This should be easy! Right? Part - 3

Geometry Level 5

The figure above represents two circles where A B = 5 AB=5 units, ∠ B A C = 3 0 ∘ \angle BAC=30^\circ , ∠ B M D = 13 5 ∘ \angle BMD=135^\circ , M D = 3.3 MD = 3.3 units and N N is the mid point of M D MD .

C D C N × C M = a c − d + e b e b − d + e \dfrac{CD}{CN}\times{CM}=\frac{a\sqrt{c-\sqrt{d}+\sqrt{e}}}{b\sqrt{e}\sqrt{b-\sqrt{d}+\sqrt{e}}}

The equation above holds true for positive integers a , b , c , d a,b,c,d and e e with b , d b,d and e e square-free, a > b a>b , and a , b a,b are coprime.

Find a + b + c + d + e a+b+c+d+e .

Clarification : M D MD a diameter of the smaller circle. M M is the centre of the larger circle. The two circles tangential at D D .

Try Part-1 and Part-2 also.


The answer is 50.

Akshay Yadav by Akshay Yadav , 20, India

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1 solution

First let the radius of the bigger circle to be r and let the angle between sides MD and MC be θ \theta .

Then,

C D 2 = 2 r 2 − 2 r 2 cos ⁡ θ \displaystyle { CD }^{ 2 }=2{ r }^{ 2 }-2{ r }^{ 2 }\cos { \theta }

C N 2 = r 2 + r 2 4 − r 2 cos ⁡ θ \displaystyle { CN }^{ 2 }={ r }^{ 2 }+\frac { { r }^{ 2 } }{ 4 } -{ r }^{ 2 }\cos { \theta }

C M 2 = r 2 \displaystyle { CM }^{ 2 }={ r }^{ 2 }

C D 2 C N 2 × C M 2 = 2 r 2 ( 1 − cos ⁡ θ ) r 2 ( 5 4 − cos ⁡ θ ) × r 2 \displaystyle \frac { { CD }^{ 2 } }{ { CN }^{ 2 } } \times { CM }^{ 2 }=\frac { 2{ r }^{ 2 }\left( 1-\cos { \theta } \right) }{ { r }^{ 2 }\left( \frac { 5 }{ 4 } -\cos { \theta } \right) } \times { r }^{ 2 }

θ = 135 − 2 × 30 = 75 = 30 + 45 \displaystyle \theta = 135-2 \times 30=75=30+45 . Because BMC is twice BAC by the Euclidean geometry theorem.

cos ⁡ θ = cos ⁡ 45 + 30 = 3 − 1 2 2 \displaystyle \cos { \theta } =\cos { 45+30 } =\frac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } }

C D 2 C N 2 × C M 2 = 2 r 2 ( 1 − cos ⁡ θ ) r 2 ( 5 4 − cos ⁡ θ ) × r 2 = 2 ( 1 − ( 3 − 1 ) 2 2 ) ( 5 4 − ( 3 − 1 ) 2 2 ) × r 2 \displaystyle \frac { { CD }^{ 2 } }{ { CN }^{ 2 } } \times { CM }^{ 2 }=\frac { 2{ r }^{ 2 }\left( 1-\cos { \theta } \right) }{ { r }^{ 2 }\left( \frac { 5 }{ 4 } -\cos { \theta } \right) } \times { r }^{ 2 }=\frac { 2\left( 1-\frac { \left( \sqrt { 3 } -1 \right) }{ 2\sqrt { 2 } } \right) }{ \left( \frac { 5 }{ 4 } -\frac { \left( \sqrt { 3 } -1 \right) }{ 2\sqrt { 2 } } \right) } \times { r }^{ 2 }

= 2 ( 4 − 6 − 2 ) ( 5 − 6 − 2 ) × r 2 \displaystyle =\frac { 2\left( 4- \sqrt { 6 } - \sqrt { 2 } \right) }{ \left( 5 -\sqrt { 6 } - \sqrt { 2 } \right) } \times { r }^{ 2 }

r = 3.3 = 33 10 \displaystyle r=3.3=\frac{33}{10}

= 2 ( 4 − 6 − 2 ) ( 5 − 6 − 2 ) × ( 33 10 ) 2 \displaystyle =\frac { 2\left( 4- \sqrt { 6 } - \sqrt { 2 } \right) }{ \left( 5 -\sqrt { 6 } - \sqrt { 2 } \right) } \times { \left( \frac { 33 }{ 10 } \right) }^{ 2 }

C D C N × C M = = 2 ( 4 − 6 − 2 ) ( 5 − 6 − 2 ) × ( 33 10 ) 2 \displaystyle \frac { { CD } }{ { CN } } \times { CM }==\sqrt { \frac { 2\left( 4- \sqrt { 6 } - \sqrt { 2 } \right) }{ \left( 5 -\sqrt { 6 } - \sqrt { 2 } \right) } \times { \left( \frac { 33 }{ 10 } \right) }^{ 2 } }

= 2 ( 4 − 6 − 2 ) ( 5 − 6 − 2 ) × ( 33 10 ) \displaystyle =\sqrt { \frac { 2\left( 4- \sqrt { 6 } - \sqrt { 2 } \right) }{ \left( 5 -\sqrt { 6 } - \sqrt { 2 } \right) } } \times { \left( \frac { 33 }{ 10 } \right) }

= ( 4 − 6 − 2 ) ( 5 − 6 − 2 ) × ( 33 5 2 ) \displaystyle =\sqrt { \frac { \left( 4- \sqrt { 6 } - \sqrt { 2 } \right) }{ \left( 5 -\sqrt { 6 } - \sqrt { 2 } \right) } } \times { \left( \frac { 33 }{ 5\sqrt { 2 } } \right) }

= ( 33 5 2 ) × ( 4 − 6 − 2 ) ( 5 − 6 − 2 ) \displaystyle = { \left( \frac { 33 }{ 5\sqrt { 2 } } \right) }\times \sqrt { \frac { \left( 4- \sqrt { 6 } - \sqrt { 2 } \right) }{ \left( 5 -\sqrt { 6 } - \sqrt { 2 } \right) } }

= ( 33 5 2 ) × ( 4 − 6 − 2 ) ( 5 − 6 − 2 ) = ( a b e ) × ( c − d − e ) ( b − d − e ) \displaystyle = { \left( \frac { 33 }{ 5\sqrt { 2 } } \right) }\times \sqrt { \frac { \left( 4- \sqrt { 6 } - \sqrt { 2 } \right) }{ \left( 5 -\sqrt { 6 } - \sqrt { 2 } \right) } } = { \left( \frac { a }{ b\sqrt { e } } \right) }\times \sqrt { \frac { \left( c- \sqrt { d } - \sqrt { e } \right) }{ \left( b -\sqrt { d } - \sqrt { e } \right) } }

So,

a = 33 , b = 5 , c = 4 , d = 6 , e = 2 \displaystyle a=33,b=5,c=4,d=6,e=2

a + b + c + d + e = 33 + 5 + 4 + 6 + 2 = 50 \displaystyle a+b+c+d+e=33+5+4+6+2=\boxed{50}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Thanks for posting a solution to my problem. I thought no will be willing to do so and in the end I will have to do it myself.

Akshay Yadav - 5 years, 3 months ago

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Is there a better / more direct approach to this problem? I quite like it, but this solution makes it look really painful

Calvin Lin Staff - 5 years, 3 months ago

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Sir, when I was making this problem, even I used the same method as given in the solution of @คลุง แจ็ค

Akshay Yadav - 5 years, 3 months ago

Is this your own problem ? It is really cool.

Venkata Karthik Bandaru - 5 years, 3 months ago

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Yeah! Its 100% original problem. Thanks.

Akshay Yadav - 5 years, 3 months ago

Great problem. But isn't there any approach that uses less computations?

A Former Brilliant Member - 5 years, 2 months ago

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Even I did it the same way as given in the solution, you can say that I deliberately made it so that its calculations are hard to do.

Akshay Yadav - 5 years, 2 months ago

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