BdMO(Regional) 2012 Problem

BdMO

In $\triangle BOC, \sin OBC = \frac{OC}{OB} \rightarrow \sin 60^\circ = \frac{OC}{2} \rightarrow OC = \sqrt{3}$

Draw a perpendicular $CM$ from point $C$ on $AB$ .

In $\triangle OCM,\sin COM = \frac{CM}{OC} \rightarrow \sin 30^\circ = \frac{CM}{\sqrt{3}} \rightarrow CM = \frac{\sqrt{3}}{2}$

$\cos COM = \frac{OM}{OC} \rightarrow \cos 30^\circ = \frac{OM}{\sqrt{3}} \rightarrow OM = \frac{3}{2}$

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In $\triangle ACM, AC^2 = AM^2 + CM^2 = (OA+OM)^2 + CM^2 = (2+\frac{3}{2})^2 + (\frac{\sqrt{3}}{2})^2 = \frac{49}{4} + \frac{3}{4} = 13$

$AC = \sqrt{13}$

$$

In $\triangle AOC, \frac{\sin x}{\sin AOC} = \frac{OC}{AC} \rightarrow \sin x = \frac{\sqrt{3} \times \sin 150^\circ}{\sqrt{13}} \rightarrow \sin x = \frac{\sqrt{3}}{2\sqrt{13}}$

$\cos x = \sqrt{1-sin^{2}x} = \sqrt{1 - \frac{3}{52}} = \sqrt{\frac{49}{52}} = \frac{7}{2\sqrt{13}}$

$\tan x = \frac{\sin x}{\cos x} = \frac{\sqrt{3}}{7}$

$\tan x = \frac{OE}{OA} \rightarrow OE = OA\tan x \rightarrow OE = 2\frac{\sqrt{3}}{7}$

$EF = 2OE = 4\frac{\sqrt{3}}{7}$

$4+3+7=\boxed{14}$

(Please consider Fahin points names above to follow my solution)

Triangle $BCO$ is a "half-equilateral-triangle" with "side" $BO = 2$ (radius). So, $BC = 1$ ("half-side") and $CO = \sqrt { 3 }$ .

Because $CM$ is a height of right triangle $BOC$ , we know $BO\cdot CM=CB\cdot CO$ and then we get $CM = \frac { \sqrt { 3 } }{ 2 }$ . Applying Pythagorean Theorem at triangle $COM$ , we get $OM = \frac { 3 }{ 2 }$ . Thus, $AM = \frac { 7 }{ 2 }$ .

Triangles AEO and ACM are similar, so $\frac { EO }{ CM } =\frac { AO }{ AM }$ and we get $EO = \frac { 2\sqrt { 3 } }{ 7 }$ . Since the symmetry of given construction, $EF = 2EO$ , therefore, $EF = \frac { 4\sqrt { 3 } }{ 7 }$ and the answer is $4 + 3 + 7 = \boxed{14}$ .

Try my geometry classic question Find the area . I'm sure u will enjoy this

For convenience, I would also like to borrow Shakkhor's diagram.

If $\angle COB = 30^\circ$ and $\angle OCB = 90^\circ$ , then $\triangle COB$ is a $30^\circ - 60^\circ - 90^\circ$ triangle with $\overline{OB}$ as its hypotenuse and $\overline{CB}$ as the short leg. $30^\circ - 60^\circ - 90^\circ$ triangles are a special kind of right triangle in that the measurement of the short leg is half of that of the hypotenuse and the measure of the longer leg is that of the short leg multiplied by $\sqrt{3}$ . Since $\overline{OB}$ is a radius of $\bigcirc O$ which has a diameter of $4$ , $\overline{OB} = 2$ , and since $\overline{OB}$ is the hypotenuse of $\triangle COB$ , $m\overline{CB} = \frac{m\overline{OB}}{2} = \frac{2}{2} = 1$ and $m\overline{OC} = m\overline{CB} \times \sqrt{3} = (1)(\sqrt{3}) = \sqrt{3}$ .

Let us construct point $M$ such that $\overline{OB} \perp \overline{CM}$ . Now, $\triangle COM$ is also a $30^\circ - 60^\circ - 90^\circ$ triangle with $\overline{OC}$ as the hypotenuse and $\overline{CM}$ as the short leg. Since $m\overline{OC} = \sqrt{3}$ , $m\overline{CM} = \frac{m\overline{OC}}{2} = \frac{\sqrt{3}}{2}$ and $m\overline{OM} = m\overline{CM} \times \sqrt{3} = (\frac{\sqrt{3}}{2})(\sqrt{3}) = \frac{(\sqrt{3})^2}{2} = \frac{3}{2}$ .

Consider triangles $\triangle AMC$ and $\triangle AOE$ . Since $\angle A \cong \angle A$ and $\angle AMC \cong \angle AOE$ , $\triangle AMC$ and $\triangle AOE$ are similar triangles. We can assume that the ratios of $\overline{AM}$ to $\overline{MC}$ and $\overline{AO}$ to $\overline{OE}$ are the same as they are corresponding sides of similar triangles. $\overline{AO}$ is another radius of $\bigcirc O$ so $\overline{AO} = \overline{OB} = 2$ . From the above, we can do the following:

$\begin{aligned} \frac{m\overline{MC}}{m\overline{AM}} &= \frac{m\overline{OE}}{m\overline{AO}} \\ \frac{m\overline{MC}}{(m\overline{AO} + m\overline{OM})} &= \frac{m\overline{OE}}{m\overline{AO}} \\ \frac{\frac{\sqrt{3}}{2}}{2 + \frac{3}{2}} &= \frac{m\overline{OE}}{2} \\ \frac{\frac{\sqrt{3}}{2}}{\frac{4}{2} + \frac{3}{2}} &= \frac{m\overline{OE}}{2} \\ \frac{\frac{\sqrt{3}}{2}}{\frac{7}{2}} &= \frac{m\overline{OE}}{2} \\ (\frac{\sqrt{3}}{2})(\frac{2}{7}) &= \frac{m\overline{OE}}{2} \\ \frac{ \sqrt{3}}{7} &= \frac{m\overline{OE}}{2} \\ \frac{2\sqrt{3}}{7} &= m\overline{OE} \\ \end{aligned}$

We now know how long $\overline{OE}$ is. But to find the length of $\overline{EF}$ , we have to find first $m \overline{OF}$ . Looking at the diagram, we can get a hunch that $\overline{OE} \cong \overline{OF}$ . Consider triangles $\triangle AOC$ and $\triangle BOD$ . $\angle AOC$ and $\angle BOD$ are vertical angles formed by the intersection of $\overline{AB}$ and $\overline{CD}$ and thus $\angle AOC \cong \angle BOD$ . Also, we already know that $\overline{CO} \cong \overline{DO}$ and $\overline{AO} \cong \overline{BO}$ . By that, $\triangle AOC$ and $\triangle BOD$ are congruent (by $SAS$ congruence) and $\angle ACO \cong \angle BDO$ . Further, consider triangles $\triangle EOC$ and $\triangle FOD$ . Since the sets of points $C$ , $O$ , and $D$ , and $E$ , $O$ , $F$ are collinear, then $\angle EOC$ and $\angle FOD$ are vertical angles and are congruent to each other. It is already known that $\overline{CO} \cong \overline{DO}$ . Finally, $\angle ACO \cong \angle ECO$ , $\angle BDO \cong \angle FDO$ and $\angle ACO \cong \angle FDO$ . Thus, $\triangle EOC$ and $\triangle FOD$ are congruent (by $ASA$ congruence) and $\overline{EO} \cong \overline{FO}$ .

Thus, $m\overline{EF} = m\overline{EO} + m\overline{OF} = m\overline{EO} + m\overline{EO} = \frac{2 \sqrt{3}}{7} + \frac{2 \sqrt{3}}{7} = \frac{4 \sqrt{3}}{7}$ . Finally, $a = 4$ , $b = 3$ , $c = 7$ and $a + b + c = 4 + 3 + 7 = \boxed{14}$ .

EF=(4√3)/7 a + b + c = 4 + 3 + 7 = 14

1. Join D to A. Since, ∠COB = ∠AOD with DO = OC and AO = OB; AD = CB and AC = DB.
2. Thus, △AOD and △COB are congruent triangles. Also, △AOC and △DOB are congruent. So, ∠ODA = ∠BCO = 90° or AD||BC. Also, ∠CAO = ∠OBD. This indicates that AC and BD are parallel. Hence, ACBD is a parallelogram. Thus, AD=BC and AC=BD. From ∟△AOD, ∠OAD = 60° (Since, ∠AOD = ∠COB = 30°). In a 30°:60°:90° triangle, the sides are in the ratio 1:2:√3. Given AB=4, AO=OB=2. Also given, OC=OD = √3. Thus, AD = CB = 1.
3. Now, drop a perpendicular from D to AB say DX. From ∟△ADX, cos30°= DX/1 or DX = (√3)/2. By Pythagorean theorem, AX = 1/2. Thus, XB = 3.5 or 7/2.
4. From ∟△DBX, by Pythagorean theorem, DB = √((7/2)^2+((√3)/2)^2) = √13.
5. Since, OF||XD, by AAA theorem, ∟△BOF and ∟△BXD are similar triangles. Thus, BO/BX = BF/BD = OF/XD or 2/3.5 = BF/√13 = OF/(√3/2). This gives, OF = (2√3)/7.
6. Since ACBD is a parallelogram, with points E, O and F being collinear, OE=OF.
7. Hence, EF = 2OE = 2OF = 2((2√3)/7) = (4√3)/7 = (a√b)/c. a, b, c are integers and b, c are primes.
8. Or, a+b+c = 14.

EF=(4√3)/7 a + b + c = 4 + 3 + 7 = 14

\boxed{14}

Good Question :)

Easy solution: Find CO through pitagoras in COB then find AC through cossine law in AOC. Call the area from EOC as A1 and the area of EOA as A2. A1 is the product of EOxOCxsin60º/2, and that makes 3/4xY, beeing Y the half of EF. A2 is AOxY/2, and that makes Y. We know now that the area of the triangle AOC is 7xY/4, but this same area can be found by multiplying AOxOCxsin(150)/2, and that makes sqrt(3)/2. So, we know that 7xY/4 is equal to sqrt(3)/2. Doing the math we find that Y = 2xsqrt(3)/7, and EF is 2xY. So EF is 4xsqrt(3)/7. a = 4, b = 3 and c = 7. The sum makes 14.