take $a=rsinp, b=rcosp$ and $ c=rsinq, b=rcosq ( 0<p,q<90)$ putting in 2nd and simplifying we get $sin(p-q)=1/2$ now the value which we have to find can be simplified to $cos(p-q)$ which is $\sqrt{3}/2$

Construct a quadrilateral with sides $a,b,c,d$ and one diagonal of Length $\sqrt{b^2+c^2+bc}$ Then the quadrilateral has one angle $60$ and the opposite $120$ .Then use Ptolemy and Areas to get answer $\sqrt{3}$/$2$

phymaths wrote: Construct a quadrilateral with sides $a,b,c,d$ and one diagonal of Length $\sqrt{b^2+c^2+bc}$ Then the quadrilateral has one angle $60$ and the opposite $120$ .Then use Ptolemy and Areas to get answer $\sqrt{3}$/$2$ motivation??

This idea has been used here:http://mathometer.weebly.com/solving-problems-with-geometric-ideas.html Basically the conditions look a lot like Law of Cosines, so here we go. EDIT: Turns out this was exactly the same problem in Baltic way.

phymaths wrote: Construct a quadrilateral with sides $a,b,c,d$ and one diagonal of Length $\sqrt{b^2+c^2+bc}$ Then the quadrilateral has one angle $60$ and the opposite $120$ .Then use Ptolemy and Areas to get answer $\sqrt{3}$/$2$ Same solution!!

From the given condition, we have $a^2-c^2+b^2-d^2=0$ and $a^2-c^2-(b^2-d^2)=ad+bc$ $\Longleftrightarrow a^2-c^2=\frac 12(ad+bc)$ and $b^2-d^2=-\frac 12(ad+bc).$ $\therefore (a^2-c^2)(b^2-d^2)=(ac+bd)^2-(ad+bc)^2=-\frac 14(ad+bc)^2$ $\therefore (ab+cd)^2=\frac 34(ad+bc)^2.$ If $a,\ b,\ c,\ d$ are real numbers, for $ad+bc\neq 0$, we have $\left(\frac{ab+cd}{ad+bc}\right)^2=\frac 34$, yielding $\frac{ab+cd}{ad+bc}=\pm \frac{\sqrt{3}}{2}.$

@above, they are +ve, so it cannot be -ve

tarzanjunior wrote: @above, they are +ve, so it cannot be -ve As I have already said, if a, b, c, d are any real numbers such that $ab+cd \neq 0.$ Needless to say, if a, b, c, d are positive, then the answer to the original problem is definitely, $\frac{ab+cd}{ad+bc}=\frac{\sqrt{3}}{2}.$ I just showed my proof, we could solved the problem without constructing a quadrilateral even if a, b, c, d are positive. My solution is for Japanese Kids at age of 10, 11, 12 aiming at IMO.

I had also done the trigonometry solution when i solved it at home. Unfortunately was not able to give nmtc this year. The quadrilateral solution was awesome.

$$a^2+b^2=c^2+d^2$$$$\iff (c+d)^2-(a-b)^2 = 2(ab+cd) \ \textendash \ \boxed{1}$$$$(a+b)^2-(c-d)^2 = 2(ab+cd) \ \textendash \ \boxed{2}$$$$\boxed{1} \times \boxed{2} \Rightarrow 4(ab+cd)^2 = (-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d) \ \textendash \ \boxed{3}$$$$a^2+d^2-ad=b^2+c^2+bc$$$$\iff (a+d)^2-(b-c)^2 = 3(ad+bc) \ \textendash \ \boxed{4}$$$$(b+c)^2-(a-d)^2=(ad+bc) \ \textendash \ \boxed{5}$$$$\boxed{4}\times\boxed{5} \Rightarrow 3(ad+bc)^2 = (-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)$$$$ \boxed{3} \ \& \ \boxed{6} \Rightarrow 3(ad+bc)^2 = 4(ab+cd)^2 \Rightarrow \frac{ab+cd}{ad+bc} = \frac{\sqrt{3}}{2}$$ My proof is not as good as others though. Sadly I couldn't give Stage - II, I botched up 3-4 easy questions in Stage - I.

Source of problem revealed! https://artofproblemsolving.com/community/c6h613429p3649212

TheDarkPrince wrote: If $a,b,c,d$ are positive reals such that $a^2+b^2=c^2+d^2$ and $a^2+d^2-ad=b^2+c^2+bc$, find the value of $\frac{ab+cd}{ad+bc}$ here: $ad+bc=a^2-c^2+d^2-b^2=2(a^2-c^2)=2(d^2-b^2)$, $\frac{ab+cd}{ad+bc}=\sqrt{1-\frac{(a^2-c^2)(d^2-b^2)}{(ad+bc)^2}}=\frac{\sqrt{3}}{2}.$