Kinda Familiar... Ah! A Theorem

Geometry Level 5

In the diagram, a circle ( I ; r ) (I; r) is internally tangent to a circle ( O ; R ) (O; R) .

Point M M lies on ( O ) (O) . Draw tangents M D , M E MD, ME to circle ( I ) (I) , which intersects ( O ) (O) at B , C B,C , respectively.

D E DE intersects I M IM at K K .

If we have R = 7 , r = 3 , B M C = 4 5 R = 7, r = 3, \angle BMC = 45^\circ , then 2 O K 2 2OK^2 can be written as:

2 O K 2 = a b c 2OK^2 = a-b\sqrt{c}

where a , b a, b and c c are positive integers with c c square-free.

Find a + b + c a+b+c .

Details and assumptions:

This problem is based on a geometric theorem, but it hasn't been wiki-ed yet.


The answer is 79.

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1 solution

Euler's theorem in geometry... K is the intersection of triangle BCM's bisectors.

But why K K is the incenter of triangle B C M BCM ?

Tran Quoc Dat - 5 years, 1 month ago

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You could start the wiki on the theorem :)

Calvin Lin Staff - 5 years, 1 month ago

Because it is well known that the tangential chord of the mixtilinear circle has its midpoint as incenter.

Vishwash Kumar ΓΞΩ - 3 years, 10 months ago

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