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Easy Geometry (Again)

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Source: 2017 KMO Problem 3

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1721 posts

#1 h Nov 12, 2017, 10:07 AM • 5 Y

Y by CZRorz, tenplusten, HoRI_DA_GRe8, samrocksnature, Adventure10

Let there be a scalene triangle $ABC$ , and its incircle hits $BC, CA, AB$ at $D, E, F$ . The perpendicular bisector of $BC$ meets the circumcircle of $ABC$ at $P, Q$ , where $P$ is on the same side with $A$ with respect to $BC$ . Let the line parallel to $AQ$ and passing through $D$ meet $EF$ at $R$ . Prove that the intersection between $EF$ and $PQ$ lies on the circumcircle of $BCR$ .

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1594 posts

#2 h Nov 12, 2017, 10:18 AM • 4 Y

Y by tenplusten, Pluto1708, Adventure10, Mango247

Let $K=EF\cap PQ$ . Obviously $A, I, Q$ are colinear where $I$ is the incenter. This implies that $DR\perp EF$ and $DRKM$ is cyclic.

Now let $T=EF\cap BC$ . Then $TR\cdot TK=TD\cdot TM$ and it suffices to show that $TB\cdot TC = TD\cdot TM$ . To do that, let $\Delta DB_1C_1$ be the orthic triangle of $\Delta BIC$ . Recall that $B_1, C_1\in EF$ (Iran) and quadrilaterals $BB_1C_1C$ and $B_1C_1DM$ are cyclic. Hence $TB\cdot TC=TB_1\cdot TC_1=TD\cdot TM$ as desired.

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1721 posts

#3 h Nov 12, 2017, 10:23 AM • 2 Y

Y by Adventure10, Mango247

To show $TB \cdot TC = TD \cdot TM$ , just notice the harmonic bundle.

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1594 posts

#4 h Nov 12, 2017, 10:29 AM • 1 Y

Y by Adventure10

Oops, I am a bit hasty

. Note that $(B,C;D,T)=-1$ , therefore $MD\cdot MT=MB^2$ hence
$TD\cdot TM = TM^2-MD\cdot MT = (TM+MB)\cdot(TM-MB)=TB\cdot TC$ as desired.

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1979 posts

#5 h Nov 12, 2017, 10:41 AM • 3 Y

Y by vsathiam, HoRI_DA_GRe8, Adventure10

It is clear that $\overline{DR} \perp \overline{EF}$ . Now we just need $\overline{EF}$ to be the external bisector of $\angle BRC$ ; however that follows since $(B,C,D, \overline{EF} \cap \overline{BC})=-1$ .

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PARISsaintGERMAIN

246 posts

PARISsaintGERMAIN

#6 h Nov 12, 2017, 7:25 PM • 3 Y

Y by benstein, vsathiam, Adventure10

Always sad that Korean geo problems are much easier than US ones.
Also sad that I cannot take KMO anymore since I moved to the US.

I tried this problem and it was really easy to prove by length bash. Not even a bash.

Outline: Let $X=PQ\cap EF, Y=EF\cap PQ.$

WTS: $YR\cdot YX=YB\cdot YC$

Since $(R,D,M,X)$ is cyclic where $M$ is the midpoint of the $BC,$ we want to prove that $YM\cdot YD=YB\cdot YC.$

Let $BC=a, CA=b, AB=c.$ Then by Menelaus on $\triangle ABC$ and line $YRX$ , we get $YB=\frac{a(c+a-b)}{2(b-c)}.$

$\Leftrightarrow YD\cdot YM=YB YC$
$\Leftrightarrow (YB+BD)(YC-CM)=YB\cdot YC$
$\Leftrightarrow BD\cdot YC=YB\cdot CM+ BD\cdot CM$
$\Leftrightarrow BD(YB+YC)=YB\cdot CM+ BD\cdot CM$
$\Leftrightarrow BD\cdot BC=YB(CM-BD)+ BD\cdot CM$
$\Leftrightarrow BD\cdot BC=YB\cdot DM+ BD\cdot BM$
$\Leftrightarrow BD\cdot BM=YB\cdot DM$
$\Leftrightarrow \frac{(c+a-b)}{2}\cdot \frac{a}{2}=\frac{a(c+a-b)}{2(b-c)}\cdot \frac{b-c}{2}$

And we are done.

This post has been edited 1 time. Last edited by PARISsaintGERMAIN, Nov 12, 2017, 7:26 PM

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68 posts

#7 h Nov 13, 2017, 5:38 AM • 2 Y

Y by Adventure10, Mango247

Without Projective

This post has been edited 1 time. Last edited by MeineMeinung, Nov 13, 2017, 5:39 AM
Reason: Changing some terms

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91 posts

#8 h Jan 15, 2018, 9:28 PM • 2 Y

Y by Adventure10, Mango247

Lemma 9.17 EGMO kills it

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756 posts

#9 h Jul 7, 2019, 4:57 PM • 2 Y

Y by Adventure10, Mango247

Sweet problem

rkm0959 wrote:

Let there be a scalene triangle $ABC$ , and its incircle hits $BC, CA, AB$ at $D, E, F$ . The perpendicular bisector of $BC$ meets the circumcircle of $ABC$ at $P, Q$ , where $P$ is on the same side with $A$ with respect to $BC$ . Let the line parallel to $AQ$ and passing through $D$ meet $EF$ at $R$ . Prove that the intersection between $EF$ and $PQ$ lies on the circumcircle of $BCR$ .

Denote $\odot (BCR)$ by $\Omega$ . Also, let $T = \overline{RD} \cap \Omega$ .Consider the following claim:
Lemma: $T$ is the mid-point of arc $\widehat{BC}$ of $\Omega$ not containing $R$ .
Proof: Let $\overline{EF} \cap \overline{BC} = K$ . Then it is obvious that $(KD;BC) = -1$ . Also the given condition $\overline{DR} \parallel \overline{AQ} \implies \overline{DR} \perp \overline{EF}$ . Now both these combined imply that $\overline{DT}$ is the internal angle bisector of $\angle BRC \iff T$ is the mid-point of arc $\widehat{BC}$ of $\Omega$ not containing $R$ .
$\text{[math]}$
Hence, $TB = TC \implies T \in \overline{PQ}$ . Now $\angle SRT = 90^{\circ} \implies S$ is the antipode of $T$ wrt $\Omega$ .

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1915 posts

#10 h Dec 31, 2019, 6:56 PM • 4 Y

Y by GeoMetrix, Kamran011, strawberry_circle, Adventure10

As this is my first solve after 2020 , so I'll post my solution even if it is same as others.

2017 KJMO P3 wrote:

Let there be a scalene triangle $ABC$ , and its incircle hits $BC, CA, AB$ at $D, E, F$ . The perpendicular bisector of $BC$ meets the circumcircle of $ABC$ at $P, Q$ , where $P$ is on the same side with $A$ with respect to $BC$ . Let the line parallel to $AQ$ and passing through $D$ meet $EF$ at $R$ . Prove that the intersection between $EF$ and $PQ$ lies on the circumcircle of $BCR$ .

Note that $AD,BE,CF$ are concurrent at the Gregonne Point of $\triangle ABC$ . Hence, if $EF\cap BC=K$ . Then $(K.D;B,C)=-1$ . Let $EF\cap PQ=T$ also notice that $PQ$ bisects $BC$ let $PQ\cap BC=M$ . Also $\overline{A-I-Q}$ where $I$ is the incenter of $\triangle ABC$ . So, $\angle DRT=90^\circ$ . So Combining Maclaurin and PoP we get that $KB.KC=KD.KM=KR.KT\implies T=EF\cap PQ\in\odot(BCR).\blacksquare$

This post has been edited 6 times. Last edited by amar_04, Dec 31, 2019, 8:35 PM

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678 posts

#11 h Dec 31, 2019, 7:53 PM • 3 Y

Y by amar_04, Adventure10, Mango247

It's still 2019 in Azerbaijan !

And it's about 15-20 minutes left to $\text{2020}$ , so I hurry up

(Sorry for having a similar solution , HnY

)

Let $T$ be the midpoint of $BC$ and let $PQ\cap{EF}=X$
$EF\cap{BC}=Z$
$(\xi)$ $ZD.ZT=ZB.ZC$
Proof
Also clearly $Q$ is the midpoint of the arc $\overarc{BC}$ , so $\overline{A-I-Q}$
And $AQ\perp{EF}$ , thus $DR\perp{EF}\rightarrow{DRXT}$ is cyclic .

$\text{PoP}\rightarrow{ZD.ZT=ZR.ZX}$ $(\omega)$
Combining $(\xi)$ and $(\omega)$
$ZR.ZX=ZB.ZC$ $\blacksquare$

This post has been edited 2 times. Last edited by Kamran011, Jan 1, 2020, 6:52 AM

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1080 posts

#12 h Nov 23, 2021, 11:02 PM • 1 Y

Y by centslordm

Let $M$ be the midpoint of $\overline{BC},$ and $X,Y,$ and $Z$ the intersection of $\overline{EF}$ with $\overline{PQ},\overline{BC},$ and $\overline{AQ},$ respectively. Notice that $\triangle AFZ\cong\triangle AEZ$ so $\measuredangle XMD=90=\measuredangle RZQ=\measuredangle XRD$ and $XRDM$ is cyclic. By the Midpoint Lengths Lemma and Power of a Point, $YB\cdot YC=YD\cdot YM=YR\cdot YX$ and $BRXC$ is cyclic. $\square$

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