Evaluate Quotient: $ab+cd \over ad+bc$

[anemone]

Given positive real numbers $a,\,b,\,c,$ and $d$ that satisfy the system below:

$a^2+ d^2-ad = b^2+ c^2+ bc$ and

$a^2+ b^2= c^2+ d^2$.

Find all possible values of the expression $\dfrac{ab + cd}{ad + bc}$.

 

[anemone]

Given positive real numbers $a,\,b,\,c,$ and $d$ that satisfy the system below:

$a^2+ d^2-ad = b^2+ c^2+ bc$ and

$a^2+ b^2= c^2+ d^2$.

Find all possible values of the expression $\dfrac{ab + cd}{ad + bc}$.

Hint:

Spoiler

This problem can easily be solved geometrically...(Wink)(Happy)

 

[Albert1]

Hint:

Spoiler

This problem can easily be solved geometrically...(Wink)(Happy)

Spoiler

$\dfrac {\sqrt 3}{2}\approx 0.87$

 

[Albert1]

Given positive real numbers $a,\,b,\,c,$ and $d$ that satisfy the system below:

$a^2+ d^2-ad = b^2+ c^2+ bc---(1)$ and

$a^2+ b^2= c^2+ d^2--(2)$.

Find all possible values of the expression $\dfrac{ab + cd}{ad + bc}---(3)$.

my solution:

Spoiler

$A,B,C,D$ are four ponts on a circle with radius 1, $\overline{AC}$ is a diameter=2)
let:$\angle BCD=60^o,\overline{AB}=c,\overline{BC}=d,\overline{CD}=a,\overline{AD}=b$
then use law of cosine and pythagorean theorem ,it is easy to see that both (1) and (2) satisfy
for simple calculation we may let :$a=d=\sqrt 3,b=c=1$
we get (3)=$\dfrac {\sqrt 3}{2}\approx 0.87$

 

[kaliprasad]

Spoiler

$a^2+ d^2-ad = b^2+ c^2+ bc=>a^2-b^2 = c^2 -d^2 + ad +bc \cdots(1)$ and
$a^2+ b^2= c^2+ d^2\cdots(2)$.
we can take $a = r\cos\,t, b= r\sin\,t, c = r\sin\, x, d= r\cos\,x$ to satisfy (2) so we get from (1)
$r^2(\cos^2 t - \sin ^2 t) = r^2(sin ^2 x - \cos ^2 x) + r^2(\cos\,t \cos\, x + \sin\,t \sin\, x)$
or $\cos 2t - \cos 2x = \cos(t-x)$
or $2\cos (t +x)\cos (t-x) = \cos(t-x)$
or $\cos (t +x) = \frac{1}{2}$
or $t+x = 60^\circ\cdots(3)$
$\frac{ab + cd}{ad + bc}= \frac{\sin t \cos t + \sin x \cos x}{\sin t \cos x + \cos t \sin x}$
$= \frac{\sin 2 t + \sin 2x}{2\sin (t +x)}$.
$= \frac{2(\sin (t + x)\cos(t-x)}{2\sin (t +x)}=\cos(t-x)$
from (3) we get $t =60^\circ-x$
so $\cos(t-x) = \cos(60^\circ- 2x)$ as as x is $> 0$ to $< 60^\circ$ we get the range of the value say A satisfying $1 >=A>\frac{1}{2}$

 

[Albert1]

Spoiler

$a^2+ d^2-ad = b^2+ c^2+ bc=>a^2-b^2 = c^2 -d^2 + ad +bc \cdots(1)$ and
$a^2+ b^2= c^2+ d^2\cdots(2)$.
we can take $a = r\cos\,t, b= r\sin\,t, c = r\sin\, x, d= r\cos\,x$ to satisfy (2) so we get from (1)
$r^2(\cos^2 t - \sin ^2 t) = r^2(sin ^2 x - \cos ^2 x) + r^2(\cos\,t \cos\, x + \sin\,t \sin\, x)$
or $\cos 2t - \cos 2x = \cos(t-x)$
or $2\cos (t +x)\cos (t-x) = \cos(t-x)$
or $\cos (t +x) = \frac{1}{2}$
or $t+x = 60^\circ\cdots(3)$
$\frac{ab + cd}{ad + bc}= \frac{\sin t \cos t + \sin x \cos x}{\cos t \cos x + \sin t \sin x}$
$= \frac{\sin 2 t + \sin 2x}{2\sin (t +x)}$.
$= \frac{2(\sin (t + x)\cos(t-x)}{2\sin (t +x)}=\cos(t-x)$
from (3) we get $t =60^\circ-x$
so $\cos(t-x) = \cos(60^\circ- 2x)$ as as x is from 0 to $< 60^\circ$ we get the range of the value say A satisfying $1 >=A>\frac{1}{2}$

Spoiler

if $x=0$ then $c=0$
but we are given $a,b,c,d>0$
if $t+x=60^o$ then the answer seemed always remain the same =$\dfrac{\sqrt 3}{2}$ (fixed)
I tried many pairs of angles $t$ and $x$ (with $t+x=60^o$ )
(I chceked the result using $"GSP"$)
the result should be :
$\frac{ab + cd}{ad + bc}= \frac{\sin t \cos t + \sin x \cos x}{\cos t \cos x + \sin t \sin x}$
$= \frac{\sin 2 t + \sin 2x}{2\cos (t -x)}$.
$= \frac{2(\sin (t + x)\cos(t-x)}{2\cos (t -x)}=\sin(t+x)=sin 60^o=\dfrac {\sqrt 3}{2}$

 

[anemone]

Spoiler

$a^2+ d^2-ad = b^2+ c^2+ bc=>a^2-b^2 = c^2 -d^2 + ad +bc \cdots(1)$ and
$a^2+ b^2= c^2+ d^2\cdots(2)$.
we can take $a = r\cos\,t, b= r\sin\,t, c = r\sin\, x, d= r\cos\,x$ to satisfy (2) so we get from (1)
$r^2(\cos^2 t - \sin ^2 t) = r^2(sin ^2 x - \cos ^2 x) + r^2(\cos\,t \cos\, x + \sin\,t \sin\, x)$
or $\cos 2t - \cos 2x = \cos(t-x)$
or $2\cos (t +x)\cos (t-x) = \cos(t-x)$
or $\cos (t +x) = \frac{1}{2}$
or $t+x = 60^\circ\cdots(3)$
$\frac{ab + cd}{ad + bc}= \frac{\sin t \cos t + \sin x \cos x}{\sin t \cos x + \cos t \sin x}$
$= \frac{\sin 2 t + \sin 2x}{2\sin (t +x)}$.
$= \frac{2(\sin (t + x)\cos(t-x)}{2\sin (t +x)}=\cos(t-x)$
from (3) we get $t =60^\circ-x$
so $\cos(t-x) = \cos(60^\circ- 2x)$ as as x is $> 0$ to $< 60^\circ$ we get the range of the value say A satisfying $1 >=A>\frac{1}{2}$

Hi kaliprasad! It took me a while to realize you have a mistake when you are to find the expression of the intended quotient in terms of sine and cosine functions...

 

[kaliprasad]

Hi kaliprasad! It took me a while to realize you have a mistake when you are to find the expression of the intended quotient in terms of sine and cosine functions...

Thanks anemone here is the solution

Spoiler

$a^2+ d^2-ad = b^2+ c^2+ bc=>a^2-b^2 = c^2 -d^2 + ad +bc \cdots(1)$ and
$a^2+ b^2= c^2+ d^2\cdots(2)$.
we can take $a = r\cos\,t, b= r\sin\,t, c = r\sin\, x, d= r\cos\,x$ to satisfy (2) so we get from (1)
$r^2(\cos^2 t - \sin ^2 t) = r^2(sin ^2 x - \cos ^2 x) + r^2(\cos\,t \cos\, x + \sin\,t \sin\, x)$
or $\cos 2t - \cos 2x = \cos(t-x)$
or $2\cos (t +x)\cos (t-x) = \cos(t-x)$
or $\cos (t +x) = \frac{1}{2}$
or $t+x = 60^\circ\cdots(3)$
correct till this
now for the second part which is corrected as below

$\frac{ab + cd}{ad + bc}= \frac{\cos t \sin t + \sin x \cos x}{\cos t \cos x + \sin t \sin x}$
$= \frac{\sin 2 t + \sin 2x}{2\cos (t-x)}$.
$= \frac{2(\sin (t + x)\cos(t-x)}{2\cos (t -x)}=\sin(t+x)= \sin(60^\circ) = \frac{\sqrt3}{2}$