+88 Let triangle ABC have side lengths AB=13, AC=14, and BC=15. 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.
Kind of confusedon how to even start?...
THANKS FOR ANY HELP!
+26367 Let triangle ABC have side lengths AB=13, AC=14, and BC=15.
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.
\(\begin{array}{|rcll|} \hline s &=& \dfrac{a+b+c}{2} \\ s &=& \dfrac{13+15+14}{2} \\ \mathbf{s} &=& \mathbf{ 21 } \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline r &=& \sqrt{\dfrac{(s-a)(s-b)(s-c)}{s}} \\ &=& \sqrt{\dfrac{(21-13)(21-15)(21-14)}{21}} \\ &=& \sqrt{\dfrac{(8)(6)(7)}{21}} \\ &=& \sqrt{\dfrac{336}{21}} \\ &=& \sqrt{16} \\ \mathbf{r} &=& \mathbf{ 4 } \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline r_b &=& \sqrt{\dfrac{s(s-a)(s-c)}{s-b}} \\ &=& \sqrt{\dfrac{21(21-13)(21-14)}{21-15}} \\ &=& \sqrt{\dfrac{21(8)(7)}{6}} \\ &=& \sqrt{ 196 } \\ \mathbf{r_b} &=& \mathbf{ 14 } \\ \hline \end{array}\)
\(\mathbf{AU=\ ?}\)
\(\begin{array}{|rcll|} \hline AU &=& AV \\ AU+AV &=& (a+BU)+(c+CV) \\ &=& a+c+BU+CV \quad | \quad BU = BU',\ CV=CV' \\ &=& a+c+BU'+CV' \quad | \quad BU'+CV' = b \\ &=& a+c+ b \\ AU+ AV&=& 2s \quad | \quad AV = AU \\ 2AU &=& 2s \\ \mathbf{ AU } &=& \mathbf{ s } \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline x &=& \sqrt{\Big(s-(s-b)\Big)^2 + (r_b-r)^2 } \\ &=& \sqrt{b^2 + (14-4)^2 } \\ &=& \sqrt{15^2 + 10^2 } \\ &=& \sqrt{(3*5)^2 + (2*5)^2 } \\ &=& \sqrt{5^2*(3^2+2^2) } \\ &=& \sqrt{5^2*13 } \\ \mathbf{x } &=& \mathbf{ 5\sqrt{ 13 } } \\ x &\approx& 18 \\ \hline \end{array} \)
The distance between the centers of these two circles is \(\mathbf{\approx 18}\)
+36916 Strong work as always, Heureka......but after you found 'r' and 'rb' , why not just add them together to find the distance 'x' between the circle centers?
