IMO 2010 Problem 4

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15142 posts

#1 h Jul 8, 2010, 8:13 AM • 24 Y

Y by Batman007, Davi-8191, tenplusten, FlakeLCR, test20, AlastorMoody, Purple_Planet, vsamc, ilovepizza2020, megarnie, pog, EthanTAoPS, jhu08, THEfmigm, Jupiter_is_BIG, sleepypuppy, EpicBird08, Adventure10, Mango247, Aopamy, ItsBesi, cubres, Rounak_iitr, and 1 other user

Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$ ). The lines $AP$ , $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$ , $L$ , respectively $M$ . The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$ . Show that from $SC = SP$ follows $MK = ML$ .

Proposed by Marcin E. Kuczma, Poland

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m.candales

186 posts

m.candales

#2 h Jul 8, 2010, 9:03 AM • 7 Y

Y by jhu08, THEfmigm, mathmax12, Adventure10, sximoz, cubres, UnearthedCyclone

$SA \cdot SB=SC^2=SP^2$ . Then $SP$ is tangent to the circuncircle of $\triangle{ABP}$ . Then $\angle{SPA}=\angle{SBP}$ .
Let $R, T$ be the intersections of $SP$ with $\Gamma$ ( $R$ is in the $\widehat{LA}$ , and $T$ is in the $\widehat{KB}$ )
$\frac{1}{2}(\widehat{RA}+\widehat{KT})=\angle{RPA}=\angle{LBA}=\frac{1}{2}\widehat{LA}=\frac{1}{2}(\widehat{LR}+\widehat{RA})$
Then $\widehat{KT}=\widehat{LR}$
Then $\angle{LPR}=\frac{1}{2}(\widehat{LR}+\widehat{TB})=\frac{1}{2}(\widehat{KT}+\widehat{TB})=\frac{1}{2}KB=\angle{PAB}$

$\angle{SPC}=\angle{LPR}+\angle{LPC}=\angle{PAB}+\angle{BPM}=\angle{PAB}+\angle{MCB}+\angle{PBC}$

$\angle{SCP}=\angle{SCA}+\angle{ACM}=\angle{SBC}+\angle{ACM}=\angle{SBL}+\angle{PBC}+\angle{ACM}$

But $\angle{SPC}=\angle{SCP}$ . Then $\angle{SBL}+\angle{ACM}=\angle{MCB}+\angle{PAB}$

Then $\widehat{LM}=\widehat{KM}$
Then $LM=KM$

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swaqar

208 posts

swaqar

#3 h Jul 8, 2010, 9:19 AM • 3 Y

Y by jhu08, Adventure10, cubres

Solution (if it's correct please tell me)
Call the circles $(S,SC)$ $\Omega$ . Now the circles $\Omega$ and $\Gamma$ are orthogonal so the point of intersection of $AB$ and $KL$ lies on the circle $\Omega$ and call this point $S'$ . Assume that $A$ belong to $[S,B]$ . Now the line $SP$ is tangent to the circumcircle of $\triangle APB$ since $SA\cdot SB=SC^{2}=SP^{2}$ . Now $SC$ is the external symmedian of the vertex $C$ . Calling the intersection of the angle bisector of $C$ with the line $AB$ , $P'$ , we have the points $(S',P',A,B)$ forming a harmonic range and this implies that $SC = SP'$ . So the circle $\Omega$ is is the appollonian circle of the vertex $C$ of the $\triangle ABC$ . Define the point $N$ on the line segment $KL$ so that the line $CN$ is the bisector of the $\angle LCM$ . Let the external symmedian meet the the $S'C$ in $J$ . We have $JN=JC$ and $SC = SP$ so the traingle $SCP$ and $JCN$ are homothetic so the points $A , P$ and $N$ are collinear and we have $M$ being the midpoint of the arc $\overarc{KMN}$ .

This post has been edited 11 times. Last edited by swaqar, Jul 8, 2010, 2:12 PM

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leejho20

1 post

leejho20

#4 h Jul 8, 2010, 9:24 AM • 3 Y

Y by jhu08, Adventure10, cubres

here's my proof

Since, angle BPC = angle BAK + angle CAK + angle ABL + angle ACL and angle SPC = angle SCP

To see angle MLK = angle ACM + angle ABL = angle BCM + angle BAK = angle MKL

it is sufficient to prove angle BPS = angle BPK.

But we have SA * SB = SC^2 = SP^2, from this it is clear that triangle BPS is similar to triangle PAS

this proves angle BPS = angle APS.

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Justanotherone

10 posts

Justanotherone

#5 h Jul 8, 2010, 9:48 AM • 3 Y

Y by jhu08, Adventure10, cubres

we know,as SC is tangent to the circle SC^2 =SA*SB
so by the problem SP^2=SA*SB
so SP/SA=SB/SP and as triangle SPA and triangle SPB have one angle in common,they are similar.so angleSPA and anglePBA(both equal to a) are equal.
extending SP to meet the circle at x,we also have angleXPM=anglePAB(both equal to b)
also,angleSCA=angleCBA(both equal to c)
again ,being on the same arc, angleACM=angleABM(both equal to d)
similarly,angleCAK=angleCMK(both equal to e)
now in triangle APM, angleAPM=angleCPK=angleCAK+angleACM=e+d
also, angleSPC=angleSCP=c+d,and angleSPA=a
so adding,180=a+c+2d+e -(1)
now, angleKLM=angleKLB+angleBLM=b+angleBLM
also, angleBLM=angleBAM=180-b-c-d-e,after calculating from triangle CAM
From (1)this turns out to be equal to a+d-b
so angleKLM=a+d
also angleLKM=angleLKA+angleAKM=a+d
so in triangle KLM we have angleKLM=angleLKM
so ML=MK................

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Pirshtuk

17 posts

Pirshtuk

#6 h Jul 8, 2010, 10:52 AM • 9 Y

Y by DHu, Sush0, jhu08, Adventure10, Mango247, cubres, Ulapu_Kanama, and 2 other users

My proof:

Because triangles $APC$ and $MPK$ are similar, we have that $\frac{MK}{MP}=\frac{AC}{AP}.$ Analogously we have that $\frac{ML}{MP}=\frac{BC}{BP}.$ So it's enough to prove that $\frac{AP}{BP}=\frac{AC}{BC}$ .

We have that $SP^2=SC^2=SA \cdot SB,$ so $\frac{SP}{SA}=\frac{SB}{SP}$ . Using this result and because angle $CSA$ is common for triangles $SPA$ and $SBP$ , we conclude that these triangles are similar. So $\frac{AP}{BP}=\frac{SA}{SP}=\frac{SA}{SC}.$ But triangles $SCA$ and $SBC$ are similar too. So $\frac{SA}{SC}=\frac{AC}{BC}.$ Therefore $\frac{AP}{BP} = \frac{AC}{BC},$ and we have done!

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silouan

3952 posts

silouan

#7 h Jul 8, 2010, 11:04 AM • 10 Y

Y by UK2019Project, AllanTian, jhu08, ericxyzhu, Adventure10, Mango247, cubres, and 3 other users

Let me add my solution too. I think it is a little different.

Well by the powe point theorem we have that $SP^2=SA\cdot SB$ which means that the triangles
SPB, PAS are similar. Using this fact a bit angle chasing shows that $\angle{SPL}=\angle{BAP}=\angle{BLK}$ so $SP//LK$ (1). But $OC\bot CS$ and $\angle{SPC}=\angle{SCP}$ , so $OM\bot SP$ (2)
From (1) and (2) we have that $OM\bot LK$ and we are done.

Attachments:

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KoolNerd

3 posts

KoolNerd

#8 h Jul 8, 2010, 11:22 AM • 4 Y

Y by jhu08, Adventure10, Mango247, cubres

As SA.SB = SC^2=SP^2, SP is tangent to the circumcircle of triangle ABP.

=> angle BPS = angle BAK = angle BLK.
Let KL intersect BC at R and SP intersect BC at Q.

In triangles BRL and BQP, angle RBL is common. angle BPQ = angle BPR. So, angle PQB = angle LRB.
=> SP is parallel to KL.

As SP = SC, angle SPC = angle SCP = angle MLC = angle NPM ( N is the intersection point of SP and the circle, N is on smaller are LA)
So,
In triangles MLC and MPD where D is the intersection point of ML and SR,
angle LMC is common, angle MPD = angle MLC. So, angle MDP = angle LCM and we know angle LCM = angle MKL.

As, SR is parallel to KL, angle MDP = angle MLK = angle MKL

We Are Done!

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YamoSky

209 posts

YamoSky

#9 h Jul 8, 2010, 11:33 AM • 3 Y

Y by jhu08, Adventure10, cubres

KoolNerd wrote:

As SA.SB = SC^2=SP^2, SP is tangent to the circumcircle of triangle ABP.

=> angle BPS = angle BAK = angle BLK.
Let KL intersect BC at R and SP intersect BC at Q.

In triangles BRL and BQP, angle RBL is common. angle BPQ = angle BPR. So, angle PQB = angle LRB.
=> SP is parallel to KL.

As SP = SC, angle SPC = angle SCP = angle MLC = angle NPM ( N is the intersection point of SP and the circle, N is on smaller are LA)
So,
In triangles MLC and MPD where D is the intersection point of ML and SR,
angle LMC is common, angle MPD = angle MLC. So, angle MDP = angle LCM and we know angle LCM = angle MKL.

As, SR is parallel to KL, angle MDP = angle MLK = angle MKL

We Are Done!

My solution is quite similar to yours...It's really an easy problem...

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April

1270 posts

April

#10 h Jul 8, 2010, 1:14 PM • 4 Y

Y by jhu08, Adventure10, Mango247, cubres

mavropnevma wrote:

Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$ ). The lines $AP$ , $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$ , $L$ , respectively $M$ . The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$ . Show that from $SC = SP$ follows $MK = ML$ .

We have $MK=ML\iff \angle PCA + \angle PBA = \angle PCB + \angle PAB\iff \angle PAB - \angle PBA = \angle PCA - \angle PCB$

From $SC=SP$ we conclude that $SP^2= SC^2=SA\cdot SB$ $\Longrightarrow SP$ is tangent to the circumcircle of triangle $ABP$ at $P$ . Therefore $\angle PAB - \angle PBA = \angle PAB - \angle APS = \angle PSB\quad (1)$ .

On the other hand, $\angle PCA - \angle PCB=\angle SCP - \angle SCA-\angle PCB=\angle SPC-\angle SBC - \angle PCB=\angle PSB\quad (2)$ .

From $(1)$ and $(2)$ we have the conclusion.

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Mithril

28 posts

Mithril

#11 h Jul 8, 2010, 3:38 PM • 4 Y

Y by jhu08, Adventure10, Mango247, cubres

Let the second tangent from $S$ meet $\Gamma$ at $C'$ . Notice that $(A,B;C,C')$ is harmonic, because projecting from $S$ fixes $C$ and $C'$ , and swaps $A$ and $B$ .

Now, let $C'P$ meet $\Gamma$ again at $M'$ . Projecting from $P$ gives $(A,B;C,C') = (K,L;M,M') = -1$ , thus $LM/MK = LM'/M'K$ . Now, if we prove that $M$ and $M'$ are antipodal, we're done, because the only way for those fractions to be equal is that $M$ is the midpoint of arc $LK$ (if we move $M$ , one fraction grows while the other is smaller).

That is the same as saying $\angle C'CP + \angle CC'M' = 90^{o} - \angle C'BC$ . But we know that $SC' = SP = SC$ , thus $S$ is the circumcenter of $C'PC$ , and $\angle C'CP + \angle CC'M = \angle C'SC / 2$ .

Now, if $O$ is the circumcenter of $ABC$ , we have $\angle C'SC / 2 = (180^{o} - \angle C'OC)/2 = 90^{o} - \angle C'BC$ , as we wanted.

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m.candales

186 posts

m.candales

#12 h Jul 8, 2010, 5:19 PM • 5 Y

Y by jhu08, Adventure10, Mango247, cubres, and 1 other user

An even simpler solution:

$SA \cdot SB=SC^2=SP^2$ . Then $SP$ is tangent to the circuncircle of $\triangle{ABP}$ . Then $\angle{SPA}=\angle{SBP}$ .
Let $R, T$ be the intersections of $SP$ with $\Gamma$ ( $R$ is in $\widehat{LA}$ , and $T$ is in $\widehat{KB}$ )

$\frac{1}{2}(\widehat{LR}+\widehat{RA})=\frac{1}{2}\widehat{LA}=\angle{LBA}=\angle{RPA}=\frac{1}{2}(\widehat{RA}+\widehat{KT})$ . Then $\widehat{LR}=\widehat{KT}$

$\frac{1}{2}(\widehat{RC}+\widehat{RM})=\frac{1}{2}\widehat{CM}=\angle{SCP}=\angle{SPC}=\frac{1}{2}(\widehat{RC}+\widehat{TM})$ . Then $\widehat{RM}=\widehat{TM}$

Then $\widehat{LM}=\widehat{KM}$ . Then $LM=KM$

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SnowEverywhere

801 posts

SnowEverywhere

#13 h Jul 8, 2010, 5:41 PM • 6 Y

Y by ayan_mathematics_king, jhu08, Adventure10, Mango247, cubres, and 1 other user

Yay. I liked this problem. I think that my solution is also a little bit different. Looks like there were quite a number of different solutions to this problem...

Solution

Let the tangent at $M$ to $\Gamma$ intersect $SC$ at $X$ . We now have that since $\triangle{XMC}$ and $\triangle{SPC}$ are both isosceles, $\angle{SPC}=\angle{SCP}=\angle{XMC}$ . This yields that $MX \| PS$ .

Now consider the power of point $S$ with respect to $\Gamma$ .

$SC^2 = SP^2 =SA \cdot SB \quad \Rightarrow \quad \frac{SP}{SA}=\frac{SB}{SP}$
Hence $\triangle{SPA} \sim \triangle{SBP}$ . Combining this with the arc angle theorem yields that $\angle{SPA}=\angle{SBP}=\angle{PKL}$ . Hence $PS \| LK$ .

This implies that the tangent at $M$ is parallel to $LK$ and therefore that $M$ is the midpoint of arc $LK$ . Hence $MK=ML$ .

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JG

19 posts

JG

#14 h Jul 8, 2010, 8:22 PM • 4 Y

Y by jhu08, Adventure10, Mango247, cubres

Let $SP$ cut the circumcircle at $X$ and $Y$ , then $SC^2=SP^2=SX\cdot SY$ so if $Q$ is the reflection of $P$ over $S$ then $X$ , $Y$ are harmonic conjugates of $P$ , $Q$ . Notice also that $\angle QCP=90^{\circ}$ then $\angle XCP=\angle PCY$ , this means $MX=MY$ , but since $SP$ is tangent to the circumcircle of $PAB$ ( $SP^2=SC^2=SA\cdot SB$ ) we have $\angle SPA=\angle PBA=\angle PKL$ so $XY$ is parallel to $KL$ and the result follows.

This post has been edited 1 time. Last edited by JG, Jan 16, 2013, 5:00 PM

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Jorge Miranda

157 posts

Jorge Miranda

#15 h Jul 8, 2010, 8:32 PM • 4 Y

Y by jhu08, Adventure10, Mango247, cubres

A straightforward angle chase

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Shreyasharma

667 posts

Shreyasharma

#120 h Nov 11, 2023, 6:31 AM • 1 Y

Y by cubres

Otis Walkthrough:
Let $D$ denote the second tangency point from $S$ . Let $DP$ meet $(ABC)$ at $N \neq D$ . Now note that,

$-1 = (AB,DC) \overset{P}{=} (KL,MN)$

and thus we have $KMLN$ is harmonic. Now it suffices to show $KMLN$ is a kite, which is equivalent to showing that $MN$ is a diameter. We will now use phantom points. Let $N'$ be the $M$ -antipode and let $D' = N'P \cap (ABC)$ . It then suffices to show $D' \equiv D$ .

Begin by noting that from the Three Tangents Lemma applied to $MNX$ , where $X = MD' \cap NC$ , we find that $OD'$ and $OC'$ are tangents to $(CD'P)$ where $O$ is the center of $(ABC)$ . Therefore $(CD'P)$ and $(ABC)$ are orthogonal. Now by noting $SC = SP$ and $SC$ tangent to $(ABC)$ we can conclude $S$ is the circumcenter of $CD'P$ . From the orthogonality we then must have $SD'$ tangent to $(ABC)$ from which $D' = D$ . Thus $MN$ is a diameter and we must have $NL = NK$ from harmonic $KMLN$ .

Attachments:

10IMO4.pdf (416kb)

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shendrew7

792 posts

shendrew7

#121 h Nov 26, 2023, 2:04 AM • 1 Y

Y by cubres

Our given condition implies $\frac{SC}{SB} = \frac{SA}{SC} \implies \frac{SP}{SB} = \frac{SA}{SP}$ , or $\triangle SPB \sim \triangle SAP$ . Thus we get
$\angle SPB = \angle PAS = \angle KLB,$ or $LK \parallel PS$ . Notice the desired implies that $LK$ is parallel to the tangent at $M$ , so we want to show this tangent is also parallel to $PS$ .

Suppose the tangent at $M$ meets the tangent at $C$ at point $T$ . Since $TM = TC$ and $SP = SC$ , we know $PS \parallel MT$ , which concludes our proof. $\blacksquare$

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AngeloChu

470 posts

AngeloChu

#122 h Jan 31, 2024, 12:46 AM • 1 Y

Y by cubres

let $MC$ and $KL$ intersect at $Q$ , and let $SP$ intersect $MK$ at $R$
we get that $MCK$ and $MKQ$ are similar, and that $MLQ$ and $MCL$ are similar
we have $BAC=BLC=BCS$
extend $CK$ to meet $SP$ at $T$
we easily get $TKR=MLC$ , and that $TKR$ is similar to $TPC$
we then get that $SP||KL$ , and that $SPQ=PQL=MLC=MCS$ , so $SC=SP$

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blueberryfaygo_55

339 posts

blueberryfaygo_55

#123 h May 12, 2024, 3:59 AM • 8 Y

Y by megarnie, deduck, heheman, KevinYang2.71, dedapperdoop, smellyman, think4l, cubres

Solved with deduck.
Claim. The desired condition is equivalent to $\angle PBA + \angle ACP = \angle PCB + \angle PAB$ .
Proof. $MK=ML$ is equivalent to $\angle MLK = \angle MKL$ , so by angle chasing, we have
$\begin{align*} \angle MKL &= \angle MKA + \angle AKL \\ &= \angle MCA + \angle ABL \\ &= \angle PCA + \angle ABP \end{align*}$ Further, we also have,
$\begin{align*} \angle MLK &= \angle MLB + \angle BLK \\ &= \angle MCB + \angle BAK \\ &= \angle PCB + \angle PAB \end{align*}$ so the claim follows. $\blacksquare$

Thus, we could remove points $M,L,K$ from the diagram. Now, we note that $SB \cdot SA = SC^2 = SP^2$ which leads to $\Delta SAP \sim \Delta SPB$ , and $\angle SAP = \angle SPB$ . To finish, we have, $\begin{align*} \angle PCB + \angle PAB &= \angle PCB + \angle BPS \\ &= \angle PCS + \angle BPS - \angle BCS \\ &= \angle BPC - \angle BCS \\ &= \angle BPC - \angle BAC \end{align*}$ Therefore, it is left to show that $\angle PBA + \angle ACP + \angle BAC = \angle BPC$ . However, this is true since,
$\begin{align*} \angle PBA + \angle ACP + \angle BAC &= \angle PBA + \angle ACP + \angle BAP + \angle PAC \\ &= (180^{\circ} - \angle APB) + (180^{\circ} - \angle APC) \\ &= 360^{\circ} - \angle APB - \angle APC \\ &= \angle BPC \end{align*}$ as desired. $\blacksquare$

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dolphinday

1318 posts

dolphinday

#124 h May 16, 2024, 5:22 AM • 1 Y

Y by cubres

We use PoP to get $SA \cdot SB = SC^2$ and also $SC^2 = SP^2$ from which we get that $SP$ is tangent to $(PAB)$ . This then implies that $\angle PAB = \angle BPS$ and $ABKL$ cyclic implies $\angle PAB = \angle KAB = \angle KLB \implies KL \parallel SP$ . So then we apply Pascal on $BMCCKA$ to get that $BM \cap CK$ lies on $SP$ . Then apply the same on $MMCKLB$ to get that $MM \cap KL$ lies on $SP$ which implies $MM$ is parallel to $KL$ , so $MK = ML$ as desired.

This post has been edited 1 time. Last edited by dolphinday, May 16, 2024, 5:22 AM

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SenorSloth

37 posts

SenorSloth

#125 h Jun 30, 2024, 3:31 PM • 1 Y

Y by cubres

We WLOG assume $CA<CB$ . Draw the other tangent to $\Gamma$ from $S$ , and let the tangency point be $D$ . Let the intersection of $DP$ and $\Gamma$ distinct from $D$ be $N$ . Then we have:

$-1=(AB;CD) \overset{P}{=} (KL;MN)$ .

Now we angle chase to show that $MN$ is a diameter. Since $C$ , $D$ , and $P$ lie on a circle centered at $S$ , we have that $\angle CPD=180-\frac12 \angle CSD$ , and thus $\angle MPD = \frac12 \angle CSD$ . Then we can calculate that the measure of arc $MN$ equals $\angle COD+2\angle MPD = 180 - \angle CSD + 2\left(\frac12 \angle CSD \right) = 180$ .

Since $MN$ is a diameter of $\Gamma$ and $LMKN$ is harmonic, by symmetry we must have that $MK=ML$ , as desired.

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Duka

2 posts

Duka

#126 h Aug 3, 2024, 9:09 AM • 1 Y

Y by cubres

Since in shortlist says iff I will show that from $MK=ML$ $\implies$ $SP = SC$
Let $Q=CL\cap MA$ , from Pascals theorem on $CCMABL$ we see that $S$ , $P$ and $Q$ are coliniear.
$\angle QSP = \angle LCM = \angle LKM = \angle KLM = \angle MAK = \angle MAP$
$\implies CQAP$ is cyclic.
Then $\angle SPA = \angle QPA = \angle QCA = \angle LCA = \angle SBP$
Which implies that $SP$ is tangent to $(APB)$ $\implies SP\cdot SP= SA\cdot SB=SC\cdot SC$
$\implies SC=SP$
You can also go backwards to get $SP = SC$ $\implies$ $MK=ML$

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SomeonesPenguin

123 posts

SomeonesPenguin

#127 h Sep 7, 2024, 9:58 AM • 1 Y

Y by cubres

We will show that $SC=SP \iff MK=ML$ .

Notice that $SC=SP \iff P$ lies on the $C$ -Apollonian circle so $SC=SP \iff \frac{PA}{PB}=\frac{CA}{CB}$ . Now $MK=ML \iff \angle LCP=\angle KCP$ .

From LOS in $\triangle LCP$ and $\triangle KCP$ we have $\frac{\sin(\angle LCP)}{\sin(\angle KCP)}=\frac{LP\sin(\angle LCP)}{KP\sin(\angle CKP)}$
From PoP we have $\frac{PA}{PB}=\frac{PL}{PK}$ and from LOS we have $\frac{\sin(\angle LCP)}{\sin(\angle CKP)}=\frac{CB}{CA}$ . Hence we get $\frac{\sin(\angle LCP)}{\sin(\angle KCP)}=\frac{CB}{CA}\cdot \frac{PA}{PB}$ Now just plug in either condition in this equation to get the other one. $\square$
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.202896110170563, xmax = 19.213125730938394, ymin = -10.497987045605893, ymax = 3.8696816774635145; /* image dimensions */ pen qqzzcc = rgb(0.,0.6,0.8); pen qqzzqq = rgb(0.,0.6,0.); draw((5.,-6.)--(13.,-6.)--(5.738284588040686,1.9324528510591215)--cycle, linewidth(0.8) + qqzzcc); draw((5.738284588040686,1.9324528510591215)--(3.8138853212849435,-0.8654634846866291)--(8.117756722334056,-3.271092593269745)--cycle, linewidth(0.8) + qqzzqq); draw((5.738284588040686,1.9324528510591215)--(13.126060669122644,1.112571352554749)--(8.117756722334056,-3.271092593269745)--cycle, linewidth(0.8) + qqzzqq); /* draw figures */ draw((5.,-6.)--(13.,-6.), linewidth(0.8) + qqzzcc); draw((13.,-6.)--(5.738284588040686,1.9324528510591215), linewidth(0.8) + qqzzcc); draw((5.738284588040686,1.9324528510591215)--(5.,-6.), linewidth(0.8) + qqzzcc); draw(circle((9.,-2.3717026281975766), 5.4004205223508635), linewidth(0.8)); draw((5.738284588040686,1.9324528510591215)--(-4.729370032483704,-6.), linewidth(0.8)); draw((-4.729370032483704,-6.)--(5.,-6.), linewidth(0.8)); draw((5.738284588040686,1.9324528510591215)--(10.12208915873399,-7.65426429258731), linewidth(0.8)); draw((5.,-6.)--(13.126060669122644,1.112571352554749), linewidth(0.8)); draw((3.8138853212849435,-0.8654634846866291)--(13.,-6.), linewidth(0.8)); draw((3.8138853212849435,-0.8654634846866291)--(5.738284588040686,1.9324528510591215), linewidth(0.8)); draw((5.738284588040686,1.9324528510591215)--(13.126060669122644,1.112571352554749), linewidth(0.8)); draw((5.738284588040686,1.9324528510591215)--(3.8138853212849435,-0.8654634846866291), linewidth(0.8) + qqzzqq); draw((3.8138853212849435,-0.8654634846866291)--(8.117756722334056,-3.271092593269745), linewidth(0.8) + qqzzqq); draw((8.117756722334056,-3.271092593269745)--(5.738284588040686,1.9324528510591215), linewidth(0.8) + qqzzqq); draw((5.738284588040686,1.9324528510591215)--(13.126060669122644,1.112571352554749), linewidth(0.8) + qqzzqq); draw((13.126060669122644,1.112571352554749)--(8.117756722334056,-3.271092593269745), linewidth(0.8) + qqzzqq); draw((8.117756722334056,-3.271092593269745)--(5.738284588040686,1.9324528510591215), linewidth(0.8) + qqzzqq); /* dots and labels */ dot((5.,-6.),dotstyle); label("$A$", (4.7006540756095925,-6.6791480835613585), NE * labelscalefactor); dot((13.,-6.),dotstyle); label("$B$", (13.211848649521938,-6.169988292520035), NE * labelscalefactor); dot((5.738284588040686,1.9324528510591215),dotstyle); label("$C$", (5.694720674902673,2.299868060960004), NE * labelscalefactor); dot((-4.729370032483704,-6.),linewidth(4.pt) + dotstyle); label("$S$", (-5.0011423857687545,-5.864293213305793), NE * labelscalefactor); dot((8.117756722334056,-3.271092593269745),dotstyle); label("$P$", (8.143746020443698,-3.000412997509203), NE * labelscalefactor); dot((13.126060669122644,1.112571352554749),linewidth(4.pt) + dotstyle); label("$K$", (13.195759434826453,1.2471396820992215), NE * labelscalefactor); dot((3.8138853212849435,-0.8654634846866291),linewidth(4.pt) + dotstyle); label("$L$", (3.345695329361647,-0.9249043017914502), NE * labelscalefactor); dot((10.12208915873399,-7.65426429258731),linewidth(4.pt) + dotstyle); label("$M$", (10.18707628677048,-7.421482326940897), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

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onyqz

195 posts

onyqz

#128 h Oct 29, 2024, 11:52 PM • 1 Y

Y by cubres

simple one, posting for storage
solution

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Siddharthmaybe

106 posts

Siddharthmaybe

#129 h Jan 4, 2025, 7:04 PM • 1 Y

Y by cubres

swaqar wrote:

Solution (if it's correct please tell me)
Call the circles $(S,SC)$ $\Omega$ . Now the circles $\Omega$ and $\Gamma$ are orthogonal so the point of intersection of $AB$ and $KL$ lies on the circle $\Omega$ and call this point $S'$ . Assume that $A$ belong to $[S,B]$ . Now the line $SP$ is tangent to the circumcircle of $\triangle APB$ since $SA\cdot SB=SC^{2}=SP^{2}$ . Now $SC$ is the external symmedian of the vertex $C$ . Calling the intersection of the angle bisector of $C$ with the line $AB$ , $P'$ , we have the points $(S',P',A,B)$ forming a harmonic range and this implies that $SC = SP'$ . So the circle $\Omega$ is is the appollonian circle of the vertex $C$ of the $\triangle ABC$ . Define the point $N$ on the line segment $KL$ so that the line $CN$ is the bisector of the $\angle LCM$ . Let the external symmedian meet the the $S'C$ in $J$ . We have $JN=JC$ and $SC = SP$ so the traingle $SCP$ and $JCN$ are homothetic so the points $A , P$ and $N$ are collinear and we have $M$ being the midpoint of the arc $\overarc{KMN}$ .

bringing modern artillery to a snowball fight :stretcher:

Works though :10:

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optimusprime154

17 posts

optimusprime154

#130 h Jan 6, 2025, 4:32 PM • 2 Y

Y by SorPEEK, cubres

pascal on $CCMABL$ gives $S,P,T$ collinear where T = $MA \cap CL$ now i claim that the quadrilateral $PATC$ is cyclic , that ends the problem because, it means $\angle PCA = \angle MAK$ which means arc $MK$ = arc $ML$ so it finishes.
now i prove $\angle TPL = \angle BAK$ . notice $\angle TPL = \angle BPS = \angle BAK$ where the last one follows from the fact that $SP$ is tangent to the circumcircle of $\triangle BPA$ since $BS * AS = SC^2 = SP^2$ from $\angle TPL = \angle BAK = \angle BLK$ we obtain $TL \parallel ST$ which indicates $\angle KAC = \angle KLC = \angle PTC$ giving $PATC$ cyclic so we're done!

This post has been edited 1 time. Last edited by optimusprime154, Jan 6, 2025, 4:35 PM

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HasnatFarooq

32 posts

HasnatFarooq

#132 h Jan 13, 2025, 12:01 PM • 3 Y

Y by Jam777, Mahad_Arif, cubres

Complex Bash

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Mquej555

15 posts

Mquej555

#133 h Jan 13, 2025, 1:29 PM • 1 Y

Y by cubres

My first IMO Upload.

Attachments:

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cj13609517288

1878 posts

cj13609517288

#134 h Feb 17, 2025, 5:04 PM • 1 Y

Y by cubres

Assume $CA<CB$ .

It turns out that the circle centered at $S$ passing through $C$ is exactly the $C$ -Apollonius circle of triangle $ABC$ . Indeed, it suffices to show that $D$ , the point on $AB$ such that $CD$ bisects $\angle ACB$ , satisfies $SC=SD$ , since the reflection of $C$ over $AB$ also lies on this circle. Indeed, $\angle SCD=\angle SCA+\angle ACD=\angle B+\angle BCD=\angle CDS$ .

Now note that
$\angle PAB+\angle PCB=\angle PDB-\angle APD+\angle PCB=\angle CPD-\frac12 \angle APB-\angle ABC.$ But since
$\angle CPD=180^{\circ}-\frac12 \angle CSD=90^{\circ}+\angle CDS=90^{\circ}+\frac12 \angle C+\angle ABC,$ we just get that
$\angle PAB+\angle PCB=90^{\circ}+\frac12\angle C-\frac12 \angle APB=\angle PBA+\angle PCA,$ which implies the result. $\blacksquare$

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kotmhn

56 posts

kotmhn

#135 h Mar 3, 2025, 6:46 AM

Y by

Let $\angle PAS = x$ and $\angle PBC = y$
Now
$\angle KML = \angle CML + \angle CMK = \angle CBL + \angle CAM =y+A+C-x$
Next observe that
$SP^2 = SA^2 = SB\cdot SC$ Hence we have that $SP$ is tangent to $(PBC)$ .
Next
$\angle KLM = \angle BLK + \angle BLM = \angle BAK + \angle BCM = \angle BAK + \angle BPS = \angle BAK +\angle PBC -\angle BSP = \angle BAK + \angle PBC - \angle BSA +\angle PSA = y+A+C-x$ Hence $\angle KML = \angle KLM$ so $KM = KL$ .
Hence proved.

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