IMO ShortList 2002, geometry problem 8

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IMO ShortList 2002, geometry problem 8

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Source: IMO ShortList 2002, geometry problem 8

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3647 posts

orl

#1 h Sep 28, 2004, 12:59 PM • 2 Y

Y by Adventure10, Mango247

Let two circles $S_{1}$ and $S_{2}$ meet at the points $A$ and $B$ . A line through $A$ meets $S_{1}$ again at $C$ and $S_{2}$ again at $D$ . Let $M$ , $N$ , $K$ be three points on the line segments $CD$ , $BC$ , $BD$ respectively, with $MN$ parallel to $BD$ and $MK$ parallel to $BC$ . Let $E$ and $F$ be points on those arcs $BC$ of $S_{1}$ and $BD$ of $S_{2}$ respectively that do not contain $A$ . Given that $EN$ is perpendicular to $BC$ and $FK$ is perpendicular to $BD$ prove that $\angle EMF=90^{\circ}$ .

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orl

#2 h Sep 28, 2004, 1:00 PM • 2 Y

Y by Adventure10, Mango247

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

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#3 h Sep 29, 2004, 5:12 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Let $G=NE\cap S_1,\ X=NE\cap MK,\ Y=KF\cap MN$ . The quadrilateral $XNKY$ is ccylic because $\angle MXN=\angle MYK=\frac \pi 2$ , so $\angle MNE=\angle FKM\ (*)$ . If we show that the triangles $MNE,FKM$ are similar, then we would have $\angle NMK=\pi-\angle YMK=\frac \pi 2+\angle MKY=\frac \pi 2+\angle NME+\angle KMF$ , which is exactly what we want.

We thus need to show that $\frac{MN}{NE}=\frac{FK}{KM}\ (**)$ , which, together with $(*)$ , proves the similarity of $MNE,FKM$ .

$(**)\iff NE\cdot KF=BN\cdot BK\ (***)$ , because $BN=MK,\ BK=MN$ . Since all triangles $BCD$ are similar, no matter which line $CD$ we choose, there is a spiral similarity $\mathcal S$ cantered at $B$ which maps $D\to C$ . $\mathcal S$ maps $BD\to BC$ , so if $K'=\mathcal S(K)$ , then $K'\in BC$ and $\frac{BK'}{K'C}=\frac{BK}{KC}=\frac{CN}{NB}$ , meaning that $K'$ is the symmetric of $N$ wrt the perpendicular bisector of $BC$ .

We then have $NE\cdot NG=BN\cdot NC=BN\cdot BK'$ , and this proves $(***)$ because $\frac{K'F'}{KF}=\frac{BK'}{BK}$ and $K'F'=NG$ , so $\frac{NG}{KF}=\frac{BK'}{BK}$ .

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The QuattoMaster 6000

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The QuattoMaster 6000

#4 h Nov 21, 2008, 2:48 AM • 2 Y

Y by Adventure10 and 1 other user

Alternate Solution

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77ant

#5 h Nov 21, 2008, 4:16 AM • 2 Y

Y by Adventure10, Mango247

Strangely, $\angle EMF\neq 90^{\circ}$ in my figure.
Why does it not hold in my figure? please help me

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The QuattoMaster 6000

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The QuattoMaster 6000

#6 h Nov 21, 2008, 5:44 AM • 1 Y

Y by Adventure10

77ant wrote:

Strangely, $\angle EMF\neq 90^{\circ}$ in my figure.
Why does it not hold in my figure? please help me

I think that $E$ has to be on the major arc $CB$ .

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9782 posts

jayme

#7 h Jul 8, 2010, 12:25 PM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
this problem appears also many time on Mathlinks.
I research a synthetic proof without calculation or transformations...
Is there one?
Sincerely
Jean-Louis

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4302 posts

#8 h Dec 28, 2010, 12:28 AM • 2 Y

Y by Adventure10, Mango247

Hmm.... the problem seems to require $A$ to lie in between $C$ and $D$ ; I think 77ant's diagram satisfies all the conditions.

Here's a more straightforward way to establish (***) from grobber's solution...

The relation is equivalent to $\triangle{ENB}\sim\triangle{BKF}$ , or $\angle{EBN}+\angle{FBK}=90^\circ$ . By the Law of Sines on $\triangle{BDF}$ , we have
$\begin{align*} \frac{BD}{\sin\angle{BAD}}=\frac{BF}{\sin\angle{BDF}} &=\frac{BF}{\sin(\angle{BAD}-\angle{FBK})}\\ &=\frac{BK}{\cos\angle{FBK}\sin(\angle{BAD}-\angle{FBK})}. \end{align*}$ By symmetry,
$\begin{align*} \frac{BC}{\sin\angle{BAD}}=\frac{BC}{\sin\angle{BAC}} &=\frac{BN}{\cos\angle{EBN}\sin(\angle{BAC}-\angle{EBN})}\\ &=\frac{BN}{\cos\angle{EBN}\sin(\angle{BAD}+\angle{EBN})}. \end{align*}$ Thus $NM\|BD$ gives us
$1=\frac{CN}{CB}+\frac{BN}{BC}=\frac{NM}{BD}+\frac{BN}{BC}=\frac{BK}{BD}+\frac{BN}{BC},$ whence plugging in and simplifying yields
$\cos(\angle{FBK}+\angle{EBN})\sin(\angle{BAD}-\angle{FBK}+\angle{EBN})=0.$ But
$0=\sin(\angle{BAD}-\angle{FBK}+\angle{EBN})=\sin(180^\circ-\angle{BCE}-\angle{KBF})$ is impossible since $0<\angle{BCE},\angle{KBF}<90^\circ$ . Thus $\angle{FBK}+\angle{EBN}=90^\circ$ , as desired.

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2043 posts

Zhero

#9 h Apr 17, 2011, 7:24 PM • 2 Y

Y by Adventure10, Mango247

All lengths and angles in this proof will be directed.

Lemma 1: $\triangle CEN \sim \triangle FDK$ .
Proof: Let $FK$ intersect $S_2$ at $F'$ . Note that $F'K \perp BD$ , $\frac{DK}{KB} = \frac{DM}{MC} = \frac{BN}{NC}$ , and $\angle BF'D = \angle BAD = \angle BEC$ , so $\triangle ECB \sim \triangle F'BD$ . Hence, $\angle ECN = \angle ECB = \angle F'BD = \angle F'FD = \angle KFD$ , and $\angle FKD = \angle CNE = 90^{\circ}$ , so $\triangle CEN \sim \triangle FDK$ .

Lemma 2: $\frac{DM}{MC} \cdot \frac{CE}{EB} \cdot \frac{BF}{FD} = 1$ .
Proof: Let $\theta = \angle CAB$ and $\alpha = \angle NEC$ . It is easy to see that $\angle ECB = 90^{\circ} - \alpha$ , $\angle BEN = 180^{\circ} - \theta - \alpha$ , $\angle CBE = \theta + \alpha - 90^{\circ}$ , $\angle BDF = \alpha$ (by lemma 1), and $\angle FBD = 180^{\circ} - \theta - \alpha$ . Hence,
$\begin{align*} \frac{DM}{MC} \cdot \frac{CE}{EB} \cdot \frac{BD}{FD} &= \frac{DM}{MC} \cdot \frac{\sin(\theta + \alpha - 90^{\circ})}{\sin(90^{\circ} - \alpha)} \cdot \frac{\sin \alpha}{\sin(180^{\circ} - \theta - \alpha)} \\ &= \frac{DM}{MC} \cdot \frac{- \cos(\theta + \alpha) \sin \alpha }{ \sin(\theta + \alpha) \cos \alpha }\\ &= \frac{DM}{MC} \cdot \frac{-\cot(\alpha + \theta)}{\cot \alpha} \\ &= \frac{DM}{MC} \cdot \frac{ \cot(180^{\circ} - \theta - \alpha) }{\cot \alpha} \\ &= \frac{DM}{MC} \cdot \frac{ \cot \angle BEN }{ \cot \angle NEB } \\ &= \frac{DM}{MC} \cdot \frac{CN}{NB} \\ &= 1. \end{align*}$

We have $\angle EAF = \angle EAB + \angle BAF = \angle ECB + \angle BDF = \angle ECN + \angle KDF = 90^{\circ}$ , since $\triangle CEN \sim \triangle FDK$ . Hence, it is sufficient to show that $AMFE$ is cyclic.

Consider an inversion about $A$ with radius 1. For any point $X$ , let $X'$ denote the image of $X$ under this inversion. We wish to show that $E'$ , $F'$ , and $M'$ are collinear. $E'$ lies on $B'C'$ and $F'$ lies on $B'D'$ , so by Menelaus's theorem, it is sufficient to show
$\frac{C'E'}{E'B'} \cdot \frac{B'F'}{F'D'} \cdot \frac{D'M'}{M'C'} = -1.$
Note that
$\begin{align*} \frac{C'E'}{E'B'} = \frac{CE \cdot C'A \cdot E'A }{EB \cdot E'A \cdot B'A} = \frac{CE}{EB} \cdot \frac{BA}{CA} \tag{1} \end{align*}$
and
$\begin{align*} \frac{B'F'}{F'D'} = \frac{BF \cdot B'A \cdot F'A }{FD \cdot F'A \cdot D'A} = \frac{BF}{FD} \cdot \frac{DA}{BA} \tag{2} \end{align*}$
Let $CD$ intersect the line at infinity at $\infty$ . We have $(M, \infty; C, D) = \frac{MC}{MD}$ . Because cross-ratios are preserved under inversion, we have
$\frac{MC}{MD} = (M', A; C', D') = \frac{\frac{M'C'}{M'D'}}{\frac{AC'}{AD'}},$
or
$\begin{align*} \frac{D'M'}{M'C'} = \frac{DM}{MC} \cdot \frac{AC}{AD}. \tag{3} \end{align*}$
Combining (1), (2), (3), and lemma 2, we have
$\begin{align*} \frac{C'E'}{E'B'} \cdot \frac{B'F'}{F'D'} \cdot \frac{D'M'}{M'C'} &= \left( \frac{CE}{EB} \cdot \frac{BA}{BC} \right) \cdot \left(\frac{BF}{FD} \cdot \frac{DA}{BA} \right) \cdot \left( \frac{DM}{MC} \cdot \frac{AC}{AD} \right) \\ &= \left( \frac{CE}{EB} \cdot \frac{BF}{FD} \cdot \frac{DM}{MC} \right) \cdot \left( \frac{BA}{BC} \cdot \frac{DA}{BA} \cdot \frac{AC}{AD} \right) \\ &= -1, \end{align*}$
as desired.

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1343 posts

jgnr

#10 h Jun 18, 2011, 6:44 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

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#11 h Jul 9, 2011, 5:51 AM • 1 Y

Y by Adventure10

Solution

Let $X$ be the point such that $E$ and $X$ are on opposite sides of $BC$ and $\triangle{CXB} \sim \triangle{DFB}$ . First note that since $C$ , $A$ and $D$ are collinear, $\angle{CXB}=\angle{DFB}=\angle{CAB}=180^\circ - \angle{CEB}$ . Therefore $XBEC$ is cyclic. Let $X'$ be the point on $S_1$ such that $XX' \| BC$ and let $Y$ and $Y'$ be the projections of $X$ and $X'$ , respectively, onto $BC$ .

Now note that since the sides of $\triangle{MNC}$ and $\triangle{DKM}$ are parallel, it follows that

$\frac{CN}{NB}=\frac{BK}{KD}=\frac{BY}{YC}=\frac{CY'}{BY'}$
since $BXX'C$ is an isosceles trapezoid and $DFBK$ , $CXBY$ and $BX'CY'$ are similar. The above equality implies that $Y'=N$ and therefore that $X'$ lies on $EN$ . Since $BXX'C$ is an isosceles trapezoid, it follows that $BX=X'C$ and that $EX$ is the reflection of $EX'$ around the bisector of $\angle{CEB}$ . Since the orthocenter of $\triangle{CEB}$ , which lies on $EX'$ since $EX' \perp BC$ , and the circumcenter of $\triangle{CEB}$ are isogonal conjugates, $EX$ is a diameter of $S_1$ . This implies that

$\angle{DBF}=\angle{CBX}=90^\circ - \angle{CBE}=\angle{NEB}$
Therefore $\triangle{NEB} \sim \triangle{KBF}$ which implies that

$\frac{EN}{KB} = \frac{BK}{KF} \quad \Rightarrow \quad \frac{EN}{NM}=\frac{MK}{KF}$
since $MNBK$ is a parallelogram. This also implies that $\angle{MNE}=\angle{MKF}=90^\circ +\angle{CBD}$ . Hence $\triangle{MNE} \sim \triangle{FKM}$ which implies that

$\angle{EMF}=\angle{NMK}+\angle{EMN}+\angle{MFK}=\angle{CBD}+\angle{EMN}+\angle{MEN}=90^\circ$
Hence $\angle{EMF}=90^\circ$ as desired. $\blacksquare$

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1415 posts

#12 h Dec 30, 2011, 3:38 AM • 2 Y

Y by Adventure10, Mango247

Here is my proof. I hope I didn't make mistakes in using directed angles.

My Solution

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#13 h Aug 21, 2012, 4:47 PM • 2 Y

Y by Adventure10, Mango247

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#14 h Jun 3, 2013, 11:31 AM • 2 Y

Y by Adventure10, Mango247

We first prove that $\angle EBC + \angle FBD = 90^{\circ}$ , and $\angle ECB + \angle FDB = 90^{\circ}$ .
Proof of the previous assertion

Now, this directly implies that $\angle EAF = \angle EAB + \angle BAF = \angle ECB + \angle FDB = 90^{\circ}$ . So, we only need to prove that $E,M,A,F$ are concyclic.

Now, let $X = MK \cap EB$ . Then $\angle XMA = \angle ACB = \angle AEB = \angle AEX$ , hence $M,A,X,E$ are concyclic.

Now, using the fact that $\triangle EBN \sim \triangle BFK$ , we get $\frac{EB}{BF} = \frac{BN}{FK} = \frac{MK}{KF}$ Combining with $\angle EBF = \angle MKF$ , we get $\triangle MKF \sim \triangle EBF$ .

Thus, $\angle XEF = \angle XMF$ , which gives $X,A,M,E,F$ concyclic, and we're done.

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2079 posts

pi37

#15 h May 18, 2016, 7:13 PM • 1 Y

Y by Adventure10

Let $E_1$ , $F_1$ be the antipodes of $E,F$ with respect to their corresponding circles. It's not hard to see that the spiral similarity centered about $B$ sending $BC\cup S_1$ to $BD\cup S_2$ also sends $E_1$ to $F$ and $E$ to $F_1$ . Therefore $A,E_1,F$ are collinear and $\angle EAF=90^\circ$ . Now note that
$\frac{BN}{CN} = \frac{\cot\angle EBN}{\cot \angle ECN} = \frac{\sin \angle ECN}{\sin \angle EBN}\cdot \frac{\sin \angle CBE_1}{\sin \angle BCE_1} = (E,E_1;C,B)=(AE,AF;CD,AB)$ Now $\frac{BN}{CN}=\frac{DM}{CM}$ . Consider an inversion about $A$ , which maps $E,F$ to points $E',F'$ on lines $B'C'$ and $B'D'$ . We want to show that $E',F',$ and $M'$ are collinear. But
$(C',D';A,M')=\frac{DM}{CM}=(AE,AF;CD,AB)=(AE',AF';C'D',AB')=(E',F';AB' \cap E'F', C'D' \cap E'F') = (C',D';A,C'D' \cap E'F')$ so $C'D'$ meets $E'F'$ at $M'$ , as desired.

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ABCDE

#16 h May 19, 2016, 2:12 AM • 2 Y

Y by Adventure10, Mango247

Probably the easiest G8

Let $EN$ intersect the circumcircle of $BCE$ again at $E'$ , $EF'$ be a diameter of the circumcircle of $BCE$ , and let $K'$ be the foot of the altitude from $F'$ to $BC$ . Note that $BN=CK'$ and $E'N=F'K'$ . Note that $\angle BF'C=\angle BAC=180-\angle BAD=\angle BFD$ . Since $\frac{BK'}{K'C}=\frac{CN}{NB}=\frac{CM}{MD}=\frac{BK}{KD}$ , $BF'CK'$ and $BFDK$ are similar. Now, note that $\frac{EN}{NC}=\frac{BN}{NE'}=\frac{CK'}{K'F'}=\frac{DK}{KF}$ , and multiplying both sides by $\frac{BC}{BD}$ we have that $\frac{EN}{NM}=\frac{MK}{KF}$ . Since $\angle ENM=\angle MKF=90+\angle CBD$ , triangles $NEM$ and $KMF$ are similar. But since $MK\parallel BC\perp EN$ and $MN\parallel BC\perp FK$ , $ME\perp MF$ as desired.

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#17 h May 19, 2016, 2:26 AM • 2 Y

Y by Adventure10, Mango247

Let $EN$ intersect $S_1$ again at $E'$ . Note that $\angle CE'B=\angle CAB=\angle DFB$ , and $\dfrac{DK}{KB}=\dfrac{BN}{NC}$ . This implies that $\triangle DFB\sim \triangle BE'C$ , which you can see by fixing $B$ and $C$ on a circle and varying $A$ along an arc; clearly $\dfrac{BH}{HC}$ determines the location of $A$ where $H$ is the foot of the altitude from $A$ . Then, $\angle FBD=\angle E'CB=90-\angle CBE\implies \dfrac{FK}{KB}=\dfrac{BN}{NE}\implies \dfrac{FK}{MN}=\dfrac{KM}{NE}$ . Since $\angle FKM=90+\angle DKM=90+\angle MNC=\angle MNE$ , so $\triangle ENM\sim \triangle MKE$ . Then $\angle FME=\angle FMK+\angle KMN+\angle NME=\angle FMK+\angle DKM+\angle NFK=180-\angle FND=90$ .

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290 posts

jred

#18 h Nov 20, 2018, 3:40 AM • 1 Y

Y by Adventure10

The problem itself is not hard. However, we must point out that the claim is true only when $A$ lies on segment $CD$ . Otherwise, either $E$ or $F$ should lie on the arc containing $A$ to keep the claim true.

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#19 h Dec 29, 2019, 3:25 PM • 1 Y

Y by Adventure10

Solution

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awesomeming327.

1698 posts

awesomeming327.

#20 h Mar 20, 2022, 3:50 AM

Y by

2001 G8 was easier

https://media.discordapp.net/attachments/855230423172907008/954945615563989062/Screen_Shot_2022-03-19_at_8.32.51_PM.png

Note that $\frac{DK}{KM}=\frac{MN}{NC}$ so $\frac{DK}{KB}=\frac{BN}{NC}.$ Extend $FK$ to intersect $S_2$ at $H.$ Note that $\angle BHD=\angle BAD=\angle CEB.$ Let the transformation $S$ be from $C,B$ to $B,D$ respectively consisting of one reflection, preserving all angles and, by extension, length ratios. $N$ goes to $K$ and so the line $EN$ goes to the line $HK.$ Since $\angle CEB=\angle BHD,$ $E$ goes to $H.$

Thus, $\angle NBE=\angle HDB=\angle KFB.$ Therefore, $\angle NBE+\angle KBF=90^\circ.$ This implies that $\triangle NBE\sim\triangle KFB$ so $\frac{NE}{MK}=\frac{MK}{KF}.$ Additionally, by angle chasing $\angle MNE=\angle FKM$ so $\triangle MNE\sim\triangle FKM.$ Now, $\angle NME+\angle KMF=90^\circ-\angle MKB=\angle NMK-90^\circ,$ so we are done.

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#21 h Jul 21, 2022, 1:21 PM • 1 Y

Y by Mango247

Let $FK$ meet $S_2$ again at $G,G'$ is reflection of $G$ across perpendicular bisector of $BD.$
Obviously $\measuredangle MNE=\measuredangle FKM$ and $|BN|:|NC|=|DM|:|MC|=|DK|:|KB|,$ so by spiral similarity $BEC\stackrel{+}{\sim} BG'D\stackrel{-}{\sim} BGD\implies BNE\stackrel{+}{\sim} FKB\implies \frac{|MN|}{|FK|}=\frac{|BK|}{|FK|}=\frac{|EN|}{|BN|}=\frac{|EN|}{|KM|}\implies$ $\implies MNE\stackrel{+}{\sim} FKM\implies \measuredangle EMF=\angle (EN,MK)=\angle (EN,BC)=90^\circ \text{ } \blacksquare$

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#22 h Mar 6, 2024, 12:00 PM

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nice problem
solution
remark

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