Looking for non-trig solution

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Absur Khan Siam
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Location: Bashaboo , Dhaka

Looking for non-trig solution

Unread post by Absur Khan Siam » Mon Feb 20, 2017 12:19 pm

In a circle , $AB = 4$ is the diameter and $O$ is the centre.On the diameter that makes an angle of $30^{\circ}$ with $AB$ at the centre, two points $C$ and $D$ are chosen that $OC = OD$ and $\angle BCO = 90^{\circ}$.The perpendicular on $AB$ through $O$ meets $AC$ at $E$ and $BD$ at $F$.If $EF = \frac{a\sqrt{b}}{c}$ where $a,b,c$
are integers and $b,c$ are primes, find the value of $a + b + c$.
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

Absur Khan Siam
Posts: 61
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

Re: Looking for non-trig solution

Unread post by Absur Khan Siam » Mon Feb 20, 2017 12:38 pm

$AB = 4 \Rightarrow OA = OB = 2$.
$BC = OB\sin30 = 1$ , $OC = \sqrt{2^2 - 1^2} = \sqrt{3}$ , $CD = 2\sqrt{3}$ and $BD = \sqrt{13}$

Let , $\angle EAO = \alpha$
From the sine rule ,
$\frac{AC}{\sin AOC} = \frac{OC}{\sin\alpha} \Rightarrow \sin\alpha = \frac{\sqrt{3}}{2\sqrt{13}}$

From the cosine rule,
$OC^2 = AC^2 + OA^2 - 2AC.OA.\cos\alpha \Rightarrow \cos\alpha = \frac{7}{2\sqrt{13}}$

Thus,$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{\sqrt{3}}{7}$

Thus,$OE = OA \times \tan\alpha = \frac{2\sqrt{3}}{7}$

Similarly , $OF = \frac{2\sqrt{3}}{7}$

$EF = OE + OF = \frac{4\sqrt{3}}{7}$

$\therefore a + b + c = 14$
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

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Thanic Nur Samin
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Re: Looking for non-trig solution

Unread post by Thanic Nur Samin » Mon Feb 20, 2017 2:16 pm

Here is a solution that doesn't use trig.

Let $AB$ be the $x$-axis and the center $O$ the origin. So, $B$ is $(2,0)$ and $A$ is $(-2,0)$. Reflect $C$ through the $x$-axis. Since $\angle COG=30^{\circ}$, $\angle COC'=60^{\circ}$, and since $OC=OC'$, it $\triangle COC'$ is equilateral. If we let $G$ be the foot of perpendicular from $C$ to $x$-axis, then $OC=2CG$, and $OG^2=OC^2-CG^2\Rightarrow OG^2=4CG^2-CG^2=3CG^2\Rightarrow OG=\sqrt{3}CG$. So the equation of $OC$ is $y=x/\sqrt{3}\Rightarrow x-\sqrt{3}y=0$.

However, $BC$ is perpendicular to $OC$, so for some real $k$, the equation of $BC$ is $\sqrt{3}x+y+k=0$. Since $B\equiv (2,0)$, we get $k=-2\sqrt{3}$. So the equation is $\sqrt{3}x+y-2\sqrt{3}=0$. Solving, the coordinates of $C$ is $(3/2,\sqrt{3}/2)$.

Now, $C\equiv (3/2,\sqrt{3}/2)$ and $A\equiv (-2,0)$. $E$ is on the $y$ axis, and so $E\equiv (0,t)$. Also, $E,C,A$ are collinear. Therefore, from the collinearity criteria, we get $-2(t-\sqrt{3}/2)+3/2(-t)=0\Rightarrow t=2\sqrt{3}/7$.

This implies $OE=\dfrac{2\sqrt{3}}{7}$, and due to symmetry $OF=\dfrac{2\sqrt{3}}{7}$. So $EF=OE+OF=\dfrac{4\sqrt{3}}{7}$. So the answer is $4+3+7=14$.
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Absur Khan Siam
Posts: 61
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

Re: Looking for non-trig solution

Unread post by Absur Khan Siam » Mon Feb 20, 2017 2:46 pm

Thanks to @Thanic nur samin .Can anyone give a synthetic solution?
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

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