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August 26th, 2011, 04:02 PM   #1
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Hard geometry

Let H, I, O be orthocenter/incenter/circumcenter of triangle ABC respectively. Let L be intersection of line CI and circumcircle such that L is not C.

Given: AH = OH and IL = AB. Find all internal angles of ABC.

This problem has kept my attention for at least two hours now.. I find it very hard to use the given information. All I have so far is: and

Here denotes measurement of angle, the symbol for latex is not so good.

Any hints are very appreciated
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August 30th, 2011, 04:26 AM   #2
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Re: Hard geometry

Quote:
Originally Posted by proglote
Let L be intersection of line CI and circumcircle such that L is not C.
makes no sense to me. Can you draw the picture? is the only parameter that gives us the radius of the circumcircle; otherwise will not be unique. If we have with and , we will have a triangle, and the rest is easy.
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August 30th, 2011, 07:56 AM   #3
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Re: Hard geometry

Quote:
Originally Posted by Math Dreamer
Quote:
Originally Posted by proglote
Let L be intersection of line CI and circumcircle such that L is not C.
makes no sense to me. Can you draw the picture? is the only parameter that gives us the radius of the circumcircle; otherwise will not be unique. If we have with and , we will have a triangle, and the rest is easy.
the line CI intersects the circumcircle on two points, C and L.

I'll soon upload picture.

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August 30th, 2011, 07:26 PM   #4
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Re: Hard geometry

Toughy!
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August 31st, 2011, 08:01 AM   #5
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Re: Hard geometry

Quote:
Originally Posted by Math Dreamer
Toughy!
I got it from a hint in another website, you need only to proove:

1) AL = IL = BL;
2) Notice that AOBH is cyclic (which was the hardest part to me, I never see these things).

Then it is trivial, I will not post answer so you can try it

Here goes another one, just as hard and just as nice. I couldn't do this yet.

is a triangle with and . The incircle touches and at and respectively. Let be the circumference with diameter . The line intersects at points: and . Find the distance between the midpoint of and the midpoint of .

Geometry problems like these remind me of P = NP.. it is very hard to find solution, but very easy to check them
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September 1st, 2011, 04:42 AM   #6
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Re: Hard geometry

Thanks.
I am trying it by using vectors. It seems a bit cleaner.
Example: Let AC intersect BH at P








Then look for one more equation related to and

Goodness! Math by nature is hard. I mean it can get very, very messy.
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