Hosam's Incircles

I used coordinate geometry here; it's a bit messy but I'll outline the approach and some other findings.

The vertices of this particular triangle can be written as $A(5,12)$ , $B(0,0)$ , $C(14,0)$ . Let $P$ have coordinates $(x,y)$ .

Since the area of any triangle is the product of its inradius with its semiperimeter, we have $2[\Delta APB] = r(AP+BP+AB)$ and similarly for the other triangles. We can calculate the areas in terms of $x$ and $y$ using the shoelace theorem: $2[\Delta BPC]=14y\;\;\;\;2[\Delta CPA]=168-12x-9y\;\;\;\;2[\Delta APB]=12x-5y$

and the sidelengths involving $P$ by Pythagoras: $AP=\sqrt{(x-5)^2+(12-y)^2}\;\;\;\;BP=\sqrt{x^2+y^2}\;\;\;\;CP=\sqrt{(14-x)^2+y^2}$

Putting all these together, we have $\frac{14y}{\sqrt{x^2+y^2}+\sqrt{(14-x)^2+y^2}+14}=\frac{168-12x-9y}{\sqrt{(14-x)^2+y^2}+\sqrt{(x-5)^2+(12-y)^2}+15}=\frac{12x-5y}{\sqrt{(x-5)^2+(12-y)^2}+\sqrt{x^2+y^2}+13}=r$

This can be solved numerically to give $r=1.85754\ldots$ and an answer to the question of $\boxed{185754}$ .

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With a bit of hunting on the Encyclopedia of Triangle Centres, this turns out to be centre $X(5394)$ , the "congruent incircles point". According to the ETC, $r$ is the smallest positive root of this equation: $585 x^8 - 7044 x^7 - 243752 x^6 + 6203408 x^5 - 56957216 x^4 + 261213120 x^3 - 638822016 x^2 + 796594176 x - 398297088=0$

I'd be interested in a proof of that one!

$P$ is the Congruent Incircles Point and is listed in as $X_{5394}$ in Kimberling's Encyclopedia of Triangle Centers . Interestingly, no coordinates are known for point. However, Kimberling does offer an eight degree polynomial for finding the congruent inradius, as well as Mathematica code. The polynomial yielded an incorrect answer for me, but that was probably a careless transcription error on my part. So I resorted to numerical methods. I hope someone finds a more aesthetic solution.