Question Number 84381 by behi83417@gmail.com last updated on 12/Mar/20
Commented by behi83417@gmail.com last updated on 12/Mar/20
$$\mathrm{AB}=\mathrm{9},\mathrm{BC}=\mathrm{10},\mathrm{CA}=\mathrm{12} \\ $$$$\mathrm{FI},\mathrm{divides}\:\mathrm{triangle}\:\mathrm{in}\:\mathrm{two}\:\mathrm{parts}\:\mathrm{that}\:\mathrm{equail}\:\mathrm{in}\: \\ $$$$\mathrm{area}\:\mathrm{and}\:\mathrm{perimeter}. \\ $$$$\mathrm{find}:\mathrm{CI},\mathrm{CF},\mathrm{FI}. \\ $$
Commented by mr W last updated on 12/Mar/20
$${nice}\:{question}\:{sir}!\:{but}\:{it}'{s}\:{impossible} \\ $$$${with}\:{given}\:{geometry}.\:{AB}\:{should}\:{be} \\ $$$${at}\:{least}\:\mathrm{8}\sqrt{\mathrm{15}}−\mathrm{22}\approx\mathrm{8}.\mathrm{98}\:{such}\:{that}\:{the} \\ $$$${triangle}\:{can}\:{be}\:{divided}\:{into}\:{such}\:{two} \\ $$$${parts}. \\ $$
Commented by behi83417@gmail.com last updated on 12/Mar/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{master}. \\ $$$$\mathrm{now}\:\mathrm{it}\:\mathrm{is}\:\mathrm{fixed}! \\ $$
Answered by mr W last updated on 12/Mar/20
![ΔCFI=[ABFI]=((ΔABC)/2) ((CF×CI)/(BC×AC))=((ΔCFI)/(ΔABC))=(1/2) ⇒CF×CI=(1/2)×BC×AC=(1/2)×10×12=60 CF+CI+FI=FB+BA+AI+FI CF+CI=FB+BA+AI=CB−CF+BA+AC−CI ⇒CF+CI=(1/2)(CB+BA+AC)=((10+9+12)/2)=((31)/2) CF and CI are roots of x^2 −((31)/2)x+60=0 x=(1/2)[((31)/2)±(√((((31)/2))^2 −4×60))]=8, 7.5 ⇒CF=8, CI=7.5 or ⇒CF=7.5, CI=8. FI^2 =8^2 +7.5^2 −2×8×7.5×((10^2 +12^2 −9^2 )/(2×10×12))=((155)/4) ⇒FI=((√(155))/2)≈6.225](https://www.tinkutara.com/question/Q84397.png)
$$\Delta{CFI}=\left[{ABFI}\right]=\frac{\Delta{ABC}}{\mathrm{2}} \\ $$$$\frac{{CF}×{CI}}{{BC}×{AC}}=\frac{\Delta{CFI}}{\Delta{ABC}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{CF}×{CI}=\frac{\mathrm{1}}{\mathrm{2}}×{BC}×{AC}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\mathrm{12}=\mathrm{60} \\ $$$${CF}+{CI}+{FI}={FB}+{BA}+{AI}+{FI} \\ $$$${CF}+{CI}={FB}+{BA}+{AI}={CB}−{CF}+{BA}+{AC}−{CI} \\ $$$$\Rightarrow{CF}+{CI}=\frac{\mathrm{1}}{\mathrm{2}}\left({CB}+{BA}+{AC}\right)=\frac{\mathrm{10}+\mathrm{9}+\mathrm{12}}{\mathrm{2}}=\frac{\mathrm{31}}{\mathrm{2}} \\ $$$${CF}\:{and}\:{CI}\:{are}\:{roots}\:{of} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{31}}{\mathrm{2}}{x}+\mathrm{60}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{31}}{\mathrm{2}}\pm\sqrt{\left(\frac{\mathrm{31}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{60}}\right]=\mathrm{8},\:\mathrm{7}.\mathrm{5} \\ $$$$\Rightarrow{CF}=\mathrm{8},\:{CI}=\mathrm{7}.\mathrm{5}\:{or} \\ $$$$\Rightarrow{CF}=\mathrm{7}.\mathrm{5},\:{CI}=\mathrm{8}. \\ $$$${FI}^{\mathrm{2}} =\mathrm{8}^{\mathrm{2}} +\mathrm{7}.\mathrm{5}^{\mathrm{2}} −\mathrm{2}×\mathrm{8}×\mathrm{7}.\mathrm{5}×\frac{\mathrm{10}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} }{\mathrm{2}×\mathrm{10}×\mathrm{12}}=\frac{\mathrm{155}}{\mathrm{4}} \\ $$$$\Rightarrow{FI}=\frac{\sqrt{\mathrm{155}}}{\mathrm{2}}\approx\mathrm{6}.\mathrm{225} \\ $$
Commented by behi83417@gmail.com last updated on 12/Mar/20
$$\mathrm{thanks}\:\mathrm{in}\:\mathrm{advance}\:\mathrm{dear}\:\mathrm{master}.\mathrm{perfect}. \\ $$