In $\Delta ABC$ , $m\angle B = 45^\circ$ , $m\angle C=54^\circ$ . Point $O$ is the circumcenter. Let $\overline{OE}\perp \overline{AC} \ , \ \overline{OF}\perp \overline{AB}$ . If the ratio $\dfrac{OE}{OF}$ can be expressed as $\dfrac{a}{\sqrt{b-\sqrt{b}}}$ where $a,b \in \mathbb{Z}^+$ and $b$ is square free,

Find the value of $\displaystyle \left \lfloor\frac{1500}{a+b} \right \rfloor$ .

The answer is 214.

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Method 1:
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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}

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Method 2:
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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 $1$ . :)

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It is not true that $OE/OF = HL/HK$ ; since you are dropping the perpendiculars from $H$ , you have to do the same with $O$ . If $P$ is the foot of the perpendicular from $O$ to $AC$ , and $Q$ is the foot of the perpendicular from $O$ to $AB$ , then $OP/OQ = HL/HK$ .

Jon Haussmann
- 5 years, 9 months ago

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Sorry , my misconception. I read the page of Isogonal Conjugates from Wolfram Mathworld which did not mention the data properly. Thanks , let me edit the question :)

Nihar Mahajan
- 5 years, 9 months ago

Did you construct $H$ for the fun of it? because it is really not necessary lol

btw as you have experienced, online resources even the likes of Wolfram may contain incorrect information. For this "property" you learned, you should have tried to prove it yourself first. The topic of Isogonal Conjugates is very interesting and still considered modern and fresh. It is very probable that one could discover some unknown properties involving its configuration(It happened to me).

Xuming Liang
- 5 years, 9 months ago

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Haha , I think I did construct it for fun xD ;P .I have added method 2 also.

I have already proved this property of perpendiculars (Using cyclicity and similarity). Yeah , I got mislead due to inadequate information on Wolfram Mathworld.

Nihar Mahajan
- 5 years, 9 months ago

What are isogonal conjugates?? @Nihar Mahajan

Harsh Shrivastava
- 5 years, 9 months ago

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In
$\Delta ABC$
, cevians
$\overrightarrow{AP_1} \ and \ \overrightarrow{AP_2}$
are said to be
**
isogonal conjugates
**
of each other
**
if and only if
**
bisector of
$\angle A$
also bisects angle between them.

Nihar Mahajan
- 5 years, 9 months ago

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@Harsh Shrivastava – Actually I am not much concentrating on NTSE bcoz its much based on school sylabi. I am concentrating on RMO , JEE , NSEA.

Nihar Mahajan
- 5 years, 9 months ago

@Harsh Shrivastava – How's your preparation going on?

Nihar Mahajan
- 5 years, 9 months ago

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@Nihar Mahajan – I am also not focussing on NTSE... :P

Harsh Shrivastava
- 5 years, 9 months ago

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Hey @Nihar Mahajan

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

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Yes I am 14 years old. I am studying in Std. 10. What about you?

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Nihar Mahajan
- 5 years, 9 months ago

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So awesome in geometry at the age of 14;You are a Genius!

Pulkit Deshmukh
- 5 years, 9 months ago

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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 $\frac { OE }{ OF } =\frac { \sin { 45 } }{ \sin { 36 } } =\frac { \frac { \sqrt { 2 } }{ 2 } }{ \frac { \sqrt { 10-2\sqrt { 5 } } }{ 4 } } =\frac { 2 }{ \sqrt { 5-\sqrt { 5 } } }$ .

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