AoPS - Problem

Problem

Source: Mumbai Region RMO 2014 Problem 3

Tags: trigonometry, geometry, angle bisector

YESMAths

07.12.2014 13:10

Let $ABC$ be an acute-angled triangle in which $\angle ABC$ is the largest angle. Let $O$ be its circumcentre. The perpendicular bisectors of $BC$ and $AB$ meet $AC$ at $X$ and $Y$ respectively. The internal angle bisectors of $\angle AXB$ and $\angle BYC$ meet $AB$ and $BC$ at $D$ and $E$ respectively. Prove that $BO$ is perpendicular to $AC$ if $DE$ is parallel to $AC$ .

TelvCohl

07.12.2014 13:52

My solution: $DE \parallel AC \Longrightarrow \frac{AD}{DB}=\frac{CE}{EB} \Longrightarrow \frac{AX}{XB}=\frac{CY}{YB}$ $\Longrightarrow \frac{AX}{XC}=\frac{CY}{YA} \Longrightarrow AY=CX \Longrightarrow \frac{\sin \angle CAB}{\sin \angle BCA}=\frac{BC}{BA}=\frac{\cos \angle BCA}{\cos \angle CAB}$ $\Longrightarrow \sin 2 \angle CAB=\sin 2 \angle BCA \Longrightarrow \angle CAB=\angle BCA$ ( $\because$ $\triangle ABC$ is an acute-angled triangle) $\Longrightarrow BO \perp AC$ Q.E.D

PROF65

07.12.2014 14:23

the same as TelvCohl until AY=CX after note that BXC and BYA are isometric

aditya21

09.12.2014 08:08

my proof = as $XD,YE$ are angle bisectors thus we have $(AX/XB) = (AD/DB)$ and $( CY/YB) = (CE/EB)$ now $DE\parallel AC\Longrightarrow\frac{AD}{DB}=\frac{CE}{EB}\Longrightarrow\frac{AX}{XB}=\frac{CY}{YB} = (AX/XC) = (CY/YA)$ now triangle $AXB , CYB$ triangle. are similar thus, $(AX/CY) = (XB/BY) = ( AB/BC)$ and $\angle BAX = \angle BCY$ and $\angle AXB = \angle CYB$ also triangle $AXC , CYA$ triangle are similar but , $AC$ is common side opp. to common angle. thus , both triangles are congruent. thus, $AY = CX \Longrightarrow\ AB = BC$ thus triangle $BOY$ and $BOX$ are congruent, thus $BO$ is angle bisector and hence as $ABC$ is isosceles thus $\Longrightarrow\ BO\perp AC$

Levinay

10.12.2014 07:39

aditya21 wrote: also triangle $AXC , CYA$ triangle are similar A-X-C are collinear i think..

Levinay

10.12.2014 07:50

PROF65 wrote: the same as TelvCohl until AY=CX after note that BXC and BYA are isometric isometric means isosceles?

aditya21

10.12.2014 09:33

Levinay wrote: aditya21 wrote: also triangle $AXC , CYA$ triangle are similar A-X-C are collinear i think.. can you show me your diagram. i dont think so still please show your diagram.

Levinay

10.12.2014 10:41

aditya21 wrote: Levinay wrote: aditya21 wrote: also triangle $AXC , CYA$ triangle are similar A-X-C are collinear i think.. can you show me your diagram. i dont think so still please show your diagram. its there in the problem itself...bisector meets AC at X....

PROF65

10.12.2014 16:22

Levinay wrote: PROF65 wrote: the same as TelvCohl until AY=CX after note that BXC and BYA are isometric isometric means isosceles? means have the same angles and side lengths note that BXY is isoceles

guptaamitu1

10.10.2020 22:40

aditya21 wrote: my proof = as $XD,YE$ are angle bisectors thus we have $(AX/XB) = (AD/DB)$ and $( CY/YB) = (CE/EB)$ now $DE\parallel AC\Longrightarrow\frac{AD}{DB}=\frac{CE}{EB}\Longrightarrow\frac{AX}{XB}=\frac{CY}{YB} = (AX/XC) = (CY/YA)$ now triangle $AXB , CYB$ triangle. are similar thus, $(AX/CY) = (XB/BY) = ( AB/BC)$ and $\angle BAX = \angle BCY$ and $\angle AXB = \angle CYB$ also triangle $AXC , CYA$ triangle are similar but , $AC$ is common side opp. to common angle. thus , both triangles are congruent. thus, $AY = CX \Longrightarrow\ AB = BC$ thus triangle $BOY$ and $BOX$ are congruent, thus $BO$ is angle bisector and hence as $ABC$ is isosceles thus $\Longrightarrow\ BO\perp AC$ can you please explain why triangles AXB, CYB are similar ?

guptaamitu1

10.10.2020 22:42

in my solution, I simply proved that BO is perpendicular to DE (though, I agree that the official solution is more elegant and faster)

NTistrulove

12.01.2022 13:27

Another Solution. Edit: The second attachment is the first part of the solution and the first attachment is the last part..

Attachments:

Jishnu4414l

25.09.2024 20:01

It suffices to show that

$\triangle ABC$ is

$B$ -isosceles. Clearly,

$\angle BED=\angle BCA$ . Also,

$\angle BXD=\frac{1}{2}\angle AXB=\angle BCA$ . This gives us that

$\angle BXD=\angle BED\implies$

$B,E,X,D$ are concyclic. Analogously, we get that

$B,E,Y,D$ are concyclic. Thus,

$X,Y,E,D$ are concyclic. Since

$XY\parallel DE$ , we must have that

$XYDE$ is an isosceles trapezoid. So,

$\angle XDE=\angle YED$ . Notice that

$\angle XDE=\angle DXA=\angle BCA$ , and that

$\angle YED=\angle EYC=\angle BAC$ . Thus, we have that

$\angle BAC=\angle BCA$ , and we are done!