Given a triangle $ABC$ with $\angle ACB = 120^{\circ}.$ A point $L$ is marked in the side $AB$ such that $CL$ bisects $\angle ACB.$ Points $N$ and $K$ are chosen in the sides $AC$ and $BC $ such that $CK+CN=CL.$ Prove that the triangle $KLN$ is equilateral.
Problem
Source: RMO 2023 P1
Tags: geometry
29.10.2023 16:03
Which state?
29.10.2023 16:08
TheHimMan wrote: Which state? Idk It’s of KV
29.10.2023 16:09
Ok,do you have the other problems as well?
29.10.2023 16:10
You need not latex them r.n.
29.10.2023 16:17
Let $CK = x$ and $CN = y.$ Then $CL = x+y.$ Now, applying Law of Cosines in $\Delta CNK$ gives \begin{align*} NK &= \sqrt{CK^2+CN^2-2\cdot CK\cdot CN\cdot \cos{\angle NCK}} \\ &= \sqrt{x^2+y^2-2xy\cos{120^{\circ}}} \\ &= \sqrt{x^2+y^2+xy} \end{align*}Similarly, by applying Law of Cosines in $\Delta CLK$ and $\Delta NCL$ we can show that $$ KL=LN=\sqrt{x^2+y^2+xy}.$$Hence, $\Delta KLN$ is equilateral. $QED.$
29.10.2023 16:45
Let LS parallel AC and intersects BC at S.It’s not difficult to see the congruence of triangle LNC and triangle LKS.Triangle CLS is an equilateral so triangle KLN is also an equilateral.
29.10.2023 16:49
very easy
29.10.2023 20:30
Let $P \in CL \ni CK = CP.$ Then $$PL=CL-CP=CL-CK=CN.$$And the triangle $CPK$ is equilateral. So, $PK=CK$ and $\angle LPK = 120^{\circ}.$ Thus, by ASA congruence, $$\Delta LPK \cong \Delta NCK.$$So, $NK=LK$ and $\angle NKL = \angle NKP + \angle PKL = \angle NKP + \angle CKN = \angle CKP = 60^{\circ}.$ Hence, the triangle $KLN$ is equilateral. $QED.$
05.11.2023 13:15
Need to prove $NLKC$ is a cyclic quad. Introduce point $X$ on $CL$ such that $CX = CK$. Then by ASA congruency, $$\Delta CNK \cong \Delta XLK.$$Hence, $ \angle CLK = \angle CNK = 60^{\circ} $. So $NLKC$ is cyclic.
12.04.2024 20:30
Synthetic solution: In the figure, let $K'$ be a point on side $BC$ such that $CNLK'$ is cyclic. Therefore, $\angle K'CL=\angle K'NL=60^{\circ}$. $\angle NCL=\angle NK'L=60^{\circ}\implies K'NL$ is equilateral. Now, we prove that $K=K'$. Firstly, let $N'$ be a point on line $BC$ such that $CN=CN'$. Also, $\angle NCN'=60^{\circ}$. Therefore, $NCN'$ is equilateral. $\implies \angle CN'N=60^{\circ}=\angle NCL$. Also, as $CNLK'$ is cyclic, $\angle NLC=\angle NK'C= \angle NK'N'$. Therefore, by $AAS$ congruency criterion, $\Delta NN'K' \cong \Delta NCL \implies CN=N'K' = N'C+CK'= NC+CK'$. Therefore, $K=K'$, and we're done!
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17.09.2024 17:44
Let $CK=a$, and $CN=b$. Using cosine law, we see that $LK=LN=NK=\sqrt{a^2+ab+b^2}$