Indian RMO 2011: Question 5

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Indian RMO 2011: Question 5

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366 posts

Sampro

#1 h Dec 4, 2011, 11:48 AM • 3 Y

Y by nguyenvanthien63, Adventure10, Mango247

Let $ABC$ be a triangle and let $BB_1,CC_1$ be respectively the bisectors of $\angle{B},\angle{C}$ with $B_1$ on $AC$ and $C_1$ on $AB$ , Let $E,F$ be the feet of perpendiculars drawn from $A$ onto $BB_1,CC_1$ respectively. Suppose $D$ is the point at which the incircle of $ABC$ touches $AB$ . Prove that $AD=EF$

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Potla

#2 h Dec 4, 2011, 12:23 PM • 1 Y

Y by Adventure10

Let $I$ be the incentre of $\triangle ABC.$
Note that from simple angle chasing, we have $\angle FAI=\frac{B}{2}$ and $\angle EAI=\frac{C}{2}.$
So, $EF\parallel BC \implies \frac{EF}{BC}=\frac{IE}{IB}=\frac{AI\sin\frac{C}{2}}{IB}=\frac{\sin \frac B2\sin\frac C2}{\sin\frac A2}.$
Using the half-angle formulae, we get
$EF=a\cdot\frac{s-a}{a}=s-a=AD,$ and we are done. $\Box$
Too disappointed at the leakout again.

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634 posts

#3 h Dec 4, 2011, 1:24 PM • 2 Y

Y by Adventure10, Mango247

Geometric proof : (although i'm also disappointed for the same reason )

Let $BB_1 \cap CC_1= I$ . Note that the points $E,F,A,I,D$ are cyclic, with $AI$ as diameter and $O$ as center (assume). Simple angle chasing will give you that $\triangle ADO \cong \triangle EFO$ . So $AD = EF$

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Sahil

#4 h Dec 4, 2011, 1:51 PM • 1 Y

Y by Adventure10

Potla wrote:

Let $I$ be the incentre of $\triangle ABC.$
Note that from simple angle chasing, we have $\angle FAI=\frac{B}{2}$ and $\angle EAI=\frac{C}{2}.$
So, $EF\parallel BC \implies \frac{EF}{BC}=\frac{IE}{IB}=\frac{AI\sin\frac{C}{2}}{IB}=\frac{\sin \frac B2\sin\frac C2}{\sin\frac A2}.$
Using the half-angle formulae, we get
$EF=a\cdot\frac{s-a}{a}=s-a=AD,$ and we are done. $\Box$
Too disappointed at the leakout again.

I think u meant $\angle FAI=\angle \frac{C}{2}$ coz otherwise u wud end up proving that ABC is isosceles

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Potla

#5 h Dec 4, 2011, 3:35 PM • 2 Y

Y by Adventure10, Mango247

sahilsharma94 wrote:

I think u meant $\angle FAI=\angle \frac{C}{2}$ coz otherwise u wud end up proving that ABC is isosceles

I wrote what I meant.
$\angle ABE=\frac B2\implies \angle EAB=\frac{\pi}{2}-\frac {B}{2},$ and $\angle BAI=\frac A2;$ so that
$\angle EAI=\frac{\pi}{2}-\frac{B}{2}-\frac{A}{2}=\frac{C}{2}.$

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#6 h Dec 4, 2011, 5:37 PM • 3 Y

Y by Adventure10 and 2 other users

Take $M, N$ the reflections of $A$ in $E$ and $F$ respectively; as $BM=AB, CN=AC \implies NM=AB+AC-BC$ . See that $EF$ is midline of $\triangle AMN \implies EF=AD=\frac{AB+AC-BC}{2}$ .

Best regards,
sunken rock

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#7 h Dec 5, 2011, 1:18 AM • 1 Y

Y by Adventure10

Amusingly enough, this is pretty much AIME I 2011 #4.

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yilun

#8 h Dec 5, 2011, 1:33 AM • 1 Y

Y by Adventure10

What do you mean by leakout?

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#9 h Dec 5, 2011, 1:35 AM • 2 Y

Y by Adventure10, Mango247

Someone managed to find the papers with the questions, and posted the questions for the RMO on AoPS a couple days ago.

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Sahil

#10 h Dec 5, 2011, 3:07 AM • 2 Y

Y by Adventure10, Mango247

bzprules wrote:

Amusingly enough, this is pretty much AIME I 2011 #4.

It has been know for ages , that olympiad papers in india are rarely original and mostly copied from american british or russian olympiads/olympiad books.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2209675&sid=46703032b64b104a1469d375d78a8db4#p2209675

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#11 h Dec 5, 2011, 2:57 PM • 5 Y

Y by Sayan, Sampro, Adventure10, Mango247, and 1 other user

another solution,

in $\triangle ADI, \frac{AD}{cos(\frac{A}{2})} = AI ......(1)$

as $AFIE$ is a cyclic quadrilateral ,

let $\angle AEF = \angle AIF = \alpha ...(2)$

in $\triangle AFE$ ,

$\frac{EF}{cos(\frac{A}{2})} = \frac{AF}{sin\alpha} ...(3)$

also in $\triangle AFI, \frac{AF}{sin\alpha} = AI ...(4)$

from $(1),(2),(3)\&(4)$

$\boxed{AD=EF}$

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2125 posts

#12 h Dec 24, 2018, 5:27 PM • 2 Y

Y by Adventure10, Mango247

:mad:

Pretty angry at myself after looking at other people's solution
Here's my try........
$AD=s-a$ and $EF=\frac{AF \cos \tfrac{A}{2}}{\cos \tfrac{B}{2}}=\frac{b\sin \tfrac{C}{2} \cos \tfrac{A}{2}}{\cos \tfrac{B}{2}}$
Let's assume $AD=EF$ , therefore, we then have on a bit simplification, $b \left( \sin C \tan \tfrac{B}{2} -\cos C \right)=c-a \implies \sin C \frac{\sin \tfrac{B}{2} }{\cos \tfrac{B}{2}} + \frac{\sin A}{\sin B}= \cos C +\frac{\sin C}{\sin B} \implies \cos \tfrac{B}{2} (\sin A -\sin C)=\sin B \cos \left( \frac{B}{2}+C \right)$
Simplifying further, $2 \sin \tfrac{B}{2} \cos \tfrac{B}{2} \sin \left( \frac{A-C}{2} \right)= \sin B \cos \left( \frac{B}{2} +C \right) \implies \sin \left( \frac{A-C}{2} \right) =\cos \left( \frac{B}{2} +C \right)=\cos \left( 90^{\circ}- \left( \frac{A-C}{2} \right) \right)= \sin \left( \frac{A-C}{2} \right)$ Which is absolutely true, hence, $AD=EF$ is also true!!

@below
Thanks! I will make sure this doesn't happen next time!!

This post has been edited 2 times. Last edited by AlastorMoody, Dec 24, 2018, 6:28 PM

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#13 h Dec 24, 2018, 6:06 PM • 3 Y

Y by AlastorMoody, PhysicsMonster_01, Adventure10

@above: that's fallacious reasoning! You can't start with the thing you want to prove and simplify it to an expression that's trivially true, as this is circular reasoning! In order for this proof to be valid, you must start with $\sin(\tfrac{A-C}2) = \sin(\tfrac{A-C}2)$ and work your way up to $b \left( \sin C \tan \tfrac{B}{2} -\cos C \right)=c-a$ . (Fortunately, all of your steps seem to be reversible, so this doesn't really affect the validity of your proof.)

This may seem like a pedantic point, but it can lead to incorrect proofs when the steps you take are not reversible. For example, I can write
$-1 = 1 \Rightarrow (-1)^2 = 1^2\Rightarrow 1 = 1$ and conclude that since $1 = 1$ is true, it must follow that $-1 = 1$ is true.

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Ilove_mathematics

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Ilove_mathematics

#14 h Dec 24, 2018, 6:47 PM • 4 Y

Y by AlastorMoody, Mathcat1234, Adventure10, Mango247

Only ANGLE CHASING:

Let $I$ be $CC_1 \cap BB_1$ , and let $\omega$ be the circle with diameter $AI.$ As, $\angle AFI = \angle AEI = 90º$ , then $\omega$ passes through $A, I, E, F.$ Now, just see that $AB$ is tangent to the incircle of $\triangle ABC$ at $D$ and the incircle has center $I$ , then $ID \perp AB \implies \angle ADI = 90º.$ Thus, $\omega$ passes through $A, I, D, E, F \implies ADFE$ is cyclic, we want $AD = EF \iff AE || DF \iff \angle BDF = \angle BAE.$

For other side, $\angle BDF = \angle C_1DF = \angle AIF (AIFD$ is cyclic $),$ make $2\alpha = \angle A$ , $2\beta = \angle B$ and $2\gamma = \angle C,$ therefore, $\angle IAC = \angle IAB = \alpha$ and $\angle ICA = \angle ICB = \gamma \implies \angle AIF = \angle IAC + \angle ICA = \alpha+\gamma,$ too $\angle BAE = 90º - \angle ABE = 90º-\beta$ , but $2\alpha + 2\beta + 2\gamma = 180º \implies \beta = \frac{180º-2\alpha-2\gamma}{2} = 90º-\alpha-\gamma$ , so $\angle BAE = 90º-\beta = 90º-(90º-\alpha-\gamma) = \alpha + \gamma.$ Hence, $\angle BAE = \angle BDF.$ As we desired.

This post has been edited 1 time. Last edited by Ilove_mathematics, Dec 25, 2018, 12:54 AM

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#15 h Dec 24, 2018, 6:50 PM • 2 Y

Y by Adventure10, Mango247

realy lovely,,very nice Ilove_mathematique

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#16 h Dec 25, 2018, 7:57 AM • 1 Y

Y by Adventure10

Let the line through A parallel to BC meet line BM at J. Let the line through J parallel to AB meet line BC at K. Let MN hit AB at X and AC at Y.

We will show that MX = AB/2, XY = BC/ 2, and YN = AC/2, from which the desired result follows.

Lemma 1: JKBA is a rhombus.
Proof: Since JK || AB and AJ || BK, JKBA is a parallelogram. Hence, we need only prove that a pair of consecutive sides are equal to conclude JKBA is a rhombus.

Since

<ABM = (<A + <C)/2.

From right triangle BAM, we find

<BAM = 90 - <ABM = <B/2.

Since AJ || BC, we have <JAB = <B, so MA bisects <BAJ. Thus, <JAM = <BAM and triangles BAM and JAM are congruent by ASA. Thus, AJ = AB and parallelogram JKBA is a rhombus.

end lemma

Lemma 2: MN || BC and MN is equidistant from lines AJ and BC.
Proof: The diagonals of a rhombus bisect each other, so triangles AMJ and KMB are congruent. Thus, the altitudes from M to AJ and BK are equal and M is equidistant from lines AJ and line BK. By symmetry, this is also true for N. Thus, MN || BC and MN is equidistant from AJ and BC.

end lemma

Lemma 3 : MX = AB/2.
Since MX || BK, triangle AMX ~ triangle AKB. Since AM = AK/2, we have MX = KB/2. Since JKBA is a rhombus (Lemma 1), KB = AB, so MX = AB/2, as desired.

end lemma

By symmetry, we also have NY = AC/2 from Lemma 3. Since XY is the midline of ABC parallel to side BC, we have XY = BC/2. Thus,

MN = MX + XY + YN = AB/2 + AC/2 + BC/2,

as desired.

(Solution method found by community member fanzha in the Olympiad Geometry class)

https://artofproblemsolving.com/articles/images/fangeom3.gif

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jayme

#17 h Dec 25, 2018, 10:23 AM • 2 Y

Y by AlastorMoody, Adventure10

Dear Mathlinkers,

1. (U) the circle with diameter AB
2. EF is also a diameter line of (U) and intersect (U) at X
3. (V) the circle with diameter AI
4. Reim's theorem : DE// BX which is also // to AE
5. AFDE is a cyclic trapez and we are done...

Sincerely
Merry Christmas
Jean-Louis

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#18 h Oct 14, 2019, 6:20 AM • 1 Y

Y by Adventure10

Solution:

Let $I$ be the incenter of $\Delta ABC$

Now, $\angle EAI = 90 - \frac{A+C}{2} = \frac{B}{2}$ and $\angle FAI = 90 - \frac{A+B}{2} = \frac{C}{2}$ .

So, $AE = AI \cos {\frac{C}{2}}$ and $AF = AI \cos {\frac{B}{2}}$
Also, $EI = AI \sin {\frac{C}{2}}$ and $FI = AI \sin {\frac{B}{2}}$

As $AEIF$ is cyclic, we can use Ptolmey Theorem to get
$AI \cdot EF = AE \cdot FI + AF \cdot EI$
$EF = \frac{ AI \sin {\frac{B}{2}} \cdot AI \cos {\frac{C}{2}} + AI \sin {\frac{C}{2}} \cdot AI \cos {\frac{B}{2}}}{AI}$
$EF = AI \cdot \sin{\frac{B+C}{2}} = AI \cos {\frac{A}{2}}$
Now, $AI \cos {\frac{A}{2}}$ is the projection of $AI$ on $AB$ which is equal to $AD$ as $I$ is the incenter $\blacksquare$

This post has been edited 1 time. Last edited by Megamind123, Oct 14, 2019, 6:21 AM

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