Let triangle ABC have side lengths AB=13, AC=14, and BC=16. There are two circles located inside angle BAC which are tangent to rays AB, AC, and segment BC. Compute the distance between the centers of these two circles.
+14903 Compute the distance between the centers of these two circles.
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\(s=\frac{a+b+c}{2}=\frac{16+14+13}2=21.5\\ \rho=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}=\sqrt{\frac{(21.5-16)(21.5-14)(21.5-13)}{21.5}}=4.0383\\ \cos\alpha=\frac{b^2+c^2-a^2}{2bc}=\frac{14^2+13^2-16^2}{2\cdot 14\cdot 13}\\ \alpha =\arccos\frac{14^2+13^2-16^2}{2\cdot 14\cdot 13}\\ \alpha=72.5754^o\)
\(m_{f(x)}=\tan\frac{\alpha}{2}=\tan\frac{72.5754}{2}=0.7342425\)
\(\color{blue}f(x)=0.7342425\cdot x\\ 0.7342425\cdot x_1=\rho\\ 0.7342425\cdot x_1=4.0383\\ x_1=5.5\)
\(P_1(5.5,\ 4.0383)\)
\(r=0.7342425\cdot x_2\\ x_2=\frac{r}{m}\\ (\rho+r)^2=(r-\rho)^2+(x_2-x_1)^2\\ (\rho+r)^2=(r-\rho)^2+(\frac{r}{m}-x_1)^2\\ \color{blue} (4.0383+r)^2=(r-4.0383)^2+(\frac{r}{0.7342425}-5,5)^2\\ {\color{blue}r=15.75}\ _{w olfram\ alpha}\)
The distance between the centers of these two circles is
4.0383 + 15.75 = 19.788
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