Question Number 16754 by ajfour last updated on 26/Jun/17
Commented by ajfour last updated on 26/Jun/17
$$\mathrm{Q}.\mathrm{16748}\:\left(\mathrm{solution}\right) \\ $$$$\:\mathrm{by}\:\mathrm{fault}\:\mathrm{it}\:\mathrm{gets}\:\mathrm{uploaded}\:\mathrm{as}\: \\ $$$$\mathrm{question}. \\ $$
Answered by ajfour last updated on 26/Jun/17
![equation of RC: y=−(h/k)(x−a) equation if AP: x=h H is their intersection, so x_H =h, y_H =−(h/k)(h−a) therefore H≡[h, (h/k)(a−h)] equation of OL: x=a/2 equation of ON: y−(k/2)=−(h/k)(x−(h/2)) circumcenter (x_0 , y_0 )lies on both; so x_0 =a/2 y_0 =(k/2)−(h/k)((a/2)−(h/2)) hence O≡[(a/2), (k/2)−(h/(2k))(a−h)] OH^(→) =r_H ^→ −r_O ^→ =[hi^ +(h/k)(a−h)j^ ]−[(a/2)i^ +((k/2)−(h/(2k))(a−h)] =(h−(a/2))i^ +[((3h)/(2k))(a−h)−(k/2)]j^ ..(i) OA^(→) +OB^(→) +OC^(→) = r_A ^→ +r_B ^→ +r_C ^→ −3r_O ^→ = (hi^ +kj^ )+(0i^ +0j^ )+(ai^ +0j) −3[(a/2)i^ +((k/2)−(h/(2k))(a−h)] =(h−(a/2))i^ +[((3h)/(2k))(a−h)−(k/2)]j^ OA^(→) +OB^(→) +OC^(→) = OH^(→) [see (i)].](Q16759.png)
$$\:\mathrm{equation}\:\mathrm{of}\:\mathrm{RC}: \\ $$$$\:\mathrm{y}=−\frac{\mathrm{h}}{\mathrm{k}}\left(\mathrm{x}−\mathrm{a}\right)\:\:\: \\ $$$$\:\mathrm{equation}\:\mathrm{if}\:\mathrm{AP}:\:\:\mathrm{x}=\mathrm{h} \\ $$$$\mathrm{H}\:\mathrm{is}\:\mathrm{their}\:\mathrm{intersection},\:\mathrm{so} \\ $$$$\:\mathrm{x}_{\mathrm{H}} =\mathrm{h},\:\:\:\:\mathrm{y}_{\mathrm{H}} =−\frac{\mathrm{h}}{\mathrm{k}}\left(\mathrm{h}−\mathrm{a}\right) \\ $$$$\:\:\mathrm{therefore}\:\:\:\mathrm{H}\equiv\left[\mathrm{h},\:\frac{\mathrm{h}}{\mathrm{k}}\left(\mathrm{a}−\mathrm{h}\right)\right] \\ $$$$\:\mathrm{equation}\:\mathrm{of}\:\mathrm{OL}:\:\:\:\mathrm{x}=\mathrm{a}/\mathrm{2} \\ $$$$\:\mathrm{equation}\:\mathrm{of}\:\mathrm{ON}: \\ $$$$\:\:\:\:\mathrm{y}−\frac{\mathrm{k}}{\mathrm{2}}=−\frac{\mathrm{h}}{\mathrm{k}}\left(\mathrm{x}−\frac{\mathrm{h}}{\mathrm{2}}\right) \\ $$$$\:\mathrm{circumcenter}\:\left(\mathrm{x}_{\mathrm{0}} ,\:\mathrm{y}_{\mathrm{0}} \right)\mathrm{lies}\:\mathrm{on}\:\mathrm{both}; \\ $$$$\mathrm{so}\:\:\mathrm{x}_{\mathrm{0}} =\mathrm{a}/\mathrm{2} \\ $$$$\:\:\:\:\:\:\mathrm{y}_{\mathrm{0}} =\frac{\mathrm{k}}{\mathrm{2}}−\frac{\mathrm{h}}{\mathrm{k}}\left(\frac{\mathrm{a}}{\mathrm{2}}−\frac{\mathrm{h}}{\mathrm{2}}\right) \\ $$$$\mathrm{hence}\:\:\:\mathrm{O}\equiv\left[\frac{\mathrm{a}}{\mathrm{2}},\:\frac{\mathrm{k}}{\mathrm{2}}−\frac{\mathrm{h}}{\mathrm{2k}}\left(\mathrm{a}−\mathrm{h}\right)\right] \\ $$$$\:\overset{\rightarrow} {\mathrm{OH}}\:=\overset{\rightarrow} {\mathrm{r}}_{\mathrm{H}} −\overset{\rightarrow} {\mathrm{r}}_{\mathrm{O}} \\ $$$$\:\:\:\:=\left[\mathrm{h}\hat {\mathrm{i}}+\frac{\mathrm{h}}{\mathrm{k}}\left(\mathrm{a}−\mathrm{h}\right)\hat {\mathrm{j}}\right]−\left[\frac{\mathrm{a}}{\mathrm{2}}\hat {\mathrm{i}}+\left(\frac{\mathrm{k}}{\mathrm{2}}−\frac{\mathrm{h}}{\mathrm{2k}}\left(\mathrm{a}−\mathrm{h}\right)\right]\right. \\ $$$$\:\:\:=\left(\mathrm{h}−\frac{\mathrm{a}}{\mathrm{2}}\right)\hat {\mathrm{i}}+\left[\frac{\mathrm{3h}}{\mathrm{2k}}\left(\mathrm{a}−\mathrm{h}\right)−\frac{\mathrm{k}}{\mathrm{2}}\right]\hat {\mathrm{j}}\:\:..\left(\mathrm{i}\right) \\ $$$$\overset{\rightarrow} {\mathrm{OA}}+\overset{\rightarrow} {\mathrm{OB}}+\overset{\rightarrow} {\mathrm{OC}}\:=\:\overset{\rightarrow} {\mathrm{r}}_{\mathrm{A}} +\overset{\rightarrow} {\mathrm{r}}_{\mathrm{B}} +\overset{\rightarrow} {\mathrm{r}}_{\mathrm{C}} −\mathrm{3}\overset{\rightarrow} {\mathrm{r}}_{\mathrm{O}} \\ $$$$\:=\:\left(\mathrm{h}\hat {\mathrm{i}}+\mathrm{k}\hat {\mathrm{j}}\right)+\left(\mathrm{0}\hat {\mathrm{i}}+\mathrm{0}\hat {\mathrm{j}}\right)+\left(\mathrm{a}\hat {\mathrm{i}}+\mathrm{0j}\right) \\ $$$$\:\:\:\:−\mathrm{3}\left[\frac{\mathrm{a}}{\mathrm{2}}\hat {\mathrm{i}}+\left(\frac{\mathrm{k}}{\mathrm{2}}−\frac{\mathrm{h}}{\mathrm{2k}}\left(\mathrm{a}−\mathrm{h}\right)\right]\right. \\ $$$$\:=\left(\mathrm{h}−\frac{\mathrm{a}}{\mathrm{2}}\right)\hat {\mathrm{i}}+\left[\frac{\mathrm{3h}}{\mathrm{2k}}\left(\mathrm{a}−\mathrm{h}\right)−\frac{\mathrm{k}}{\mathrm{2}}\right]\hat {\mathrm{j}}\: \\ $$$$\:\:\overset{\rightarrow} {\mathrm{OA}}+\overset{\rightarrow} {\mathrm{OB}}+\overset{\rightarrow} {\mathrm{OC}}\:=\:\overset{\rightarrow} {\mathrm{OH}}\:\:\:\left[\mathrm{see}\:\left(\mathrm{i}\right)\right]. \\ $$$$\:\:\: \\ $$
Commented by ajfour last updated on 26/Jun/17
$$\mathrm{please}\:\mathrm{do}\:\mathrm{so}. \\ $$
Answered by Tinkutara last updated on 26/Jun/17
Commented by Tinkutara last updated on 26/Jun/17
$$\mathrm{Since}\:{A}'{BHC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{parallelogram},\:\mathrm{so} \\ $$$$\overset{\rightarrow} {{HC}}\:+\:\overset{\rightarrow} {{HB}}\:=\:\overset{\rightarrow} {{HA}'} \\ $$$$\mathrm{and}\:\overset{\rightarrow} {{HA}'}\:+\:\overset{\rightarrow} {{HA}}\:=\:\mathrm{2}\overset{\rightarrow} {{HO}} \\ $$$$\Rightarrow\:\overset{\rightarrow} {{HA}}\:+\:\overset{\rightarrow} {{HB}}\:+\:\overset{\rightarrow} {{HC}}\:=\:\mathrm{2}\overset{\rightarrow} {{HO}} \\ $$$$\mathrm{But}\:\mathrm{then}\:\overset{\rightarrow} {{OA}}\:+\:\overset{\rightarrow} {{OB}}\:+\:\overset{\rightarrow} {{OC}}\:= \\ $$$$\overset{\rightarrow} {{OH}}\:+\:\overset{\rightarrow} {{HA}}\:+\:\overset{\rightarrow} {{OH}}\:+\:\overset{\rightarrow} {{HB}}\:+\:\overset{\rightarrow} {{OH}}\:+\:\overset{\rightarrow} {{HC}} \\ $$$$=\:\mathrm{3}\overset{\rightarrow} {{OH}}\:+\:\mathrm{2}\overset{\rightarrow} {{HO}}\:=\:\overset{\rightarrow} {{OH}}, \\ $$$$\mathrm{as}\:\mathrm{desired}. \\ $$