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Source: Saudi Arabia JTST -2015,P4

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#1 h Oct 7, 2016, 12:49 PM • 3 Y

Y by buratinogigle, tarzanjunior, Adventure10

Let $ABC$ be a right triangle with the hypotenus $BC.$ Let $BE$ be the bisector of the angle $\angle ABC.$ The circumcircle of the triangle $BCE$ cuts the segment $AB$ again at $F.$ Let $K$ be the projection of $A$ on $BC.$ The point $L$ lies on the segment $AB$ such that $BL=BK.$ Prove that $\frac{AL}{AF}=\sqrt{\frac{BK}{BC}}.$

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#2 h Oct 7, 2016, 3:45 PM • 1 Y

Y by Adventure10

Any idea?

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1574 posts

#3 h Oct 7, 2016, 4:00 PM • 1 Y

Y by Adventure10

What do you mean by projection?Are you taking component?

This post has been edited 1 time. Last edited by Googlu15, Oct 7, 2016, 4:02 PM

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#4 h Oct 7, 2016, 4:10 PM • 2 Y

Y by Adventure10, Mango247

Please can someone explain me?

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849 posts

#5 h Oct 7, 2016, 4:12 PM • 2 Y

Y by Adventure10, Mango247

Projection means the foot of perpendicular

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#6 h Oct 7, 2016, 4:18 PM • 2 Y

Y by Adventure10, Mango247

@above thanks

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#7 h Oct 7, 2016, 4:24 PM • 2 Y

Y by Adventure10, Mango247

I am getting the intersection of FE and AK as the circumcenter of triangle AFE.

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849 posts

#8 h Oct 7, 2016, 4:33 PM • 2 Y

Y by Adventure10, Mango247

Length bashing kills the problem, $BK=c^2/a$ , $BC=a$ , $AL=(ac-c^2)/a$ , $AF=b^2/(a+c)$

This post has been edited 1 time. Last edited by tarzanjunior, Oct 7, 2016, 4:34 PM
Reason: typo

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#9 h Oct 7, 2016, 4:35 PM • 2 Y

Y by Adventure10, Mango247

I think similarity also kills the problem.

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#10 h Oct 7, 2016, 4:36 PM • 2 Y

Y by Adventure10, Mango247

tarzanjunior wrote:

Length bashing kills the problem, $BK=c^2/a$ , $BC=a$ , $AL=(ac-c^2)/a$ , $AF=b^2/(a+c)$

CAn you show your full solution?

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#11 h Oct 7, 2016, 4:36 PM • 2 Y

Y by Adventure10, Mango247

Length basing is much easier, just write all lengths in terms of the sides $a,b,c$

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#12 h Oct 7, 2016, 4:38 PM • 2 Y

Y by Adventure10, Mango247

tarzanjunior wrote:

Length basing is much easier, just write all lengths in terms of the sides $a,b,c$

Please show your solution(500th post).

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Orkhan-Ashraf_2002

299 posts

Orkhan-Ashraf_2002

#13 h Oct 7, 2016, 4:49 PM • 2 Y

Y by Adventure10, Mango247

Ferid.---. wrote:

Let $ABC$ be a right triangle with the hypotenus $BC.$ Let $BE$ be the bisector of the angle $\angle ABC.$ The circumcircle of the triangle $BCE$ cuts the segment $AB$ again at $F.$ Let $K$ be the projection of $A$ on $BC.$ The point $L$ lies on the segment $AB$ such that $BL=BK.$ Prove that $\frac{AL}{AF}=\sqrt{\frac{BK}{BC}}.$

We should use law of sines, $AF \cdot AB=AE \cdot AC$ and angle-bisector theorem.

This post has been edited 1 time. Last edited by Orkhan-Ashraf_2002, Oct 7, 2016, 4:51 PM

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#14 h Oct 7, 2016, 5:02 PM • 2 Y

Y by Adventure10, Mango247

Ferid.---. wrote:

Let $ABC$ be a right triangle with the hypotenus $BC.$ Let $BE$ be the bisector of the angle $\angle ABC.$ The circumcircle of the triangle $BCE$ cuts the segment $AB$ again at $F.$ Let $K$ be the projection of $A$ on $BC.$ The point $L$ lies on the segment $AB$ such that $BL=BK.$ Prove that $\frac{AL}{AF}=\sqrt{\frac{BK}{BC}}.$

$AL = c - BL = c - BK = c - \frac{{{c^2}}}{a} \Leftrightarrow$ $\boxed{AL = \frac{{c(a - c)}}{a}}$ $(1)$

$AF \cdot c = AE \cdot b = \frac{{bc}}{{a + c}} \cdot b \Leftrightarrow$ $\boxed{AF = \frac{{{b^2}}}{{a + c}}}$ $(2)$

$(1), (2)$ : $\frac{{AL}}{{AF}} = \frac{{c(a - c)(a + c)}}{{a{b^2}}} = \frac{{c({a^2} - {c^2})}}{{a{b^2}}} = \frac{{c{b^2}}}{{a{b^2}}} = \frac{c}{a} = \sqrt {\frac{{{c^2}}}{{{a^2}}}} = \sqrt {\frac{{{c^2}}}{{{b^2} + {c^2}}}} = \sqrt {1 + \frac{{{c^2}}}{{{b^2}}}} = \sqrt {1 + \frac{{BK}}{{KC}}} = \sqrt {\frac{{BK}}{{BC}}}$

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#15 h Nov 1, 2022, 9:47 PM • 1 Y

Y by buratinogigle

proposed by Tran Quang Hung, Vietnam

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