Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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

$${im}\:{most}\:{grateful}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com