IGO 2016 Elementary/4

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Thamim Zahin
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IGO 2016 Elementary/4

Unread post by Thamim Zahin » Tue Jan 10, 2017 2:19 pm

4. In a right-angled triangle $ABC (\angle A = 90)$, the perpendicular bisector of $BC$ intersects the line $AC$ in $K$ and the perpendicular bisector of $BK$ intersects the line $AB$ in $L$. If the line $CL$ be the internal bisector of $\angle C$, find all possible values for angles $\angle B$ and $\angle C$.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

aritra barua
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Re: IGO 2016 Elementary/4

Unread post by aritra barua » Sat Feb 25, 2017 12:33 pm

Isn't BC the diameter of the circumcircle?

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ahmedittihad
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Re: IGO 2016 Elementary/4

Unread post by ahmedittihad » Sat Feb 25, 2017 4:50 pm

Yes. It's a very trivial observation. Try to solve the problem.
Frankly, my dear, I don't give a damn.

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Thamim Zahin
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Re: IGO 2016 Elementary/4

Unread post by Thamim Zahin » Sun Feb 26, 2017 12:04 am

The fun thing(or cruel) about this problem is: You don't even have to do angle chasing in this problem. But when it is saying something about $90^o$, all we want is diameter. Through, use everything you learnt in class $6,7,8$. By that, you might solve the problem.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

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ahmedittihad
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Re: IGO 2016 Elementary/4

Unread post by ahmedittihad » Sun Feb 26, 2017 8:24 am

Why couldn't you? Thamim
Frankly, my dear, I don't give a damn.

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Thamim Zahin
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Re: IGO 2016 Elementary/4

Unread post by Thamim Zahin » Wed Mar 01, 2017 7:20 pm

It is a very nasty problem. By the way, thanks me if you read the solution. It was a hard time to $LateX$ all of them. It might look big, but, it is not very hard.
Case 1: $AC > AB$.

Claim 1: In $\triangle ABC$, $\angle LBK=\angle LKB$.
Proof: Let $X$ be the midpoint of $KB$. We know that, $LX \perp BK$. So. by $SAS$, $\triangle LXK \cong \triangle LXB$.

We denote $\angle LKB=\angle LBK=2a$. that means $\angle KLA=4a$.

Claim 2: $\angle KBC= 45^o-a$
Proof: Like Claim 1, we can easily show that $\triangle KBC$ is isosceles and $KB=KC$. Now $\angle BKC= \angle KBA+\angle BAK= 90^o+2a$. So, $\angle KBC= \frac{180^o-90^o-2a}{2}=45-a$.

Let $A'$ be the altitude on $BC$ from $L$.

Claim 3: $LA=LA'$
Proof: $CL=CL$, $\angle A'CL=\angle ACL$ and $\angle LAC=\angle LA'C$. So, by $ASA$, $\triangle ACL \cong \triangle A'CL$. So, $LA=LA'$.

Claim 4: $\angle LKA=\angle LBA'$.
Proof: $LA=LA'$[from claim 3],$LB=LK$[from claim 1], $\angle LAK=LA'B=90^o$. So, by $SSA$, $\triangle LKA \cong \triangle LA'B$. So,$\angle LKA=\angle LBA'$.


Now, $\angle LKA= 90^o-4a$, $$\angle LBA=45^o-a+21^o=45+a$$
$$\Rightarrow 90^o-4a=45^o+a$$
$$\Rightarrow a=9^o$$.
So, $\angle B= 45^o+a=54^o$. So. $\angle C=90^o-54^o=36^o$.

Case 2: $AC < AB$

Claim 5: $\angle LKB = \angle LBK$.
Proof: Same as Claim 1.

Let denote $\angle LKB = \angle LBK = a$. So, $\angle KLA=2a$.

Let $A'$ be the altitude of $L$ on $BC$.

Claim 6: $LA=LA'$
Proof: $CL=CL$, $\angle LCA=\angle LCA'$. $\angle CA'L=\angle CAL$. So, $\triangle CLA \cong \triangle CLA'$.So, $LA=LA'$.

Claim 7: $\angle AKL=\angle A'BL$.
Proof: $LK=LB$[Claim 5], $LA=LA'$[CLaim 6], $\angle LA'B=\angle LAK=90^o$. So, by $SSA$, $\triangle LAK \cong \triangle LA'B$.By that, $\angle AKL=\angle A'BL$.

Claim 8: $\triangle KBC$ is equilateral.

From [Claim 2] and [Claim 7], we got, $\angle LKB = \angle LBK$ and angle $AKL=\angle A'BL$. By combining them, $\angle CBK=\angle BKC$. Or, $CB=CK$. And we have, $KC=KB$[the perpendicular bisector $BC$ intersect the line $AC$ at $K$].
From that we get $CB=BK=KC$. Or, $\triangle KBC$ is equilateral.

That means. $\angle C=60^o$. So. $\angle B=90^o-60^o=30^o$.

Case 3: $AC=AB$.

Claim 9: It is impossible.
In this case, $K \equiv A$ and $L$ is the midpoint of $AB$. Let $T$ be a point on $BC$ such that $LT\perp BC$. We know that the line $CL$ is the internal bisector of $\angle C$, so $LT = LA = LB$ which is impossible.

So, all possible solutions are: $(\angle B, \angle C)=(54^o,36^o), (30^o,60^o)$.
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Last edited by Thamim Zahin on Thu Mar 02, 2017 12:51 am, edited 1 time in total.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

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Thamim Zahin
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Re: IGO 2016 Elementary/4

Unread post by Thamim Zahin » Wed Mar 01, 2017 9:44 pm

ahmedittihad wrote:Why couldn't you? Thamim
DONE. Didn't use any angle chasing as I said :D :D :D
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

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Thamim Zahin
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Re: IGO 2016 Elementary/4

Unread post by Thamim Zahin » Thu Mar 02, 2017 8:08 pm

I more solution Case 1:$AC>AB$
By a well know lemma we know that $BLKC$ is cyclic.

Now denote, $\angle LBK=\angle LCK=\angle LCB=\angle LKB=a$.
And we have that $KC=KB$. So, $\angle KBC=2a$. So, we have $a+2a+a+a=90^o \Rightarrow a=18^o$.
That means $2a=\angle C=18^o\times 2=36^o$. So, $\angle B= 90^o-36^o=54^o$.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

aritra barua
Posts: 50
Joined: Sun Dec 11, 2016 2:01 pm

Re: IGO 2016 Elementary/4

Unread post by aritra barua » Fri Mar 03, 2017 8:01 pm

It is just the official solution....is there any unique technique to solve that?

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