Question-199358

Question Number 199358 by ajfour last updated on 01/Nov/23

Commented by ajfour last updated on 01/Nov/23

$${Find}\:{R}. \\ $$

Answered by mr W last updated on 01/Nov/23

(R+R+1+(√((R+1)^2 +R^2 )))×1=R(R+1) (√((R+1)^2 +R^2 ))=R^2 −R−1 R^2 −2R−3=0 (R−3)(R+1)=0 ⇒R=3

$$\left({R}+{R}+\mathrm{1}+\sqrt{\left({R}+\mathrm{1}\right)^{\mathrm{2}} +{R}^{\mathrm{2}} }\right)×\mathrm{1}={R}\left({R}+\mathrm{1}\right) \\ $$$$\sqrt{\left({R}+\mathrm{1}\right)^{\mathrm{2}} +{R}^{\mathrm{2}} }={R}^{\mathrm{2}} −{R}−\mathrm{1} \\ $$$${R}^{\mathrm{2}} −\mathrm{2}{R}−\mathrm{3}=\mathrm{0} \\ $$$$\left({R}−\mathrm{3}\right)\left({R}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{R}=\mathrm{3} \\ $$

Answered by Frix last updated on 01/Nov/23

There′s exactly one rectangular triangle with sides a, b=a+1 and incircle of radius equal to 1: a=R=3, b=R+1=4, c=5, r_I =1

$$\mathrm{There}'\mathrm{s}\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{rectangular}\:\mathrm{triangle} \\ $$$$\mathrm{with}\:\mathrm{sides}\:{a},\:{b}={a}+\mathrm{1}\:\mathrm{and}\:\mathrm{incircle}\:\mathrm{of}\:\mathrm{radius} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{1}: \\ $$$${a}={R}=\mathrm{3},\:{b}={R}+\mathrm{1}=\mathrm{4},\:{c}=\mathrm{5},\:{r}_{\mathrm{I}} =\mathrm{1} \\ $$

Commented by ajfour last updated on 01/Nov/23

$${yes},\:{thanks}.\:{simple}\:{one}\:{for}\:{you}. \\ $$

Answered by cortano12 last updated on 02/Nov/23

(2R−1)^2 = (R+1)^2 +R^2 4R^2 −4R+1=2R^2 +2R+1 2R^2 −6R=0 2R(R−3)=0 R=3

$$\left(\mathrm{2R}−\mathrm{1}\right)^{\mathrm{2}} =\:\left(\mathrm{R}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{R}^{\mathrm{2}} \\ $$$$\:\mathrm{4R}^{\mathrm{2}} −\mathrm{4R}+\mathrm{1}=\mathrm{2R}^{\mathrm{2}} +\mathrm{2R}+\mathrm{1} \\ $$$$\:\:\mathrm{2R}^{\mathrm{2}} −\mathrm{6R}=\mathrm{0} \\ $$$$\:\mathrm{2R}\left(\mathrm{R}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\:\:\mathrm{R}=\mathrm{3} \\ $$

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