Let ABC be a triangle with AB=1,AC=3,and angl | vedclass

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(B) Place $A$ at $(0,0)$,$B$ at $(1,0)$,and $C$ at $(0,3)$.The circumcircle of $	riangle ABC$ has diameter $BC$. The center $C_1$ is the midpoint of

Vedclass · Accepted Answer

$0.84$

Vedclass · Answer

(B) Place $A$ at $(0,0)$,$B$ at $(1,0)$,and $C$ at $(0,3)$.The circumcircle of $	riangle ABC$ has diameter $BC$. The center $C_1$ is the midpoint of $BC$,so $C_1 = (\frac{1}{2}, \frac{3}{2})$.The radius $R$ of the circumcircle is $R = \frac{BC}{2} = \frac{\sqrt{1^2+3^2}}{2} = \frac{\sqrt{10}}{2}$.The small circle has radius $r$ and touches $AB$ $(y=0)$ and $AC$ $(x=0)$ in the first quadrant,so its center is $C_2 = (r, r)$.Since the small circle touches the circumcircle internally,the distance between centers $C_1 C_2$ must be $R - r$.$C_1 C_2^2 = (r - \frac{1}{2})^2 + (r - \frac{3}{2})^2 = (R - r)^2 = (\frac{\sqrt{10}}{2} - r)^2$.$r^2 - r + \frac{1}{4} + r^2 - 3r + \frac{9}{4} = \frac{10}{4} - \sqrt{10}r + r^2$.$r^2 - 4r + \sqrt{10}r = 0$.Since $r > 0$,$r = 4 - \sqrt{10} \approx 4 - 3.162 = 0.838$.Rounding to two decimal places,$r \approx 0.84$.

Let $ABC$ be a triangle with $AB=1$,$AC=3$,and $\angle BAC=\frac{\pi}{2}$. If a circle of radius $r>0$ touches the sides $AB$,$AC$ and also touches the circumcircle of triangle $ABC$ internally,then the value of $r$ is:

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