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Question-64469

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Question Number 64469 by Tawa1 last updated on 18/Jul/19
Answered by MJS last updated on 18/Jul/19
the radius of the incircle of a rectangular triangle is given by r=((a+b−(√(a^2 +b^2 )))/2) ((a+b−(√(a^2 +b^2 )))/2)=703 a+b−(√(a^2 +b^2 ))=1406 (√(a^2 +b^2 ))=a+b−1406 a^2 +b^2 =(a+b−1406)^2 2812a−2ab+2812b−1976836=0 ⇒ b=1406+((988418)/(a−1406)) put a=1406+t with t∈N^★ b=((988418)/t)+1406 988418=2×19^2 ×37^2 b∈N ⇒ t∣988418 we must try t∈{1, 2, 19, 37, 38, 74, 361, 703, 722, 1369, 1406, 2738, 13357, 26011, 26714, 52022, 494209, 988418} ⇒ with a<b<c [we can change a⇄b] a=1407 b=989724 c=989825 a=1408 b=495615 c=495617 a=1425 b=53428 c=53447 a=1443 b=28120 c=28156 a=1444 b=27417 c=27455 a=1480 b=14763 c=14837 a=1767 b=4144 c=4505 a=2109 b=2812 c=3515 a=2128 b=2775 c=3497
$$\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{incircle}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rectangular} \\ $$$$\mathrm{triangle}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${r}=\frac{{a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\frac{{a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mathrm{2}}=\mathrm{703} \\ $$$${a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{1406} \\ $$$$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }={a}+{b}−\mathrm{1406} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left({a}+{b}−\mathrm{1406}\right)^{\mathrm{2}} \\ $$$$\mathrm{2812}{a}−\mathrm{2}{ab}+\mathrm{2812}{b}−\mathrm{1976836}=\mathrm{0} \\ $$$$\Rightarrow\:{b}=\mathrm{1406}+\frac{\mathrm{988418}}{{a}−\mathrm{1406}} \\ $$$$\mathrm{put}\:{a}=\mathrm{1406}+{t}\:\mathrm{with}\:{t}\in\mathbb{N}^{\bigstar} \\ $$$${b}=\frac{\mathrm{988418}}{{t}}+\mathrm{1406} \\ $$$$\mathrm{988418}=\mathrm{2}×\mathrm{19}^{\mathrm{2}} ×\mathrm{37}^{\mathrm{2}} \\ $$$${b}\in\mathbb{N}\:\Rightarrow\:{t}\mid\mathrm{988418} \\ $$$$\mathrm{we}\:\mathrm{must}\:\mathrm{try} \\ $$$${t}\in\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{19},\:\mathrm{37},\:\mathrm{38},\:\mathrm{74},\:\mathrm{361},\:\mathrm{703},\:\mathrm{722},\:\mathrm{1369},\:\mathrm{1406},\:\mathrm{2738},\:\mathrm{13357},\:\mathrm{26011},\:\mathrm{26714},\:\mathrm{52022},\:\mathrm{494209},\:\:\mathrm{988418}\right\} \\ $$$$\Rightarrow \\ $$$$\mathrm{with}\:{a}<{b}<{c}\:\left[\mathrm{we}\:\mathrm{can}\:\mathrm{change}\:{a}\rightleftarrows{b}\right] \\ $$$${a}=\mathrm{1407}\:\:{b}=\mathrm{989724}\:\:{c}=\mathrm{989825} \\ $$$${a}=\mathrm{1408}\:\:{b}=\mathrm{495615}\:\:{c}=\mathrm{495617} \\ $$$${a}=\mathrm{1425}\:\:{b}=\mathrm{53428}\:\:{c}=\mathrm{53447} \\ $$$${a}=\mathrm{1443}\:\:{b}=\mathrm{28120}\:\:{c}=\mathrm{28156} \\ $$$${a}=\mathrm{1444}\:\:{b}=\mathrm{27417}\:\:{c}=\mathrm{27455} \\ $$$${a}=\mathrm{1480}\:\:{b}=\mathrm{14763}\:\:{c}=\mathrm{14837} \\ $$$${a}=\mathrm{1767}\:\:{b}=\mathrm{4144}\:\:{c}=\mathrm{4505} \\ $$$${a}=\mathrm{2109}\:\:{b}=\mathrm{2812}\:\:{c}=\mathrm{3515} \\ $$$${a}=\mathrm{2128}\:\:{b}=\mathrm{2775}\:\:{c}=\mathrm{3497} \\ $$
Commented by Prithwish sen last updated on 18/Jul/19
great!
$$\mathrm{great}! \\ $$
Commented by MJS last updated on 18/Jul/19
I will post Heron′s formula and its conclusions as soon as I can find the time
$$\mathrm{I}\:\mathrm{will}\:\mathrm{post}\:\mathrm{Heron}'\mathrm{s}\:\mathrm{formula}\:\mathrm{and}\:\mathrm{its}\:\mathrm{conclusions} \\ $$$$\mathrm{as}\:\mathrm{soon}\:\mathrm{as}\:\mathrm{I}\:\mathrm{can}\:\mathrm{find}\:\mathrm{the}\:\mathrm{time} \\ $$
Commented by Prithwish sen last updated on 18/Jul/19
Sir how could one be good at geometry . What is the path one should follow ?
$$\mathrm{Sir}\:\mathrm{how}\:\mathrm{could}\:\mathrm{one}\:\mathrm{be}\:\mathrm{good}\:\mathrm{at}\:\mathrm{geometry}\:.\:\mathrm{What} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{path}\:\mathrm{one}\:\mathrm{should}\:\mathrm{follow}\:? \\ $$
Commented by Prithwish sen last updated on 18/Jul/19
MJS Sir please don′t ignore. Suggest something.
$$\mathrm{MJS}\:\mathrm{Sir}\:\mathrm{please}\:\mathrm{don}'\mathrm{t}\:\mathrm{ignore}.\:\mathrm{Suggest}\:\mathrm{something}. \\ $$
Commented by Tawa1 last updated on 18/Jul/19
God bless you sirs, i appreciate your effort
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sirs},\:\mathrm{i}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort} \\ $$
Commented by mr W last updated on 18/Jul/19
Area of rectangular triangle A A=((ab)/2) A=((r(a+b+c))/2)=((r(a+b+(√(a^2 +b^2 ))))/2) ((r(a+b+(√(a^2 +b^2 ))))/2)=((ab)/2) ⇒r=((ab)/(a+b+(√(a^2 +b^2 )))) ⇒r=((ab(a+b−(√(a^2 +b^2 ))))/((a+b+(√(a^2 +b^2 )))(a+b−(√(a^2 +b^2 ))))) ⇒r=((ab(a+b−(√(a^2 +b^2 ))))/(2ab)) ⇒r=((a+b−(√(a^2 +b^2 )))/2)
$${Area}\:{of}\:{rectangular}\:{triangle}\:{A} \\ $$$${A}=\frac{{ab}}{\mathrm{2}} \\ $$$${A}=\frac{{r}\left({a}+{b}+{c}\right)}{\mathrm{2}}=\frac{{r}\left({a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)}{\mathrm{2}} \\ $$$$\frac{{r}\left({a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)}{\mathrm{2}}=\frac{{ab}}{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{{ab}}{{a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\Rightarrow{r}=\frac{{ab}\left({a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)}{\left({a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)\left({a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)} \\ $$$$\Rightarrow{r}=\frac{{ab}\left({a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)}{\mathrm{2}{ab}} \\ $$$$\Rightarrow{r}=\frac{{a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$
Commented by MJS last updated on 18/Jul/19
sorry no time at the moment
$$\mathrm{sorry}\:\mathrm{no}\:\mathrm{time}\:\mathrm{at}\:\mathrm{the}\:\mathrm{moment} \\ $$
Commented by Prithwish sen last updated on 18/Jul/19
ok thank you sir.
$$\mathrm{ok}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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