Consider the circle $(S)$ of radius r = 1 and the line $l$ directed. Let $k$ be a directed line parallel to the line $l$ at a distance $h$ and A a point on this line outside of the circle $(S)$. Let A be a point on the line $k$ outside of the circle $(S)$ and H the foot of a normal from the point A to the line $l$. Let $b, c$ be the 2 directed tangents from the point A to the circle $(S)$ intersecting the line $l$ at points B, C and let $\beta = \widehat{(b, k)} = \widehat{(b, l)} = \measuredangle ABH$ $\gamma = \widehat{(c, k)} = \widehat{(c, l)} = 180^\circ - \measuredangle ACH$ be the directed angles formed by the directed tangents $b, c$ with the directed lines $k, l$. (A directed tangent must have the same direction as the directed circle at their tangency point. As opposed to 2 undirected lines, which form 2 different complementary angles, 2 directed lines form a unique angle.) Then $\frac{r}{PB} = \frac{PS}{PB} = \tan \widehat{SBP} = \tan \left(90^\circ - \frac{\widehat{ABH}}{2}\right) = \tan \left(90^\circ - \frac{\beta}{2}\right) = \frac{1}{\tan \frac \beta 2}$ ${\frac{r}{PC} = \frac{PS}{PC} = \tan \widehat{SCP}} = \tan \frac{\widehat{ACH}}{2} = \tan \left(90^\circ - \frac{\gamma}{2}\right) = \frac{1}{\tan \frac \gamma 2}$ The power of the directed line $k$ to the directed circle $(S)$ is equal to $\tan \frac{\widehat{(b, k)}}{2} \cdot \tan \frac{\widehat{(c, k)}}{2} = \tan \frac{\beta}{2} \cdot \tan \frac{ \gamma}{2} = \frac{PB \cdot PC}{r^2}$ and it does not depend on the position of the point A on the directed line $k$. Obviously, this is a dual concept of the power of a point to an undirected circle, which does not depend on the angle of the circle secant from this point. I guess it is not considered a pure geometry, because it contains trigonometric functions . To calculate the power of the directed line $k$ to the directed circle $(S)$, we can move the point A into the foot $A_0$ of a normal from the tangency point P to the line $k$, because then $\gamma_0 = \beta_0$ and $\tan \frac{\beta}{2} \cdot \tan \frac{ \gamma}{2} = \tan \frac{\beta_0}{2} \cdot \tan \frac{ \gamma_0}{2} = \tan^2 \frac{\beta_0}{2}$ Let $U_0, V_0$ be the tangency points of the directed tangents from the point $A_0$ to the directed circle $(S)$ intersecting the directed line $l$ at points $B_0, C_0$. $\frac{r}{h - r} = \frac{U_0S}{A_0S} = \cos \widehat{A_0SU_0} = \cos \widehat{A_0BP} = \cos (180^\circ - \beta_0) = -\cos \beta_0$ $\cos \beta_0 = \cos^2 \frac{\beta_0}{2} - \sin^2 \frac{\beta_0}{2} = \cos^2 \frac{\beta_0}{2} \left(1 - \tan^2 \frac{\beta_0}{2}\right) = \frac{1 - \tan^2 \frac{\beta_0}{2}}{1 + \tan^2 \frac{\beta_0}{2}} = \frac{1 - x}{1 + x}$ Solving for x, we get $x = \tan^2 \frac{\beta_0}{2} = \frac{1 - \cos \beta_0}{1 + \cos \beta_0} = \frac{1 + \frac{r}{h - r}}{1 - \frac{r}{h - r}} = \frac{h - r + r}{h - r - r} = \frac{h}{h - 2r}$ $PB \cdot PC = r^2 \tan^2 \frac{\beta_0}{2} = \frac{r^2h}{h - 2r}$ and for a unit circle r = 1, $PB \cdot PC = \frac{h}{h - 2}$

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HI yetti , your solution is nice but it is very long. Let AB=c and AC=b and BC=a and (a+b+c)/2 =k In ABC we know that S(the area of triangle) = k*r r=1 ===> S=k=a*h/2 ==>k-a = a*(h-2)/2 PC=k-c and PB=k-b we khow S = [(k-a)*(k-b)*(k-c)*k]^(1/2) ===>PC*PB = k /k-a ===>PB*PC=(a*h/2)/(a*(h-2)/2) so PB*PC = h/h-2

Length of a solution depends mainly on how many details and comments is the solver willing to put down and perhaps how general the solution is. Still, if a problem was phrased "Find the shortest possible solution of ...", I would not touch it, because I do not search for brevity, but for illumination. Yetti

Naser yetties solutions are not just for finishing the probleme but to find most possible facts about the argumentsd generlising it as more as possible .our dear yetti is one of the best guys in calculating and his solutions are just elegant in forms of calculating . if you want to see some nice peices of art of problem solving then read the solution of yetti for the most hardest problems with elgant calculatings.

I agree with you, Lupin! But such Yeti often solved a lot of the difficult problems. In the triangle ABC there is the relation: $h_a=\frac{2r(s-b)(s-c)}{|r^2-(s-b)(s-c)|}$ In the forum http://www.mathlinks.ro/Forum/viewtopic.php?t=42518 I proposed the following problem: "the ecuation $h_a=\frac{2(s-b)(s-c)x}{|x^2-(s-b)(s-c)|}$ has the solutions r,$r_a$ and the ecuation ... has the solutions $r_b,r_c.$" This problem even Yeti solved on one-two pages. From r=1, $h_a=h\ge 2.$ results $PB.PC=(s-b)(s-c)=\frac{h}{h-2}.$ because $r^2\le (s-b)(s-c)$.