In the figure above (not drawn to scale), the triangle ABC is inscribe

Question

https://gmatclub.com/forum/download/file.php?id=27294 In the figure above (not drawn to scale), the triangle ABC is inscribed in the circle with the center O and the diameter AB. AC is 4 inches shorter than AB. The ratio of the angle COB’s measure to the measure of angle OBC is 2:1. What is the length of OB? A. 2√2 B. 2√3 C. 4 D. 4 + 2√2 E. 4 + 2√3 Kudos for a correct solution . circle.gif

ENGRTOMBA2018 · Answer

\angle{COB}=2\angle{OBC}  ----> In triangle OBC,  \angle{obc}+\angle{ocb}+\angle{cob} = 180  ---> x+x+2x = 180 ---> x=45 and  \angle{cob} = {90}^\circ

Similarly, in Triangle AOC,  \angle{OCA} = \angle{OAC} = {45}^\circ

Let R be the radius of the circle and thus AC = 2R-4 (given) and OC = AC= R

We can proceed forward in 2 ways:

1. Trigonometric way (not recommended for GMAT)

In triangle ACO, AO / AC =cos (45)---> \frac{R}{2R-4} = \frac{1}{\sqrt{2}}  --->  R = 4+2\sqrt{2} , D is the correct answer.

2. Simpler and straighforward:

After you calculate AC=2R-4, in right triangle AOC (right angled at  \angle{AOC} ), apply the Pythagorean rule (Hyp^2 = Height ^2 + Base ^2) to get the following equation:

(2R-4)^2 = R^2+R^2 ---> R = 4+2\sqrt{2} , D is the correct answer.

Alternately, for this method after you calculate AC = 2R-4, in a 45-45-90 degree right triangle , the sides are in the ratio  1:1:\sqrt{2}----> R\sqrt{2} = 2R-4 ---> R =4+2\sqrt{2}

kelvind13 · Answer

Sorry if this sounds like a stupid question although I'm not very clear on how you managed to get Ang ocb = Ang Obc => x

from the question it states that COB and OBC are 2:1, couldn't OCB be equal to a value other than OBC?

ENGRTOMBA2018 · Answer

No question is a "stupid question".

Look below for answers to your doubts.

In triangle OCB, you have been given that  \angle {cob} = 2* \angle {obc} . So now assume  \angle {obc} = x ---> \angle{cob} = 2x

now as OC and OB are radii of the circle, thus OC = OB ---> In triangle OCB as OC = OB, angles opposite to equal sides are always equal (remember this, this is true for ALL triangles!)--->  \angle {OCB} = \angle{OBC} = x

Thus in triangle OCB, we get x+x+2x = 180...etc

GMATinsight · Answer

Given : COB = 2*OBC

But angle OBC = angle OCB

Also Angle OBC + Angle BOC + Angle OCB = 180
therefore if angle OBC = a
then 2a+a+a = 180
i.e. a = 45 and angle COB = 90
i.e Triangle OBC is an Isosceles Right angle Triangle
Let Side OA =   OB = Radius = r  
i.e. AB = 2r
i.e. AC = 2r-4
Now, Triangle OAC also is an Isosceles Right angle Triangle in which Hypotenuse AC = r√2 = 2r-4 (given)

2r - r√2 = 4
i.e.  r  = 4/(2-√2) = 2(2+√2) =  4+2√2

Answer: Option D

I strictly urge GMAT test takers to stay away from Trigonometrical relations if you are not already good at them

GMATinsight · Answer

Relatively easier explanation would be

Line OC = Line OB

So angle OCB = angle OBC  
 
I hope it helps!

ENGRTOMBA2018 · Answer

GMATinsight , that is what I have mentioned in my post above

chetan2u · Answer

Hi,
another straight way to find that the angle CAO =45..
an angle at the circumference and the angle at the center on  the same arc are in the ratio 1:2 ...
therefore angle CAO:angle COB =1:2...
so angle CAO and CBO are equal and 45..

chetan2u · Answer

Hi,
an angle at the circumference and the angle at the center on the same arc are in the ratio 1:2...
therefore angle CAO:angle COB =1:2...
so angle CAO and CBO are equal and 45, as the third angle is 90..
now two sides are x each and the hyp/dia =x+4..
therefore in the right angle triangle, x \sqrt{2} =x+4...
therefore  x\sqrt{2}-x=4 ...
 x=4(\sqrt{2}+1) ..
radius = \frac{(x+4)}{2} ...
 4+2\sqrt{2} ..
ans D

Mo2men · Answer

Hi Engr2012,
 \angle{COB}=2\angle{OBC}  ----> In triangle OBC,  \angle{obc}+\angle{ocb}+\angle{cob} = 180  ---> x+x+2x = 180 ---> x=45 and  \angle{cob} = {90}^\circ

Similarly, in Triangle AOC,  \angle{OCA} = \angle{OAC} = {45}^\circ

Let R be the radius of the circle and thus AC = 2R-4 (given) and OC = AC= R

We can proceed forward in 2 ways:

1. Trigonometric way (not recommended for GMAT)

In triangle ACO, AO / AC =cos (45)---> \frac{R}{2R-4} = \frac{1}{\sqrt{2}}  --->   R = 4+4\sqrt{2}  , D is the correct answer.

2. Simpler and straighforward:

After you calculate AC=2R-4, in right triangle AOC (right angled at  \angle{AOC} ), apply the Pythagorean rule (Hyp^2 = Height ^2 + Base ^2) to get the following equation:

(2R-4)^2 = R^2+R^2 --->  R =4+4\sqrt{2}  , D is the correct answer.

Alternately, for this method after you calculate AC = 2R-4, in a 45-45-90 degree right triangle , the sides are in the ratio  1:1:\sqrt{2}----> R\sqrt{2} = 2R-4 ---> R = 4+4\sqrt{2}

Hi Engr2012,
Thanks for all alternatives. However it seems you small type mistakes. The correct answer is  R = 4+2\sqrt{2}  .

ENGRTOMBA2018 · Answer

Thanks  Mo2men , I have corrected the typos.

Bunuel · Answer

800score Official Solution:

Let us use the fact that the ratio of the angle COB’s measure to the measure of the angle OBC is 2 to 1. Denote the measure of the angle OBC as x degrees. Then, the measure of the angle COB is 2x degrees. OB = OC, as the both are radii. Therefore the triangle BOC is isosceles and angle OCB = angle OBC = x degrees. Since the sum of all the angle measures in any triangle is 180 degrees, we can write the equation:
x + x + 2x = 180
4x = 180
x = 45

So, the measure of the angle BOC is 2x = 90 degrees. The angles AOC and COB are contiguous so angle AOC = 180° – 90° = 90°. Also AO = OC as radii. Therefore, triangle AOC is an isosceles right triangle (90°-45°-45°). The sides of such triangle have 1 : 1 : √2 ratio.

The question statement defines AC = AB – 4. Since AB is a diameter and AO is a radius AC = 2AO – 4. When we plug it in AO : AC = 1 : √2 we get AO : (2AO – 4) = 1 : √2 .
√2AO = 2AO – 4
4 = 2AO – √2AO
4 = (2 – √2)AO
AO = 4/(2 – √2)
AO = 4(2 + √2) / ((2 – √2)(2 + √2))
AO = (8 + 4√2) / 2
AO = 4 + 2√2

The correct answer is D.

If you don't remember the property of a 90°-45°-45° triangle, there is an alternative ending of the solution. Let us use the fact that AC is 4 inches shorter than AB. Denote radii by y inches. So, we can deduce three things:
1. AO = OB = OC = y
2. AB = 2y
3. AC = 2y – 4

Then we use the Pythagorean Theorem for triangle AOC and get the following equation:
(2y – 4)² = y² + y²
4y² – 16y + 16 = 2y²
2y² – 16y + 16 = 0
y² – 8y + 8 = 0
y = 4 + 2√2 and y = 4 – 2√2 are two solutions to this equation. But, the length of AC (2y – 4) must be a positive number, so y = 4 + 2√2 is the only option.

The correct answer is D.

LaxAvenger · Answer

\angle{COB}=2\angle{OBC}  ----> In triangle OBC,  \angle{obc}+\angle{ocb}+\angle{cob} = 180  ---> x+x+2x = 180 ---> x=45 and  \angle{cob} = {90}^\circ

Similarly, in Triangle AOC,  \angle{OCA} = \angle{OAC} = {45}^\circ

Let R be the radius of the circle and thus AC = 2R-4 (given) and OC = AC= R

We can proceed forward in 2 ways:

1. Trigonometric way (not recommended for GMAT)

In triangle ACO, AO / AC =cos (45)---> \frac{R}{2R-4} = \frac{1}{\sqrt{2}}  --->  R = 4+2\sqrt{2} , D is the correct answer.

2. Simpler and straighforward:

After you calculate AC=2R-4, in right triangle AOC (right angled at  \angle{AOC} ), apply the Pythagorean rule (Hyp^2 = Height ^2 + Base ^2) to get the following equation:

(2R-4)^2 = R^2+R^2 ---> R = 4+2\sqrt{2} , D is the correct answer.

Can you show how the above formula is solved to get 4+2 *root of 2 ??

ENGRTOMBA2018 · Answer

Sure, look below for the explanation:

(2R-4)^2 = R^2+R^2  ---->   ^2 = 2R^2

4(R-2)^2 = 2R^2  --->  2(R-2)^2 = R^2  --->  2(R^2+4-4R) = R^2  (applied the relation  (a+b)^2 = a^2+b^2+2ab )

Rearranging the terms you get,

R^2 - 8R + 8 =0  ----> For a quadratic equation, ax^2+bx+c = 0  , the roots of the equations are :  (x1,x2) = \frac{-b\pm \sqrt{b^2-4ac}}{2a} . Thus for the given equation, we get,

R = \frac{8\pm \sqrt{64-32}}{2}  ---->  R = 4 \pm 2\sqrt{2}  . Now as the length of AC ( = 2R – 4) must be a positive number, so R = 4 + 2√2 is the only option.

Hope this helps.

bumpbot · Answer

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In the figure above (not drawn to scale), the triangle ABC is inscribe

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