Doctor A wrote: (F.Nilov) Given right triangle $ ABC$ with hypothenuse $ AC$ and $ \angle A = 50^{\circ}$. Points $ K$ and $ L$ on the cathetus $ BC$ are such that $ \angle KAC = \angle LAB = 10^{\circ}$. Determine the ratio $ CK/LB$. See that $ |BC| = sin(50)*|AC|$, $ |BK| = sin(40)*|AC|$, $ |BL| = sin(10)*|AC|$. So $ \frac{|CK|}{|BL|} = \frac{sin(50)-sin(40)}{sin(10)}$

i think you're wrong... we have that $ \frac{CK}{BL}=\frac{\frac{AC \sin 10}{\sin 80}}{\frac{AB \sin 10}{\sin 100}}=\frac{\sin 100}{\sin 50\sin 40}=2$...

I don't see anything's wrong with Mathias_DK's, but we can further simplify the expression into $ \frac1{\sqrt2\cos5^{\circ}}$. I think campos's solutions is incorrect, because $ CK=\frac{AC\sin10^{\circ}}{\sin130^{\circ}}$ and $ BL=\frac{AB\sin10^{\circ}}{\sin80^{\circ}}$.

yes, it has a typo, however $ \frac{\frac{AC\sin 10}{\sin 50}}{\frac{AB\sin 10}{\sin 100}}=\frac{\sin100}{\sin 40\sin 50}=2$...

Sorry.. Must have been sleeping or something.. :S $ |BL| = tan(10) * |AB|$, $ |BK| = tan(40) * |AB|$, $ |BC| = tan(50) * |AB|$, $ |CK| = |BC| - |BK| = (tan(50) - tan(40))*|AB|$. So $ \frac {|CK|}{|BL|} = \frac {tan(50) - tan(40)}{tan(10)} = 2$

Seems to me that F. Nilov got to this problem working on Langley's 80-80-20 triangle. Here is PWW figure. M.T.

There is a very nice and short geometrical solution!

1) See that ALC is an isosceles triangle with AL = LC ( 1 ) 2) Take L' the symmetrical of L w.r.t. B. AL = AL' ( 2 ) The triangle AL'K is isosceles, with AL' = KL' ( 3 ). From (1), (2) and (3) we get CL = KL', hence CK = LL', but LL' = 2 BL, so the ratio is 2. Best regards, sunken rock

I think that the configuration of the regular 9-gon, help us to solve this problem without words. Kostas vittas.

This is my first post in this forum. I`m sorry because my english is not good. Let N the midpoint of AC. Note that triangles KAC and BNL are similar(*) and the BD/AC=BL/KC=1/2 and hence CK/LB=2. (* it`s not hard to prove this)[/code]

Nice solutions guys.

Draw KQ perpendicular to AK,KQ meet BC at Q;we have :△AKQ∽△ABL, Thus,BL/KQ=cos40;By angles,we also have :KQ=QC, Hence CK=2QK=2cos40*QK=2cos40*(BL/cos40)=2BL;obviously,CK/BL=2..

Doctor A wrote: (F.Nilov) Given right triangle $ ABC$ with hypothenuse $ AC$ and $ \angle A = 50^{\circ}$. Points $ K$ and $ L$ on the cathetus $ BC$ are such that $ \angle KAC = \angle LAB = 10^{\circ}$. Determine the ratio $ CK/LB$. $2sin(x)sin(90 - x) = 2sin(x)cos(x) = sin(2x)$ and $sin(x) = sin(180 - x)$. We apply the $sinus$ $theorem$ in triangle $ALB$ and we get $\frac{LB}{sin(10)} = \frac{AB}{sin(80)}$.Now we apply $sinus$ $theorem$ in triangle $KCA$ and we get $\frac{KC}{sin(10)} = \frac{AC}{sin(50)}$ but $AC = \frac{AB}{sin(40)}$ $\implies$ $\frac{KC}{sin(10)} = \frac{AB}{sin(40)*sin(50)} = \frac{2AB}{sin(100)} = \frac{2AB}{sin(80)} = \frac{2LB}{sin(10)}$ and now we get that $KC = 2LB$ $\implies$ $\frac{KC}{LB} = 2$.

In ∆ ABC, sin ACB=AB/AC AC/ AB=1 / sin 40° In ∆AKC, sin KAC/ CK= sin LKC/ AC sin 10°/ CK= sin 130°/ AC sin 10°= (CK sin 50° )/AC In ∆ ABL, sin BAL/ BL= sin ALB/ AC sin 10°/ BL= sin 80°/ AB sin 10°= (LB sin 80° )/AB CK/ LB=( AC/ AB)* (sin 80°/ sin 50°) = 2 sin 40° cos 40°/ sin 40° . sin 50° = 2 cos 40°/ sin 50° = 2 CK/ LB =2