Prove-that-if-A-B-and-C-are-the-midpoints-of-the-sides-BC-CA-and-AB-respectively-then-AA-BB-CC-lt-AB-BC-CA-

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

Prove that if A^{'}, B^{'} and C^{'} are the

midpoints of the sides B C, C A and A B,

respectively, then

{A A}^{'} + {B B}^{'} + {C C}^{'} < A B + B C + C A

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

A G + B G > A B \Rightarrow \frac{2}{3} ({A A}^{^{'}} + {B B}^{^{'}}) > A B

B G + G C > B C \Rightarrow \frac{2}{3} ({B B}^{^{'}} + {C C}^{^{'}}) > B C

A G + G C > A C \Rightarrow \frac{2}{3} ({A A}^{^{'}} + {C C}^{^{'}}) > A C

\Rightarrow 1) A B + B C + C A < \frac{4}{3} ({A A}^{^{'}} + {B B}^{^{'}} + {C C}^{^{'}})

2) \frac{∣ A B - A C ∣}{2} < {A A}^{^{'}} < \frac{A B + A C}{2}

3) a (m_{a}^{2} + \frac{a}{2} . \frac{a}{2}) = b^{2} . \frac{a}{2} + c^{2} . \frac{a}{2} ({E s t e w a r t}^{'} s t e o r e m)

\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} - 2 b c . c o s A)}{4} = \frac{1}{4} (b^{2} + c^{2} + 2 b c . c o s A) =

= \frac{1}{4} (b^{2} + c^{2} + 2 b c . c o s A) < \frac{1}{4} (b^{2} + c^{2} + 2 b c) =

= \frac{1}{4} {(b + c)}^{2} \Rightarrow m_{a} < \frac{1}{2} (b + c)

s i m i l a r l y :\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} (2 a + 2 b + 2 c) = a + b + c

\Rightarrow m_{a} + m_{b} + m_{c} < a + b + c .

4) i f : m_{a} = \frac{b + c}{2} \Rightarrow \frac{2 (b^{2} + c^{2}) - a^{2}}{4} = \frac{b^{2} + c^{2} + 2 b c}{4}

\Rightarrow b^{2} + c^{2} - 2 b c = a^{2} \Rightarrow {(b - c)}^{2} = a^{2} \Rightarrow b - c = a (i m p o s s i b l e)

o r : b^{2} + c^{2} - 2 b c = b^{2} + c^{2} - 2 b c . c o s A

\Rightarrow c o s A = 1 \Rightarrow ∡ A = 0 (i m p o s s i b l e)

s o : m_{a} ⪇ \frac{1}{2} (b + c) a n d : m_{a} < \frac{1}{2} (b + c) .

5) A = 90^{°} \Rightarrow 4 m_{a}^{2} = b^{2} + c^{2} = a^{2} \Rightarrow m_{a} = \frac{a}{2} .

A = 120^{°} \Rightarrow 4 m_{a}^{2} = b^{2} + c^{2} - b c

A = 60 \Rightarrow 4 m_{a}^{2} = b^{2} + c^{2} + b c

6) A = 90, B = 60, C = 30

\Rightarrow 4 (m_{a}^{2} + m_{b}^{2} + m_{c}^{2}) = a^{2} + a^{2} + c^{2} + a c + a^{2} + b^{2} + \sqrt{3} a b =

= 4 a^{2} + \sqrt{3} a (b + c) .

7) A = 90, b = c \Rightarrow Σ m_{a}^{2} = 4 a^{2} + 2 \sqrt{2} a b \overset{}{.}

8) {B C}^{2} = {B G}^{2} + {G C}^{2} - 2 B G . G C . c o s B \overset{}{G C}

∡ B G C = φ \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} c o s φ

\Rightarrow \frac{9}{4} a^{2} = m_{b}^{2} + m_{c}^{2} - 2 m_{b} . m_{c} . c o s φ

φ = 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 (i f : m_{b} ⊥ m_{c}) \Rightarrow (5 a^{2} = b^{2} + c^{2}), (m_{a} = \frac{3}{2} a)

9) A B ⊥ A C \Rightarrow m_{b}^{2} = \frac{2 (a^{2} + c^{2}) - b^{2}}{4} = \frac{b^{2} + 2 c^{2}}{4}

m_{c}^{2} = \frac{2 (a^{2} + b^{2}) - c^{2}}{4} = \frac{2 b^{2} + c^{2}}{4}

\Rightarrow m_{b}^{2} + m_{c}^{2} = \frac{3}{4} (b^{2} + c^{2}) = \frac{3}{4} a^{2} = 3 m_{a}^{2}

\Rightarrow {\begin{cases} b ⊥ c \Rightarrow m_{b}^{2} + m_{c}^{2} = 3 m_{a}^{2}, m_{a} = \frac{1}{2} a . \\ m_{b} ⊥ m_{c} \Rightarrow b^{2} + c^{2} = 5 a^{2}, m_{a} = \frac{3}{2} a . \end{cases}

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

i n c a s e o f : ∡ A G B = ∡ B G C = ∡ A G C

s h o w t h a t :

1) S_{A B C} = \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} = a l t i t u d e f r o m \overset{}{A} t o B C)

Answered by mrW1 last updated on 24/May/17

{A A}^{'} = \frac{1}{2} \sqrt{{A B}^{2} + {A C}^{2} + 2 \times A B \times A C \times \cos ∠ A}

< \frac{1}{2} \sqrt{{A B}^{2} + {A C}^{2} + 2 \times A B \times A C}

= \frac{1}{2} \sqrt{{(A B + A C)}^{2}} = \frac{1}{2} (A B + A C)

{A A}^{'} < \frac{1}{2} (A B + A C)

{B B}^{'} < \frac{1}{2} (B A + B C)

{C C}^{'} < \frac{1}{2} (C B + C A)

\Rightarrow {A A}^{'} + {B B}^{'} + {C C}^{'} < A B + B C + C A

Commented by mrW1 last updated on 24/May/17

Y o u c a n a l s o p r o v e l i k e t h i s :

s i n c e {B A}^{'} = A^{'} C

\Rightarrow A^{'} A^{″} = A^{'} A^{‴}

{A A}^{″} < A C

{A A}^{‴} < A B

{A A}^{'} = \frac{1}{2} ({A A}^{″} + {A A}^{‴}) < \frac{1}{2} (A C + A B)

Commented by mrW1 last updated on 24/May/17

U s i n g l a w o f c o s i n e s w e g e t :

{A B}^{2} = {B A}^{' 2} + {A A}^{' 2} - 2 \times {B A}^{'} \times {A A}^{'} \times \cos ∠ {A A}^{'} B \dots (i)

{A C}^{2} = {C A}^{' 2} + {A A}^{' 2} - 2 \times {C A}^{'} \times {A A}^{'} \times \cos ∠ {A A}^{'} C

{A C}^{2} = {C A}^{' 2} + {A A}^{' 2} - 2 \times {C A}^{'} \times {A A}^{'} \times \cos (180 ° - ∠ {A A}^{'} B)

{A C}^{2} = {C A}^{' 2} + {A A}^{' 2} + 2 \times {C A}^{'} \times {A A}^{'} \times \cos ∠ {A A}^{'} B

{A C}^{2} = {C A}^{' 2} + {A A}^{' 2} + 2 \times {B A}^{'} \times {A A}^{'} \times \cos ∠ {A A}^{'} B \dots (i i)

(i) + (i i) :

{A B}^{2} + {A C}^{2} = {B A}^{' 2} + {C A}^{' 2} + 2 \times {A A}^{' 2}

{A B}^{2} + {A C}^{2} = {(\frac{B C}{2})}^{2} + {(\frac{B C}{2})}^{2} + 2 \times {A A}^{' 2}

{A B}^{2} + {A C}^{2} = \frac{{B C}^{2}}{2} + 2 \times {A A}^{' 2}

{A A}^{' 2} = \frac{1}{2} ({A B}^{2} + {A C}^{2} - \frac{{B C}^{2}}{2}) \dots (i i i)

{B C}^{2} = {A B}^{2} + {A C}^{2} - 2 \times A B \times A C \times \cos ∠ A

i n t o (i i i) :

{A A}^{' 2} = \frac{1}{4} ({A B}^{2} + {A C}^{2} + 2 \times A B \times A C \times \cos ∠ A) \dots (i i i)

\Rightarrow {A A}^{'} = \frac{1}{2} \sqrt{{A B}^{2} + {A C}^{2} + 2 \times A B \times A C \times \cos ∠ A}

Commented by ajfour last updated on 24/May/17

h o w n i c e!

Commented by mrW1 last updated on 24/May/17

Commented by ajfour last updated on 24/May/17

a l l t b e m o r e b e t t e r!

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

How did u get

{A A}^{'} = \frac{1}{2} \sqrt{{A B}^{2} + {A C}^{2} + 2 \times A B \times A C \times \cos ∠ A}

Commented by Tinkutara last updated on 25/May/17

Thanks .