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Question-214876

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Question Number 214876 by mr W last updated on 22/Dec/24
Commented by TonyCWX08 last updated on 22/Dec/24
Want a clarification Does the tangent point at Blue circle cuts the line into 2 equal parts? If no, then I have no idea.
$${Want}\:{a}\:{clarification} \\ $$$${Does}\:{the}\:{tangent}\:{point}\:{at}\:{Blue}\:{circle}\:{cuts}\:{the}\:{line}\:{into}\:\mathrm{2}\:{equal}\:{parts}? \\ $$$${If}\:{no},\:{then}\:{I}\:{have}\:{no}\:{idea}. \\ $$
Commented by mr W last updated on 22/Dec/24
yes. everything is as it looks like: blue triangle is right−angled. yellow circle is inscribed in the blue triangle and touches the big circle. the blue circle touches the big circle and the blue side at the middle point.
$${yes}. \\ $$$${everything}\:{is}\:{as}\:{it}\:{looks}\:{like}: \\ $$$${blue}\:{triangle}\:{is}\:{right}−{angled}. \\ $$$${yellow}\:{circle}\:{is}\:{inscribed}\:{in}\:{the} \\ $$$${blue}\:{triangle}\:{and}\:{touches}\:{the}\:{big} \\ $$$${circle}. \\ $$$${the}\:{blue}\:{circle}\:{touches}\:{the}\:{big}\:{circle} \\ $$$${and}\:{the}\:{blue}\:{side}\:{at}\:{the}\:{middle}\:{point}. \\ $$
Commented by TonyCWX08 last updated on 22/Dec/24
Oh. Then I know what to do!
$${Oh}. \\ $$$${Then}\:{I}\:{know}\:{what}\:{to}\:{do}! \\ $$
Answered by TonyCWX08 last updated on 22/Dec/24
AC=13 a=((13+12+5)/2)=2 CG=6.5 GH=12+5+13−3−6.5=3.5 OC^2 =OG^2 +6.5^2 OX=OP−2 OX^2 =XF^2 +OF^2 (OP−2)^2 =3.5^2 +(2+OG)^2 OP^2 −4OP+4=12.25+4+4OG+OG^2 OG^2 +42.25−4OP=12.25+4OG+OG^2 42.25−4OP=12.25+4OG −4OP−4OG=−30 OP+OG=7.5 Observe that 2b=OP+OH 2b=7.5 b=3.75 (b/a)=((3.75)/2)=1.875
$${AC}=\mathrm{13} \\ $$$${a}=\frac{\mathrm{13}+\mathrm{12}+\mathrm{5}}{\mathrm{2}}=\mathrm{2} \\ $$$${CG}=\mathrm{6}.\mathrm{5} \\ $$$${GH}=\mathrm{12}+\mathrm{5}+\mathrm{13}−\mathrm{3}−\mathrm{6}.\mathrm{5}=\mathrm{3}.\mathrm{5} \\ $$$$ \\ $$$${OC}^{\mathrm{2}} ={OG}^{\mathrm{2}} +\mathrm{6}.\mathrm{5}^{\mathrm{2}} \\ $$$$ \\ $$$${OX}={OP}−\mathrm{2} \\ $$$${OX}^{\mathrm{2}} ={XF}^{\mathrm{2}} +{OF}^{\mathrm{2}} \\ $$$$\left({OP}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{3}.\mathrm{5}^{\mathrm{2}} +\left(\mathrm{2}+{OG}\right)^{\mathrm{2}} \\ $$$${OP}^{\mathrm{2}} −\mathrm{4}{OP}+\mathrm{4}=\mathrm{12}.\mathrm{25}+\mathrm{4}+\mathrm{4}{OG}+{OG}^{\mathrm{2}} \\ $$$${OG}^{\mathrm{2}} +\mathrm{42}.\mathrm{25}−\mathrm{4}{OP}=\mathrm{12}.\mathrm{25}+\mathrm{4}{OG}+{OG}^{\mathrm{2}} \\ $$$$\mathrm{42}.\mathrm{25}−\mathrm{4}{OP}=\mathrm{12}.\mathrm{25}+\mathrm{4}{OG} \\ $$$$−\mathrm{4}{OP}−\mathrm{4}{OG}=−\mathrm{30} \\ $$$${OP}+{OG}=\mathrm{7}.\mathrm{5} \\ $$$$ \\ $$$${Observe}\:{that}\: \\ $$$$\mathrm{2}{b}={OP}+{OH} \\ $$$$\mathrm{2}{b}=\mathrm{7}.\mathrm{5} \\ $$$${b}=\mathrm{3}.\mathrm{75} \\ $$$$ \\ $$$$\frac{{b}}{{a}}=\frac{\mathrm{3}.\mathrm{75}}{\mathrm{2}}=\mathrm{1}.\mathrm{875} \\ $$
Commented by TonyCWX08 last updated on 23/Dec/24
Thanks sir! I finally able to utilize what I had learnt!
$${Thanks}\:{sir}! \\ $$$${I}\:{finally}\:{able}\:{to}\:{utilize}\:{what}\:{I}\:{had}\:{learnt}! \\ $$
Commented by TonyCWX08 last updated on 22/Dec/24
Commented by mr W last updated on 22/Dec/24
right sir!
$${right}\:{sir}! \\ $$
Answered by A5T last updated on 31/Dec/24
R^2 =6.5^2 +(2b−R)^2 ⇒0=42.25+4b^2 −4bR ⇒0=6.5^2 +4b^2 −4bR⇒R=((4b^2 +42.25)/(4b)) a(((5+12+13)/2))=((5×12)/2)⇒a=((60)/(30))=2 (a−R+2b)^2 +(6.5−3)^2 =(R−a)^2 ⇒4b^2 +8b−4bR+12.25=0 ⇒−30+8b=0⇒8b=30⇒b=((15)/4)=3.75 ⇒(b/a)=((3.75)/2)=1.875
$$\mathrm{R}^{\mathrm{2}} =\mathrm{6}.\mathrm{5}^{\mathrm{2}} +\left(\mathrm{2b}−\mathrm{R}\right)^{\mathrm{2}} \Rightarrow\mathrm{0}=\mathrm{42}.\mathrm{25}+\mathrm{4b}^{\mathrm{2}} −\mathrm{4bR} \\ $$$$\Rightarrow\mathrm{0}=\mathrm{6}.\mathrm{5}^{\mathrm{2}} +\mathrm{4b}^{\mathrm{2}} −\mathrm{4bR}\Rightarrow\mathrm{R}=\frac{\mathrm{4b}^{\mathrm{2}} +\mathrm{42}.\mathrm{25}}{\mathrm{4b}} \\ $$$$\mathrm{a}\left(\frac{\mathrm{5}+\mathrm{12}+\mathrm{13}}{\mathrm{2}}\right)=\frac{\mathrm{5}×\mathrm{12}}{\mathrm{2}}\Rightarrow\mathrm{a}=\frac{\mathrm{60}}{\mathrm{30}}=\mathrm{2} \\ $$$$\left(\mathrm{a}−\mathrm{R}+\mathrm{2b}\right)^{\mathrm{2}} +\left(\mathrm{6}.\mathrm{5}−\mathrm{3}\right)^{\mathrm{2}} =\left(\mathrm{R}−\mathrm{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4b}^{\mathrm{2}} +\mathrm{8b}−\mathrm{4bR}+\mathrm{12}.\mathrm{25}=\mathrm{0} \\ $$$$\Rightarrow−\mathrm{30}+\mathrm{8b}=\mathrm{0}\Rightarrow\mathrm{8b}=\mathrm{30}\Rightarrow\mathrm{b}=\frac{\mathrm{15}}{\mathrm{4}}=\mathrm{3}.\mathrm{75} \\ $$$$\Rightarrow\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{3}.\mathrm{75}}{\mathrm{2}}=\mathrm{1}.\mathrm{875} \\ $$

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