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Question Number 199940 by cortano12 last updated on 11/Nov/23

Commented by Frix last updated on 11/Nov/23

∣AD∣=3 ∣BD∣=(9/2) r=((√(15))/4)

$$\mid{AD}\mid=\mathrm{3} \\ $$$$\mid{BD}\mid=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${r}=\frac{\sqrt{\mathrm{15}}}{\mathrm{4}} \\ $$

Commented by cortano12 last updated on 11/Nov/23

Find AD

$$\:\:\:\mathrm{Find}\:\mathrm{AD}\: \\ $$

Commented by cortano12 last updated on 11/Nov/23

how get AD=3 ?

$$\mathrm{how}\:\mathrm{get}\:\mathrm{AD}=\mathrm{3}\:? \\ $$

Answered by Frix last updated on 11/Nov/23

∣AD∣^2 =∣BD∣^2 +p∣BD∣+q We know 1. ∣BD∣=0 ⇔ ∣AD∣=6 2. ∣BD∣=8 ⇔ ∣AD∣=4 ⇒ ∣AD∣=(√(∣BD∣^2 −((21)/2)∣BD∣+36)) We have now the sides of both triangles Δ_1 : 6, ∣BD∣, ∣AD∣ Δ_2 : 4, 8−∣BD∣, ∣AD∣ Now use the formula for the incircle r_I (a, b, c)=((√(2(a^2 b^2 +b^2 c^2 +c^2 a^2 )−(a^4 +b^4 +c^4 )))/(2(a+b+c))) to find ∣BD∣

$$\mid{AD}\mid^{\mathrm{2}} =\mid{BD}\mid^{\mathrm{2}} +{p}\mid{BD}\mid+{q} \\ $$$$\mathrm{We}\:\mathrm{know} \\ $$$$\mathrm{1}.\:\mid{BD}\mid=\mathrm{0}\:\Leftrightarrow\:\mid{AD}\mid=\mathrm{6} \\ $$$$\mathrm{2}.\:\mid{BD}\mid=\mathrm{8}\:\Leftrightarrow\:\mid{AD}\mid=\mathrm{4} \\ $$$$\Rightarrow \\ $$$$\mid{AD}\mid=\sqrt{\mid{BD}\mid^{\mathrm{2}} −\frac{\mathrm{21}}{\mathrm{2}}\mid{BD}\mid+\mathrm{36}} \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{now}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{both}\:\mathrm{triangles} \\ $$$$\Delta_{\mathrm{1}} :\:\mathrm{6},\:\mid{BD}\mid,\:\mid{AD}\mid \\ $$$$\Delta_{\mathrm{2}} :\:\mathrm{4},\:\mathrm{8}−\mid{BD}\mid,\:\mid{AD}\mid \\ $$$$\mathrm{Now}\:\mathrm{use}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{the}\:\mathrm{incircle} \\ $$$${r}_{{I}} \left({a},\:{b},\:{c}\right)=\frac{\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)−\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)}}{\mathrm{2}\left({a}+{b}+{c}\right)} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mid{BD}\mid \\ $$

Answered by mr W last updated on 11/Nov/23

AD=x BD=y, DC=8−y ((6+x+y)/y)=((4+x+8−y)/(8−y)) ⇒y=((4(x+6))/(5+x)) cos B=((6^2 +y^2 −x^2 )/(2×6y))=((6^2 +8^2 −4^2 )/(2×6×8)) ((6^2 +y^2 −x^2 )/y)=((21)/2) x^4 +10x^3 +15x^2 −90x−216=0 (x−3)(x+3)(x+4)(x+6)=0 ⇒x=3=AD

$${AD}={x} \\ $$$${BD}={y},\:{DC}=\mathrm{8}−{y} \\ $$$$\frac{\mathrm{6}+{x}+{y}}{{y}}=\frac{\mathrm{4}+{x}+\mathrm{8}−{y}}{\mathrm{8}−{y}} \\ $$$$\Rightarrow{y}=\frac{\mathrm{4}\left({x}+\mathrm{6}\right)}{\mathrm{5}+{x}} \\ $$$$\mathrm{cos}\:{B}=\frac{\mathrm{6}^{\mathrm{2}} +{y}^{\mathrm{2}} −{x}^{\mathrm{2}} }{\mathrm{2}×\mathrm{6}{y}}=\frac{\mathrm{6}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} }{\mathrm{2}×\mathrm{6}×\mathrm{8}} \\ $$$$\frac{\mathrm{6}^{\mathrm{2}} +{y}^{\mathrm{2}} −{x}^{\mathrm{2}} }{{y}}=\frac{\mathrm{21}}{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +\mathrm{10}{x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} −\mathrm{90}{x}−\mathrm{216}=\mathrm{0} \\ $$$$\left({x}−\mathrm{3}\right)\left({x}+\mathrm{3}\right)\left({x}+\mathrm{4}\right)\left({x}+\mathrm{6}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{3}={AD} \\ $$

Commented by cortano12 last updated on 11/Nov/23

very simple

$$\mathrm{very}\:\mathrm{simple} \\ $$

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