Trig kills this

TheDarkPrince wrote: Trig kills this Indeed. Short chase gives us, \[OB_i = \frac{\cos A_i}{\cos (A_{i + 1} - A_{i - 1})}.\] Using this we have $OB_1 \ge OB_2 \ge OB_3 \iff A_1 \le A_2 \le A_3$. WLOG assume this condition to be true. Now $OB_1$ will be minimum when $\cos A_1$ is minimum and $\cos (A_2 - A_3)$ is maximum. \[\max\{\cos(A_2 - A_3)\} = 1\text{ when }A_2 = A_3.\] Note that $A_1 \le \frac{A_1 + A_2 + A_3}{3} = 60^{\circ}$. Since, $\cos$ is decreasing in the interval $[0, \tfrac12\pi]$ \[\min_{0 < A_1 \le 60} \{ \cos A_1 \} = \frac{1}{2} \text{ when } A_1 = 60^{\circ}.\] Thus, $\min\{\max_{1 \le i \le 3} \{OB_i\}\} = \frac{1}{2}$ achieved when $A_i = 60^{\circ}$ and $A_{i - 1} = A_{i + 1}$. ~~~~~ Really messed this part up ~~~~ ayan.nmath wrote: Am I correct? Indeed

@above thnx Vrangr wrote: Let $A_1A_2A_3$ be an acute-angled triangle inscribed into a unit circle centered at $O$. The cevians from $A_i$ passing through $O$ meet the opposite sides at points $B_i$ $(i = 1, 2, 3)$ respectively. Find the minimal possible length of the longest of three segments $B_iO$. Find the maximal possible length of the shortest of three segments $B_iO$. Solution . The answer to both the parts is $\boxed{\tfrac12}.$ Without loss of generality assume $OB_1\ge OB_2\ge OB_3.$ Using Gergonne-Euler Theorem we get, \begin{align*} &\tfrac{OB_1}{1+OB_1}+\tfrac{OB_2}{1+OB_2}+\tfrac{OB_3}{1+OB_3}=1 \implies \sum_{i=1}^{3}\left(1-\tfrac{1}{1+OB_i}\right)=1 \implies \sum_{i=1}^3\tfrac{1}{1+OB_i}=2\end{align*}Now, \[2=\sum_{i=1}^3\tfrac{1}{1+OB_i}\ge \tfrac{3}{1+OB_1}\implies 1+OB_1\ge \tfrac32\implies OB_1\ge \tfrac12.\]Also, \[2=\sum_{i=1}^3\tfrac{1}{1+OB_i}\le \tfrac{3}{1+OB_3}\implies 1+OB_3\le \tfrac32\implies OB_3\le \tfrac12.\]It can be easily seen that one case of equality is when $\triangle A_1A_2A_3$ is equilateral. So we are done. $\blacksquare$

Even I got $\frac{1}{2}$ and $\frac{1}{2}$ for both.

Attachments:
Problem 12.pdf (134kb)

Oh, wow $\cos(60^{\circ}) = \tfrac12$, I feel so... I should have proofread my solutions.

No trig!

Attachments:
Sharygin_P12_Pranjal.pdf (411kb)

Problem 3.18 Evan Chen. Open and close. But dunno why I proved it. My god, ayan and I keep thinking the same. Magic!

Attachments:
L.pdf (144kb)

AREA CHASING!!!

Attachments:
P12.pdf (382kb)