+0

help geometry

0
59
1

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.

Feb 15, 2022

#1
+13581
+2

Compute the distance between the centers of these two circles.

Hello Guest!

$$s=\frac{a+b+c}{2}=\frac{16+14+13}2=21.5\\ \rho=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}=\sqrt{\frac{(21.5-16)(21.5-14)(21.5-13)}{21.5}}=4.0383\\ \cos\alpha=\frac{b^2+c^2-a^2}{2bc}=\frac{14^2+13^2-16^2}{2\cdot 14\cdot 13}\\ \alpha =\arccos\frac{14^2+13^2-16^2}{2\cdot 14\cdot 13}\\ \alpha=72.5754^o$$

$$m_{f(x)}=\tan\frac{\alpha}{2}=\tan\frac{72.5754}{2}=0.7342425$$

$$\color{blue}f(x)=0.7342425\cdot x\\ 0.7342425\cdot x_1=\rho\\ 0.7342425\cdot x_1=4.0383\\ x_1=5.5$$

$$P_1(5.5,\ 4.0383)$$

$$r=0.7342425\cdot x_2\\ x_2=\frac{r}{m}\\ (\rho+r)^2=(r-\rho)^2+(x_2-x_1)^2\\ (\rho+r)^2=(r-\rho)^2+(\frac{r}{m}-x_1)^2\\ \color{blue} (4.0383+r)^2=(r-4.0383)^2+(\frac{r}{0.7342425}-5,5)^2\\ {\color{blue}r=15.75}\ _{w olfram\ alpha}$$

The distance between the centers of these two circles is

4.0383 + 15.75 = 19.788

!

Feb 16, 2022
edited by asinus  Feb 16, 2022