I know someone who decided that this Olympiad, he a friend of mine. Here it is sufficient to prove that the quadrilateral ABCD is tangential.

@samariddin: but that is already given (the quadrilateral 4 is tangential), hence you must mean the outer one!! Best regards, sunken rock

Does anyone have a complete solution? If yes, please, post it.

The most interesting (from my point of view) is the following solution. Lemma 1 Given the triangle $ABC$. Each of two circles $w_1,w_2$ touches the sides $AB,AC$ and intersects the side $BC$. If arcs of these circles located inside the triangle have equal angle measures, then $w_1, w_2$ are the same. Proof. $w_1$ and $w_2$ are homotetic with center $A$ and WLOG $k>1$ and both $D,F$ lie closer to $A$ than $E,G$. Thus, $\stackrel{\frown}{DF}=\stackrel{\frown}{EG},\stackrel{\frown}{DH}=\stackrel{\frown}{EI},\stackrel{\frown}{FJ}=\stackrel{\frown}{GK}$. Since $k>1 \quad I,K$ lie outside of a triangle and $\stackrel{\frown}{LEM} <\stackrel{\frown}{HDJ}$. A contradiction. Lemma 2 Given a convex quadrilateral $ABCD$. Two circles $y_1,y_2$ are tangent to the sides $AB,AD$ and $CB,CD$ correspondingly. $ABCD$ is tangential iff arcs of the circles that are lying in the same half-plane with respect to $AC$ have equal angle measures. Proof. 1) Build two new circles $y_3$ homotetic to $y_1$ and tangent to $BC$ and $y_4$ homotetic to $y_2$ and tangent to $AB$ (it is always possible). If $ABCD$ is tangential $y_3$ and $y_4$ are the same and it's homotetic images $y_1$ and $y_2$ have equal arcs in each half-plane. 2) The other way, if $y_1,y_2$ have equal arcs, then $y_3,y_4$ have equal arcs inside triangle $ABC$. Lemma 1 tells us $y_3,y_4$ must be the same and $ABCD$ is tangential. Now to the main problem. 1) Circle $z_4$ (see figure) is homotetic to $z_3$ (with center $E$), thus upper arc of $z_4$ (with respect to $AC$) equals lower arc of $z_3$. From the same argument lower arc of $z_3$ equals upper arc of $z_2$. From Lemma 2 we can deduce that $ABCD$ is tangential. 2) Using Lemma 2 again we get upper arc of $z_1$ (with respect to $BD$) equals upper arc of $z_6$. But from homotetic transformations upper arc of $z_1$ equals lower arc of $z_3$ equals upper arc of $z_5$. Corresponding arcs of $z_5$ and $z_6$ are equal $\Rightarrow$ quadrilateral 5 is tangential. P.S. This solution was found by Igor Voronovich during the jury discussion. It's different from official solution. Also several interesting solutions were found by the participants during competition.

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This problem easily follows from the following lemma: a convex quadrilateral $ ABCD $ has an incircle if and only if $ \tan{\frac{\angle ABD}{2}}\cdot\tan{\frac{\angle BDC}{2}}=\tan{\frac{\angle ADB}{2}}\cdot\tan{\frac{\angle DBC}{2}} $ Official solution was the same.

Yeah! dizzy I have solved it by using this lemma. We obtain from "quadrilaterals $2,3,4$" that quadrilateral $ABCD$ has an inscribed circle. And If quadrilaterals $ABCD$, $1$ and $3$ have inscribed circles, then we obtain quadrilateral $5$ also has an inscribed circle!

dizzy wrote: This problem easily follows from the following lemma: a convex quadrilateral $ ABCD $ has an incircle if and only if $ \tan{\frac{\angle ABD}{2}}\cdot\tan{\frac{\angle BDC}{2}}=\tan{\frac{\angle ADB}{2}}\cdot\tan{\frac{\angle DBC}{2}} $ Official solution was the same. It is actually the Iosifescu's theorem

Is it possible to solve the problem of applying 2 nd variant monges theorem?

Dear Mathlinkers, you can see also http://perso.orange.fr/jl.ayme vol. 20 Equal incircles theorem p. 18-20 Sincerely Jean-Louis

imomyrat wrote: Is it possible to solve the problem of applying 2 nd variant monges theorem? Yes, we can solve this problem by apply Monge d'Alembert theorem several times. First we prove that the largest quadrilateral is tangential. We will show that $AI_1,BI_2,DI_4$ are concurrent. By applying the Monge d'Alembert theorem one can prove that if $X$ is the intersection of two diagonals then $AI_1, BI_2, DI_4$ concurrent on $XI_3$. Next, we will prove that the insimilicenter of $(I)$ and the circle with the circle lies on $I_3C_1$ that touches $BC, DC$ is actually $C_1$. So we have quadrilateral 5 is tangential. Sorry for vague notations, but I'm just bloody lazy now.

How to prove this: $ \tan{\frac{\angle ABD}{2}}\cdot\tan{\frac{\angle BDC}{2}}=\tan{\frac{\angle ADB}{2}}\cdot\tan{\frac{\angle DBC}{2}} $

How to prove this: $ \tan{\frac{\angle ABD}{2}}\cdot\tan{\frac{\angle BDC}{2}}=\tan{\frac{\angle ADB}{2}}\cdot\tan{\frac{\angle DBC}{2}} $, if ABCD has incircle?

Abubakir wrote: How to prove this: $ \tan{\frac{\angle ABD}{2}}\cdot\tan{\frac{\angle BDC}{2}}=\tan{\frac{\angle ADB}{2}}\cdot\tan{\frac{\angle DBC}{2}} $, if ABCD has incircle? Proof: Let $AB=a,\ BC=b,\ CD=c,\ DA=d,\ AC=e,\ BD=f.$ Let $r_1$ and $r_2$ are the radii of the incircles of $\triangle ABD$ and $\triangle CBD$ respectively. Then, $tg\frac{\angle ABD}{2}=\frac{2r_1}{a+f-d},\ tg\frac{\angle CDB}{2}=\frac{2r_2}{c+f-b},$ $tg\frac{\angle CBD}{2}=\frac{2r_2}{b+f-c},\ tg\frac{\angle ADB}{2}=\frac{2r_1}{d+f-a}.$ So, $tg\frac{\angle ABD}{2}tg\frac{\angle CDB}{2}=tg\frac{\angle CBD}{2}tg\frac{\angle ADB}{2}\Leftrightarrow$ $\Leftrightarrow (a+f-d)(c+f-b)=(b+f-c)(d+f-a)\Leftrightarrow$ $\Leftrightarrow 2af+2cf=2bf+2df\Leftrightarrow$ $\Leftrightarrow a+c=b+d.$ $Q.E.D.$

I didn't understand this problem was the same as the problem in PRASOLOV(Geometry book). This book was printed before than IZHO 2014. There was 1,2,3,4 and 5 quadrilaterals are inscribed circles and it have to been proven that given the greater quadrilateral also inscribed circle. !!!

After several years I learned Monge's theorem, and now I can post it here . It is very sad story, that I did not solve it during contest Nevertheless, this is my solution. Call vertices of greatest quadlirateral ABCD, such that A is the vertice of 4th quadlirateral too, B is the vertice of 1st qualdirateral, C is the vertice of 2nd quadlirateral. By Monge's theorem for 2,3 and 4 we have that exsimilicenter of 2 and 4 lies on diagonal AC. Then consider circle(call it 6th circle) that touches sides AB, BC and AD. Then by Monge's theorem for 2,4, and 6th we have that C is the exsimilicenter of 2 and 6, since A is the exsimilicenter of 4 and 6, the exsimilicenter of 2 and 4 lies on AC, so the exsimilicenter of 2 and 6 lies on AC, on the other hand we have that BC is common ex-tangent line of 2 and 6. It means that CD also tangent to 6th circle. Similiarly using that 1st, 3rd and 6th, and circle tangent to 3 sides of 5th circle(without CD), we can prove that it is also tangent to CD. imomyrat wrote: Is it possible to solve the problem of applying 2 nd variant monges theorem?

Abubakir wrote: After several years I learned Monge's theorem, and now I can post it here . It is very sad story, that I did not solve it during contest Nevertheless, this is my solution. Call vertices of greatest quadlirateral ABCD, such that A is the vertice of 4th quadlirateral too, B is the vertice of 1st qualdirateral, C is the vertice of 2nd quadlirateral. By Monge's theorem for 2,3 and 4 we have that exsimilicenter of 2 and 4 lies on diagonal AC. Then consider circle(call it 6th circle) that touches sides AB, BC and AD. Then by Monge's theorem for 2,4, and 6th we have that C is the exsimilicenter of 2 and 6, since A is the exsimilicenter of 4 and 6, the exsimilicenter of 2 and 4 lies on AC, so the exsimilicenter of 2 and 6 lies on AC, on the other hand we have that BC is common ex-tangent line of 2 and 6. It means that CD also tangent to 6th circle. Similiarly using that 1st, 3rd and 6th, and circle tangent to 3 sides of 5th circle(without CD), we can prove that it is also tangent to CD. imomyrat wrote: Is it possible to solve the problem of applying 2 nd variant monges theorem? After here assume that circle 7 is a circle that tangent to $DA,DC$ and 1 other side of quadrilateral 5. Then Monge's theorem on circles 1,6,7 and circles 1,3,7 finishes the problem.