Question Number 189973 by mr W last updated on 25/Mar/23
Commented by mr W last updated on 25/Mar/23
$${find}\:{the}\:{area}\:{and}\:{side}\:{lengths}\:{of}\:{the} \\ $$$${shaded}\:{triangle}. \\ $$
Commented by MJS_new last updated on 25/Mar/23
$$\frac{\mathrm{1}}{\mathrm{7}}\:\mathrm{of}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\Delta{ABC} \\ $$$$\mathrm{I}'\mathrm{ll}\:\mathrm{show}\:\mathrm{my}\:\mathrm{work}\:\mathrm{later} \\ $$
Commented by MJS_new last updated on 25/Mar/23
![btw to a given triangle with sides a, b, c there are 2 possible “inner” triangles with sides o_k , p_k , q_k : o_1 =((√(6a^2 −2b^2 +3c^2 ))/7); o_2 =((√(6a^2 +3b^2 −2c^2 ))/7) [p_k , q_k by cycling a, b, c] it′s easy to solve these for a, b, c with given o, p, q: a_1 =(√(6o^2 +3p^2 −2q^2 )); a_2 =(√(6o^2 −2p^2 +3q^2 )) [again b_k , c_k by cycling o, p, q] all these triangles share the same mass center](https://www.tinkutara.com/question/Q189995.png)
$$\mathrm{btw}\:\mathrm{to}\:\mathrm{a}\:\mathrm{given}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{sides}\:{a},\:{b},\:{c} \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{2}\:\mathrm{possible}\:“\mathrm{inner}''\:\mathrm{triangles}\:\mathrm{with} \\ $$$$\mathrm{sides}\:{o}_{{k}} ,\:{p}_{{k}} ,\:{q}_{{k}} : \\ $$$${o}_{\mathrm{1}} =\frac{\sqrt{\mathrm{6}{a}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{2}} }}{\mathrm{7}};\:{o}_{\mathrm{2}} =\frac{\sqrt{\mathrm{6}{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} }}{\mathrm{7}} \\ $$$$\left[{p}_{{k}} ,\:{q}_{{k}} \:\mathrm{by}\:\mathrm{cycling}\:{a},\:{b},\:{c}\right] \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{these}\:\mathrm{for}\:{a},\:{b},\:{c}\:\mathrm{with}\:\mathrm{given} \\ $$$${o},\:{p},\:{q}: \\ $$$${a}_{\mathrm{1}} =\sqrt{\mathrm{6}{o}^{\mathrm{2}} +\mathrm{3}{p}^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} };\:{a}_{\mathrm{2}} =\sqrt{\mathrm{6}{o}^{\mathrm{2}} −\mathrm{2}{p}^{\mathrm{2}} +\mathrm{3}{q}^{\mathrm{2}} } \\ $$$$\left[\mathrm{again}\:{b}_{{k}} ,\:{c}_{{k}} \:\mathrm{by}\:\mathrm{cycling}\:{o},\:{p},\:{q}\right] \\ $$$$\mathrm{all}\:\mathrm{these}\:\mathrm{triangles}\:\mathrm{share}\:\mathrm{the}\:\mathrm{same}\:\mathrm{mass}\:\mathrm{center} \\ $$
Commented by mr W last updated on 25/Mar/23
$${thanks}\:{sir}! \\ $$
Answered by ajfour last updated on 25/Mar/23
$${B}\:{origin}. \\ $$$${vectors}\:\:{a},\:{c},\:{p} \\ $$$${r}_{{C}} ={a} \\ $$$${r}_{{A}} ={c} \\ $$$${A}\:'={P}\:\:{and}\:{so}\:{on}.. \\ $$$${r}_{{Q}} =\mathrm{2}{p} \\ $$$${r}_{{R}} ={a}+\mathrm{2}\left(\mathrm{2}{p}−{c}\right)={c}+\frac{{p}−{c}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{2}{a}+\mathrm{7}{p}=\mathrm{5}{c} \\ $$$$\mathrm{2}{Area}_{{sh}} =\mid{p}×\left\{{a}+\mathrm{2}\left(\mathrm{2}{p}−{c}\right)\right\}\mid \\ $$$$\Rightarrow\:\mathrm{98}\bar {{A}}_{{sb}} \\ $$$$=\left(\mathrm{5}{c}−\mathrm{2}{a}\right)×\left\{\mathrm{7}{a}+\mathrm{4}\left(\mathrm{5}{c}−\mathrm{2}{a}\right)−\mathrm{14}{c}\right\} \\ $$$$=\left(\mathrm{5}{c}−\mathrm{2}{a}\right)\left(\mathrm{6}{c}−{a}\right) \\ $$$$=\mathrm{7}\mid{a}×{c}\mid \\ $$$$\Rightarrow\:\:{A}_{{sh}} =\frac{\mid{a}×{c}\mid}{\mathrm{14}}=\frac{\bigtriangleup_{{ABC}} }{\mathrm{7}} \\ $$$$\bigstar\:\:\:\:\mathrm{2}\bigtriangleup_{{ABC}} =\mid{a}×{c}\mid \\ $$$$ \\ $$
Commented by mr W last updated on 25/Mar/23
$${thanks}\:{sir}! \\ $$
Answered by mr W last updated on 25/Mar/23
Commented by mr W last updated on 25/Mar/23
$${we}\:{see}\:{all}\:{seven}\:{small}\:{triangles}\:{have} \\ $$$${the}\:{same}\:{area},\:{therefore}\: \\ $$$$\Delta_{{A}'{B}'{C}'} =\frac{\Delta_{{ABC}} }{\mathrm{7}}. \\ $$$${say}\:{AC}'={C}'{A}'={p},\:{BA}'={A}'{B}'={q},\:{CB}'={B}'{C}'={r} \\ $$$$\mathrm{cos}\:\beta=−\frac{\mathrm{4}{q}^{\mathrm{2}} +{r}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{qr}}=\frac{{q}^{\mathrm{2}} +{r}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{2}{qr}} \\ $$$$−\mathrm{2}{p}^{\mathrm{2}} +\mathrm{6}{q}^{\mathrm{2}} +\mathrm{3}{r}^{\mathrm{2}} ={a}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$${similarly} \\ $$$$\mathrm{3}{p}^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} +\mathrm{6}{r}^{\mathrm{2}} ={b}^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\mathrm{6}{p}^{\mathrm{2}} +\mathrm{3}{q}^{\mathrm{2}} −\mathrm{2}{r}^{\mathrm{2}} ={c}^{\mathrm{2}} \:\:\:…\left({iii}\right) \\ $$$$\Rightarrow{p}^{\mathrm{2}} =\frac{−\mathrm{2}{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} +\mathrm{6}{c}^{\mathrm{2}} }{\mathrm{49}}\:\Rightarrow{p}=\frac{\sqrt{−\mathrm{2}{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} +\mathrm{6}{c}^{\mathrm{2}} }}{\mathrm{7}} \\ $$$$\Rightarrow{q}^{\mathrm{2}} =\frac{\mathrm{6}{a}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{2}} }{\mathrm{49}}\:\Rightarrow{q}=\frac{\sqrt{\mathrm{6}{a}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{2}} }}{\mathrm{7}} \\ $$$$\Rightarrow{r}^{\mathrm{2}} =\frac{\mathrm{3}{a}^{\mathrm{2}} +\mathrm{6}{b}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} }{\mathrm{49}}\:\Rightarrow{r}=\frac{\sqrt{\mathrm{3}{a}^{\mathrm{2}} +\mathrm{6}{b}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} }}{\mathrm{7}} \\ $$
Answered by manxsol last updated on 26/Mar/23
$$\mathrm{2}{A}={acsinx}={bcsiny}={basinz} \\ $$$$\mathrm{2}{A}_{\mathrm{1}} =\mathrm{2}{cbsiny}=\mathrm{2}\left(\mathrm{2}{A}\right)=\mathrm{4}{A} \\ $$$$\mathrm{2}{A}_{\mathrm{2}} =\mathrm{2}{acsinx}=\mathrm{2}\left(\mathrm{2}{A}\right)=\mathrm{4}{A} \\ $$$$\mathrm{2}{A}_{\mathrm{3}} =\mathrm{2}{basinz}=\mathrm{2}\left(\mathrm{2}{A}\right)=\mathrm{4}{A} \\ $$$$\frac{\Delta{blue}}{\Delta{total}}=\frac{{A}}{\mathrm{7}{A}}=\frac{\mathrm{1}}{\mathrm{7}} \\ $$
Commented by manxsol last updated on 26/Mar/23
