![\"\"](\"https:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2021\/05\/Screen-Shot-2021-05-17-at-4.19.55-PM-1024x789.png\")
<\/figure>\n\n\n\nLet the radius of $\u03c9_1$ as $r_1$, and we have $$r_1=\\dfrac{Area\\ of\\ \\triangle{ABM}}{\\dfrac{AB+BM+MA}{2}}=\\dfrac{Area\\ of\\ \\triangle{ABC}}{AB+BM+MA}=\\dfrac{\\dfrac{1}{2}\\cdot 12\\cdot 16}{12+10+10}=3$$ Similarly, we have the radius of $\u03c9_2$ as $r_2=\\dfrac{8}{3}$.<\/p>\n\n\n\n
$$\\because OP=MO+MP=ML+MN=MN+LN+MN=2MN+LN$$ $$JK=QK+QJ=QL+QN=QL+QL+LN=2QL+LN$$ $$OP=JK$$ <\/p>\n\n\n\n
$$\\therefore 2MN+LN=2QL+LN, QL=MN$$<\/p>\n\n\n\n
$$\\because MN=MP=MC-CP=MC-CH=\\dfrac{1}{2}AC-\\dfrac{1}{2}BC=2$$ $$\\therefore KQ=LQ=2, LN=ML-LN=\\dfrac{ML}{DL}\\cdot DL-2=\\dfrac{4}{3}\\cdot 3-2=2$$<\/p>\n\n\n\n
Method 1<\/strong> – Using Trigonometry<\/strong>: $$\\tan{\\angle{KDL}}=\\tan{(2\\angle{KDQ})}=\\dfrac{2\\tan{\\angle{KDQ}}}{1-\\tan^2{\\angle{KDQ}}}=\\dfrac{2\\cdot\\dfrac{2}{3}}{1-(\\dfrac{2}{3})^2}=\\dfrac{12}{5}$$<\/p>\n\n\n\n $$\\tan{\\angle{IDK}}=\\tan{(180^\\circ-\\angle{KDM})}=-\\tan{\\angle{KDM}} =-\\tan{(\\angle{KDL}+\\angle{LDM})}$$<\/p>\n\n\n\n $$=-\\dfrac{\\tan{\\angle{KDL}}+\\tan{\\angle{LDM}}}{1-\\tan\\angle{KDL}\\cdot\\tan\\angle{LDM}}=-\\dfrac{\\dfrac{12}{5}+\\dfrac{4}{3}}{1-\\dfrac{12}{5}\\cdot\\dfrac{4}{3}}=\\dfrac{56}{33}$$<\/p>\n\n\n\n $$\\sin{\\angle{IDK}}=\\dfrac{56}{\\sqrt{56^2+33^2}}=\\dfrac{56}{65}, \\cos{\\angle{IDK}}=\\dfrac{33}{\\sqrt{56^2+33^2}}=\\dfrac{33}{65}$$<\/p>\n\n\n\n $$\\tan{\\angle{IDX}}=\\tan{\\dfrac{\\angle{IDK}}{2}}=\\dfrac{\\sin{\\angle{IDK}}}{1+\\cos{\\angle{IDK}}}=\\dfrac{\\dfrac{56}{65}}{1+\\dfrac{33}{65}}=\\dfrac{4}{7}$$<\/p>\n\n\n\n $$BX=BI-IX=\\dfrac{1}{2}\\cdot AB-ID\\cdot\\tan{\\angle{IDX}}=6-3\\cdot\\dfrac{4}{7}=\\dfrac{30}{7} \\tag{1}$$<\/p>\n\n\n\n $$\\because \\angle{IDK}+\\angle{KDF}+\\angle{FDM}=180^\\circ\\ , \\angle{HFJ}+\\angle{JFD}+\\angle{DFM}=180^\\circ$$<\/p>\n\n\n\n $$\\therefore (\\angle{IDK}+\\angle{KDF}+\\angle{FDM})+(\\angle{HFJ}+\\angle{JFD}+\\angle{DFM})=360^\\circ$$<\/p>\n\n\n\n $$\\therefore (\\angle{IDK}+\\angle{HFJ})+(\\angle{KDF}+\\angle{JFD})+(\\angle{FDM}+\\angle{DFM})=360^\\circ$$<\/p>\n\n\n\n $$\\because \\angle{KDF}+\\angle{JFD}=360^\\circ-(\\angle{DKJ}+\\angle{FJK})=360^\\circ-(90^\\circ+90^\\circ)=180^\\circ, \\angle{FDM}+\\angle{DFM}=90^\\circ$$<\/p>\n\n\n\n $$\\therefore \\angle{IDK}+\\angle{HFJ}=90^\\circ$$<\/p>\n\n\n\n $$\\therefore \\angle{IDX}+\\angle{HFY}=45^\\circ$$<\/p>\n\n\n\n $$\\therefore \\tan{\\angle{HFY}}=\\tan(45^\\circ-\\angle{IDX})=\\dfrac{\\tan{45^\\circ}-\\tan{\\angle{IDX}}}{1+\\tan{45^\\circ}\\cdot\\tan{\\angle{IDX}}}=\\dfrac{1-\\dfrac{4}{7}}{1+1\\cdot \\dfrac{4}{7}}=\\dfrac{3}{11}$$<\/p>\n\n\n\n $$\\therefore BY=BH-BY=\\dfrac{1}{2}BC-FH\\cdot\\tan{\\angle{HFY}}=8-\\dfrac{8}{3}\\cdot\\dfrac{3}{11}=\\dfrac{80}{11} \\tag{2}$$<\/p>\n\n\n\n Based on $(1)$ and $(2)$, the area of $\\triangle{BXY}$ is: $\\dfrac{1}{2}\\cdot BX\\cdot BY=\\dfrac{1}{2}\\cdot\\dfrac{30}{7}\\cdot\\dfrac{80}{11}=\\dfrac{1200}{77}$.<\/p>\n\n\n\n Therefore, $m+n=1200+77=\\boxed{1277}$.<\/p>\n\n\n\n Method 2 – Using Menelaus’s Theorem<\/strong>: Extend both tangent lines $AC$ and $XY$ so that they intersect at $Z$. Draw line $DZ$, and $D$, $F$ and $Z$ are co-line, as $D$ and $F$ are centers of circles $\u03c9_1$ and $\u03c9_2$, respectively, and $Z$ is the intersection point of two external tangent lines of both circles $\u03c9_1$ and $\u03c9_2$. <\/p>\n\n\n\n $$\\because \\angle{DKZ}=\\angle{FJZ}=90^\\circ, \\triangle{DKZ}\\sim \\triangle{FJZ}$$<\/p>\n\n\n\n $$\\therefore \\dfrac{KZ}{JZ}=\\dfrac{DK}{FJ}=\\dfrac{r_1}{r_2}=\\dfrac{3}{\\dfrac{8}{3}}=\\dfrac{9}{8}$$<\/p>\n\n\n\n $$\\because KZ=KJ+JZ=KQ+QJ+YZ=KQ+QL+LN+JZ=2+2+2+JZ=JZ+6$$<\/p>\n\n\n\n $$\\therefore \\dfrac{JZ+6}{JZ}=\\dfrac{9}{8}, JZ=48, KZ=OZ=JZ+6=54$$<\/p>\n\n\n\n $$\\therefore MZ=OZ-OM=54-4=50, AZ=MZ+AM=50+10=60, CZ=MZ-MC=50-10=40$$<\/p>\n\n\n\n According to Menelaus’s Theorem, we have:<\/p>\n\n\n\n $$\\dfrac{AZ}{MZ}\\cdot\\dfrac{MQ}{QM}\\cdot\\dfrac{BX}{XA}=1, \\dfrac{MZ}{CZ}\\cdot\\dfrac{CY}{BY}\\cdot\\dfrac{BQ}{QM}=1$$<\/p>\n\n\n\n $$\\therefore \\dfrac{60}{50}\\cdot\\dfrac{6}{4}\\cdot\\dfrac{BX}{12-BX}=1, \\dfrac{50}{40}\\cdot\\dfrac{16-BY}{BY}\\cdot\\dfrac{4}{6}=1$$<\/p>\n\n\n\n $$\\therefore AX=\\dfrac{30}{7}, BY=\\dfrac{80}{11}$$<\/p>\n\n\n\n Therefore, the area of $\\triangle{BXY}$ is: $\\dfrac{1}{2}\\cdot BX\\cdot BY=\\dfrac{1}{2}\\cdot\\dfrac{30}{7}\\cdot\\dfrac{80}{11}=\\dfrac{1200}{77}$.<\/p>\n\n\n\n Therefore, $m+n=1200+77=\\boxed{1277}$.<\/p>\n","protected":false},"excerpt":{"rendered":" $\\triangle{ABC}$ has a right angle at $B$, $AB = 12$, and $BC = 16$. Let $M$ be the midpoint of $AC$. Let $\u03c9_1$ be the incircle of $\\triangle{ABM}$ and $\u03c9_2$ be the incircle of $\\triangle{BCM}$. The line externally tangent to … Continue reading ![\"\"](\"https:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2021\/06\/Screen-Shot-2021-06-07-at-12.31.48-AM-1024x762.png\")
<\/figure>\n\n\n\n