/3.... SOOoooo?? you ask.
If you divide the transverse axis into two segments on each side of one foci, their lengths will be \(3+\sqrt{5} \) and \(3 - \sqrt{b}\)
Using the formula \( \frac{2ab}{a+b} \) gives the harmonic mean of these two values as 4/3.
The solution may remind students that if a diameter of a circle is divided into two segments, the perpendicular to the diameter will be the geometric mean of the two lengths. We might remind them that that works for ANY division of the diameter, where this applies only to the division at a focus.
Hopefully I have omitted my usual characteristic mis-calculations and typos. And I still want to share other approaches to solutions of these.
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