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

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The discussion centers on evaluating the expression \( \frac{ab + cd}{ad + bc} \) under specific conditions involving positive real numbers \( a, b, c, \) and \( d \). Two equations are provided: \( a^2 + d^2 - ad = b^2 + c^2 + bc \) and \( a^2 + b^2 = c^2 + d^2 \). Participants identify a mistake in the approach to express the quotient in terms of sine and cosine functions. The conversation emphasizes the need to correctly manipulate the given equations to derive the desired expression. Ultimately, the focus remains on finding all possible values for the quotient based on the established conditions.
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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 $\dfrac{ab + cd}{ad + bc}$.
 
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anemone said:
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:

This problem can easily be solved geometrically...(Wink)(Happy)
 
anemone said:
Hint:

This problem can easily be solved geometrically...(Wink)(Happy)
$\dfrac {\sqrt 3}{2}\approx 0.87$
 
Last edited:
anemone said:
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:
$A,B,C,D$ 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)=$\dfrac {\sqrt 3}{2}\approx 0.87$
 
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$a^2+ d^2-ad = b^2+ c^2+ bc=>a^2-b^2 = c^2 -d^2 + ad +bc \cdots(1)$ 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 $1 >=A>\frac{1}{2}$
 
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kaliprasad said:
$a^2+ d^2-ad = b^2+ c^2+ bc=>a^2-b^2 = c^2 -d^2 + ad +bc \cdots(1)$ 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 $1 >=A>\frac{1}{2}$
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 :
$\frac{ab + cd}{ad + bc}= \frac{\sin t \cos t + \sin x \cos x}{\cos t \cos x + \sin t \sin x}$
$= \frac{\sin 2 t + \sin 2x}{2\cos (t -x)}$.
$= \frac{2(\sin (t + x)\cos(t-x)}{2\cos (t -x)}=\sin(t+x)=sin 60^o=\dfrac {\sqrt 3}{2}$
 
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kaliprasad said:
$a^2+ d^2-ad = b^2+ c^2+ bc=>a^2-b^2 = c^2 -d^2 + ad +bc \cdots(1)$ 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...
 
anemone said:
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

$a^2+ d^2-ad = b^2+ c^2+ bc=>a^2-b^2 = c^2 -d^2 + ad +bc \cdots(1)$ 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

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

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