IMO 2014 Problem 4

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2 posts

#1 h Jul 9, 2014, 11:38 AM • 19 Y

Y by codyj, Davi-8191, Abdollahpour, itslumi, jhu08, mathematicsy, megarnie, Lamboreghini, Adventure10, Mango247, Rounak_iitr, Shadow6885, and 7 other users

Let $P$ and $Q$ be on segment $BC$ of an acute triangle $ABC$ such that $\angle PAB=\angle BCA$ and $\angle CAQ=\angle ABC$ . Let $M$ and $N$ be the points on $AP$ and $AQ$ , respectively, such that $P$ is the midpoint of $AM$ and $Q$ is the midpoint of $AN$ . Prove that the intersection of $BM$ and $CN$ is on the circumference of triangle $ABC$ .

Proposed by Giorgi Arabidze, Georgia.

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McItran

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McItran

#2 h Jul 9, 2014, 11:50 AM • 13 Y

Y by mathtastic, TinTin028, LeonhardEuler0, jhu08, megarnie, Lamboreghini, Adventure10, Mango247, and 5 other users

Just consider points $B'$ and $C'$ which are symmetric to $B$ and $C$ with respect to $A$ and find two pairs of similar triangles $C'BC$ , $ABM$ and $BCB'$ , $ACN$ .

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m.candales

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m.candales

#3 h Jul 9, 2014, 12:23 PM • 17 Y

Y by Mahi, Pirkuliyev Rovsen, might_guy, MeowX2, TinTin028, AllanTian, jhu08, Lamboreghini, xiaofengfeng, Adventure10, Mango247, Rounak_iitr, and 5 other users

$QP$ and $NM$ are parallel.

$\angle{ANM} = \angle{AQC} = \angle{A}$

$\angle{AMN} = \angle{APB} = \angle{A}$

Let $R$ be the intersection of $BM$ and $CN$ .
Let $X, Y$ be points on $AB, AC$ such that $A$ is the midpoint of $BX$ and $A$ is the midpoint of $YC$
$\triangle{ABC}$ and $\triangle{QAC}$ are similar, then $\triangle{XBC}$ and $\triangle{NAC}$ are similar. Then $\angle{ANC} = \angle{BXC}$
$\triangle{ACB}$ and $\triangle{PAB}$ are similar, then $\triangle{YCB}$ and $\triangle{MAB}$ are similar. Then $\angle{AMB} = \angle{CYB}$
$XCBY$ is a parallelogram. Take $H$ be the midpoint of $BC$ . Then $AH$ is parallel to $YB$ and $XC$ . Then $\angle{A} = \angle{BXC} + \angle{CYB}$

$\angle{RNM} = \angle{ANM} - \angle{ANC} = \angle{A} - \angle{BXC}$

$\angle{RMN} = \angle{RMN} - \angle{AMB} = \angle{A} - \angle{CYB}$

$\angle{RNM} + \angle{RMN} = 2\angle{A} - \angle{BXC} - \angle{CYB} = \angle{A}$

$\angle{BRC} + \angle{A} = 180$

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tchebytchev

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tchebytchev

#4 h Jul 9, 2014, 12:24 PM • 11 Y

Y by sunny2000, Lamboreghini, Adventure10, Mango247, Idontlikecreativenames, and 6 other users

Let $x=<QCN$ and $y=<PBM$ , by similitude we have $\frac{AQ}{AB}=\frac{AC}{BC}=\frac{QC}{AC}$ so $QN=\frac{AB.AC}{BC}$ and $QC=\frac{AC^2}{BC}$ and $<CQN=180-<AQC=180-A$ so in the triangle $QCN$ we have $\frac{\sin(x-A)}{\sin(x)}=\frac{AC}{AB}$ and by the same way $\frac{\sin(y-A)}{\sin(y)}=\frac{AB}{AC}$
so $\sin(x-A)\sin(y-A)=\sin(x)\sin(y)$ so $-\cos(x+y-2A)+\cos(x-y)=2\sin(x)\sin(y)$ and therefore $\cos(x+y-2A)=\cos(x+y)$ which gives $x+y=A$ and if $R$ is the intersection point $<BRC+<BCR=A$

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nima1376

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nima1376

#5 h Jul 9, 2014, 12:25 PM • 10 Y

Y by nzzr2017, Springles, Lamboreghini, Adventure10, Mango247, and 5 other users

very easy problem
$\widehat{BPM}=\widehat{CQN},\frac{BP}{PN}=\frac{BP}{AN}=\frac{\sin C}{\sin B}=\frac{AQ}{QC}=\frac{QN}{QC}$
so $\widehat{CNQ}=\widehat{CBM}$
let the intersection of $BM$ and $CN$ is $S$ .
so $BQSN$ is cycle
so $\widehat{BAC}=\widehat{BQN}=\widehat{BSN}.$
done!

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YS0819

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YS0819

#6 h Jul 9, 2014, 12:50 PM • 11 Y

Y by AstrapiGnosis, myh2910, Lamboreghini, Adventure10, Mango247, and 6 other users

Barycentric coordinates overkill this problem, yet straightfoward and not messy, took me about 5 minutes.
We have similar triangles $PBA$ , $QAC$ , $ABC$ , with area ratios $\frac{c^2}{a^2}, \frac{b^2}{a^2}, 1$ respectively.
Therefore we have the barycentric coordinates $P=(0:1-\frac{c^2}{a^2}:\frac{c^2}{a^2}), Q=(0:\frac{b^2}{a^2}:1-\frac{b^2}{a^2}).$
Now note that the area of $MAB$ is the double of $PAB$ , and area of $MCA$ is the double of $PCA$ . $MBC$ has the same area as $ABC$ , but negative, and do the similar for point $N$ .
One then obtains $M=(-1:2(1-\frac{c^2}{a^2}):\frac{2c^2}{a^2}), N=(-1:\frac{2b^2}{a^2}:2(1-\frac{c^2}{a^2})).$
Lines $BM$ and $CN$ have equations $z=-\frac{2c^2x}{a^2}$ and $y=-\frac{2b^2x}{a^2}$ respectively.
The intersection point has un-homogenized coordinates $(1:-\frac{2b^2x}{a^2}:-\frac{2c^2x}{a^2})$ , which can be easily verified to satisfy the circumcircle equation $a^2yz+b^2xz+c^2xy=0$ .

Yay!

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ThirdTimeLucky

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ThirdTimeLucky

#7 h Jul 9, 2014, 1:12 PM • 6 Y

Y by Adventure10 and 5 other users

Another approach.

Let $O$ be the circumcenter of $\triangle ABC$ . Observe that $OB \perp AP$ , $OC \perp AQ$ . Let $OB \cap AP = L$ , $OC \cap AQ = K$ .

Let $X,Y$ be the midpoints of $AB,AC$ . Let $XP \cap YQ = S$ . Then, we want to show that $S$ lies on circle $(AXY)$ , i.e, the circle $\Gamma$ with diameter $OA$ .

Note that $X,K,L,Y$ lie on $\Gamma$ , so by Pascal's Theorem on $KOLYAX$ , we get $KO \cap YA = C, LO \cap XA = B$ and $LY \cap KX := D$ are collinear, i.e $D$ is on $\overline{BC} = \overline{PQ}$ . Considering the hexagon $KALYSX$ , we have the intersections, $KA \cap SY = Q, LA \cap SX = P$ and $LY \cap KX = D$ are collinear, so by converse of Pascal's Theorem, $S$ is on the unique conic passing through $A,K,L,X,Y$ which is precisely circle $\Gamma$ .

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Davi Medeiros

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Davi Medeiros

#8 h Jul 9, 2014, 1:22 PM • 8 Y

Y by Adventure10, Mango247, and 6 other users

Let $X$ and $Y$ be the midpoints of $AB$ and $AC$ , respectively, and let $Z$ be the intersection between $PX$ and $QY$ . Also, let $\gamma = \angle ACB$ and $\beta = \angle ABC$ . A quick angle chasing show us that $\angle APQ = \angle AQP = 180 -\alpha -\beta$ , and the triangles $ABP$ and $CAQ$ are similar, where we get $\angle YQC = \angle APX = \alpha$ (because both $QY$ and $PX$ are medians of the same corresponding size). Thus, $\angle YPQ=180- \alpha - \beta - \gamma$ and finally:

$\angle XZY = \angle QZP = 180 - \angle ZQP - \angle ZPQ = 180 -\alpha - (180 - \alpha - \beta - \gamma)= \beta + \gamma = 180 - \angle A$ .

So $AXZY$ is cyclic.

Now, let $R$ be the intersection of $BM$ and $CN$ . A homotety of center $A$ and ratio 2 takes $X$ to $B$ , $Y$ to $C$ , $Q$ to $N$ and $P$ to $M$ . So it takes $Z$ to $R$ , which means that $ABRC$ is cyclic as well. This ends the proof.

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Sardor

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Sardor

#9 h Jul 9, 2014, 1:35 PM • 7 Y

Y by Adventure10, Mango247, and 5 other users

It's very nice and easy problem.Let $(BAQ) \cap (CAP) = X$ , and let $AX \cap (ABC)=Y$ , $BM \cap CN=Y$ .We have by angles chasing $AQ=AP$ , $MY || PX$ and $NY || QX$ so the triangles $QXP$ and $NYM$ are similar hamotetic (with ratio 2) .On the other $\angle NYM = \angle QXP=180 - \angle BAC$ so we have $Y$ lies on the circumcircle of the triangle $ABC$ . DONE !

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mathuz

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mathuz

#10 h Jul 9, 2014, 1:59 PM • 9 Y

Y by OlympusHero, Complete_quadrilateral, Adventure10, DroneChaudhary, and 5 other users

we have $ABP$ be similar to $ACQ$ . So $BPM$ is similar to $CQN$ and we are done!

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silouan

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silouan

#11 h Jul 9, 2014, 2:00 PM • 7 Y

Y by OlympusHero, Adventure10, and 5 other users

If you knew the following problem, then you had a big advantage.
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=73134&p=422851#p422851

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thkim1011

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thkim1011

#12 h Jul 9, 2014, 3:23 PM • 8 Y

Y by Creeper1612, Adventure10, and 6 other users

An interesting note to make: Draw the circumcircle and the tangents to it at $B$ and $C$ . Suppose these intersect at $K$ . Obviously $\angle ACK = \angle ABC = \angle CAQ$ . Thus $AQ || KC$ . Similarly, $AP || KB$ . Also, $MN || BC$ so there exists a homothety mapping $KBC$ to $AMN$ , which implies that $BM$ and $CN$ intersect on the symmedian through $A$ .

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v_Enhance

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v_Enhance

#13 h Jul 9, 2014, 3:38 PM • 31 Y

Y by Mewto55555, InCtrl, AstrapiGnosis, aayush-srivastava, Durjoy1729, sa2001, e_plus_pi, TinTin028, char2539, hsiangshen, myh2910, Abidabi, mijail, OlympusHero, HamstPan38825, WGM_RhuanSA, rayfish, RP3.1415, Creeper1612, EpicBird08, Lamboreghini, Adventure10, Bryan0224, and 8 other users

What an excellent barycentric tutorial problem!

Since $PB = c^2/a$ we have $P = (0 : a^2-c^2 : c^2)$ , so $M = (-a^2 : {-} : 2c^2)$ . Similarly $N = (-a^2 : 2b^2 : {-})$ . Thus $\overline{BM} \cap \overline{CN}$ is $(-a^2 : 2b^2 : 2c^2)$ which clearly lies on the circumcircle.

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manuel153

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manuel153

#14 h Jul 9, 2014, 4:04 PM • 9 Y

Y by tove.lo, Creeper1612, Adventure10, Mango247, and 5 other users

A good and easy IMO problem. I saw the cyclic quadrilateral and then the proof is already done. There are many different ways of approaching the problem. A purely analytical solution might be doable but gory.

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AnonymousBunny

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AnonymousBunny

#15 h Jul 9, 2014, 4:25 PM • 9 Y

Y by biomathematics, Adventure10, Rounak_iitr, and 6 other users

Let $BM$ and $CN$ meet at $J.$ From similar triangles $\triangle ABP$ and $\triangle ACQ,$ we have $\dfrac{AP}{BP} = \dfrac{AQ}{QC} \implies \dfrac{PM}{BP} = \dfrac{QM}{QC}.$ Combined with the fact that $\angle BPM = \angle ABC + \angle BCA = \angle QCN,$ we deduce that $\triangle BPM \sim \triangle QCN,$ implying $\angle PBM = \angle QCN.$ Let $\angle PBM = \angle QCN = \theta.$

Again, from $\triangle ABP \sim \triangle ACQ,$ $\dfrac{PM}{BM} = \dfrac{AP}{BP}= \dfrac{AB}{AC}.$ We have that $\angle BPM = \angle ABC + \angle BCA = 180^{\circ} - \angle BAC.$ We then have $\angle BMP = \theta - \angle BAC .$ Similarly, $\angle CNQ =\theta - \angle BAC.$ From sine rule in $\triangle BPM,$ $\dfrac{PM}{\sin \theta} = \dfrac{BP}{\sin (\theta - \angle BAC)}$ and from sine rule in $\triangle CQN,$ $\dfrac{QN}{\sin \theta } = \dfrac{QC}{\sin (\theta - \angle BAC )}.$ Combining them, we see that
$\sin^2(\theta - \angle BAC) = \sin^2 \theta \implies \cos (2(\theta - \angle BAC)) + 1 = \cos^2 \theta \\ \implies 2 \theta = \angle BAC \implies \theta = \dfrac{\angle BAC}{2}.$
Let $BM$ and $CN$ meet at $J.$ From isoceles $\triangle JBC,$ we have that $\angle BJC = 180^{\circ} - 2 \theta = 180^{\circ} - \angle BAC,$ so $(ABJC)$ is cyclic, as desired. $\blacksquare$

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gracemoon124

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gracemoon124

#163 h Aug 9, 2024, 4:42 AM • 1 Y

Y by Danielzh

storage :p

(a) We want to prove that the tangent to $B$ is parallel to $\overline{APM}$ , so we need the obtuse angle formed between the tangent and $BC$ to equal $\angle APC$ .

Because $\angle CAP = \angle A-\angle C$ , it's easy to see that $\angle APC = 180-\angle A$ . It's trivial that the obtuse angle between the tangent and $BC$ is $180-\angle A$ , so we're done with this part.

(b) It's well-known that $(AM; P\infty)$ is a harmonic bundle, where $P_\infty$ is the point of infinity on the tangent at $B$ .

Here, we let $\overline{BM}$ meet the circumcircle at $X$ .

(c) We have $-1=(AM; P\infty)\stackrel{B}{=}(AX; BC)$ , so $ACXB$ is a harmonic quadrilateral.

(d) If $CN$ intersects the circumcircle at $Y$ , we can get that $(AY; BC)=-1$ too -- therefore, $X$ and $Y$ have to be the same point, implying the conclusion. $\square$

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ezpotd

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ezpotd

#164 h Sep 14, 2024, 2:30 AM

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We desire to prove that $BM \cap CN,A , B, C$ cyclic, so it suffices to prove $\angle ABM + \angle ACN = 180^{\circ}$ . We write $\angle ABM + \angle ACN = \angle ABC + \angle ACB + \angle CBM + \angle BCN = 180^{\circ} - \angle BAC + \angle CBM + \angle BCN$ so it suffices to prove $\angle CBM + \angle BCN = \angle BAC$ .

Now we claim $MPB$ is similar to $CQN$ . We use $SAS$ similarity. Clearly angles $\angle MPB = \angle CQN = 180^{\circ} - \angle BAC$ . We see $CQ : QN = CQ : AQ = AC : AB = AP : PB = MP : PB$ , giving the desired similarity. Now we are clearly done since $\angle CQN = \angle PBM$ , then $\angle MBP + \angle PBM = \angle BAC$ as desired.

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khanhnx

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khanhnx

#165 h Sep 14, 2024, 12:50 PM

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Let $S$ be a point on $(ABC)$ such as $AS$ is $A$ - symmedian of $\triangle ABC;$ $Cx$ be tangent at $C$ of $(ABC)$ . Since $\angle{QAC} = \angle{ABC},$ we have $CA$ tangents $(ABQ)$ . So $\angle{AQB} = \angle{BAC} = \angle{BCx}$ or $Cx \parallel AQ$ . Hence $C(AN, Qx) = - 1 = C(AS, Bx)$ or $C, S, N$ are collinear. Similarly, we have $B, S, M$ are collinear

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shendrew7

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shendrew7

#166 h Oct 27, 2024, 8:11 PM

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Denote $K \in (ABC)$ with $(AK;CB) = -1$ . Notice
$(AM; PP_{\infty}) \overset{B}{=} (A, BM \cap (ABC); C, B) \implies K \in BM.$
Similarily, $K \in CN$ , giving the desired. $\blacksquare$

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peppapig_

278 posts

peppapig_

#167 h Oct 29, 2024, 11:43 PM

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First, we make the following claim.

***

Claim 1. $BM$ and $CN$ intersect at $X$ , where $X$ is the unique (by uniqueness of harmonic conjugates) point on $(ABC)$ such that $(AX;BC)=-1$ .

It now suffices to show that this point $X$ lies on both $BM$ and $CN$ . Let $N'=CX\cap AQ$ . We make the following claim.

***

Claim 2. $Q$ is the midpoint of $AN'$ . In other words, $N'=N$ .

Proof.
Let $T$ be the intersection of the tangents to $(ABC)$ at $B$ and $C$ . Notice that,
$\angle AQC=180-\angle CAQ-\angle C=180-\angle B-\angle C=\angle A=\angle BCT,$ so $CT\parallel AQ$ , which means that we can then get that
$-1=(AX;BC)\overset{B}{=}(AX;TY)\overset{C}{=}(AN';P_{\infty, CT}Q),$ so $(AN';QP_{\infty, CT})=-1$ , implying that $Q$ must be the midpoint of $N'$ , as desired. Therefore $N=N'$ .

***

Since $N'\in CX$ and $N=N'$ , this means that $X$ lies on $CN$ . Similarly, we can prove that $X$ also lies on $BM$ , which means that $BM$ and $CN$ both intersect at point $X$ , which lies on the circle $(ABC)$ . This completes our proof.

This post has been edited 4 times. Last edited by peppapig_, Oct 29, 2024, 11:49 PM
Reason: Wording

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lnzhonglp

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lnzhonglp

#168 h Nov 26, 2024, 6:11 PM

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Let $B'$ be the reflection of $A$ over $B$ and let $C'$ be the reflection of $A$ over $C$ . Let $X = BM \cap CN$ . Then $\triangle ABC \sim \triangle MB'A \sim \triangle NAC',$ and $\triangle B'BM \sim \triangle ANC$ , so $\begin{align*}\measuredangle BXC &= \measuredangle XBC + \measuredangle BCX \\ &= \measuredangle XMN + \measuredangle MNX \\&= \measuredangle CNA + \measuredangle C'NC \\&= \measuredangle C'NA = \measuredangle BAC.\end{align*}$ Therefore, $X$ lies on $(ABC)$ .

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ihatemath123

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ihatemath123

#169 h Nov 27, 2024, 12:49 AM • 1 Y

Y by OronSH

Let $J$ be the reflection of $B$ across $A$ and let $K$ be the reflection of $C$ across $A$ so that $JKBC$ is a parallelogram. By the definitions of $P$ and $Q$ , we have that $\triangle ABC \sim \triangle PBA \sim \triangle QAC$ . So, we also have
$PABM \sim ACBK, \qquad QACN \sim ABCJ.$ Now, letting $X$ be the intersection between lines $BM$ and $CN$ , we have that
$\begin{align*}\angle BXC &= 180^{\circ} - \angle MBP - \angle NCQ \\ &= 180^{\circ} - \angle KBA - \angle JCA = 180^{\circ} - \angle A,\end{align*}$ as desired.

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smileapple

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smileapple

#170 h Jan 6, 2025, 4:19 AM • 1 Y

Y by teomihai

Reflect $B$ and $C$ about $A$ to get points $X$ and $Y$ respectively. Then $\triangle ANC\sim\triangle BXC$ , so that $\angle ACN=\angle BCX$ and thus $\angle XCY=\angle BCN$ . Similarly, we also have that $\angle XBY=\angle CBM$ . Letting $R$ be the intersection of $BM$ and $CN$ , we find that $\angle BRC=180^\circ-\angle BCN-\angle CBM=180^\circ-\angle XCY-\angle CBY=180^\circ-\angle BAC$ , so $R$ lies on the circumcircle of $\triangle ABC$ as desired. $\blacksquare$

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Ilikeminecraft

277 posts

Ilikeminecraft

#171 h Feb 17, 2025, 1:53 AM

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Let $X = BM\cap(ABC).$ Consider the tangent at $B$ . Clearly, it is parallel to $AP.$ Hence, $-1 = (AM; P\infty) \stackrel=B (AX;BC),$ which finishes.

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Maximilian113

485 posts

Maximilian113

#172 h Mar 1, 2025, 5:12 AM

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Let $R=BM \cap CN.$ Observe that from our conditions $\triangle BPA \sim \triangle AQC \implies \frac{BP}{PM} = \frac{BP}{AP} = \frac{AQ}{CQ} = \frac{NQ}{QC}.$ But $\angle APQ = \angle AQP \implies \angle BPM = \angle NQC$ so by SAS $\triangle BPM \sim \triangle NQC \implies \angle PCR = \angle PMR,$ so $PCMR$ is cyclic. Hence, $\angle ACR = \angle ACB + \angle BCR = \angle BAP + \angle AMB = 180^\circ - \angle ABM,$ so $ABRC$ is cyclic. QED

This post has been edited 1 time. Last edited by Maximilian113, Mar 1, 2025, 5:52 AM

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Retemoeg

47 posts

Retemoeg

#173 h Mar 2, 2025, 12:49 PM

Y by

Interesting problem..

Let $BM$ and $CN$ intersect at $T$ . Denote $C’$ the reflection of $C$ in $Q$ . Note that triangles $AQC$ and $BPA$ are similar, so triangles $AC’C$ and $BMA$ are similar, implying that $\angle CAC’ = \angle ABM$ . Now, as $C’ACN$ is a parallelogram, we should have:
$\angle ABT + \angle ACT = \angle ABM + \angle ACN = \angle CAC’ + 180^{\circ} - \angle CAC’ = 180^{\circ}$ Thus providing that $T$ lies on $(ABC)$ , as desired.

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Sadigly

118 posts

Sadigly

#174 h Mar 2, 2025, 1:47 PM • 1 Y

Y by ihatemath123

Seems bashable,will solve it tmrw

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eg4334

597 posts

eg4334

#175 h Mar 8, 2025, 1:58 AM

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Use bary on $\triangle ABC$ . By tangent circles and pop, $BP = \frac{c^2}{a}$ so then its immediate that $P = (0, \frac{a^2-c^2}{a^2}, \frac{c^2}{a^2})$ and similarly $Q = (0, \frac{b^2}{a^2}, \frac{a^2-b^2}{a^2})$ . Then, $M = (-1, \frac{2(a^2-c^2)}{a^2}, \frac{2c^2}{a^2})$ and $N = (-1, \frac{2b^2}{a^2}, \frac{2(a^2-b^2)}{a^2})$ . If we let the intersection be $(-1, t, \frac{2c^2}{a^2})$ by parameterizing $BM$ then we need
$\begin{align*} \begin{vmatrix} 0 & 0 & 1\\ -1 & t & \frac{2c^2}{a^2} \\ -1 & \frac{2b^2}{a^2} & \frac{2(a^2-b^2)}{a^2} \end{vmatrix} = 0 \end{align*}$ whicih gives $t = \frac{2b^2}{a^2}$ . Now its trivial to confirm that indeed $a^2yz+b^2xz+c^2xy=0$ .

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Marcus_Zhang

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Marcus_Zhang

#176 h Mar 9, 2025, 1:22 AM

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Target practice for Bary.

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hgomamogh

29 posts

hgomamogh

#177 h Yesterday at 4:22 AM

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Let $X$ be the intersection of $BM$ and the circumcircle of $ABC$ , which we will denote as $\Gamma$ . Eyeballing seems to suggest that $X$ is on the $A$ -symmedian. We will prove this using projective geometry.

By an angle chase, we observe that the tangent to $\Gamma$ at $B$ is parallel to $AP$ . This is because both lines make an angle of $\angle BAC$ with the segment $BC$ . Furthermore, observe that $\begin{align*} (A, M; P, \infty_{AM}) = -1. \end{align*}$
Hence, when we take perspectivity at $B$ onto $\Gamma$ , we obtain $\begin{align*} (A, X; B, C) = -1. \end{align*}$
Therefore, $X$ is on the $A$ -symmedian.

We can similarly show that if $X'$ is the intersection of $CN$ and $\Gamma$ , then $X'$ also lies on the $A$ -symmedian. It follows that $X$ and $X'$ are the same point, so we are done.

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