Post Number:#6 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: $\angle KBC= 45^oa$
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, $\angle KBC= \frac{180^o90^o2a}{2}=45a$.
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, $\angle LKA= 90^o4a$, $$\angle LBA=45^oa+21^o=45+a$$
$$\Rightarrow 90^o4a=45^o+a$$
$$\Rightarrow a=9^o$$.
So, $\angle B= 45^o+a=54^o$. So. $\angle C=90^o54^o=36^o$.
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. $\angle B=90^o60^o=30^o$.
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: $(\angle B, \angle C)=(54^o,36^o), (30^o,60^o)$.
 Attachments

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Last edited by
Thamim Zahin on Thu Mar 02, 2017 12:51 am, edited 1 time in total.