Question-15567

Question Number 15567 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 11/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 11/Jun/17

A B C D & M N P Q, a r e s q u a r e s .

s h o w t h a t :

a^{2} + b^{2} = c^{2} + d^{2}

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/Jun/17

a^{2} = s^{2} + {(p + n \frac{\sqrt{2}}{2})}^{2}, b^{2} = q^{2} + {(r + n \frac{\sqrt{2}}{2})}^{2}

c^{2} = p^{2} + {(q + n \frac{\sqrt{2}}{2})}^{2}, d^{2} = r^{2} + {(s + n \frac{\sqrt{2}}{2})}^{2}

a^{2} + b^{2} = s^{2} + q^{2} + p^{2} + r^{2} + n^{2} + 2 n \sqrt{2} (p + r) (i)

c^{2} + d^{2} = s^{2} + q^{2} + p^{2} + r^{2} + n^{2} + 2 n \sqrt{2} (q + s) (i i)

q + s + + n \sqrt{2} = m, p + r + n \sqrt{2} = m

\Rightarrow p + r = q + s

(i) - (i i) \Rightarrow (a^{2} + b^{2}) - (c^{2} + d^{2}) = 0

\Rightarrow a^{2} + b^{2} = c^{2} + d^{2} .

Answered by ajfour last updated on 12/Jun/17

Commented by ajfour last updated on 12/Jun/17

l e t s \cos θ = s_{x}, s \sin θ = s_{y}

l i s t h e s i d e o f b i g g e s t s q u a r e

a l l c o m p o n e n t s o f l e n g t h s a, b,

c, d, a n d s a r e p o s i t i v e .

f r o m f i g u r e a b o v e,

c_{x} = b_{x} + s_{y}; c_{y} = a_{y} - s_{y}

d_{x} = a_{x} + s_{y}; d_{y} = b_{y} - s_{y}

c^{2} + d^{2} = c_{x}^{2} + c_{y}^{2} + d_{x}^{2} + d_{y}^{2}

= {(b_{x} + s_{y})}^{2} + {(a_{y} - s_{y})}^{2}

+ {(a_{x} + s_{y})}^{2} + {(b_{y} - s_{y})}^{2}

= (a_{x}^{2} + a_{y}^{2}) + (b_{x}^{2} + b_{y}^{2}) + 4 s_{y}^{2}

+ 2 s_{y} [(a_{x} + b_{x}) - (a_{y} + b_{y})]

N o w (a_{x} + b_{x}) = l - (s_{x} + s_{y})

a n d (a_{y} + b_{y}) = l - (s_{x} - s_{y})

\Rightarrow (a_{x} + b_{x}) - (a_{y} + b_{y}) = - 2 s_{y}

c o n t i n u i n g, w e h a v e

c^{2} + d^{2} = a^{2} + b^{2} + 4 s_{y}^{2}

+ 2 s_{y} (- 2 s_{y})

o r c^{2} + d^{2} = a^{2} + b^{2} .

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 12/Jun/17

e x c e l l e n t! m r A j c o u r . t h a n k y o u v e r y

m u c h .

Commented by ajfour last updated on 12/Jun/17

t h a n k y o u s i r, i s h a l l y e t l o o k f o r

a s h o r t e r w a y . .

Answered by ajfour last updated on 12/Jun/17

Commented by ajfour last updated on 12/Jun/17

{\bar{l}}_{x} + \bar{c} = \bar{a} + {\bar{s}}_{1} \dots 1 (i)

{\bar{l}}_{y} + \bar{b} = \bar{c} + {\bar{s}}_{2} \dots . . 1 (i i)

{\bar{l}}_{x} + \bar{b} = \bar{d} + {\bar{s}}_{1} \dots . . 1 (i i i)

{\bar{l}}_{y} + \bar{d} = \bar{a} + {\bar{s}}_{2} \dots . . 1 (i v)

{\bar{l}}_{x} . {\bar{l}}_{y} = {\bar{s}}_{1} . {\bar{s}}_{2} = 0

{\bar{l}}_{x} . {\bar{s}}_{2} = - l s (\sin θ), {\bar{l}}_{y} . {\bar{s}}_{1} = l s (\sin θ)

s u b t r a c t i n g t h e t o p l e f t t w o e q n s .

\bar{c} - \bar{b} = \bar{a} - \bar{d}

\Rightarrow \bar{c} + \bar{d} = \bar{a} + \bar{b}

S o

(\bar{c} + \bar{d}) . (\bar{c} + \bar{d}) = (\bar{c} + \bar{d}) . (\bar{a} + \bar{b})

c^{2} + d^{2} + 2 \bar{c} . \bar{d} = (\bar{c} + \bar{d}) . (\bar{a} + \bar{b}) \dots (2)

f r o m (i) :

\bar{c} - \bar{a} = {\bar{s}}_{1} - {\bar{l}}_{x} a n d f r o m (i v)

\bar{d} - \bar{a} = {\bar{s}}_{2} - {\bar{l}}_{y}

t a k i n g d o t p r o d u c t,

(\bar{c} - \bar{a}) . (\bar{d} - \bar{a}) = - {\bar{s}}_{1} . {\bar{l}}_{y} - {\bar{s}}_{2} . {\bar{l}}_{x}

= - l s (\sin θ) + l s (\sin θ) = 0 . . (3)

r e a r r a n g i n g (i i) a n d (i i i) :

\bar{c} - \bar{b} = {\bar{l}}_{y} - {\bar{s}}_{2}

\bar{b} - \bar{d} = {\bar{s}}_{1} - {\bar{l}}_{x}

d o t p r o d u c t y i e l d s,

(\bar{c} - \bar{b}) . (\bar{b} - \bar{d}) = {\bar{s}}_{1} . {\bar{l}}_{y} + {\bar{s}}_{2} . {\bar{l}}_{x}

= l s (\sin θ) - l s (\sin θ) = 0 \dots (4)

s u b t r a c t i n g (4) a n d (3) :

\bar{c} . \bar{d} - \bar{c} . \bar{a} - \bar{a} . \bar{d} + a^{2}

- \bar{c} . \bar{b} + \bar{c} . \bar{d} + b^{2} - \bar{b} . \bar{d} = 0

o r

2 \bar{c} . \bar{d} - (\bar{c} + \bar{d}) . (\bar{a} + \bar{b}) + a^{2} + b^{2} = 0

\dots . (5)

s u b t r a c t i n g (5) f r o m (2) :

(c^{2} + d^{2}) - (a^{2} + b^{2}) = 0

c^{2} + d^{2} = a^{2} + b^{2} .

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 12/Jun/17

s o n i c e d e a r m r A j f o u r . t h i s p r o o f i s

v e r y b e a u t i f u l . i l o v e t h i s . t h a n k s .