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Question Number 125337 by ajfour last updated on 10/Dec/20

Commented by ajfour last updated on 10/Dec/20

If all coloured areas are 10 units  each, find coordinates of G.  (FGE   and CGD are straight  segments)

$${If}\:{all}\:{coloured}\:{areas}\:{are}\:\mathrm{10}\:{units} \\ $$$${each},\:{find}\:{coordinates}\:{of}\:{G}. \\ $$$$\left({FGE}\:\:\:{and}\:{CGD}\:{are}\:{straight}\right. \\ $$$$\left.{segments}\right) \\ $$

Answered by ajfour last updated on 10/Dec/20

let  G(p,q)  ,  C(0,c)  ,  F(0, f)  eq. of AB :    x+y=10  E(2p, 2q−f)  ⇒   2p+2q−f=10       ...(i)  D(d, 0)  (q/(d−p))=(c/d)      ....(ii)  p(c−f)=20      ...(iii)  (10−c)(2p)=20     ....(iv)         cd=40           ....(v)  ⇒   q=(c^2 /(40))(((40)/c)−p)    ....(I)           f=c−((20)/p)      2p+(c^2 /(20))(((40)/c)−p)=10+c−((20)/p)   ..(II)  from  (iv)     c=10−((10)/p)     substituting for c in (II)    2p+10−((10)/p)−(p/(20))(10−((10)/p))^2 =10−((20)/p)  ⇒  2p+((10)/p)−5p(1−(2/p)+(1/p^2 ))=0  ⇒   2p+((10)/p)−5p+10−(5/p)=0  ⇒   3p−(5/p)=10  ⇒    3p^2 −10p−5 = 0  p=(5/3)+(√(((25)/9)+((15)/9)))      p=((5+2(√(10)))/3)        c=10(1−(3/(5+2(√(10)))))=10(((2+2(√(10)))/(5+2(√(10)))))       =4(1+(√(10)))(2(√(10))−5)/3       =(4/3)(2(√(10))−5+20−5(√(10)))    c =4(5−(√(10)))   And from (I) :  q=(c^2 /(40))(((40)/c)−p)    q= c−((c^2 p)/(40))    =20−4(√(10))−((16(5−(√(10)))^2 (5+2(√(10))))/(40×3))   q= 20−4(√(10))−((2(7−2(√(10)))(5+2(√(10))))/3)    =((60−12(√(10))−2(7−2(√(10)))(5+2(√(10))))/3)    =((60−12(√(10))−2(35+14(√(10))−10(√(10))−40))/3)    =((70−20(√(10)))/3) = ((10)/3)(7−2(√(10)))    G[((5+2(√(10)))/3) , ((10)/3)(7−2(√(10)) )] .

$${let}\:\:{G}\left({p},{q}\right)\:\:,\:\:{C}\left(\mathrm{0},{c}\right)\:\:,\:\:{F}\left(\mathrm{0},\:{f}\right) \\ $$$${eq}.\:{of}\:{AB}\::\:\:\:\:{x}+{y}=\mathrm{10} \\ $$$${E}\left(\mathrm{2}{p},\:\mathrm{2}{q}−{f}\right) \\ $$$$\Rightarrow\:\:\:\mathrm{2}{p}+\mathrm{2}{q}−{f}=\mathrm{10}\:\:\:\:\:\:\:...\left({i}\right) \\ $$$${D}\left({d},\:\mathrm{0}\right) \\ $$$$\frac{{q}}{{d}−{p}}=\frac{{c}}{{d}}\:\:\:\:\:\:....\left({ii}\right) \\ $$$${p}\left({c}−{f}\right)=\mathrm{20}\:\:\:\:\:\:...\left({iii}\right) \\ $$$$\left(\mathrm{10}−{c}\right)\left(\mathrm{2}{p}\right)=\mathrm{20}\:\:\:\:\:....\left({iv}\right) \\ $$$$\:\:\:\:\:\:\:{cd}=\mathrm{40}\:\:\:\:\:\:\:\:\:\:\:....\left({v}\right) \\ $$$$\Rightarrow\:\:\:{q}=\frac{{c}^{\mathrm{2}} }{\mathrm{40}}\left(\frac{\mathrm{40}}{{c}}−{p}\right)\:\:\:\:....\left({I}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{f}={c}−\frac{\mathrm{20}}{{p}} \\ $$$$\:\:\:\:\mathrm{2}{p}+\frac{{c}^{\mathrm{2}} }{\mathrm{20}}\left(\frac{\mathrm{40}}{{c}}−{p}\right)=\mathrm{10}+{c}−\frac{\mathrm{20}}{{p}}\:\:\:..\left({II}\right) \\ $$$${from}\:\:\left({iv}\right) \\ $$$$\:\:\:{c}=\mathrm{10}−\frac{\mathrm{10}}{{p}} \\ $$$$\:\:\:{substituting}\:{for}\:{c}\:{in}\:\left({II}\right) \\ $$$$\:\:\mathrm{2}{p}+\mathrm{10}−\frac{\mathrm{10}}{{p}}−\frac{{p}}{\mathrm{20}}\left(\mathrm{10}−\frac{\mathrm{10}}{{p}}\right)^{\mathrm{2}} =\mathrm{10}−\frac{\mathrm{20}}{{p}} \\ $$$$\Rightarrow\:\:\mathrm{2}{p}+\frac{\mathrm{10}}{{p}}−\mathrm{5}{p}\left(\mathrm{1}−\frac{\mathrm{2}}{{p}}+\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{2}{p}+\frac{\mathrm{10}}{{p}}−\mathrm{5}{p}+\mathrm{10}−\frac{\mathrm{5}}{{p}}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{3}{p}−\frac{\mathrm{5}}{{p}}=\mathrm{10} \\ $$$$\Rightarrow\:\:\:\:\mathrm{3}{p}^{\mathrm{2}} −\mathrm{10}{p}−\mathrm{5}\:=\:\mathrm{0} \\ $$$${p}=\frac{\mathrm{5}}{\mathrm{3}}+\sqrt{\frac{\mathrm{25}}{\mathrm{9}}+\frac{\mathrm{15}}{\mathrm{9}}} \\ $$$$\:\:\:\:{p}=\frac{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{10}}}{\mathrm{3}}\:\:\: \\ $$$$\:\:\:{c}=\mathrm{10}\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{10}}}\right)=\mathrm{10}\left(\frac{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{10}}}{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{10}}}\right) \\ $$$$\:\:\:\:\:=\mathrm{4}\left(\mathrm{1}+\sqrt{\mathrm{10}}\right)\left(\mathrm{2}\sqrt{\mathrm{10}}−\mathrm{5}\right)/\mathrm{3} \\ $$$$\:\:\:\:\:=\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{2}\sqrt{\mathrm{10}}−\mathrm{5}+\mathrm{20}−\mathrm{5}\sqrt{\mathrm{10}}\right) \\ $$$$\:\:{c}\:=\mathrm{4}\left(\mathrm{5}−\sqrt{\mathrm{10}}\right) \\ $$$$\:{And}\:{from}\:\left({I}\right)\::\:\:{q}=\frac{{c}^{\mathrm{2}} }{\mathrm{40}}\left(\frac{\mathrm{40}}{{c}}−{p}\right) \\ $$$$\:\:{q}=\:{c}−\frac{{c}^{\mathrm{2}} {p}}{\mathrm{40}}\: \\ $$$$\:=\mathrm{20}−\mathrm{4}\sqrt{\mathrm{10}}−\frac{\mathrm{16}\left(\mathrm{5}−\sqrt{\mathrm{10}}\right)^{\mathrm{2}} \left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{10}}\right)}{\mathrm{40}×\mathrm{3}} \\ $$$$\:{q}=\:\mathrm{20}−\mathrm{4}\sqrt{\mathrm{10}}−\frac{\mathrm{2}\left(\mathrm{7}−\mathrm{2}\sqrt{\mathrm{10}}\right)\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{10}}\right)}{\mathrm{3}} \\ $$$$\:\:=\frac{\mathrm{60}−\mathrm{12}\sqrt{\mathrm{10}}−\mathrm{2}\left(\mathrm{7}−\mathrm{2}\sqrt{\mathrm{10}}\right)\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{10}}\right)}{\mathrm{3}} \\ $$$$\:\:=\frac{\mathrm{60}−\mathrm{12}\sqrt{\mathrm{10}}−\mathrm{2}\left(\mathrm{35}+\mathrm{14}\sqrt{\mathrm{10}}−\mathrm{10}\sqrt{\mathrm{10}}−\mathrm{40}\right)}{\mathrm{3}} \\ $$$$\:\:=\frac{\mathrm{70}−\mathrm{20}\sqrt{\mathrm{10}}}{\mathrm{3}}\:=\:\frac{\mathrm{10}}{\mathrm{3}}\left(\mathrm{7}−\mathrm{2}\sqrt{\mathrm{10}}\right) \\ $$$$\:\:{G}\left[\frac{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{10}}}{\mathrm{3}}\:,\:\frac{\mathrm{10}}{\mathrm{3}}\left(\mathrm{7}−\mathrm{2}\sqrt{\mathrm{10}}\:\right)\right]\:. \\ $$

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