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Question Number 13888 by Tinkutara last updated on 24/May/17

$$\mathrm{Prove}\mathrm{that}\mathrm{if}{A}^{\prime },B^{\prime }\mathrm{and}{C}^{\prime }\mathrm{are}\mathrm{the}$$$$\mathrm{midpoints}\mathrm{of}\mathrm{the}\mathrm{sides}BC,CA\mathrm{and}AB,$$$$\mathrm{respectively},\mathrm{then}$$$${AA}^{\prime }+{BB}^{\prime }+{CC}^{\prime }<AB+BC+CA$$

Commented by Tinkutara last updated on 24/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/May/17

$$AG+BG>AB\Rightarrow \frac{2}{3}({AA}^{{}^{\prime }}+{BB}^{{}^{\prime }})>AB$$$$BG+GC>BC\Rightarrow \frac{2}{3}({BB}^{{}^{\prime }}+{CC}^{{}^{\prime }})>BC$$$$AG+GC>AC\Rightarrow \frac{2}{3}({AA}^{{}^{\prime }}+{CC}^{{}^{\prime }})>AC$$$$\Rightarrow 1)AB+BC+CA<\frac{4}{3}({AA}^{{}^{\prime }}+{BB}^{{}^{\prime }}+{CC}^{{}^{\prime }})$$$$2)\frac{\mid AB-AC\mid }{2}<{AA}^{{}^{\prime }}<\frac{AB+AC}{2}$$$$3)a(m_a^2+\frac{a}{2}.\frac{a}{2})=b^2.\frac{a}{2}+c^2.\frac{a}{2}\left({Estewart}^{\prime }steorem\right)$$$$\Rightarrow m_a^2+\frac{a^2}{4}=\frac{b^2+c^2}{2}\Rightarrow m_a^2=\frac{2(b^2+c^2)-a^2}{4}$$$$=\frac{2(b^2+c^2)-(b^2+c^2-2bc.cosA)}{4}=\frac{1}{4}(b^2+c^2+2bc.cosA)=$$$$=\frac{1}{4}(b^2+c^2+2bc.cosA)<\frac{1}{4}(b^2+c^2+2bc)=$$$$=\frac{1}{4}{(b+c)}^2\Rightarrow m_a<\frac{1}{2}(b+c)$$$$similarly:\Rightarrow m_b<\frac{1}{2}(a+c),m_c<\frac{1}{2}(a+b)$$$$\Rightarrow m_a+m_b+m_c<\frac{1}{2}(2a+2b+2c)=a+b+c$$$$\Rightarrow m_a+m_b+m_c<a+b+c.◼$$$$4)if:m_a=\frac{b+c}{2}\Rightarrow \frac{2(b^2+c^2)-a^2}{4}=\frac{b^2+c^2+2bc}{4}$$$$\Rightarrow b^2+c^2-2bc=a^2\Rightarrow {(b-c)}^2=a^2\Rightarrow b-c=a\left(impossible\right)$$$$or:b^2+c^2-2bc=b^2+c^2-2bc.cosA$$$$\Rightarrow cosA=1\Rightarrow \measuredangle A=0\left(impossible\right)$$$$so:m_a⪇\frac{1}{2}(b+c)and:m_a<\frac{1}{2}(b+c).$$$$5)A={90}^{°}\Rightarrow 4m_a^2=b^2+c^2=a^2\Rightarrow m_a=\frac{a}{2}.$$$$A={120}^{°}\Rightarrow 4m_a^2=b^2+c^2-bc$$$$A=60\Rightarrow 4m_a^2=b^2+c^2+bc$$$$6)A=90,B=60,C=30$$$$\Rightarrow 4(m_a^2+m_b^2+m_c^2)=a^2+a^2+c^2+ac+a^2+b^2+\sqrt{3}ab=$$$$=4a^2+\sqrt{3}a(b+c).$$$$7)A=90,b=c\Rightarrow \mathrm{\Sigma }{m}_a^2=4a^2+2\sqrt{2}ab\stackrel{}{.}$$$$8){BC}^2={BG}^2+{GC}^2-2BG.GC.cosB\stackrel{}{GC}$$$$\measuredangle BGC=\phi \Rightarrow a^2=\frac{4}{9}{m}_b^2+\frac{4}{9}{m}_c^2-2\frac{2}{3}{m}_b.\frac{2}{3}{m}_{c}cos\phi$$$$\Rightarrow \frac{9}{4}{a}^2=m_b^2+m_c^2-2m_b.m_c.cos\phi$$$$\phi =90\Rightarrow \frac{9}{4}{a}^2=\frac{2(a^2+c^2)-b^2}{4}+\frac{2(a^2+b^2)-c^2}{4}\Rightarrow$$$$\Rightarrow (if:m_b\mathrm{\perp }{m}_c)\Rightarrow (5a^2=b^2+c^2),(m_a=\frac{3}{2}a)$$$$9)AB\mathrm{\perp }AC\Rightarrow m_b^2=\frac{2(a^2+c^2)-b^2}{4}=\frac{b^2+2c^2}{4}$$$$m_c^2=\frac{2(a^2+b^2)-c^2}{4}=\frac{2b^2+c^2}{4}$$$$\Rightarrow m_b^2+m_c^2=\frac{3}{4}(b^2+c^2)=\frac{3}{4}{a}^2=3m_a^2$$$$\Rightarrow \{\begin{array}{l}b\mathrm{\perp }c\Rightarrow m_b^2+m_c^2=3m_a^2,m_a=\frac{1}{2}a.\\ m_b\mathrm{\perp }{m}_c\Rightarrow b^2+c^2=5a^2,m_a=\frac{3}{2}a.\end{array}$$$$$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/May/17

$$incaseof:\measuredangle AGB=\measuredangle BGC=\measuredangle AGC$$$$showthat:$$$$1)S_{ABC}=\frac{\sqrt{3}}{12}.(a^2+b^2+c^2)$$$$2)\frac{a}{h_a}+\frac{b}{h_b}+\frac{c}{h_c}=2\sqrt{3}.(h_a=altitudefrom\stackrel{}{A}toBC)$$

Answered by mrW1 last updated on 24/May/17

$${AA}^{\prime }=\frac{1}{2}\sqrt{{AB}^2+{AC}^2+2\times AB\times AC\times \cos \mathrm{\angle }A}$$$$<\frac{1}{2}\sqrt{{AB}^2+{AC}^2+2\times AB\times AC}$$$$=\frac{1}{2}\sqrt{{(AB+AC)}^2}=\frac{1}{2}(AB+AC)$$$$$$$${AA}^{\prime }<\frac{1}{2}(AB+AC)$$$${BB}^{\prime }<\frac{1}{2}(BA+BC)$$$${CC}^{\prime }<\frac{1}{2}(CB+CA)$$$$\Rightarrow {AA}^{\prime }+{BB}^{\prime }+{CC}^{\prime }<AB+BC+CA$$

Commented by mrW1 last updated on 24/May/17

$$Youcanalsoprovelikethis:$$$$since{BA}^{\prime }=A^{\prime }C$$$$\Rightarrow A^{\prime }{A}^{″}=A^{\prime }{A}^{‴}$$$${AA}^{″}<AC$$$${AA}^{‴}<AB$$$${AA}^{\prime }=\frac{1}{2}({AA}^{″}+{AA}^{‴})<\frac{1}{2}(AC+AB)$$

Commented by mrW1 last updated on 24/May/17

$$Usinglawofcosinesweget:$$$${AB}^2={BA}^{\prime 2}+{AA}^{\prime 2}-2\times {BA}^{\prime }\times {AA}^{\prime }\times \cos \mathrm{\angle }{AA}^{\prime }B...\left(i\right)$$$$$$$${AC}^2={CA}^{\prime 2}+{AA}^{\prime 2}-2\times {CA}^{\prime }\times {AA}^{\prime }\times \cos \mathrm{\angle }{AA}^{\prime }C$$$${AC}^2={CA}^{\prime 2}+{AA}^{\prime 2}-2\times {CA}^{\prime }\times {AA}^{\prime }\times \cos (180°-\mathrm{\angle }{AA}^{\prime }B)$$$${AC}^2={CA}^{\prime 2}+{AA}^{\prime 2}+2\times {CA}^{\prime }\times {AA}^{\prime }\times \cos \mathrm{\angle }{AA}^{\prime }B$$$${AC}^2={CA}^{\prime 2}+{AA}^{\prime 2}+2\times {BA}^{\prime }\times {AA}^{\prime }\times \cos \mathrm{\angle }{AA}^{\prime }B...\left(ii\right)$$$$$$$$\left(i\right)+\left(ii\right):$$$${AB}^2+{AC}^2={BA}^{\prime 2}+{CA}^{\prime 2}+2\times {AA}^{\prime 2}$$$${AB}^2+{AC}^2={\left(\frac{BC}{2}\right)}^2+{\left(\frac{BC}{2}\right)}^2+2\times {AA}^{\prime 2}$$$${AB}^2+{AC}^2=\frac{{BC}^2}{2}+2\times {AA}^{\prime 2}$$$${AA}^{\prime 2}=\frac{1}{2}({AB}^2+{AC}^2-\frac{{BC}^2}{2})...\left(iii\right)$$$$$$$${BC}^2={AB}^2+{AC}^2-2\times AB\times AC\times \cos \mathrm{\angle }A$$$$into\left(iii\right):$$$${AA}^{\prime 2}=\frac{1}{4}({AB}^2+{AC}^2+2\times AB\times AC\times \cos \mathrm{\angle }A)...\left(iii\right)$$$$\Rightarrow {AA}^{\prime }=\frac{1}{2}\sqrt{{AB}^2+{AC}^2+2\times AB\times AC\times \cos \mathrm{\angle }A}$$

Commented by ajfour last updated on 24/May/17

$$hownice!$$

Commented by mrW1 last updated on 24/May/17

Commented by ajfour last updated on 24/May/17

$$alltbemorebetter!$$

Commented by Rishabh#1 last updated on 24/May/17

$$\mathrm{How}\mathrm{did}\mathrm{u}\mathrm{get}$$$${AA}^{\prime }=\frac{1}{2}\sqrt{{AB}^2+{AC}^2+2\times AB\times AC\times \cos \mathrm{\angle }A}$$

Commented by Tinkutara last updated on 25/May/17

$$\mathrm{Thanks}.$$

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