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BdMO Online Forum • View topic - IGO 2016 Elementary/4

IGO 2016 Elementary/4

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IGO 2016 Elementary/4

Post Number:#1  Unread postby 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 $\angle C$.
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Re: IGO 2016 Elementary/4

Post Number:#2  Unread postby aritra barua » Sat Feb 25, 2017 12:33 pm

Isn't BC the diameter of the circumcircle?
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Re: IGO 2016 Elementary/4

Post Number:#3  Unread postby 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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Re: IGO 2016 Elementary/4

Post Number:#4  Unread postby 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.
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Re: IGO 2016 Elementary/4

Post Number:#5  Unread postby 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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Re: IGO 2016 Elementary/4

Post Number:#6  Unread postby 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.
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Re: IGO 2016 Elementary/4

Post Number:#7  Unread postby 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
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Re: IGO 2016 Elementary/4

Post Number:#8  Unread postby 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, $\angle B= 90^o-36^o=54^o$.
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Re: IGO 2016 Elementary/4

Post Number:#9  Unread postby 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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