1500 followers problem! For my friend Xuming!

We first construct segments $\overline { OA } ,\overline { OB } ,$ and $\overline { OC }$ . Since O is the circumcenter, we know that these three segments have the same length, and triangles $AOB, AOC,$ and $BOC$ are isosceles. Thus, $E$ and $F$ are bisectors of sides $AC$ and $AB$ , respectively.

We then assign the measure of $\angle BAO$ as $\alpha$ , and from here we angle-chase a little bit until we reach $\angle ABO=72-\alpha$ . Since $\angle ABO=\angle BAO=\alpha =72-\alpha$ , $\alpha =36$ . Thus, $\angle FAO=\alpha =36$ , and $\angle OAE=81-\alpha =45$ . We then have $\overline { OF } =\overline { OA } \times \sin { 36 }$ , and $\overline { OE } =\overline { OA } \times \sin { 45 }$ , so $\frac { OE }{ OF } =\frac { \sin { 45 } }{ \sin { 36 } } =\frac { \frac { \sqrt { 2 } }{ 2 } }{ \frac { \sqrt { 10-2\sqrt { 5 } } }{ 4 } } =\frac { 2 }{ \sqrt { 5-\sqrt { 5 } } }$ .

$\left\lfloor \frac { 1500 }{ 2+5 } \right\rfloor =\boxed { 214 }$

Method 1:

Let $H$ be orthocenter and $HL \perp AB \ , \ HK \perp AC$ . Note that $H$ and $O$ are the isogonal conjugates of each other. By one of the properties of isogonal conjugates ,

\begin{aligned}{\dfrac{OE}{OF} = \dfrac{HL}{HK} \\ = \dfrac{2R\cos A \cos B}{2R\cos A \cos C} \\ = \dfrac{\cos B}{\cos C} \\ = \dfrac{\cos 45^\circ}{\cos 54^\circ} \\ = \dfrac{2}{\sqrt{5-\sqrt{5}}}}\end{aligned} \\ \Rightarrow a+b=2+5=7 \Rightarrow \left \lfloor \dfrac{1500}{7} \right \rfloor = \lfloor 214.28 \rfloor = \boxed{214}

Method 2:

We know that $OE = \dfrac{1}{2}(BH) = \dfrac{2R\cos B}{2} \ , \ OF = \dfrac{1}{2}(CH) = \dfrac{2R \cos C}{2}$

Clearly , it follows that $\dfrac{OE}{OF} = \dfrac{\cos B}{\cos C}$ and compute the required quantities same as in method $1$ . :)

$Note ~that ~AF=FB,~~AE=EC,~~AO=R=\dfrac{AF}{Sin54},~~\\ Sin54=\dfrac{1+\sqrt5} 4,~~Sin45=\dfrac 1 {\sqrt2},\\ by~Sin~Law~AE=\dfrac{\sqrt2}{\frac{1+\sqrt5} 4}*AF=\dfrac{4\sqrt2}{1+\sqrt5}*AF\\ \therefore~ AE^2=\dfrac 4 {3+\sqrt5}*AF^2\\$ $AO^2=\dfrac 8 {3+\sqrt5}*AF^2. \\ \therefore \dfrac{OE}{OF}= \sqrt{ \dfrac { AO^2 - AE^2 } {AO^2-AF^2} }\\ =\sqrt{ \left \{ \dfrac{ \frac 8 {3+\sqrt5 } -\frac 4 {3+\sqrt5} } { \dfrac 8 {3+\sqrt5 }- 1 } \right\} }\\ =\sqrt{\left \{\dfrac { 8 - 4}{ 8 - (3+\sqrt5 ) } \right\} } \\ =\dfrac 2 {\sqrt{5 - \sqrt5}}=\dfrac a {\sqrt{b-\sqrt b}}.\\ \therefore \left \lfloor \dfrac{1500}{2+5} \right \rfloor \\ =\Large \color{#D61F06}{214}$

Hey @Nihar Mahajan

Which standard you are actually in,Are you 14 years old???