IMO 2009 SL(G-1)

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IMO 2009 SL(G-1)

Post Number:#1  Unread postby aritra barua » Mon Jun 19, 2017 3:02 pm

Let $\bigtriangleup$ $ABC$ be an isosceles triangle with $AB$=$AC$.The bisectors of $\angle BAC$ and $\angle ABC$ intersect $BC$ and $AC$ at $D$ and $E$ respectively.Suppose $\angle BEK$ to be $45$ degree,where $K$ is the incenter of $\bigtriangleup$ $ACD$.Determine all possible values of $\angle BAC$.
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Re: IMO 2009 SL(G-1)

Post Number:#2  Unread postby ahmedittihad » Thu Jun 22, 2017 4:53 am

Ugly problem tbh.
Let us assign $\angle BAC = 4a$. Then quick angle chases give us, $\angle CAK = a $, $\angle DCI = 45-a $, $\angle CEK = 3a $.

We can apply sine law on $\triangle IEK $ and $\triangle CEK $ to get $ \frac{IK}{KC} = \frac{sin(45)*sin (45-a)}{sin(3a)*sin(90-2a)}$ ........(relation 1).

By angle bisect or theorem and sine law again, $\frac{IK}{KC} = \frac {ID}{CD} = \frac{sin(45-a)}{sin(45+a)}$ ........(relation 2).

Using the two relations together, we get that $ sin(45)*sin(45+a) = sin(3a)*sin(90-2a)$.

Now we use the product-sum rule and get $cos(a) = cos(5a-90)$.

Which implies that $cos(5a-90) - cos(a) = 0$.

Applying the product-sum rule again yields $cos(5a-90) - cos(a) = 0 = 2 sin(3a-45)*sin(2x-45)$.

We know that $0 \leq a <45$. So the two solutions are $45/2$° and $15$°.
Frankly, my dear, I don't give a damn.
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