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 August 26th, 2011, 05:02 PM #1 Senior Member   Joined: Apr 2011 From: Recife, BR Posts: 352 Thanks: 0 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: $\angle ABH \equiv \angle CBO$ and $3 \angle ABH + 2\angle HBO - \angle OAH= 90.$ Here $\angle$ denotes measurement of angle, the symbol for latex is not so good. Any hints are very appreciated
August 30th, 2011, 05:26 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.
$L \neq C$ makes no sense to me. Can you draw the picture? $IL$ is the only parameter that gives us the radius of the circumcircle; otherwise $R$ will not be unique. If we have $R$ with $OH$ and $AH$, we will have a triangle, and the rest is easy.

August 30th, 2011, 08:56 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.
$L \neq C$ makes no sense to me. Can you draw the picture? $IL$ is the only parameter that gives us the radius of the circumcircle; otherwise $R$ will not be unique. If we have $R$ with $OH$ and $AH$, we will have a triangle, and the rest is easy.
the line CI intersects the circumcircle on two points, C and L.

 August 30th, 2011, 08:26 PM #4 Senior Member   Joined: Jun 2011 Posts: 298 Thanks: 0 Re: Hard geometry Toughy!
August 31st, 2011, 09:01 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.

$ABC$ is a triangle with $A= 120 \text{degrees}$ and $BC= 12$. The incircle touches $AB$ and $AC$ at $D$ and $E$ respectively. Let $\gamma$ be the circumference with diameter $BC$. The line $DE$ intersects $\gamma$ at points: $K$ and $L$. Find the distance between the midpoint of $KL$ and the midpoint of $BC$.

Geometry problems like these remind me of P = NP.. it is very hard to find solution, but very easy to check them

 September 1st, 2011, 05:42 AM #6 Senior Member   Joined: Jun 2011 Posts: 298 Thanks: 0 Re: Hard geometry Thanks. I am trying it by using vectors. It seems a bit cleaner. Example: Let AC intersect BH at P $\vec{AH}\cdot \vec{AP}= |AH|\cdot |AP|\cos \gamma$ $\vec{AB}\cdot \vec{AP}= |AB|\cdot |AP|\cos A$ $\vec{AP}=\frac{\vec{AH}\cdot \vec{AP}}{|AH|\cos \gamma}=\frac{\vec{AB}\cdot \vec{AP}}{|AB|\cos A}$ $\frac{\sin A}{\cos \gamma}=\tan A$ Then look for one more equation related to $\gamma$ and $IL$ Goodness! Math by nature is hard. I mean it can get very, very messy.

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