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Question Number 37751 by ajfour last updated on 17/Jun/18

Commented by ajfour last updated on 17/Jun/18

Find area of quadrilateral PQRS in terms of a, b. ABCD is a square.

$${Find}\:{area}\:{of}\:{quadrilateral}\:{PQRS} \\ $$$${in}\:{terms}\:{of}\:\boldsymbol{{a}},\:\boldsymbol{{b}}. \\ $$$${ABCD}\:{is}\:{a}\:{square}. \\ $$

Commented by MJS last updated on 17/Jun/18

ABCD must be a square? the center of the circle lies on x=y?

$${ABCD}\:\mathrm{must}\:\mathrm{be}\:\mathrm{a}\:\mathrm{square}? \\ $$$$\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{lies}\:\mathrm{on}\:{x}={y}? \\ $$

Commented by ajfour last updated on 17/Jun/18

center of circle on y=x (not given) ABCD is a square (i had missed mentioning).

$${center}\:{of}\:{circle}\:{on}\:{y}={x}\:\left({not}\:{given}\right) \\ $$$${ABCD}\:{is}\:{a}\:{square}\:\left({i}\:{had}\:{missed}\right. \\ $$$$\left.{mentioning}\right). \\ $$

Commented by MJS last updated on 17/Jun/18

ok thank you

$$\mathrm{ok}\:\mathrm{thank}\:\mathrm{you} \\ $$

Commented by ajfour last updated on 17/Jun/18

Commented by ajfour last updated on 17/Jun/18

solution part 2_(−) (refer diagram with circle center as origin) let side of square=2s=2rcos α 4s^2 =a^2 +b^2 ....(i) and radius of circle=r angles as marked in figure above. Area_(PQRS) =A A=A_(△SDC) −(A_(△PDC) +A_(△DCR) −A_(△QDC) ) =s[Rsin α−(Rsin α+y_P +Rsin α+y_R ) +Rsin α+y_Q ] ⇒ A=s(y_Q −y_P −y_R ) .....(ii) M≡(−rcos φ, −rsin φ) N≡(rcos φ, rsin φ) C≡(rcos α,−rsin α) D(−rcos α,−rsin α) to evaluate y_P : [rcos α−y_P (((cos α)/(sin α)))]tan ((α/2)−(φ/2)) = rsin α+y_P y_P =((rcos αtan ((α/2)−(φ/2))−rsin α)/(1+(((cos α)/(sin α)))tan ((α/2)−(φ/2)))) to find y_R : [rcos α−y_R (((cos α)/(sin α)))]tan ((α/2)+(φ/2)) = rsin α+y_R y_R =((rcos αtan ((α/2)+(φ/2))−rsin α)/(1+(((cos α)/(sin α)))tan ((α/2)+(φ/2)))) for y_Q : 2rcos α=(rsin α+y_Q )[cot ((α/2)−(φ/2))+cot ((α/2)+(φ/2))] y_Q =((2rcos α)/([cot ((α/2)−(φ/2))+cot ((α/2)+(φ/2))]))−rsin α ....... with rcos α = s = ((√(a^2 +b^2 ))/2) rsin α = (√(r^2 −(((a^2 +b^2 )/4)))) tan θ = (b/a) ⇒ asin θ = ((ab)/(√(a^2 +b^2 ))) acos θ = (a^2 /(√(a^2 +b^2 ))) tan 𝛅 =((rcos 𝛂−acos 𝛉)/(asin 𝛉+r(2cos 𝛂−sin 𝛂))) 𝛉−𝛗 = (𝛑/2)−𝛉+𝛅 ⇒ 𝛗=2𝛉−𝛅−(𝛑/2) . ( just a plan) ! if a=b , r=((5(√2) a)/6) ; θ=(π/4) 𝛗=𝛅 = 0 ; rcos α =(a/(√2)) ⇒ cos α = (3/5) sin α = (4/5) ; tan (α/2) = (1/2) y_P = −((10(√2) a)/(33)) =y_R y_Q = −((5a(√2))/(12)) A = (a^2 /(√2))(((20(√2))/(33))−((5(√2))/(12))) = ((75a^2 )/(33×12)) (A_(PQRS) )∣_(a=b) = ((25a^2 )/(132)) .

$$\underset{−} {{solution}\:{part}\:\mathrm{2}}\:\left({refer}\:{diagram}\:\right. \\ $$$$\left.{with}\:{circle}\:{center}\:{as}\:{origin}\right) \\ $$$${let}\:{side}\:{of}\:{square}=\mathrm{2}\boldsymbol{{s}}=\mathrm{2}{r}\mathrm{cos}\:\alpha \\ $$$$\mathrm{4}{s}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:\:\:\:\:....\left({i}\right) \\ $$$${and}\:{radius}\:{of}\:{circle}=\boldsymbol{{r}} \\ $$$${angles}\:{as}\:{marked}\:{in}\:{figure}\:{above}. \\ $$$${Area}_{{PQRS}} ={A} \\ $$$$\boldsymbol{{A}}={A}_{\bigtriangleup{SDC}} −\left({A}_{\bigtriangleup{PDC}} +{A}_{\bigtriangleup{DCR}} −{A}_{\bigtriangleup{QDC}} \right) \\ $$$$\:\:=\boldsymbol{{s}}\left[{R}\mathrm{sin}\:\alpha−\left({R}\mathrm{sin}\:\alpha+{y}_{{P}} +{R}\mathrm{sin}\:\alpha+{y}_{{R}} \right)\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{R}\mathrm{sin}\:\alpha+{y}_{{Q}} \:\right] \\ $$$$\Rightarrow\:\:\:{A}={s}\left({y}_{{Q}} −{y}_{{P}} −{y}_{{R}} \right)\:\:\:.....\left({ii}\right) \\ $$$${M}\equiv\left(−{r}\mathrm{cos}\:\phi,\:−{r}\mathrm{sin}\:\phi\right) \\ $$$${N}\equiv\left({r}\mathrm{cos}\:\phi,\:{r}\mathrm{sin}\:\phi\right) \\ $$$${C}\equiv\left({r}\mathrm{cos}\:\alpha,−{r}\mathrm{sin}\:\alpha\right) \\ $$$${D}\left(−{r}\mathrm{cos}\:\alpha,−{r}\mathrm{sin}\:\alpha\right) \\ $$$${to}\:{evaluate}\:\boldsymbol{{y}}_{\boldsymbol{{P}}} \:: \\ $$$$\left[{r}\mathrm{cos}\:\alpha−{y}_{{P}} \left(\frac{\mathrm{cos}\:\alpha}{\mathrm{sin}\:\alpha}\right)\right]\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}−\frac{\phi}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:=\:{r}\mathrm{sin}\:\alpha+{y}_{{P}} \\ $$$${y}_{{P}} =\frac{{r}\mathrm{cos}\:\alpha\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}−\frac{\phi}{\mathrm{2}}\right)−{r}\mathrm{sin}\:\alpha}{\mathrm{1}+\left(\frac{\mathrm{cos}\:\alpha}{\mathrm{sin}\:\alpha}\right)\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}−\frac{\phi}{\mathrm{2}}\right)} \\ $$$${to}\:{find}\:\boldsymbol{{y}}_{\boldsymbol{{R}}} \:: \\ $$$$\left[{r}\mathrm{cos}\:\alpha−{y}_{{R}} \left(\frac{\mathrm{cos}\:\alpha}{\mathrm{sin}\:\alpha}\right)\right]\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}+\frac{\phi}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:{r}\mathrm{sin}\:\alpha+{y}_{{R}} \\ $$$${y}_{{R}} =\frac{{r}\mathrm{cos}\:\alpha\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}+\frac{\phi}{\mathrm{2}}\right)−{r}\mathrm{sin}\:\alpha}{\mathrm{1}+\left(\frac{\mathrm{cos}\:\alpha}{\mathrm{sin}\:\alpha}\right)\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}+\frac{\phi}{\mathrm{2}}\right)} \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{y}}_{\boldsymbol{{Q}}} \:: \\ $$$$\mathrm{2}{r}\mathrm{cos}\:\alpha=\left({r}\mathrm{sin}\:\alpha+{y}_{{Q}} \right)\left[\mathrm{cot}\:\left(\frac{\alpha}{\mathrm{2}}−\frac{\phi}{\mathrm{2}}\right)+\mathrm{cot}\:\left(\frac{\alpha}{\mathrm{2}}+\frac{\phi}{\mathrm{2}}\right)\right] \\ $$$${y}_{{Q}} =\frac{\mathrm{2}{r}\mathrm{cos}\:\alpha}{\left[\mathrm{cot}\:\left(\frac{\alpha}{\mathrm{2}}−\frac{\phi}{\mathrm{2}}\right)+\mathrm{cot}\:\left(\frac{\alpha}{\mathrm{2}}+\frac{\phi}{\mathrm{2}}\right)\right]}−{r}\mathrm{sin}\:\alpha \\ $$$$....... \\ $$$${with}\:\:{r}\mathrm{cos}\:\alpha\:=\:{s}\:=\:\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{r}\mathrm{sin}\:\alpha\:=\:\sqrt{{r}^{\mathrm{2}} −\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{4}}\right)}\: \\ $$$$\:\:\mathrm{tan}\:\theta\:=\:\frac{{b}}{{a}}\:\:\Rightarrow\:\:{a}\mathrm{sin}\:\theta\:=\:\frac{{ab}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$${a}\mathrm{cos}\:\theta\:=\:\frac{{a}^{\mathrm{2}} }{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\: \\ $$$$\mathrm{tan}\:\boldsymbol{\delta}\:=\frac{\boldsymbol{{r}}\mathrm{cos}\:\boldsymbol{\alpha}−\boldsymbol{{a}}\mathrm{cos}\:\boldsymbol{\theta}}{\boldsymbol{{a}}\mathrm{sin}\:\boldsymbol{\theta}+\boldsymbol{{r}}\left(\mathrm{2cos}\:\boldsymbol{\alpha}−\mathrm{sin}\:\boldsymbol{\alpha}\right)} \\ $$$$\boldsymbol{\theta}−\boldsymbol{\phi}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{\theta}+\boldsymbol{\delta} \\ $$$$\Rightarrow\:\:\:\boldsymbol{\phi}=\mathrm{2}\boldsymbol{\theta}−\boldsymbol{\delta}−\frac{\boldsymbol{\pi}}{\mathrm{2}}\:\:. \\ $$$$\left(\:{just}\:{a}\:{plan}\right)\:! \\ $$$${if}\:{a}={b}\:\:,\:\:\:{r}=\frac{\mathrm{5}\sqrt{\mathrm{2}}\:{a}}{\mathrm{6}}\:\:;\:\:\theta=\frac{\pi}{\mathrm{4}}\: \\ $$$$\boldsymbol{\phi}=\boldsymbol{\delta}\:=\:\mathrm{0}\:;\:\: \\ $$$${r}\mathrm{cos}\:\alpha\:=\frac{{a}}{\sqrt{\mathrm{2}}}\:\Rightarrow\:\:\:\:\mathrm{cos}\:\alpha\:=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{sin}\:\alpha\:=\:\frac{\mathrm{4}}{\mathrm{5}}\:\:;\:\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}_{{P}} =\:−\frac{\mathrm{10}\sqrt{\mathrm{2}}\:{a}}{\mathrm{33}}\:\:={y}_{{R}} \\ $$$${y}_{{Q}} \:=\:−\frac{\mathrm{5}{a}\sqrt{\mathrm{2}}}{\mathrm{12}} \\ $$$${A}\:=\:\frac{{a}^{\mathrm{2}} }{\sqrt{\mathrm{2}}}\left(\frac{\mathrm{20}\sqrt{\mathrm{2}}}{\mathrm{33}}−\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{12}}\right)\:=\:\frac{\mathrm{75}{a}^{\mathrm{2}} }{\mathrm{33}×\mathrm{12}} \\ $$$$\:\:\:\:\left({A}_{{PQRS}} \right)\mid_{{a}={b}} =\:\frac{\mathrm{25}{a}^{\mathrm{2}} }{\mathrm{132}}\:. \\ $$

Commented by MJS last updated on 17/Jun/18

for b=a I get the same solution :−)

$$\left.\mathrm{for}\:{b}={a}\:\mathrm{I}\:\mathrm{get}\:\mathrm{the}\:\mathrm{same}\:\mathrm{solution}\::−\right) \\ $$

Commented by ajfour last updated on 17/Jun/18

thank you sir, for confirming.

$${thank}\:{you}\:{sir},\:{for}\:\:{confirming}. \\ $$

Answered by ajfour last updated on 17/Jun/18

Refer to diagram in question:_(−) solution part 1_(−) let circle center be (h,k) h^2 +k^2 =r^2 (as circle passes through origin) here we take O as origin; later circle center as origin in the commented diagram. C(b,a+b) ; D(a+b,a) eq. of circle : 2hx+2ky = x^2 +y^2 C and D lies on circle, So 2bh+2(a+b)k=b^2 +(a+b)^2 2(a+b)h+2ak=(a+b)^2 +a^2 2h = ((a[b^2 +(a+b)^2 ]−(a+b)[(a+b)^2 +a^2 ])/(ab−(a+b)^2 )) 2k= ((b[(a+b)^2 +a^2 ]−(a+b)[(a+b)^2 +a^2 ])/(ab−(a+b)^2 )) r^2 =h^2 +k^2 . [if a=b we have h=k= ((5a)/6) and r=((5(√2) a)/6) ].

$$\underset{−} {{Refer}\:{to}\:{diagram}\:{in}\:{question}:} \\ $$$$\underset{−} {{solution}\:\:{part}\:\mathrm{1}} \\ $$$${let}\:{circle}\:{center}\:{be}\:\left({h},{k}\right) \\ $$$${h}^{\mathrm{2}} +{k}^{\mathrm{2}} ={r}^{\mathrm{2}} \:\:\:\left({as}\:{circle}\:{passes}\:{through}\:{origin}\right) \\ $$$${here}\:{we}\:{take}\:{O}\:{as}\:{origin}; \\ $$$${later}\:{circle}\:{center}\:{as}\:{origin}\:{in} \\ $$$${the}\:{commented}\:{diagram}. \\ $$$${C}\left({b},{a}+{b}\right)\:\:\:;\:\:{D}\left({a}+{b},{a}\right) \\ $$$${eq}.\:{of}\:{circle}\:: \\ $$$$\mathrm{2}{hx}+\mathrm{2}{ky}\:=\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${C}\:{and}\:{D}\:{lies}\:{on}\:{circle},\:{So} \\ $$$$\mathrm{2}\boldsymbol{{bh}}+\mathrm{2}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)\boldsymbol{{k}}=\boldsymbol{{b}}^{\mathrm{2}} +\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)\boldsymbol{{h}}+\mathrm{2}\boldsymbol{{ak}}=\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)^{\mathrm{2}} +\boldsymbol{{a}}^{\mathrm{2}} \\ $$$$\mathrm{2}{h}\:=\:\frac{{a}\left[{b}^{\mathrm{2}} +\left({a}+{b}\right)^{\mathrm{2}} \right]−\left({a}+{b}\right)\left[\left({a}+{b}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \right]}{{ab}−\left({a}+{b}\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}{k}=\:\frac{{b}\left[\left({a}+{b}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \right]−\left({a}+{b}\right)\left[\left({a}+{b}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \right]}{{ab}−\left({a}+{b}\right)^{\mathrm{2}} } \\ $$$$\boldsymbol{{r}}^{\mathrm{2}} =\boldsymbol{{h}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} \:. \\ $$$$\left[{if}\:{a}={b}\:\:\:\:{we}\:{have}\:{h}={k}=\:\frac{\mathrm{5}{a}}{\mathrm{6}}\:\right. \\ $$$$\left.\:\:\:\:\:{and}\:\:\:{r}=\frac{\mathrm{5}\sqrt{\mathrm{2}}\:{a}}{\mathrm{6}}\:\right]. \\ $$

Answered by MJS last updated on 17/Jun/18

I took the first picture you posted... A= ((a),(0) ) B= ((0),(b) ) C= ((b),((a+b)) ) D= (((a+b)),(a) ) ∣SO∣=∣SC∣=∣SD∣ ⇒ ⇒ S= (((((a+b)^3 −ab(a+2b))/(2((a+b)^2 −ab)))),((((a+b)^3 −ab(2a+b))/(2((a+b)^2 −ab)))) ) ⇒ r=((√((a^2 +b^2 )((a+b)^4 +((a+b)^2 −ab)^2 )))/(2((a+b)^2 −ab))) ⇒ M= (((((a+b)^3 −ab(a+2b))/((a+b)^2 −ab))),(0) ) N= ((0),((((a+b)^3 −ab(2a+b))/((a+b)^2 −ab))) ) P=CM∩DS P= ((((((a+b)^2 −ab)((a+b)^2 +ab))/((a+b)^3 +a((a+b)^2 −ab)))),(((a(a+b)((a+b)^2 +b^2 ))/((a+b)^3 +a((a+b)^2 −ab)))) ) Q=CM∩DN Q= ((((b((a+b)^2 +a^2 ))/((a+b)^2 ))),(((a((a+b)^2 +b^2 ))/((a+b)^2 ))) ) R=CS∩DN R= ((((b(a+b)((a^2 +b^2 )+a^2 ))/((a+b)^3 +b((a+b)^2 −ab)))),(((((a+b)^2 −ab)((a+b)^2 +ab))/((a+b)^3 +b((a+b)^2 −ab)))) ) area(PQRS)=area(CPS)−area(CQR)= =((b(a+b)^2 (a^2 +b^2 )((a+b)^2 +a^2 ))/(4((a+b)^2 −ab)((a+b)^2 (2a+b)−a^2 b)))− −((b^3 (a^2 +b^2 )((a+b)^2 +a^2 ))/(2(a+b)^2 ((a+b)^2 (a+2b)−ab^2 )))= =((ab(a^2 +b^2 )(2(a+b)^8 +2ab(a+b)^6 −7a^2 b^2 (a+b)^4 +6a^3 b^3 (a+b)^2 −2a^4 b^4 ))/(4(a+b)^2 ((a+b)^2 −ab)((a+b)^3 +a((a+b)^2 −ab))((a+b)^3 +b((a+b)^2 −ab))))=

$$\mathrm{I}\:\mathrm{took}\:\mathrm{the}\:\mathrm{first}\:\mathrm{picture}\:\mathrm{you}\:\mathrm{posted}... \\ $$$$ \\ $$$${A}=\begin{pmatrix}{{a}}\\{\mathrm{0}}\end{pmatrix}\:{B}=\begin{pmatrix}{\mathrm{0}}\\{{b}}\end{pmatrix}\:{C}=\begin{pmatrix}{{b}}\\{{a}+{b}}\end{pmatrix}\:{D}=\begin{pmatrix}{{a}+{b}}\\{{a}}\end{pmatrix} \\ $$$$\mid{SO}\mid=\mid{SC}\mid=\mid{SD}\mid\:\Rightarrow \\ $$$$\Rightarrow\:{S}=\begin{pmatrix}{\frac{\left({a}+{b}\right)^{\mathrm{3}} −{ab}\left({a}+\mathrm{2}{b}\right)}{\mathrm{2}\left(\left({a}+{b}\right)^{\mathrm{2}} −{ab}\right)}}\\{\frac{\left({a}+{b}\right)^{\mathrm{3}} −{ab}\left(\mathrm{2}{a}+{b}\right)}{\mathrm{2}\left(\left({a}+{b}\right)^{\mathrm{2}} −{ab}\right)}}\end{pmatrix} \\ $$$$\Rightarrow\:{r}=\frac{\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\left({a}+{b}\right)^{\mathrm{4}} +\left(\left({a}+{b}\right)^{\mathrm{2}} −{ab}\right)^{\mathrm{2}} \right)}}{\mathrm{2}\left(\left({a}+{b}\right)^{\mathrm{2}} −{ab}\right)} \\ $$$$\Rightarrow\:{M}=\begin{pmatrix}{\frac{\left({a}+{b}\right)^{\mathrm{3}} −{ab}\left({a}+\mathrm{2}{b}\right)}{\left({a}+{b}\right)^{\mathrm{2}} −{ab}}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:{N}=\begin{pmatrix}{\mathrm{0}}\\{\frac{\left({a}+{b}\right)^{\mathrm{3}} −{ab}\left(\mathrm{2}{a}+{b}\right)}{\left({a}+{b}\right)^{\mathrm{2}} −{ab}}}\end{pmatrix} \\ $$$${P}={CM}\cap{DS} \\ $$$${P}=\begin{pmatrix}{\frac{\left(\left({a}+{b}\right)^{\mathrm{2}} −{ab}\right)\left(\left({a}+{b}\right)^{\mathrm{2}} +{ab}\right)}{\left({a}+{b}\right)^{\mathrm{3}} +{a}\left(\left({a}+{b}\right)^{\mathrm{2}} −{ab}\right)}}\\{\frac{{a}\left({a}+{b}\right)\left(\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{\left({a}+{b}\right)^{\mathrm{3}} +{a}\left(\left({a}+{b}\right)^{\mathrm{2}} −{ab}\right)}}\end{pmatrix} \\ $$$${Q}={CM}\cap{DN} \\ $$$${Q}=\begin{pmatrix}{\frac{{b}\left(\left({a}+{b}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}{\left({a}+{b}\right)^{\mathrm{2}} }}\\{\frac{{a}\left(\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{\left({a}+{b}\right)^{\mathrm{2}} }}\end{pmatrix} \\ $$$${R}={CS}\cap{DN} \\ $$$${R}=\begin{pmatrix}{\frac{{b}\left({a}+{b}\right)\left(\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+{a}^{\mathrm{2}} \right)}{\left({a}+{b}\right)^{\mathrm{3}} +{b}\left(\left({a}+{b}\right)^{\mathrm{2}} −{ab}\right)}}\\{\frac{\left(\left({a}+{b}\right)^{\mathrm{2}} −{ab}\right)\left(\left({a}+{b}\right)^{\mathrm{2}} +{ab}\right)}{\left({a}+{b}\right)^{\mathrm{3}} +{b}\left(\left({a}+{b}\right)^{\mathrm{2}} −{ab}\right)}}\end{pmatrix} \\ $$$$\mathrm{area}\left({PQRS}\right)=\mathrm{area}\left({CPS}\right)−\mathrm{area}\left({CQR}\right)= \\ $$$$=\frac{{b}\left({a}+{b}\right)^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\left({a}+{b}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}{\mathrm{4}\left(\left({a}+{b}\right)^{\mathrm{2}} −{ab}\right)\left(\left({a}+{b}\right)^{\mathrm{2}} \left(\mathrm{2}{a}+{b}\right)−{a}^{\mathrm{2}} {b}\right)}− \\ $$$$\:\:\:\:\:−\frac{{b}^{\mathrm{3}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\left({a}+{b}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}{\mathrm{2}\left({a}+{b}\right)^{\mathrm{2}} \left(\left({a}+{b}\right)^{\mathrm{2}} \left({a}+\mathrm{2}{b}\right)−{ab}^{\mathrm{2}} \right)}= \\ $$$$=\frac{{ab}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\mathrm{2}\left({a}+{b}\right)^{\mathrm{8}} +\mathrm{2}{ab}\left({a}+{b}\right)^{\mathrm{6}} −\mathrm{7}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}+{b}\right)^{\mathrm{4}} +\mathrm{6}{a}^{\mathrm{3}} {b}^{\mathrm{3}} \left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{4}} {b}^{\mathrm{4}} \right)}{\mathrm{4}\left({a}+{b}\right)^{\mathrm{2}} \left(\left({a}+{b}\right)^{\mathrm{2}} −{ab}\right)\left(\left({a}+{b}\right)^{\mathrm{3}} +{a}\left(\left({a}+{b}\right)^{\mathrm{2}} −{ab}\right)\right)\left(\left({a}+{b}\right)^{\mathrm{3}} +{b}\left(\left({a}+{b}\right)^{\mathrm{2}} −{ab}\right)\right)}= \\ $$

Commented by ajfour last updated on 17/Jun/18

Great solution Sir !

$$\mathcal{G}{reat}\:{solution}\:{Sir}\:! \\ $$

Commented by ajfour last updated on 17/Jun/18

was it easy to calculate P = CM∩DS and Q, R ? you show no working Sir ..

$${was}\:{it}\:{easy}\:{to}\:{calculate}\: \\ $$$${P}\:=\:{CM}\cap{DS}\: \\ $$$${and}\:{Q},\:{R}\:?\:{you}\:{show}\:{no}\:{working} \\ $$$${Sir}\:.. \\ $$

Commented by MJS last updated on 18/Jun/18

the method is simple, it′s just finding a line between two points and then intersecting the lines, the only hard work is simplifying. I usually do this by hand because I don′t have any calculator which can properly handle equations of this complexity. you won′t be able to read my scribbling... as you can see at my integral solutions I′m a correct worker and I can show step by step I just decided it′s too hard typing all the equations on my smartphone. sorry for that

$$\mathrm{the}\:\mathrm{method}\:\mathrm{is}\:\mathrm{simple},\:\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{finding}\:\mathrm{a}\:\mathrm{line} \\ $$$$\mathrm{between}\:\mathrm{two}\:\mathrm{points}\:\mathrm{and}\:\mathrm{then}\:\mathrm{intersecting} \\ $$$$\mathrm{the}\:\mathrm{lines},\:\mathrm{the}\:\mathrm{only}\:\mathrm{hard}\:\mathrm{work}\:\mathrm{is}\:\mathrm{simplifying}. \\ $$$$\mathrm{I}\:\mathrm{usually}\:\mathrm{do}\:\mathrm{this}\:\mathrm{by}\:\mathrm{hand}\:\mathrm{because}\:\mathrm{I}\:\mathrm{don}'\mathrm{t} \\ $$$$\mathrm{have}\:\mathrm{any}\:\mathrm{calculator}\:\mathrm{which}\:\mathrm{can}\:\mathrm{properly} \\ $$$$\mathrm{handle}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{this}\:\mathrm{complexity}. \\ $$$$\mathrm{you}\:\mathrm{won}'\mathrm{t}\:\mathrm{be}\:\mathrm{able}\:\mathrm{to}\:\mathrm{read}\:\mathrm{my}\:\mathrm{scribbling}... \\ $$$$\mathrm{as}\:\mathrm{you}\:\mathrm{can}\:\mathrm{see}\:\mathrm{at}\:\mathrm{my}\:\mathrm{integral}\:\mathrm{solutions}\:\mathrm{I}'\mathrm{m} \\ $$$$\mathrm{a}\:\mathrm{correct}\:\mathrm{worker}\:\mathrm{and}\:\mathrm{I}\:\mathrm{can}\:\mathrm{show}\:\mathrm{step}\:\mathrm{by}\:\mathrm{step} \\ $$$$\mathrm{I}\:\mathrm{just}\:\mathrm{decided}\:\mathrm{it}'\mathrm{s}\:\mathrm{too}\:\mathrm{hard}\:\mathrm{typing}\:\mathrm{all}\:\mathrm{the} \\ $$$$\mathrm{equations}\:\mathrm{on}\:\mathrm{my}\:\mathrm{smartphone}.\:\mathrm{sorry}\:\mathrm{for}\:\mathrm{that} \\ $$

Commented by MJS last updated on 18/Jun/18

my methods: ∣SO∣=∣SC∣=∣SD∣ put S= ((x_S ),(y_S ) ) ∣SO∣^2 =x_S ^2 +y_S ^2 =r^2 ∣SC∣^2 =(x_S −b)^2 +(y_S −(a+b))^2 =r^2 ∣SD∣^2 =(x_S −(a+b))^2 +(y_S −b)^2 =r^2 subtracting two of them to get x_S , y_S without x_S ^2 , y_S ^2 and then solving is the fastest method r=(√(x_S ^2 +y_S ^2 )) S is the center of MN ⇒ M= (((2x_S )),(0) ); N= ((0),((2y_S )) ) P=CM∩DS (and the other 2) line between 2 points: U= ((x_U ),(y_U ) ); V= ((x_V ),(y_V ) ) y=kx+d k=((y_V −y_U )/(x_V −x_U )); d=((x_V y_U −x_U y_V )/(x_V −x_U )) intersection y=k_1 x+d_1 y=k_2 x+d_2 should be obvious now we need the lenghts^2 of the lines CP, CS, PS and CQ, CR, QR ∣UV∣^2 =(x_V −x_U )^2 +(y_V −y_U )^2 the area of the triangles (the hardest part): area(ijk)=((√((i+j+k)(i+j−k)(i+k−j)(j+k−i)))/4)= =((√(2(i^2 j^2 +i^2 k^2 +j^2 k^2 )−(i^4 +j^4 +k^4 )))/4) all you need is 4P: PencilPaperPrecisionPatience

$$\mathrm{my}\:\mathrm{methods}: \\ $$$$\mid{SO}\mid=\mid{SC}\mid=\mid{SD}\mid \\ $$$$\mathrm{put}\:{S}=\begin{pmatrix}{{x}_{{S}} }\\{{y}_{{S}} }\end{pmatrix} \\ $$$$\mid{SO}\mid^{\mathrm{2}} ={x}_{{S}} ^{\mathrm{2}} +{y}_{{S}} ^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\mid{SC}\mid^{\mathrm{2}} =\left({x}_{{S}} −{b}\right)^{\mathrm{2}} +\left({y}_{{S}} −\left({a}+{b}\right)\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\mid{SD}\mid^{\mathrm{2}} =\left({x}_{{S}} −\left({a}+{b}\right)\right)^{\mathrm{2}} +\left({y}_{{S}} −{b}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\mathrm{subtracting}\:\mathrm{two}\:\mathrm{of}\:\mathrm{them}\:\mathrm{to}\:\mathrm{get}\:{x}_{{S}} ,\:{y}_{{S}} \:\mathrm{without}\: \\ $$$${x}_{{S}} ^{\mathrm{2}} ,\:{y}_{{S}} ^{\mathrm{2}} \:\mathrm{and}\:\mathrm{then}\:\mathrm{solving}\:\mathrm{is}\:\mathrm{the}\:\mathrm{fastest}\:\mathrm{method} \\ $$$$ \\ $$$${r}=\sqrt{{x}_{{S}} ^{\mathrm{2}} +{y}_{{S}} ^{\mathrm{2}} } \\ $$$$ \\ $$$${S}\:\mathrm{is}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:{MN}\:\Rightarrow\:{M}=\begin{pmatrix}{\mathrm{2}{x}_{{S}} }\\{\mathrm{0}}\end{pmatrix};\:{N}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{2}{y}_{{S}} }\end{pmatrix} \\ $$$$ \\ $$$${P}={CM}\cap{DS}\:\left(\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{line}\:\mathrm{between}\:\mathrm{2}\:\mathrm{points}: \\ $$$${U}=\begin{pmatrix}{{x}_{{U}} }\\{{y}_{{U}} }\end{pmatrix};\:{V}=\begin{pmatrix}{{x}_{{V}} }\\{{y}_{{V}} }\end{pmatrix} \\ $$$${y}={kx}+{d} \\ $$$${k}=\frac{{y}_{{V}} −{y}_{{U}} }{{x}_{{V}} −{x}_{{U}} };\:{d}=\frac{{x}_{{V}} {y}_{{U}} −{x}_{{U}} {y}_{{V}} }{{x}_{{V}} −{x}_{{U}} } \\ $$$$\mathrm{intersection} \\ $$$${y}={k}_{\mathrm{1}} {x}+{d}_{\mathrm{1}} \\ $$$${y}={k}_{\mathrm{2}} {x}+{d}_{\mathrm{2}} \\ $$$$\mathrm{should}\:\mathrm{be}\:\mathrm{obvious} \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{need}\:\mathrm{the}\:\mathrm{lenghts}^{\mathrm{2}} \:\mathrm{of}\:\mathrm{the}\:\mathrm{lines} \\ $$$${CP},\:{CS},\:{PS}\:\mathrm{and}\:{CQ},\:{CR},\:{QR} \\ $$$$\mid{UV}\mid^{\mathrm{2}} =\left({x}_{{V}} −{x}_{{U}} \right)^{\mathrm{2}} +\left({y}_{{V}} −{y}_{{U}} \right)^{\mathrm{2}} \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangles}\:\left(\mathrm{the}\:\mathrm{hardest}\:\mathrm{part}\right): \\ $$$$\mathrm{area}\left({ijk}\right)=\frac{\sqrt{\left({i}+{j}+{k}\right)\left({i}+{j}−{k}\right)\left({i}+{k}−{j}\right)\left({j}+{k}−{i}\right)}}{\mathrm{4}}= \\ $$$$=\frac{\sqrt{\mathrm{2}\left({i}^{\mathrm{2}} {j}^{\mathrm{2}} +{i}^{\mathrm{2}} {k}^{\mathrm{2}} +{j}^{\mathrm{2}} {k}^{\mathrm{2}} \right)−\left({i}^{\mathrm{4}} +{j}^{\mathrm{4}} +{k}^{\mathrm{4}} \right)}}{\mathrm{4}} \\ $$$$ \\ $$$$\mathrm{all}\:\mathrm{you}\:\mathrm{need}\:\mathrm{is}\:\mathrm{4P}: \\ $$$$\mathrm{PencilPaperPrecisionPatience} \\ $$

Commented by ajfour last updated on 18/Jun/18

Too good Sir, thanks a lot.

$${Too}\:{good}\:{Sir},\:{thanks}\:{a}\:{lot}. \\ $$

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