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Tangent problem

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Tangent problem

by hemant_rajput » Wed Feb 27, 2013 9:32 am
In the figure below AB is 12 meter in length and is tangent at point C to the inner circle of the two concentric circles. It is known that radii of two circles are integer. The radius of the inner circle is :-
a.5
b.8
c.6
d.3
e.4
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by Anurag@Gurome » Wed Feb 27, 2013 9:55 am
hemant_rajput wrote:In the figure below AB is 12 meter in length and is tangent at point C to the inner circle of the two concentric circles. It is known that radii of two circles are integer. The radius of the inner circle is
Tricky Solution:
Refer to the figure below
Image

Say, radius of the inner circle = OC = r and radius of the larger circle = OB = R
And, AC = CB = AB/2 = 6

Now, triangle OCB is a right-angled triangle with angle OCB = 90 degrees
Hence, OC, CB, and CB forms a Pythagorean triplet.
Hence, (r, 6, R) is a Pythagorean triplet in which both r and R are integers.
Now, there is only one integral Pythagorean triplet of which 6 is a part, which is (6, 8, 10)
Hence, r must be 8.

The correct answer is B.
Last edited by Anurag@Gurome on Wed Feb 27, 2013 10:15 am, edited 1 time in total.
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by Anurag@Gurome » Wed Feb 27, 2013 9:57 am
Algebraic Solution:
Triangle OCB is a right-angled triangle with angle OCB = 90 degrees
Hence, OC² + CB² = OB²
--> r² + 6² = R²
--> R² - r² = 36
--> (R - r)(R + r) = 36

As both r and R are positive integers, (R - r) and (R + r) must be integers too with (r + r) > (R - r).
36 can be expressed as the product of two integers in the following ways : 1*36, 2*18, 3*12, 4*9 and 6*6

r = [(R + r) - (R - r)]/2 = [Larger factor - Smaller factor]/2
As r is an integer, [Larger factor - Smaller factor] must be even.
Hence, either both the factors are odd or both are even.

Only possible cases are 2*18 and 6*6
Hence, r = [18 - 2]/2 = 8 or r = [6 - 6]/2 = 0
As r cannot be 0, r must be 8.

The correct answer is B.
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by subhakam » Thu Feb 28, 2013 5:47 pm
Anurag@Gurome wrote:
hemant_rajput wrote:In the figure below AB is 12 meter in length and is tangent at point C to the inner circle of the two concentric circles. It is known that radii of two circles are integer. The radius of the inner circle is
Tricky Solution:
Refer to the figure below
Image

Say, radius of the inner circle = OC = r and radius of the larger circle = OB = R
And, AC = CB = AB/2 = 6

Now, triangle OCB is a right-angled triangle with angle OCB = 90 degrees
Hence, OC, CB, and CB forms a Pythagorean triplet.
Hence, (r, 6, R) is a Pythagorean triplet in which both r and R are integers.
Now, there is only one integral Pythagorean triplet of which 6 is a part, which is (6, 8, 10)
Hence, r must be 8.

The correct answer is B.
How did you assume C is the midpoint between AB? It does not explicitly state in the question. Can you please help me understand that part?
Thanks
Subhakam
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by ananthbi » Thu Feb 28, 2013 7:10 pm
Hi Shubhakam,

AC = CB because a general rule goes like this: "A radius or diameter with an end point at the point of tangency is perpendicular to the tangent line, and , conversely, a line that is perpendicular to the radius or diameter at one of it's endpoints is tangent to the circle at the endpoint."

Therefore AC=CB=AB/2

Let me know if this is clear.
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by hemant_rajput » Thu Feb 28, 2013 8:28 pm
subhakam wrote:
Anurag@Gurome wrote:
hemant_rajput wrote:In the figure below AB is 12 meter in length and is tangent at point C to the inner circle of the two concentric circles. It is known that radii of two circles are integer. The radius of the inner circle is
Tricky Solution:
Refer to the figure below
Image

Say, radius of the inner circle = OC = r and radius of the larger circle = OB = R
And, AC = CB = AB/2 = 6

Now, triangle OCB is a right-angled triangle with angle OCB = 90 degrees
Hence, OC, CB, and CB forms a Pythagorean triplet.
Hence, (r, 6, R) is a Pythagorean triplet in which both r and R are integers.
Now, there is only one integral Pythagorean triplet of which 6 is a part, which is (6, 8, 10)
Hence, r must be 8.

The correct answer is B.
How did you assume C is the midpoint between AB? It does not explicitly state in the question. Can you please help me understand that part?
Thanks
Subhakam
There is one more theorem,"In an isosceles triangle an altitude from a vertex containing two equal side bisects the base."

Hope this helps.
Cheers,
Hemant
I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.
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by Anurag@Gurome » Thu Feb 28, 2013 9:54 pm
subhakam wrote:How did you assume C is the midpoint between AB? It does not explicitly state in the question. Can you please help me understand that part?
I think a mathematical explanation will help you to analyze this kind of problems in future than remembering zillions of theorems and formulas.

Take the two triangles OAC and OBC
  • OB = OA (As they are radii of the larger circle)
    angle OCA = angle OCB (As the tangent (AB) is perpendicular to any radius (OC) at the point of tangency (C))
    OC is a common side
Hence, the triangles OAC and OCB are congruent triangles.
Hence, AC = BC, i.e. C is the midpoint of AB.

Hope that helps.
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by subhakam » Sun Mar 03, 2013 7:51 am
Thank you folks for your replies; now i am int he clear with respect to this concept! Many thanks again!
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