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Source: China south east mathematical olympiad 2006 day2 problem 5

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jred

#1 h Jul 4, 2013, 2:26 PM • 2 Y

Y by Adventure10, Mango247

In $\triangle ABC$ , $\angle A=60^\circ$ . $\odot I$ is the incircle of $\triangle ABC$ . $\odot I$ is tangent to sides $AB$ , $AC$ at $D$ , $E$ , respectively. Line $DE$ intersects line $BI$ and $CI$ at $F$ , $G$ respectively. Prove that $FG=\frac{BC}{2}$ .

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#2 h Jul 4, 2013, 3:58 PM • 1 Y

Y by Adventure10

jred wrote:

In⊿ABC,∠A=60°. ⊙I is the incircle of ⊿ABC. ⊙I is tangent to side AB, AC at D, E respectively. Line DE intersects line BI and CI at F, G respectively. Prove that $FG=\frac{BC}{2}$ .

Using this lemma: $\Delta ABC$ , I is the incircle center of $\Delta ABC$ , ⊙I is tangent to side AB, AC at D, E respectively. Line DE intersects line BI at F. We can prove $BF \perp CF$ .
Using this lemma into this problem, we can easily prove $BF \perp CF, BG \perp CG$ .
Hence, $DGIB,BGFC$ are cyclic quadrilaterals.
We have: $\frac{GF}{BC}=\frac{GI}{IB}=\sin \widehat{GBI}=\sin \widehat{IDE}=\sin 30^{\circ}=\frac{1}{2}$ ( $\Delta GIF \sim \Delta BIC$ )
Accordingly, we have $GF=\frac{BC}{2}$ and that's what we need to prove.

This post has been edited 2 times. Last edited by JokerNVT, Jul 5, 2013, 11:48 AM

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#3 h Jul 4, 2013, 9:26 PM • 1 Y

Y by Adventure10

Obviously, $\triangle ADE$ is equilateral, also $\widehat{CIF}=\frac{\hat B+\hat C}{2}=60^\circ=\angle DEA$ , hence $F,I,C,E$ are concyclic and $CF\bot BF$ , similarly $BG\bot CI$ .
Let $B'=AC\cap BG, C'=AB\cap CF$ ; due to symmetry, $\angle B'IC=\angle BIC=\angle C'IB=120^\circ$ , so $B',C',I$ are collinear.
Now let's take $M\in AC| GM\parallel BC$ and $N\in AB|FN\parallel BC$ . $GM$ is median in right-angled $\triangle B'GC$ , hence $GM=\frac{B'C}{2}=\frac{BC}{2}$ . Similarly $FN=GM$ ; with $GM$ midline in $\triangle BB'C'$ , we easily get $\triangle GFN$ isosceles, and $FG=FN$ , done.

Best regards,
sunken rock

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#4 h Aug 5, 2013, 12:30 PM • 2 Y

Y by Adventure10, Mango247

How did the obviously equilateral come in?

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Burii

#5 h Aug 5, 2013, 12:56 PM • 1 Y

Y by Adventure10

biomathematics wrote:

How did the obviously equilateral come in?

Because this triangle is isosceles and one of it'a angle is $60^{\circ}$ .

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2568 posts

#6 h Mar 4, 2015, 1:06 AM • 2 Y

Y by Adventure10, Mango247

Burii wrote:

biomathematics wrote:

How did the obviously equilateral come in?

Because this triangle is isosceles and one of it'a angle is $60^{\circ}$ .

Oh oops. I thought D and E were given to be the intersection of the angle bisectors with the corresponding sides

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#7 h Mar 4, 2015, 2:35 AM • 2 Y

Y by Adventure10, Mango247

This problem was pretty fun to do, since I haven't been doing geometry recently.
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#8 h Mar 5, 2015, 6:07 AM • 2 Y

Y by Adventure10, Mango247

we first prove a lemma.
lemma = let $ABC$ be a triangle with incircle touching $AB,AC$ at $D,E$ . let $I$ be incentre and let $CI$ intersect $DE$ at $G$ . than $\angle CGB = 90$ .

proof = since $AD=AE$ we get $\angle AED = 90-A/2$
and hence $\angle CEG = 90+A/2$ which gives $\angle CGE =\angle IGE= B/2$
thus $BIGD$ is cyclic quad.
and hence , $\angle CGB=\angle IGB = \angle IDG = 90$
thus proving the lemma.

main proof = by the above lemma we have $\angle CGB=\angle BFC=90$
and hence $BGFC$ is cyclic quad. with $BC$ as diameter.
now let $O$ be midpoint of $BC$
than $\angle FGB=180-\angle FCB = 90+B/2$
thus $\angle FCB = 90-B/2$
similarly $\angle CBG = 90-C/2$

now $\angle OFG = 180-\angle GBC-\angle OFC = 90+C/2-\angle OFC=90+C/2-(180-2\angle FCB)=90+C/2-180-2(90-B/2) = 90-A/2$

thus by sine law in triangle $OFG$
$OB/2sin(90-A/2) =OB/2cosA/2= FG/sin A$
and hence $FG = 2OBsinA/2 = BCsin30 = BC/2$
and hence proved

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