Cool Circles Problem

$\Delta ACB$ is a right triangle with side lengths $AC:CB:BA$ in the ratio $3:4:5$ . By similarity, triangles $\Delta ADC$ and $\Delta CDB$ are right triangles with the same side length ratio as $\Delta ACB.$

Now as the inradius of a $3/4/5$ triangle is $1$ , the inradii of triangles $\Delta ADC$ and $\Delta CDB$ will scale up to values of $3$ and $4$ , respectively.

Now draw a line parallel to $AB$ passing through $P$ , and let this line intersect the perpendicular from $Q$ to $AB$ at $M$ . Then $\Delta PMQ$ is a right triangle with side lengths $PM = 3 + 4 = 7$ and $QM = 4 - 3 = 1.$ The hypotenuse $PQ$ will then have length $\sqrt{7^{2} + 1^{2}} = \boxed{\sqrt{50}}.$

The altitude divide the triangle in two rigth triangles with sides $b_1=9$ , $c_1=12$ , $a_1=15$ and $b_2=12$ , $c_2=16$ , $a_2=20$ respectively (easy to prove by using Euclid's theorems!). The inradii of these triangles are $r_1=\frac{b_1+c_1-a_1}{2}=3 \qquad r_2=\frac{b_2+c_2-a_2}{2}=4$
so the distance between the centers of these circles is $d=\sqrt{\left(r_1+r_2\right)^2+\left(r_2-r_1\right)^2}=\sqrt{2(r_1^2+ r_2^2)}=\boxed{\sqrt{50}}$

we have : AC=15 , BC=20 ; SO : AB=25 and now by using Euclid's theory : AC × BC = AB × CD →CD=12

FROM ACD→AD=9→BD=25-9=16

the radius of any circle inside any triangle is : r²={(s-a)(s-b)(s-c)/s) where s=(a+b+c)/2

by using it we got : the radius of the circle Q=3×\sqrt { 2 } as well : the radius of the circle P = 3 now the distance that is needed (PQ)=r(Q)+r(P)=3+3×\sqrt{2}} =7.2 and that result almost equal to \sqrt{50}

The distance $d$ between the centers of two circles inscribed by two parts (obtained by dropping a perpendicular from right vertex to the hypotenuse) of a right triangle having orthogonal sides $a$ & $b$ is given by the general formula

$\boxed{d=\frac{a+b-\sqrt{a^2+b^2}}{\sqrt2}}$

As per given problem, setting $a=15$ & $b=20$ in above general formula

$d=\frac{15+20-\sqrt{15^2+20^2}}{\sqrt2}=\frac{35-25}{\sqrt2}=5\sqrt2=\sqrt{50}$