Question Number 228342 by ajfour last updated on 08/Apr/26
Commented by ajfour last updated on 08/Apr/26
$${If}\:\:{for}\:{rectangles}\:{ABCD}\:{and} \\ $$$${PQRS}:\:\:\:\:\:\:{AB}=\mathrm{2}{a},\:{BC}=\mathrm{2}{b}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{PQ}={a},\:{QR}={b} \\ $$$${find}\:{lengths}\:{BE},\:{DF}. \\ $$
Commented by fantastic2 last updated on 08/Apr/26
$${is}\:{the}\:{big}\:{triangle}\:{isoceles}? \\ $$
Commented by ajfour last updated on 08/Apr/26
$${not}\:{necessarily} \\ $$
Commented by mr W last updated on 08/Apr/26
$${let}\:{DF}={p},\:{BE}={q} \\ $$$$\left(\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{q}^{\mathrm{2}} }−{a}\right)\left(\mathrm{4}{ab}−{pq}\right)={b}\left(\mathrm{4}{a}^{\mathrm{2}} +{q}^{\mathrm{2}} \right) \\ $$$${there}\:{seems}\:{to}\:{be}\:{only}\:{one}\:{single} \\ $$$${possibility}:\:{p}=\mathrm{0},\:{q}=\mathrm{0} \\ $$
Commented by mr W last updated on 09/Apr/26
Commented by mr W last updated on 09/Apr/26
$${this}\:{is}\:{the}\:{only}\:{possbility}. \\ $$
Commented by mr W last updated on 09/Apr/26
Commented by ajfour last updated on 09/Apr/26
$${Thanks}\:{sir},\:{but}\:{maybe}\:{i}'{ll}\:{check} \\ $$$${again},\:{cuz}\:{i}\:{didnot}\:{understand} \\ $$$${the}\:{logic}\:{of}\:{your}\:{equation}. \\ $$
Commented by ajfour last updated on 09/Apr/26
$${AI}\:{Gemini}\:{tells}\:{me}\:{E}\:{and}\:{F}\:{are} \\ $$$${then}\:{midpoints}\:{of}\:{those}\:{sides}. \\ $$
Answered by mr W last updated on 09/Apr/26
Commented by mr W last updated on 09/Apr/26
![AE=(√(4a^2 +q^2 )) eqn. of AE: qx−2ay=0 F(p, 2b) h=((−pq+4ab)/( (√(4a^2 +q^2 )))) (h/( (√(4a^2 +q^2 ))))=((h−b)/a) ((√(4a^2 +q^2 ))−a)((4ab−pq)/( (√(4a^2 +q^2 ))))=b(√(4a^2 +q^2 )) ⇒((√(4a^2 +q^2 ))−a)(4ab−pq)=b(4a^2 +q^2 ) ⇒(p/b)=(1/(((q/a))))[4−((4+((q/a))^2 )/( (√(4+((q/a))^2 ))−1))] ⇒λ=(1/ξ)(4−((4+ξ^2 )/( (√(4+ξ^2 ))−1))) such that λ, ξ≥0, there is only one possibility: ξ=λ=0, i.e. p=q=0.](https://www.tinkutara.com/question/Q228363.png)
$${AE}=\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{q}^{\mathrm{2}} } \\ $$$${eqn}.\:{of}\:{AE}: \\ $$$${qx}−\mathrm{2}{ay}=\mathrm{0} \\ $$$${F}\left({p},\:\mathrm{2}{b}\right) \\ $$$${h}=\frac{−{pq}+\mathrm{4}{ab}}{\:\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{q}^{\mathrm{2}} }} \\ $$$$\frac{{h}}{\:\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{q}^{\mathrm{2}} }}=\frac{{h}−{b}}{{a}} \\ $$$$\left(\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{q}^{\mathrm{2}} }−{a}\right)\frac{\mathrm{4}{ab}−{pq}}{\:\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{q}^{\mathrm{2}} }}={b}\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{q}^{\mathrm{2}} } \\ $$$$\Rightarrow\left(\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{q}^{\mathrm{2}} }−{a}\right)\left(\mathrm{4}{ab}−{pq}\right)={b}\left(\mathrm{4}{a}^{\mathrm{2}} +{q}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\frac{{p}}{{b}}=\frac{\mathrm{1}}{\left(\frac{{q}}{{a}}\right)}\left[\mathrm{4}−\frac{\mathrm{4}+\left(\frac{{q}}{{a}}\right)^{\mathrm{2}} }{\:\sqrt{\mathrm{4}+\left(\frac{{q}}{{a}}\right)^{\mathrm{2}} }−\mathrm{1}}\right] \\ $$$$\Rightarrow\lambda=\frac{\mathrm{1}}{\xi}\left(\mathrm{4}−\frac{\mathrm{4}+\xi^{\mathrm{2}} }{\:\sqrt{\mathrm{4}+\xi^{\mathrm{2}} }−\mathrm{1}}\right) \\ $$$${such}\:{that}\:\lambda,\:\xi\geqslant\mathrm{0},\:{there}\:{is}\:{only}\:{one} \\ $$$${possibility}:\:\xi=\lambda=\mathrm{0},\:{i}.{e}.\:{p}={q}=\mathrm{0}. \\ $$
Commented by ajfour last updated on 10/Apr/26
$${yeah}\:{hope}\:{so}. \\ $$$${But}\:{we}\:{can}\:{still}\:{think}\:{of}\:{more} \\ $$$${questions}\:{out}\:{of}\:{this}\:{question}, \\ $$
Commented by fantastic2 last updated on 10/Apr/26
$${right}\:{sir} \\ $$
Commented by fantastic2 last updated on 10/Apr/26
$${mr}\:{W}\:{sir}\:{is}\:{absolutely}\:{right} \\ $$
Answered by fantastic2 last updated on 10/Apr/26
Commented by fantastic2 last updated on 10/Apr/26
![AE⇒y=(p/(2a))x △AEF=(1/2)[x_1 (y_2 −y_3 )+x_2 (y_3 −y_1 )+x_3 (y_1 −y_2 )] here x_1 =0 ,y=0 x_2 =2a ,y_2 =p x_3 =q,y_3 =2b △=(1/2)(4ab−pq) FT=h (1/2)×AE×h=(1/2)(4ab−pq) h=((4ab−pq)/( (√(4a^2 +p^2 )))) eq of FT m^⊥ =−((2a)/p) y−y_1 =m^⊥ (x−x_1 ) ⇒y=−((2a)/p)(x−q)+2b x of T (p/(2a))x=−((2a)/p)x+((2aq)/p)+2b ⇒x_T =((4a(aq+pb))/(p^2 +4a^2 )) ⇒y_T =((2p(aq+pb))/(p^2 +4a^2 )) (√(x_T ^2 +y_T ^2 ))+(√((x_T −2a)^2 +(y_T −p)^2 ))=(√(4a^2 +p^2 )) simplifying we get 2(aq+pb)=4a^2 +p^2 ...I now (√(x_T ^2 +y_(Th) ^2 ))(1/h)=((AP)/b) AP=(b/h)(√(x_T ^2 +y_T ^2 )) similarly QE=(b/h)(√((x_T −2a)^2 +(y_T −p)^2 )) ∴(1/2)×b×(b/h)(√(x_T ^2 +y_T ^2 ))+(1/2)b×(b/h)(√((x_T −2a)^2 +(y_T −p)^2 ))+(1/2)×a×(h−b)+ab=(1/2)(4ab−pq) ⇒(b^2 /h)((√(x_T ^2 +y_T ^2 ))+(√((x_T −2a)^2 +(y_T −p)^2 )))+a(h−b)=2ab−pq ⇒((b^2 ×(√(4a^2 +p^2 )))/(4ab−pq))(√(4a^2 +p^2 ))+a(((4ab−pq)/( (√(4a^2 +p^2 ))))−b)=2ab−pq ⇒((b^2 (4a^2 +p^2 ))/(4ab−pq))+a((4ab−pq)/( (√(4a^2 +p^2 ))))=3ab−pq...II](https://www.tinkutara.com/question/Q228377.png)
$${AE}\Rightarrow{y}=\frac{{p}}{\mathrm{2}{a}}{x} \\ $$$$\bigtriangleup{AEF}=\frac{\mathrm{1}}{\mathrm{2}}\left[{x}_{\mathrm{1}} \left({y}_{\mathrm{2}} −{y}_{\mathrm{3}} \right)+{x}_{\mathrm{2}} \left({y}_{\mathrm{3}} −{y}_{\mathrm{1}} \right)+{x}_{\mathrm{3}} \left({y}_{\mathrm{1}} −{y}_{\mathrm{2}} \right)\right] \\ $$$${here}\:{x}_{\mathrm{1}} =\mathrm{0}\:,{y}=\mathrm{0} \\ $$$${x}_{\mathrm{2}} =\mathrm{2}{a}\:,{y}_{\mathrm{2}} ={p} \\ $$$${x}_{\mathrm{3}} ={q},{y}_{\mathrm{3}} =\mathrm{2}{b} \\ $$$$\bigtriangleup=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}{ab}−{pq}\right) \\ $$$${FT}={h} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×{AE}×{h}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}{ab}−{pq}\right) \\ $$$${h}=\frac{\mathrm{4}{ab}−{pq}}{\:\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{p}^{\mathrm{2}} }} \\ $$$${eq}\:{of}\:{FT} \\ $$$${m}^{\bot} =−\frac{\mathrm{2}{a}}{{p}} \\ $$$${y}−{y}_{\mathrm{1}} ={m}^{\bot} \left({x}−{x}_{\mathrm{1}} \right) \\ $$$$\Rightarrow{y}=−\frac{\mathrm{2}{a}}{{p}}\left({x}−{q}\right)+\mathrm{2}{b} \\ $$$${x}\:{of}\:{T} \\ $$$$\frac{{p}}{\mathrm{2}{a}}{x}=−\frac{\mathrm{2}{a}}{{p}}{x}+\frac{\mathrm{2}{aq}}{{p}}+\mathrm{2}{b} \\ $$$$\Rightarrow{x}_{{T}} =\frac{\mathrm{4}{a}\left({aq}+{pb}\right)}{{p}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} }\: \\ $$$$\Rightarrow{y}_{{T}} =\frac{\mathrm{2}{p}\left({aq}+{pb}\right)}{{p}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} }\: \\ $$$$\sqrt{{x}_{{T}} ^{\mathrm{2}} +{y}_{{T}} ^{\mathrm{2}} }+\sqrt{\left({x}_{{T}} −\mathrm{2}{a}\right)^{\mathrm{2}} +\left({y}_{{T}} −{p}\right)^{\mathrm{2}} }=\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{p}^{\mathrm{2}} } \\ $$$${simplifying}\:{we}\:{get} \\ $$$$\mathrm{2}\left({aq}+{pb}\right)=\mathrm{4}{a}^{\mathrm{2}} +{p}^{\mathrm{2}} \:…\mathrm{I} \\ $$$${now} \\ $$$$\sqrt{{x}_{{T}} ^{\mathrm{2}} +{y}_{{Th}} ^{\mathrm{2}} }\frac{\mathrm{1}}{{h}}=\frac{{AP}}{{b}} \\ $$$${AP}=\frac{{b}}{{h}}\sqrt{{x}_{{T}} ^{\mathrm{2}} +{y}_{{T}} ^{\mathrm{2}} } \\ $$$${similarly} \\ $$$${QE}=\frac{{b}}{{h}}\sqrt{\left({x}_{{T}} −\mathrm{2}{a}\right)^{\mathrm{2}} +\left({y}_{{T}} −{p}\right)^{\mathrm{2}} } \\ $$$$\therefore\frac{\mathrm{1}}{\mathrm{2}}×{b}×\frac{{b}}{{h}}\sqrt{{x}_{{T}} ^{\mathrm{2}} +{y}_{{T}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}{b}×\frac{{b}}{{h}}\sqrt{\left({x}_{{T}} −\mathrm{2}{a}\right)^{\mathrm{2}} +\left({y}_{{T}} −{p}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}×{a}×\left({h}−{b}\right)+{ab}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}{ab}−{pq}\right) \\ $$$$\Rightarrow\frac{{b}^{\mathrm{2}} }{{h}}\left(\sqrt{{x}_{{T}} ^{\mathrm{2}} +{y}_{{T}} ^{\mathrm{2}} }+\sqrt{\left({x}_{{T}} −\mathrm{2}{a}\right)^{\mathrm{2}} +\left({y}_{{T}} −{p}\right)^{\mathrm{2}} }\right)+{a}\left({h}−{b}\right)=\mathrm{2}{ab}−{pq} \\ $$$$\Rightarrow\frac{{b}^{\mathrm{2}} ×\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{p}^{\mathrm{2}} }}{\mathrm{4}{ab}−{pq}}\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{p}^{\mathrm{2}} }+{a}\left(\frac{\mathrm{4}{ab}−{pq}}{\:\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{p}^{\mathrm{2}} }}−{b}\right)=\mathrm{2}{ab}−{pq} \\ $$$$\Rightarrow\frac{{b}^{\mathrm{2}} \left(\mathrm{4}{a}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)}{\mathrm{4}{ab}−{pq}}+{a}\frac{\mathrm{4}{ab}−{pq}}{\:\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{p}^{\mathrm{2}} }}=\mathrm{3}{ab}−{pq}…\mathrm{II} \\ $$