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Zhautykov Olympiad 2015 problem 6 This links discusses the olympiad problem which none of students could solve , meaning it is very hard.

Question:

The area of a convex pentagon $ABCDE$ is $S$, and the circumradii of the triangles $ABC$, $BCD$, $CDE$, $DEA$, $EAB$ are $R_1$, $R_2$, $R_3$, $R_4$, $R_5$. Prove the inequality $$R_1^4+R_2^4+R_3^4+R_4^4+R_5^4\geq {4\over 5\sin^2 108^\circ}S^2$$

A month ago,I tried to solve this problem,because I know the following Möbius theorem(1880):

Theorem: Let $a,b,c,d,e$ are area of $\Delta ABC,\Delta BCD,\Delta CDE,\Delta DEA,\Delta EAB$,then we have $$S^2-S(a+b+c+d+e)+ab+bc+cd+de+ea=0$$ This is a result by Gauss 1880,you can see this paper (for proof):(Gauss Carl Fridrich Gauss Werke Vol 4.2ter Abdruck,1880:406-407)

Using this Theorem ,in 2002 ,Chen Ji and Xiongbin proved this result:

$$a^2+b^2+c^2+d^2+e^2\ge\dfrac{20S^2}{(5+\sqrt{5})^2}$$

I use the well known result: $$a\le\dfrac{3\sqrt{3}}{4}R^2_{1},b\le\dfrac{3\sqrt{3}}{4}R^2_{2},c\le\dfrac{3\sqrt{3}}{4}R^2_{3},d\le\dfrac{3\sqrt{3}}{4}R^2_{4},e\le\dfrac{3\sqrt{3}}{4}R^2_{5}$$ so we have $$R_1^4+R_2^4+R_3^4+R_4^4+R_5^4\ge\dfrac{16}{27}(a^2+b^2+c^2+d^2+e^2) \ge\dfrac{16\cdot 20}{27(5+\sqrt{5})^2}S^2$$

But I found $$\dfrac{16\cdot 20}{27(5+\sqrt{5})^2}<\dfrac{32}{5(5+\sqrt{5})}={4\over 5\sin^2 108^\circ}$$ so my work can't solve this contest problem. If we can solve following question,then the Olympiad problem can be solved:

Question 2:

$$a^2+b^2+c^2+d^2+e^2\ge\dfrac{54}{5(5+\sqrt{5})}S^2$$

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  • $\begingroup$ Hey , nice question! I've edited the question , I hope you don't mind. Can anyone check that my interpretation of the last sentence is correct? $\endgroup$ – A Googler Apr 27 '15 at 17:24
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Dan Schwarz (one of the major problem proposers for EGMO, RMM, Balkan...) has posted a solution at here. I'll briefly sketch it here.

First, one takes the midpoints $M_1$, ..., $M_5$ of $A_1A_2$, ..., $A_5A_1$. Then at each angle $A_i$, one takes the circumcircle of triangle $M_{i-1}A_iM_{i+1}$ (which has radius $\frac 12 R_i$) and the point diametrically opposite $A_i$ on it, say $X_i$. This gives a quadrilateral $A_iA_{i-1}X_iA_{i+1}$, with area at most $(\frac 12 R_i)^2 \cdot 2\sin A_i$. Consider all five of these quadrilaterals, as shown.

Pentagon covering

It's not too hard to show that these five quadrilaterals cover the entire pentagon (look at perpendicular bisectors). So summing gives $$ S \le \frac{1}{2} \sum_i R_i^2 \sin A_i. $$ Then, by the Cauchy-Schwarz Inequality, we have $$ 4S^2 \le \left( \sum_i R_i^2 \sin A_i \right)^2 \le \left( \sum_i R_i^4 \right) \left( \sum_i \sin^2 A_i \right). $$ So it remains to show $$\sum_i \sin^2 A_i \le 5\sin^2 (108^{\circ})$$ reducing this to a purely algebraic problem (with $0^{\circ} \le A_i \le 180^{\circ}$ and $\sum A_i = 540^{\circ}$).

It's tempting to try and apply, say, Jensen's Inequality, but the function $\sin^2 x$ has a few inflection points, so one has to proceed more delicately using a fudging argument (along the lines of the $n-1$ equal-value trick). Indeed, we do this by showing that if $A_1 \le A_2 \le A_3 \le A_4 \le A_5$ and $A_1 < 108^{\circ}$, then we can replace $A_1$ with $108^{\circ}$ and $A_5$ by $A_1+A_5-108^{\circ}$ while increasing $\sum_i \sin^2 A_i$. Repeating this process until all $A_i$ are equal completes the proof.

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  • $\begingroup$ Nice.Thank you very much!+1 $\endgroup$ – math110 Apr 29 '15 at 6:12
  • $\begingroup$ and today I have found you links this PDF can't open it in china $\endgroup$ – math110 Apr 30 '15 at 2:21

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