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Question Number 20131 by NECC last updated on 22/Aug/17
In rectangle ABCD,AB=8, BC=20.P is a point on AD so that ∠BPC=90°.If r_1 ,r_2 ,r_3 are the radii of the incircles of APB, BPC, and CPD. find r_1 +r_2 +r_3
$${In}\:{rectangle}\:{ABCD},{AB}=\mathrm{8}, \\ $$$${BC}=\mathrm{20}.{P}\:{is}\:{a}\:{point}\:{on}\:{AD}\:{so} \\ $$$${that}\:\angle{BPC}=\mathrm{90}°.{If}\:{r}_{\mathrm{1}} ,{r}_{\mathrm{2}} ,{r}_{\mathrm{3}} \:{are}\:{the} \\ $$$${radii}\:{of}\:{the}\:{incircles}\:{of}\:{APB}, \\ $$$${BPC},\:{and}\:{CPD}.\:{find}\:{r}_{\mathrm{1}} +{r}_{\mathrm{2}} +{r}_{\mathrm{3}} \\ $$
Commented by ajfour last updated on 22/Aug/17
Commented by NECC last updated on 22/Aug/17
please help
$${please}\:{help} \\ $$
Commented by ajfour last updated on 22/Aug/17
r_1 +r_2 +r_3 =r_1 (1+(√5)+2)=(3+(√5))r_1 (as shall be proved below) r_1 +r_1 cot θ=AP (from figure) OP=((BC)/2)=10 ; PQ=AB=8 OQ=(√(OP^2 −PQ^2 )) =6 BQ=AP=OB−OQ=10−6=4 BP=(√(PQ^2 +BQ^2 )) =(√(64+16)) =4(√5) CD=AB=8 tan 2θ=((PQ)/(BQ))=(8/4)=2 ((2tan θ)/(1−tan^2 θ))=2 ⇒ tan^2 θ+tan θ−1=0 ⇒ tan θ=(((√5)−1)/2) r_1 (1+cot θ)=AP=4 r_1 (1+(2/( (√5)−1)))=4 ⇒ r_1 =((4((√5)−1))/( (√5)+1)) ⇒ r_1 =6−2(√5) . The three triangles are similar, so their inradii are in the ratio of their sides; r_1 : r_2 : r_3 =AP: BP: CD = 4 : 4(√5) : 8 = 1 : (√5) : 2 so, r_1 +r_2 +r_3 =r_1 (1+(√5)+2) =(6−2(√5))(3+(√5)) =2(3−(√5))(3+(√5)) =2(9−5) =8 .
$${r}_{\mathrm{1}} +{r}_{\mathrm{2}} +{r}_{\mathrm{3}} ={r}_{\mathrm{1}} \left(\mathrm{1}+\sqrt{\mathrm{5}}+\mathrm{2}\right)=\left(\mathrm{3}+\sqrt{\mathrm{5}}\right){r}_{\mathrm{1}} \\ $$$$\:\:\:\:\left({as}\:{shall}\:{be}\:{proved}\:{below}\right) \\ $$$${r}_{\mathrm{1}} +{r}_{\mathrm{1}} \mathrm{cot}\:\theta={AP}\:\:\:\left({from}\:{figure}\right) \\ $$$${OP}=\frac{{BC}}{\mathrm{2}}=\mathrm{10}\:\:;\:{PQ}={AB}=\mathrm{8} \\ $$$${OQ}=\sqrt{{OP}^{\mathrm{2}} −{PQ}^{\mathrm{2}} }\:=\mathrm{6} \\ $$$${BQ}={AP}={OB}−{OQ}=\mathrm{10}−\mathrm{6}=\mathrm{4} \\ $$$${BP}=\sqrt{{PQ}^{\mathrm{2}} +{BQ}^{\mathrm{2}} }\:=\sqrt{\mathrm{64}+\mathrm{16}}\:=\mathrm{4}\sqrt{\mathrm{5}}\: \\ $$$${CD}={AB}=\mathrm{8} \\ $$$$\mathrm{tan}\:\mathrm{2}\theta=\frac{{PQ}}{{BQ}}=\frac{\mathrm{8}}{\mathrm{4}}=\mathrm{2} \\ $$$$\frac{\mathrm{2tan}\:\theta}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta}=\mathrm{2}\:\:\Rightarrow\:\:\:\mathrm{tan}\:^{\mathrm{2}} \theta+\mathrm{tan}\:\theta−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{tan}\:\theta=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\: \\ $$$$\:\:{r}_{\mathrm{1}} \left(\mathrm{1}+\mathrm{cot}\:\theta\right)={AP}=\mathrm{4} \\ $$$$\:\:{r}_{\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}−\mathrm{1}}\right)=\mathrm{4}\:\:\:\Rightarrow\:\:\:{r}_{\mathrm{1}} =\frac{\mathrm{4}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\:\sqrt{\mathrm{5}}+\mathrm{1}} \\ $$$$\:\:\Rightarrow\:\:\:\:\boldsymbol{{r}}_{\mathrm{1}} =\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}\:. \\ $$$${The}\:{three}\:{triangles}\:{are}\:{similar},\:{so} \\ $$$${their}\:{inradii}\:{are}\:{in}\:{the}\:{ratio}\:{of} \\ $$$${their}\:{sides}; \\ $$$$\:{r}_{\mathrm{1}} :\:{r}_{\mathrm{2}} :\:{r}_{\mathrm{3}} \:={AP}:\:{BP}:\:{CD} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4}\::\:\mathrm{4}\sqrt{\mathrm{5}}\::\:\mathrm{8}\:\:=\:\mathrm{1}\::\:\sqrt{\mathrm{5}}\::\:\mathrm{2}\: \\ $$$$\boldsymbol{{so}},\:{r}_{\mathrm{1}} +{r}_{\mathrm{2}} +{r}_{\mathrm{3}} ={r}_{\mathrm{1}} \left(\mathrm{1}+\sqrt{\mathrm{5}}+\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:=\left(\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}\right)\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)\: \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)\left(\mathrm{3}+\sqrt{\mathrm{5}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{9}−\mathrm{5}\right)\:=\mathrm{8}\:. \\ $$$$ \\ $$
Commented by NECC last updated on 22/Aug/17
wow.... you′re amazing sir. It feels good to be here. I′ll learn alot from you all.Gracias.....
$${wow}….\:{you}'{re}\:{amazing}\:{sir}.\: \\ $$$${It}\:{feels}\:{good}\:{to}\:{be}\:{here}.\:{I}'{ll}\:{learn} \\ $$$${alot}\:{from}\:{you}\:{all}.{Gracias}….. \\ $$
Answered by Tinkutara last updated on 22/Aug/17
Commented by Tinkutara last updated on 22/Aug/17
In a right triangle where c is hypotenuse, radius of incircle = (1/2)(a + b − c) So r_1 = (1/2)(x + 8 − a) r_2 = (1/2)(a + b − 20) r_3 = (1/2)(y + 8 − 20) r_1 + r_2 + r_3 = (1/2)(2×8) = 8
$$\mathrm{In}\:\mathrm{a}\:\mathrm{right}\:\mathrm{triangle}\:\mathrm{where}\:{c}\:\mathrm{is}\:\mathrm{hypotenuse}, \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{incircle}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({a}\:+\:{b}\:−\:{c}\right) \\ $$$$\mathrm{So}\:{r}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}\:+\:\mathrm{8}\:−\:{a}\right) \\ $$$${r}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({a}\:+\:{b}\:−\:\mathrm{20}\right) \\ $$$${r}_{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({y}\:+\:\mathrm{8}\:−\:\mathrm{20}\right) \\ $$$${r}_{\mathrm{1}} \:+\:{r}_{\mathrm{2}} \:+\:{r}_{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}×\mathrm{8}\right)\:=\:\mathrm{8} \\ $$
Commented by ajfour last updated on 22/Aug/17
thanks for the confirmation.
$${thanks}\:{for}\:{the}\:{confirmation}. \\ $$
Commented by NECC last updated on 22/Aug/17
im most grateful sir
$${im}\:{most}\:{grateful}\:{sir} \\ $$

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