Prove that $ab+xy=2(ay+bx)$, $0

Question

Let $a, b, x, y$ be real numbers such that $0<x<a$ and $0<y<b$ and

$a^2+y^2=b^2+x^2=2(ax+by)$.

Prove that $ab+xy=2(ay+bx)$

solution

$a^2+y^2=2(ax+by)>>(a+y)^2-2ay=2(ax+by)$

$b^2+x^2=2(ax+by)>>(b+x)^2-2bx=2(ax+by)$

Let's write sum of this two equations

$(a+y)^2-2ay+(b+x)^2-2bx=4(ax+by)$

we know that

$(a+y)^2-2ay=2(ax+by)$

$(b+x)^2-2bx=2(ax+by)$

$(a+b)^2-2ab+(x+y)^2-2xy=4(ax+by)$

$(a+y)^2-2ay+(b+x)^2-2bx=4(ax+by)$

$(a+b)^2-2ab+(x+y)^2-2xy+(a+y)^2-2ay+(b+x)^2-2bx=8(ax+by)$

$(a+b)^2-2ab+(x+y)^2-2xy+(a+y)^2+(b+x)^2-8(ax+by)=2(ay+bx)$

Since $8(ax+by)=2(a^2+y^2)+2(b^2+x^2)$

$(a+b)^2-2ab+(x+y)^2-2xy+(a+y)^2+(b+x)^2-2(a^2+y^2)-2(b^2+x^2)=2(ay+bx)$

$(a+b)^2+(x+y)^2+(a+y)^2+(b+x)^2-2(ab+xy)-2(a^2+y^2)-2(b^2+x^2)=2(ay+bx)$

I can't write anything after that

asdf · Answer 1 · 2017-12-11T10:51:12.397

2

Let $a=c\cdot \sin(m), y=c\cdot \cos(m), b=c\cdot \sin(n), x=c\cdot \cos(n)$. WLOG $c\neq0$ Otherwise it's trivial. Hence we have:

$2(ax+by)=c^2$ which means that $2\sin(m+n)=1$ $(1)$

Now, $ab+xy=2(ay+bx)$

$\Leftrightarrow$ $\sin(m)\sin(n)+\cos(m)\cos(n)=2(\sin(m)\cos(m)+\sin(n)\cos(n))$

$\Leftrightarrow$ $\cos(m-n) = \sin(2m)+\sin(2n)$

$\Leftrightarrow$ $\cos(m-n) = 2\cos(m-n)\sin(m+n)$

which holds by $(1)$

edited Dec 11 '17 at 10:51

answered May 15 '17 at 18:50

asdf

$\LaTeX$ tip: use \sin and \cos and \times or \cdot for multiplication. – mdave16 May 15 '17 at 18:59
How can we know that x=c cos(b)?? – Aytajdadashova May 17 '17 at 06:50
Well, you can say the following: Let $b^2+x^2=c$. Then $(\frac{b}{c})^2+(\frac{x}{c})^2=1$. Then you can take those 2 fractions to equal some sine and cosine – asdf May 17 '17 at 20:08
To make the solution full, it should be noted that there was a typo and it should've been $x=c \cdot cos(n)$ – asdf Dec 11 '17 at 10:52

1 Answers1