Beautiful identity! I have just uploaded a note, "Circumscribed quadrilaterals revisited", on my website ( http://www.cip.ifi.lmu.de/~grinberg/ ), where this identity is Theorem 10 and a synthetic proof is given. Hope you will like it (although you will probably find most of the stuff quite well-known). Darij

Very nice identity ! Thanks, Mekrazywong ! Indeed, this problem is totally trivial by trigonometry. It is equivalently with the following conditioned identity: $x+y+z+t=\pi\Longrightarrow\cos x\cdot \cos z +\cos y\cdot \cos t =\sin (x+y)\cdot \sin (y+z)\ \ (1)$. Really it isn't worth proving syntetically, at the most a syntetical prove for the conditioned identity (1) ! Remark. In general, $\cos x\cdot \cos z+\cos y\cdot \cos t =\sin (x+y)\cdot\sin (y+z)\Longleftrightarrow$$\cos \frac{x+y+z+t}{2}\cdot \cos y\cdot \cos \frac{x+y+z-t}{2}=0$. Mekrazywong and Darij, can give you a geometrical interpretation (for a quadrilateral, without circumscribed) of this equivalence ?

First of all, I would like to clarify here this result was not due to me. Actually this is a problem from Chinese TST 2003, proposed by a Chinese professor. By the way, there is a really simple solution via spiral similarity.

Denote by $ a=AB$, $ b=BC$, $ c=CD$, $ d=DA$ the lengths of the sides of $ ABCD$. Let $ r$ be the ray of the inscribed circle. Then $ OA =\frac{r}{\sin\frac{A}{2}},OB =\frac{r}{\sin\frac{B}{2}},OC =\frac{r}{\sin\frac{C}{2}},OD =\frac{r}{\sin\frac{C}{2}}$. Let $ M$ be the point where $ AB$ touches the circle. Then $ AM = rctg\frac{A}{2},\ MB = rctg\frac{B}{2}\Rightarrow a = AM+MB = r\cdot\frac{\sin\frac{A+B}{2}}{\sin\frac{A}{2}\sin\frac{B}{2}}$, $ b = r\cdot\frac{\sin\frac{B+C}{2}}{\sin\frac{B}{2}\sin\frac{C}{2}},c = r\cdot\frac{\sin\frac{C+D}{2}}{\sin\frac{C}{2}\sin\frac{D}{2}},d = r\cdot\frac{\sin\frac{D+A}{2}}{\sin\frac{D}{2}\sin\frac{A}{2}}$. Replacing $ OA,OB,OC,OD,a,b,c,d$, making some calculus and using the equalities $ \sin\frac{A+B}{2}=\sin\frac{C+D}{2}, sin\frac{B+C}{2}=\sin\frac{A+D}{2}$, the relation becomes $ \sin\frac{A}{2}\sin\frac{C}{2}+\sin\frac{B}{2}\sin\frac{D}{2}=\sin\frac{A+B}{2}\sin\frac{B+C}{2}$. But $ \sin\frac{A}{2}\sin\frac{C}{2}+\sin\frac{B}{2}\sin\frac{D}{2}=\frac{1}{2}\left[\cos\frac{A-C}{2}-\cos\frac{A+C}{2}+\cos\frac{B-D}{2}-\cos\frac{B+D}{2}\right] =$ $ =\frac{1}{2}\left(\cos\frac{A-C}{2}+\cos\frac{B-D}{2}\right) =\sin\frac{A+B}{2}\sin\frac{B+C}{2}$.

use this lemma : we have that $ AB.BC = OB^{2} + \frac {OA.OB.OC}{OD}$

My solution: Let $ E $ be a point satisfy $ \triangle CDE \sim \triangle BAO $ . Since $ C, D, E, O $ are concyclic , so from Ptolemy theorem we get $ DE \cdot OC+CE\cdot OD=CD\cdot OE $ . ... $ (1) $ Since $ \frac{DE}{OA}=\frac{CE}{OB}=\frac{CD}{AB} $ , so from $ (1) $ we get $ OA\cdot OC+OB\cdot OD=AB\cdot OE $ . ... $ (2) $ Since $ \angle EOD=\angle ECD=\angle OBA=\angle CBO , \angle DEO=\angle DCO=\angle OCB $ , so we get $ \triangle DOE \sim \triangle OBC $ and $ \frac{OE}{BC}=\frac{OD}{OB} $ . ... $ (3) $ Similarly, we can prove $ \triangle OCE \sim \triangle AOD $ and $ \frac{OE}{DA}=\frac{CE}{OD} $ . ... $ (4) $ From $ (3), (4) $ we get $ \frac{OE^2}{BC\cdot DA}=\frac{CE}{OB}=\frac{CD}{AB} $ . ... $ (5) $ From $ (2), (5) $ we get $ (OA\cdot OC+OB\cdot OD)^2=AB^2\cdot \frac{BC\cdot CD\cdot DA}{AB}=AB\cdot BC\cdot CD\cdot DA $ . Q.E.D

Another proof :

Attachments:

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Quadrilatere%20circonscriptible.pdf p. 7... Sincerely Jean-Louis

Just solving this for no reason as I just came for this page for a different problem but saw this and wanted to gain some fake satisfaction before sleep of solving an Click to reveal hidden text

chinese tst. Just consider the point of tangencies as $KEK'W$.Let $F$ be $KK' \cap EW$ Then consider a projective transformation that fixes the incircle and sends $F$ to the center of the incircle. Now ,we just have to prove the problem for a rhombus which is a mere Pythagorean .

Solved with Max Lu. The work here is almost entirely copy-pasted from our work on USAMO 2004/6 . Let the incircle have radius $r$, and let the lengths of the tangents from $A,B,C,D$ be $a,b,c,d$. Then, note that \[(w+ri)(x+ri)(y+ri)(z+ri) \text{is a negative real number}\]since the argument of it is $\frac{A}{2}+\frac{B}{2}+\frac{C}{2}+\frac{D}{2}=180$. Thus, we have $r\cdot \sum_{cyc} wxy - r^3 \sum_{cyc} a = 0\Longrightarrow r^2 = \frac{\sum_{cyc} wxy}{\sum_{cyc} w}$. Now, note that this gives \[a^2+r^2 = a^2 + \frac{\sum abc}{\sum a} = \frac{a^2\sum a + \sum abc}{\sum a} = \frac{(a+b)(a+c)(a+d)}{\sum a}\] Using this, we get that \[OA\cdot OC + OB\cdot OD = \sqrt{\frac{(a+b)(a+c)(a+d)}{a+b+c+d}} \cdot \sqrt{\frac{(a+c)(b+c)(c+d)}{a+b+c+d}}+ \sqrt{\frac{(a+b)(b+c)(b+d)}{a+b+c+d}} \cdot \sqrt{\frac{(a+d)(b+d)(c+d)}{a+b+c+d}} \]\[= \frac{((a+c)+(b+d)) \sqrt{(a+b)(a+d)(b+c)(c+d)}}{a+b+c+d} = \sqrt{AB\cdot BC\cdot CD\cdot DA}\]and we're done!

Let $X,Y,Z,W$ be touch points of the incircle of $ABCD$ and sides $DA,AB,BC,CA$, respectively. Let $A^*,B^*,C^*,D^*$ denote the inverses of $A,B,C,D$ of an inversion around $(XYZW)$, respectively. Note that $A^*,B^*,C^*,D^*$ are the midpoints of $XY,YZ,ZW,WX$, respectively. [asy][asy] import olympiad;import geometry; size(8cm);defaultpen(fontsize(10pt)); pair A,B,C,D,X,Y,Z,W,O,A0,B0,C0,D0; X=dir(110);Y=dir(205);Z=dir(280);W=dir(10);O=(0,0); A=intersectionpoint(perpendicular(X, line(X,O) ),perpendicular(Y, line(Y,O) )); B=intersectionpoint(perpendicular(Y, line(Y,O) ),perpendicular(Z, line(Z,O) )); C=intersectionpoint(perpendicular(Z, line(Z,O) ),perpendicular(W, line(W,O) )); D=intersectionpoint(perpendicular(W, line(W,O) ),perpendicular(X, line(X,O) )); A0=midpoint(X--Y);B0=midpoint(Y--Z);C0=midpoint(Z--W);D0=midpoint(W--X); draw(circumcircle(X,Y,Z),royalblue);draw(A--B--C--D--cycle,heavyred+1);draw(A0--B0--C0--D0--cycle, fuchsia);draw(X--Y--Z--W--cycle, red+0.5); draw(A--O--B,palered+1);draw(C--O--D,palered+1); dot("$O$",O,dir(90)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$W$",W,dir(W)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$A^*$",A0,dir(A0)); dot("$B^*$",B0,dir(B0)); dot("$C^*$",C0,dir(C0)); dot("$D^*$",D0,dir(D0)); [/asy][/asy] \begin{align*} OA\cdot OC+OB\cdot OD&=\sqrt{AB\cdot BC\cdot CD\cdot DA}\Longleftrightarrow\\ \frac{r^2}{OA^*}\cdot\frac{r^2}{OC^*}+\frac{r^2}{OB^*}\cdot \frac{r^2}{OD^*}&= \sqrt{\frac{r^2\cdot A^*B^*}{OA^*\cdot OB^*}\cdot \frac{r^2\cdot B^*C^*}{OB^*\cdot OC^*}\cdot \frac{r^2\cdot C^*D^*}{OC^*\cdot OD^*}\cdot \frac{r^2\cdot D^*A^*}{OD^*\cdot OA^*}}\Longleftrightarrow\\ OA^*\cdot OC^*+OB^*\cdot OD^*&=\sqrt{A^*B^*\cdot B^*C^*\cdot C^*D^*\cdot D^*A^*}\Longleftrightarrow\\ OA^*\cdot OC^*+OB^*\cdot OD^*&=\frac{XZ\cdot YW}{4}\Longleftrightarrow\\ OA^*\cdot OC^*+OB^*\cdot OD^*&=\frac{XY\cdot ZW+YZ\cdot WX}{4}\Longleftrightarrow\\ OA^*\cdot OC^*+OB^*\cdot OD^*&=XA^*\cdot C^*Z+B^*Y\cdot D^*W\Longleftrightarrow\\ r^2(\sin{\angle YXO}\cdot \sin{\angle WZO}+\sin{\angle ZYO}\cdot \sin{\angle XWO})&=r^2(\cos{\angle YXO}\cdot \cos{\angle WZO}+\cos{\angle ZYO}\cdot \cos{\angle XWO})\Longleftrightarrow\\ \cos{(\angle YXO+\angle WZO)}&=-\cos{(\angle ZYO+\angle XWO)}, \end{align*}which is true as $\angle YXO+\angle WZO=\frac{2\pi-\angle XOY-\angle ZOW}{2}=\frac{\angle YOZ+\angle WOX}{2}=\pi-\angle ZYO-\angle XWO$. $\blacksquare$

rafaello wrote: Let $X,Y,Z,W$ be touch points of the incircle of $ABCD$ and sides $DA,AB,BC,CA$, respectively. Let $A^*,B^*,C^*,D^*$ denote the inverses of $A,B,C,D$ of an inversion around $(XYZW)$, respectively. Note that $A^*,B^*,C^*,D^*$ are the midpoints of $XY,YZ,ZW,WX$, respectively. [asy][asy] import olympiad;import geometry; size(8cm);defaultpen(fontsize(10pt)); pair A,B,C,D,X,Y,Z,W,O,A0,B0,C0,D0; X=dir(110);Y=dir(205);Z=dir(280);W=dir(10);O=(0,0); A=intersectionpoint(perpendicular(X, line(X,O) ),perpendicular(Y, line(Y,O) )); B=intersectionpoint(perpendicular(Y, line(Y,O) ),perpendicular(Z, line(Z,O) )); C=intersectionpoint(perpendicular(Z, line(Z,O) ),perpendicular(W, line(W,O) )); D=intersectionpoint(perpendicular(W, line(W,O) ),perpendicular(X, line(X,O) )); A0=midpoint(X--Y);B0=midpoint(Y--Z);C0=midpoint(Z--W);D0=midpoint(W--X); draw(circumcircle(X,Y,Z),royalblue);draw(A--B--C--D--cycle,heavyred+1);draw(A0--B0--C0--D0--cycle, fuchsia);draw(X--Y--Z--W--cycle, red+0.5); draw(A--O--B,palered+1);draw(C--O--D,palered+1); dot("$O$",O,dir(90)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$W$",W,dir(W)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$A^*$",A0,dir(A0)); dot("$B^*$",B0,dir(B0)); dot("$C^*$",C0,dir(C0)); dot("$D^*$",D0,dir(D0)); [/asy][/asy] \begin{align*} OA\cdot OC+OB\cdot OD&=\sqrt{AB\cdot BC\cdot CD\cdot DA}\Longleftrightarrow\\ \frac{r^2}{OA^*}\cdot\frac{r^2}{OC^*}+\frac{r^2}{OB^*}\cdot \frac{r^2}{OD^*}&= \sqrt{\frac{r^2\cdot A^*B^*}{OA^*\cdot OB^*}\cdot \frac{r^2\cdot B^*C^*}{OB^*\cdot OC^*}\cdot \frac{r^2\cdot C^*D^*}{OC^*\cdot OD^*}\cdot \frac{r^2\cdot D^*A^*}{OD^*\cdot OA^*}}\Longleftrightarrow\\ OA^*\cdot OC^*+OB^*\cdot OD^*&=\sqrt{A^*B^*\cdot B^*C^*\cdot C^*D^*\cdot D^*A^*}\Longleftrightarrow\\ OA^*\cdot OC^*+OB^*\cdot OD^*&=\frac{XZ\cdot YW}{4}\Longleftrightarrow\\ OA^*\cdot OC^*+OB^*\cdot OD^*&=\frac{XY\cdot ZW+YZ\cdot WX}{4}\Longleftrightarrow\\ OA^*\cdot OC^*+OB^*\cdot OD^*&=XA^*\cdot C^*Z+B^*Y\cdot D^*W\Longleftrightarrow\\ r^2(\sin{\angle YXO}\cdot \sin{\angle WZO}+\sin{\angle ZYO}\cdot \sin{\angle XWO})&=r^2(\cos{\angle YXO}\cdot \cos{\angle WZO}+\cos{\angle ZYO}\cdot \cos{\angle XWO})\Longleftrightarrow\\ \cos{(\angle YXO+\angle WZO)}&=-\cos{(\angle ZYO+\angle XWO)}, \end{align*}which is true as $\angle YXO+\angle WZO=\frac{2\pi-\angle XOY-\angle ZOW}{2}=\frac{\angle YOZ+\angle WOX}{2}=\pi-\angle ZYO-\angle XWO$. $\blacksquare$ MOST BEAUTIFUL IDEE!(AND DIAGRAME!)