I found this problem on a website and I couldn't do anything. Do you have any ideas, hints? enter image description here

Edit: If I say $$\frac { { \partial }^{ 2 }f }{ \partial { a }^{ 2 } } +\frac { { \partial }^{ 2 }f }{ \partial { b }^{ 2 } } =0$$

am I wrong?


Nothing to do with harmonic functions :)

This is a version of the classical Problem of Apollonius, for which there are both geometric and algebraic solutions. See http://en.wikipedia.org/wiki/Problem_of_Apollonius#Ten_combinations_of_points.2C_circles.2C_and_lines and/or http://mathworld.wolfram.com/ApolloniusProblem.html. For the algebraic approach, you can use the chord lengths $a$ and $b$ to write down equations of the two lines (say, through the fixed point $(-1,0)$ on the unit circle).

Edit #3: OK, it turns out one has to be a bit more careful than I was. Given $a<b$, there are two configurations (up to congruence). So, at long last, we have $f(a,b)=$ $$ \frac{a^2 (2+\sqrt{4-b^2})+a b (\sqrt{4-a^2}-\sqrt{4-b^2})+(\sqrt{4-a^2}-2)(4 \sqrt{4-b^2}+8-b^2)}{(a+b)^2} $$ when the chords are on the same side of the diameter through the common point of the chords, and

$$\frac{a^2(2-\sqrt{4-b^2})+ab(\sqrt{4-a^2}+\sqrt{4-b^2})-(\sqrt{4-a^2}-2) (b^2+4\sqrt{4-b^2}-8)}{(a+b)^2}$$ when they are on opposite sides of the diameter.

I thank @Rahul Narain for insisting I get it right :) It is reassuring that these formulas do agree when $b=2$.

Still no harmonic functions.

  • $\begingroup$ The function of radius holds that harmonic function or not? I am just asking. What do you think? $\endgroup$ – newzad May 25 '13 at 17:35
  • 1
    $\begingroup$ Honestly, I have no idea. This is going to be difficult to work out. If you are going to try to solve it algebraically, note that two circles intersect tangentially if and only if the distance between their centers is $|r_1\pm r_2|$ (where $r_i$ are their radii), and a circle centered at $(x_0,y_0)$ with radius $r$ intersects the line $Ax+By+C=0$ tangentially if and only if $|Ax_0+By_0+C|=r$. $\endgroup$ – Ted Shifrin May 25 '13 at 17:49
  • $\begingroup$ There are indeed some sign issues. $f(a,b)$ is larger than $2$ for all the values of $a,b\in(0,2)$ I tried. $\endgroup$ – Rahul May 25 '13 at 18:31
  • $\begingroup$ Is it reasonable that now $f(0.5,b)$ crosses zero at $b\approx1.33095$? $\endgroup$ – Rahul May 26 '13 at 2:49
  • $\begingroup$ @RahulNarain, I did play with Geometer's Sketchpad this morning and have verified that the final version above is correct. I should have realized we needed two formulas depending on the configuration. Doh. $\endgroup$ – Ted Shifrin May 26 '13 at 16:17

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