geometric-construction
Exercise 8.10 at this site
http://www.euclidea.xyz/
claims that harmonic mean can be built with just 4 simple objects (lines and/or circles)
My best result is 6 objects (including perpendiculars and segment bisector).
Do you have any hint to simplify the construction?
As the page is not accessible if you have not done all the previous ones, here is a screenshot of the page 
All of these constructions are fairly complex, and I'm not going to give them in any detail. OP has asked for information that will help move the ball forward. I'll give some comments and references, and will add detail or clarification or corrections where requested. I have not vetted any of the constructions in every detail. I'll number the claims to match the numbering in the question.
Two non-intersecting circles with a point on the join of the centers. The site referred to by OP is most likely http://mathafou.free.fr/pbg_en/sol100a.html. The construction is the topic of C. Gram. A remark on the construction of the centre of a circle by means of the ruler. Math. Scand., 4:157–160, 1956.
Three non-intersecting circles. Apparently in D. Cauer. Uber die Konstruktion des Mittelpunktes eines Kreises mit dem Lineal allein (Berichtigung). Math. Ann., 73(1):90–94, 1912 and 74(3):462–464, 1913 (German). If you can find Smogorzhevskii's The ruler in geometrical constructions, the construction for three circles not in the same coaxial system is given at the end of the book.
Two congruent non-intersecting circles with an arbitrary point on their radical axis. Use the following construction to get the line $X_{11}X_{12}$, which joins the circle center. After that use the same construction as in 3. In the figure, givens are in blue, and the points are numbered in order of creation. The four chords (e.g. $X_1X_2$) are polars of $P$ and X_6.
Any arc of the circle with the center identified. See Chris Impens' Geometry without compasses
Bonus case. Two non-congruent non-intersecting circles with an arbitrary point on their radical axis. As for case 5, build the points $X_1,\dots,X_5$. Let $X_E$ be the intersection of the lines $X_2X_3$ and $X_1X_4$. Then line $X_5X_E$ goes through the centers.
Akopyan and Fedorov's Two circles and only a straightedge gives a useful overview of some of these topics.
Best Answer
$$H_\left(a,b\right) = \frac{2ab}{a+b} \Rightarrow \frac{H}{b} = \frac{2a}{a+b}$$
Mildly obfuscated solution illustrated here:
As you can see, this solution will get you a 4L/7E score. Getting 4/4 is a little tricker. :D