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 130339 by mnjuly1970 last updated on 24/Jan/21

Answered by MJS_new last updated on 25/Jan/21

A= ((0),(a) ) B= ((0),(0) ) C= ((c),(0) ) AC: y=−(a/c)x+a 0≤d∈R BD: y=dx D= ((((ac)/(a+cd))),(((acd)/(a+cd))) ) radius of incircle ABD ((ac)/(a+cd+(√(a^2 +c^2 ))+c(√(d^2 +1))))=r_(ABD) radius of incircle BCD ((acd)/(a+cd+d(√(a^2 +c^2 ))+a(√(d^2 +1))))=r_(BCD) r_(ABD) =r_(BCD) leads to (a−cd)(√(d^2 +1))=(a+cd)(d−1) ⇒ d=(((a(√2)+(√(ac))−c(√2))(√a))/( (2a−c)(√c))) this means, for any rectangular triangle we can find a d in order to get 2 part triangles with equal incircles (lim_(c→2a) d =(3/4)) now we know ∣BD∣=13 ((ac(√(d^2 +1)))/(a+cd))=13 the question is if the area is unique under these conditions indeed because of a>0∧c>0 inserting for d we get ac=338 ⇒ area is ((ac)/2)=169

$${A}=\begin{pmatrix}{\mathrm{0}}\\{{a}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{{c}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${AC}:\:{y}=−\frac{{a}}{{c}}{x}+{a} \\ $$$$\mathrm{0}\leqslant{d}\in\mathbb{R} \\ $$$${BD}:\:{y}={dx} \\ $$$${D}=\begin{pmatrix}{\frac{{ac}}{{a}+{cd}}}\\{\frac{{acd}}{{a}+{cd}}}\end{pmatrix} \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{incircle}\:{ABD} \\ $$$$\frac{{ac}}{{a}+{cd}+\sqrt{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }+{c}\sqrt{{d}^{\mathrm{2}} +\mathrm{1}}}={r}_{{ABD}} \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{incircle}\:{BCD} \\ $$$$\frac{{acd}}{{a}+{cd}+{d}\sqrt{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }+{a}\sqrt{{d}^{\mathrm{2}} +\mathrm{1}}}={r}_{{BCD}} \\ $$$${r}_{{ABD}} ={r}_{{BCD}} \:\mathrm{leads}\:\mathrm{to} \\ $$$$\left({a}−{cd}\right)\sqrt{{d}^{\mathrm{2}} +\mathrm{1}}=\left({a}+{cd}\right)\left({d}−\mathrm{1}\right) \\ $$$$\Rightarrow\:{d}=\frac{\left({a}\sqrt{\mathrm{2}}+\sqrt{{ac}}−{c}\sqrt{\mathrm{2}}\right)\sqrt{{a}}}{\:\left(\mathrm{2}{a}−{c}\right)\sqrt{{c}}} \\ $$$$\mathrm{this}\:\mathrm{means},\:\mathrm{for}\:\mathrm{any}\:\mathrm{rectangular}\:\mathrm{triangle}\:\mathrm{we} \\ $$$$\mathrm{can}\:\mathrm{find}\:\mathrm{a}\:{d}\:\mathrm{in}\:\mathrm{order}\:\mathrm{to}\:\mathrm{get}\:\mathrm{2}\:\mathrm{part}\:\mathrm{triangles} \\ $$$$\mathrm{with}\:\mathrm{equal}\:\mathrm{incircles}\:\left(\underset{{c}\rightarrow\mathrm{2}{a}} {\mathrm{lim}}\:{d}\:=\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{know}\:\mid{BD}\mid=\mathrm{13} \\ $$$$\frac{{ac}\sqrt{{d}^{\mathrm{2}} +\mathrm{1}}}{{a}+{cd}}=\mathrm{13} \\ $$$$\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{if}\:\mathrm{the}\:\mathrm{area}\:\mathrm{is}\:\mathrm{unique}\:\mathrm{under} \\ $$$$\mathrm{these}\:\mathrm{conditions} \\ $$$$\mathrm{indeed}\:\mathrm{because}\:\mathrm{of}\:{a}>\mathrm{0}\wedge{c}>\mathrm{0}\:\mathrm{inserting}\:\mathrm{for}\:{d} \\ $$$$\mathrm{we}\:\mathrm{get} \\ $$$${ac}=\mathrm{338} \\ $$$$\Rightarrow\:\mathrm{area}\:\mathrm{is}\:\frac{{ac}}{\mathrm{2}}=\mathrm{169} \\ $$

Commented by mnjuly1970 last updated on 25/Jan/21

grateful mr mjs.thanks alot...

$${grateful}\:\:{mr}\:{mjs}.{thanks}\:{alot}... \\ $$

Answered by Olaf last updated on 24/Jan/21

By symmetry AB = BC D middle of [AC] BD = AD = DC AC = 26 Area[ABC] = (1/2)AC×BD = 169

$$\mathrm{By}\:\mathrm{symmetry}\:\mathrm{AB}\:=\:\mathrm{BC} \\ $$$$\mathrm{D}\:\mathrm{middle}\:\mathrm{of}\:\left[\mathrm{AC}\right] \\ $$$$\mathrm{BD}\:=\:\mathrm{AD}\:=\:\mathrm{DC} \\ $$$$\mathrm{AC}\:=\:\mathrm{26} \\ $$$$\mathrm{Area}\left[\mathrm{ABC}\right]\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AC}×\mathrm{BD}\:=\:\mathrm{169} \\ $$

Commented by MJS_new last updated on 25/Jan/21

it′s not necessary to have ∣AB∣=∣BC∣. it′s always true for ∣AB∣×∣BC∣=338

$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{necessary}\:\mathrm{to}\:\mathrm{have}\:\mid{AB}\mid=\mid{BC}\mid.\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{always}\:\mathrm{true}\:\mathrm{for}\:\mid{AB}\mid×\mid{BC}\mid=\mathrm{338} \\ $$

Commented by mnjuly1970 last updated on 25/Jan/21

perfect.thank you mr olaf...

$${perfect}.{thank}\:{you}\:{mr}\:{olaf}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com