four points lie on a circle

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1145 posts

#1 h Jun 28, 2007, 6:32 PM • 17 Y

Y by ArefS, Davi-8191, r0518, Abdollahpour, Adventure10, jhu08, HWenslawski, mathematicsy, veirab, Mango247, Rounak_iitr, Tastymooncake2, ehuseyinyigit, cubres, and 3 other users

Let $ABCD$ be a trapezoid with parallel sides $AB > CD$ . Points $K$ and $L$ lie on the line segments $AB$ and $CD$ , respectively, so that $AK/KB=DL/LC$ . Suppose that there are points $P$ and $Q$ on the line segment $KL$ satisfying $\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.$ Prove that the points $P$ , $Q$ , $B$ and $C$ are concyclic.

Proposed by Vyacheslev Yasinskiy, Ukraine

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Yimin Ge

253 posts

Yimin Ge

#2 h Jun 29, 2007, 12:06 PM • 7 Y

Y by dwip_neel, Adventure10, jhu08, Mango247, Tastymooncake2, and 2 other users

$AD$ and $BC$ intersect at $S$ , $AP$ and $DQ$ intersect at $X$ , $BP$ and $CQ$ intersect at $Y$ , $BP$ and $CD$ intersect at $R$ .

Because $AB\parallel CD$ , there exists a homothety taking $\triangle SAB$ to $\triangle SDC$ . But $AK/KB=DL/LC$ , so this homothety also takes $K$ to $L$ , hence $S,L,K$ are collinear.

The points $A,X,P$ are collinear by definition, so by Menelaos' theorem applied to $\triangle SQD$ , we obtain
$\frac{QX}{XD}\frac{DA}{AS}\frac{SP}{PQ}=-1.\ \ \ \ \ \ \ \ (1)$
Also, $B,Y,P$ are collinear as well, so using Menelaos' theorem on $\triangle SCQ$ , we obtain
$\frac{QY}{YC}\frac{CB}{BS}\frac{SP}{PQ}=-1.\ \ \ \ \ \ \ \ (2)$
But it follows from $AB\parallel CD$ that
$\frac{DA}{AS}=\frac{CB}{BS}\ \ \ \ \ \ \ \ (3)$
so combining (1), (2), (3), we obtain
$\frac{QX}{XD}=\frac{QY}{YC}$
so $XY\parallel CD$ .

From $XY\parallel CD$ , it follows that $\angle CRY=\angle RYX$ , and since $\angle XPY=\angle APB=\angle BCD=\angle BCR$ , the triangles $XPY$ and $BCR$ are similar, so
$\angle PXY = \angle RBC=\angle PBC.\ \ \ \ \ \ \ \ (4)$
Also, $\angle XPY=\angle APB = \angle BCD = \pi-\angle ABC = \pi-\angle CQD =\pi-\angle YQX$ , so $XPYQ$ is a cyclic quadrilateral. It follows that
$\angle PXY=\angle PQY=\angle PQC.\ \ \ \ \ \ \ \ (5)$
From (4) and (5), we conclude that $\angle PBC=\angle PQC$ , so $PQBC$ is a cyclic quadrilateral. $\Box$

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watchcatno1

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watchcatno1

#3 h Jun 29, 2007, 12:40 PM • 3 Y

Y by Adventure10, jhu08, Mango247

I think the proof that XY||CD can also be done with Desargue's theorem

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Amir.S

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Amir.S

#4 h Jun 30, 2007, 7:25 AM • 11 Y

Y by Sx763_, IsaacJimenez, siddigss, Muradjl, Adventure10, jhu08, Tastymooncake2, ehuseyinyigit, and 3 other users

$\angle CQD=\angle ABC=\angle DCS$ hence circumcircle of $CQD$ is tangent to $SC$ .
$\angle APB=180^{\circ}-\angle CQD=DP'C$ hence $CP'\parallel BP\ ,\ DP'\parallel AP$ so $\angle BQP=\angle CP'Q=\angle QCB$ .
now we're done.

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TomciO

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TomciO

#5 h Jun 30, 2007, 9:56 PM • 8 Y

Y by Mobashereh, Adventure10, jhu08, Mango247, Tastymooncake2, and 3 other users

Draw line parallel to $AP$ from $D$ and line parallel to $BP$ from $C$ . Since $\frac{DL}{LC}=\frac{AK}{BK}$ they meet in point $R$ on line $KL$ . Then $\angle DRC = \angle APB = 180^\circ-\angle DQC$ so quadrilateral $QDRC$ is cyclic. So $\angle DRQ = \angle DCQ$ and since $\angle DRC = \angle APB = \angle DCB$ we also have $\angle QRC = \angle QCB$ . But $\angle QRC = \angle KPB = 180^\circ-\angle BPQ$ and we are done.

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JSteinhardt

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JSteinhardt

#6 h Jul 3, 2007, 12:42 AM • 7 Y

Y by Adventure10, jhu08, Mango247, Tastymooncake2, and 3 other users

In a similar vein to Amir, first note that the ratio condition means that $PQ$ , $BC$ , and $AD$ concur (call the point of concurrence $E$ ). Now we start with $A,B,C,D,E$ and determine where $P$ and $Q$ could be. Construct $\omega$ through $C$ and $D$ tangent to $EB$ , and $\omega '$ through $B$ and $A$ tangent to $EB$ . Note that $P$ must lie on $\omega$ , and $Q$ on $\omega '$ , because these two circles are the locuses of points such that $\angle CQD = \angle CBA$ , and such that $\angle BPA = \angle BCD$ . Since $PQ$ goes through $E$ , we can draw a line $l$ from $E$ to attain any candidate points $P$ and $Q$ by noting where $l$ intersects $\omega$ and $\omega '$ . Let $P'$ be the intersection of $l$ with $\omega$ that is not $Q$ . By power of a point, $(EP')(EQ) = (EC)^{2}$ . But $\omega$ and $\omega '$ are homothetic with ratio $\frac{BE}{CE}$ , thus $\frac{EP}{EP'}= \frac{BE}{CE}$ , so $(EP)(EQ) = (EC)(EB)$ , so that $PQBC$ is cyclic by the converse of Power of a Point.

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windrock

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windrock

#7 h Jan 20, 2009, 1:44 PM • 2 Y

Y by jhu08, Adventure10

Assume that AD intersect BC at S. Easy to see that S is center of ray (CDQ) and (APB). We get: S, P, Q is conlinear and SCB is common targent of (APB) and (CDQ). Hence, easy to finish problem
( Denote that (XYZ) is circumcenter of triangle XYZ)

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windrock

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windrock

#8 h Jan 28, 2009, 1:45 PM • 3 Y

Y by jhu08, Adventure10, Mango247

Assume that $AD$ intersect $BC$ at $S$ . Easy to see that $S$ is center of ray $(CDQ)$ and $(APB)$ . We get: $S, P, Q$ is conlinear and $SCB$ is common targent of $(APB)$ and $(CDQ)$ . Hence, easy to finish problem
( Denote that $(XYZ)$ is circumcenter of $\Delta XYZ$ )

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Zhero

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Zhero

#9 h Apr 4, 2010, 3:50 PM • 4 Y

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

Let $X$ be the intersection of lines $AD$ and $BC$ ; the problem's conditions tell us that $K$ , $L$ , and $X$ are collinear. Let $Q'$ be the intersection (distinct from $P$ ) of the circumcircle of $BPC$ and line $KL$ , and let $m \angle ABP = a$ , $m \angle PBC = b$ , $m \angle BCQ' = c$ , and $m \angle Q'CD = d$ . Clearly, showing that $Q' = Q$ finishes the proof.

Since $BPQ'C$ is cyclic, $m \angle KPB = c$ , so by definition $m \angle APK = d$ . Note that since $ABCD$ is a trapezoid, $a + b + c + d = 180^{\circ}$ , so $m \angle PAB = b$ .

Let the homothety centered at $X$ that brings $AB$ to $CD$ bring $P$ to $P'$ . We trivially have that $m \angle APK = m \angle DP'K = d$ . On the other hand, $m \angle Q'CD = d$ , so $Q'CP'D$ is cyclic. Therefore, $m \angle DQ'C = 180 - m \angle DP'C = 180 - d - c = a + b = m \angle ABC$ . Since there exists a unique point $Q$ such that $m \angle DQC = m \angle ABC$ , we have that $Q = Q'$ , as desired.

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SnowEverywhere

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SnowEverywhere

#10 h Jul 5, 2010, 7:37 PM • 11 Y

Y by Stens, Ali3085, jhu08, Jahir, Adventure10, Mango247, Tastymooncake2, and 4 other users

Solution

Let $X=AD \cap BC$ . From the condition, we have that $X$ is the center of homothety that maps $L$ to $K$ and therefore $X$ , $L$ and $K$ are collinear.

Consider the homothety that maps $D$ to $A$ , $C$ to $B$ and $Q$ to $Y$ . Because $\angle{APB}+\angle{AYB}=\angle{ABC}+\angle{BCD}=180$ , we have that $APBY$ is cyclic. Hence,

$\angle{QPB}=\angle{YPB}=\angle{YAB}=\angle{QDC}=180-\angle{DQC}-\angle{DCQ}=\angle{QCB}$
Hence $PQBC$ is cyclic.

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Rijul saini

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Rijul saini

#11 h Mar 6, 2011, 8:37 AM • 7 Y

Y by jhu08, Adventure10, Mango247, Tastymooncake2, and 3 other users

Let $AP \cap QD = M, BP \cap QC = N$ . Also, let $l$ be a line passing through $M$ , parallel to $AB,CD$ and meeting the lines $PB,QC,PQ$ at $N',N'',Z$ respectively. Finally, let $X$ be a point on $BC$ on the opposite side of $CD$ as $B$ and $Y$ be a point on $BC$ on the opposite side of $AB$ as $C$ .
We have,
$\angle APB = \angle BCD = \angle YBA$ since $AB || CD$ .
This implies that the line $BC$ is tangent to the circle $\cdot (PAB)$ at $B$ .
Therefore, $\angle PBC = \angle PAB$ .
Also, note that
$\angle MPN = \angle APB = \angle BCD = 180^{\circ} - \angle ABC = 180^{\circ} - \angle DQC = 180^{\circ} - \angle MQN$
Therefore, $M,P,N,Q$ are concyclic. Thus, $\angle PQN = \angle PMN$ .
Thus,
$BCPQ \text{ is cyclic} \iff \angle PBC = \angle PQN \iff \angle PAB = \angle PMN$
So, it suffices to prove that $MN || AB$ .
Suppose the contrary, and thus $N' \not = N''$ .
But then, since $MN'' || CD || AB$ therefore, $\frac{MZ}{ZN''}=\frac{DL}{LC} = \frac{AK}{KB} = \frac{MZ}{ZN'}$ a contradiction.

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StefanS

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StefanS

#12 h Jul 8, 2011, 8:35 PM • 4 Y

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

My Solution

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StefanS

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StefanS

#13 h Jul 9, 2011, 12:08 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Zhero wrote:

Let $X$ be the intersection of lines $AD$ and $BC$ ; the problem's conditions tell us that $K$ , $L$ , and $X$ are collinear. Let $Q'$ be the intersection (distinct from $P$ ) of the circumcircle of $BPC$ and line $KL$ , and let $m \angle ABP = a$ , $m \angle PBC = b$ , $m \angle BCQ' = c$ , and $m \angle Q'CD = d$ . Clearly, showing that $Q' = Q$ finishes the proof.

Since $BPQ'C$ is cyclic, $m \angle KPB = c$ , so by definition $m \angle APK = d$ . Note that since $ABCD$ is a trapezoid, $a + b + c + d = 180^{\circ}$ , so $m \angle PAB = b$ .

Let the homothety centered at $X$ that brings $AB$ to $CD$ bring $P$ to $P'$ . We trivially have that $m \angle APK = m \angle DP'K = d$ . On the other hand, $m \angle Q'CD = d$ , so $Q'CP'D$ is cyclic. Therefore, $m \angle DQ'C = 180 - m \angle DP'C = 180 - d - c = a + b = m \angle ABC$ . Since there exists a unique point $Q$ such that $m \angle DQC = m \angle ABC$ , we have that $Q = Q'$ , as desired.

You also need to prove that $~$ $Q'$ $~$ is on the same side of $~$ $CD$ $~$ as $~$ $A, \, B, \, P$ $~$ and $~$ $Q.$

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exmath89

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exmath89

#14 h Jul 3, 2013, 4:45 AM • 3 Y

Y by jhu08, Adventure10, Markas

Solution

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sunken rock

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sunken rock

#15 h Jul 4, 2013, 8:04 AM • 2 Y

Y by jhu08, Adventure10

Remark: If $X\in CD\cap AP, Y\in AB\cap DQ$ , then $BCXPQY$ is cyclic.

Best regards,
sunken rock

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bjump

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bjump

#65 h Nov 29, 2023, 2:41 AM

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Let $\measuredangle$ denote directed angles $\text{Mod } 180^\circ$

Let $BC \cap AD=X$ Note that a homothety about $X$ sends $K$ to $L$ . By the angle conditions $(APB)$ , and $(DQC)$ are tangent to $BC$ . Let $P \neq (ABP)\cap KL=J$ . Since a homothety sends $\triangle JAB$ to $\triangle DQC$ about $X$ .
$\measuredangle QPB = \measuredangle JPB = \measuredangle JBC = \measuredangle QCX= \measuredangle QCB$ Hence $PQBC$ is cyclic. $\square$

This post has been edited 1 time. Last edited by bjump, Nov 29, 2023, 2:49 AM
Reason: defining directed angles

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shendrew7

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shendrew7

#66 h Dec 14, 2023, 1:57 AM

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The ratio condition tells us $AD$ , $BC$ , and $KL$ concur at a point, say $O$ . The given angle condition tells us $(PAB)$ and $(QCD)$ are both tangent to $AB$ at $B$ and $C$ , respectively.

Thus we notice a homothety at $O: (PAB) \rightarrow (QCD)$ . Suppose we denote $G = (PAB) \cap KL \neq P$ . Our homothety induces parallel lines $CQ$ and $BG$ , and hence
$\measuredangle PQC = \measuredangle PGB = \measuredangle PBC. \quad \blacksquare$

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blueprimes

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blueprimes

#67 h Mar 20, 2024, 2:21 PM

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A nice problem!

Suppose $(PCB)$ intersects $KL$ at $Q'$ . It suffices to show that $\angle CQ'D = \angle ABC$ . Now observe that
$\angle Q'CK = \angle BCD - \angle BCQ' = \angle APB - \angle LPB = \angle APL.$ But $\angle CKQ' = \angle ALP$ , then by AA Similarity, $\triangle Q'CK \sim \triangle APL.$ Then we obtain the relation $\frac{Q'K}{AL} = \frac{CK}{PL}$ . But by trapezoid properties and the given side length ratio, we also have $\frac{AL}{DK} = \frac{BL}{CK}$ . Multiplying these two ratios together yields $\frac{Q'K}{DK} = \frac{BL}{PL}.$ But $\angle Q'KD = \angle BLD$ , so by SAS Similarity, $\triangle Q'KD \sim \triangle BLP$ . So
$\angle KQ'D = \angle LBP \implies \angle CQ'D = \angle CQ'P + \angle KQ'D = \angle CBP + \angle LBP = \angle ABC.$ Our proof is complete.

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john0512

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john0512

#68 h Apr 13, 2024, 4:46 AM

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Let's restate the problem in a more natural reference frame.

Let $\triangle EAB$ be a triangle and $C$ and $D$ on sides $BE$ and $AE$ respectively such that $CD\parallel AB$ . Let $\ell$ be a cevian of $\triangle EAB$ through $E$ . Let $P$ and $Q$ be on $\ell$ such that $(ABP)$ and $(CDQ)$ are both tangent to $EB$ . We wish to show that $CBPQ$ is cyclic.

Let $\ell$ intersect $(CDQ)$ againt at $P'$ and $(APB)$ again at $Q'$ . Note that a homothety at $E$ sends $P'$ to $P$ and $Q$ to $Q'$ . Thus, by this homothety, $EP\cdot EQ=EP'\cdot EQ'.$ Finally, by the circumcircle tangencies, $EC^2=EP'\cdot EQ,EB^2=EP\cdot EQ'$ $(EB\cdot EC)^2=(EP\cdot EQ)\cdot (EP'\cdot EQ')$ $EB\cdot EC=EP\cdot EQ,$ as desired.

remark: there are three main ideas used in this solution. First and most importantly, we stated the problem in a much more natural reference frame that allows us to think about it much more easily. Then, the "considering the conjugate" idea is also used. Since we have a line and a circle with one marked intersection, we also added in the other intersection to our diagram. Finally, this problem demonstrates another example of "power flipping", which in essence is multiplying one of the two lengths in a power expression by some ratio and dividing the other by the same ratio to get an equivalent power. This was really what was being done with $P,Q,P',Q'$ here.

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amogususususus

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amogususususus

#70 h May 9, 2024, 5:53 AM

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Define $AD \cap BC = E$ , $BP \cap CQ = X$ , and $AP \cap DQ=Y$ .

Claim 1. $E,L,K$ are collinear
Proof. Let $EL \cap AB = K`$ , since $AB \parallel CD$ we have $\triangle EDL \sim EAK`$ and $\triangle ECL \sim \triangle EBK`$ . Using the similarity condition, we get
$\frac{DL}{AK}=\frac{EL}{EK`}=\frac{LC}{K`B}$ Since $\frac{DL}{LC}=\frac{AK}{KB} \Rightarrow \frac{DL}{AK}=\frac{LC}{KB}$ , we must have $KB=K`B$ . Because both $K$ and $K`$ are on $\overline{BC}$ , we must have $K`=K$ . As desired.

Claim 2. $Q,X,P,Y$ are concyclic
Proof. Note that $X$ and $Y$ are on a different side wrt $QP$ . Since $ABCD$ is a trapezoid, and by using the angle condition in the problem we have
$\angle XPY = \angle APB = \angle BCD = 180^{\circ} - \angle ABC = 180^{\circ} - \angle CQD = 180^{\circ} - \angle XQY$ $\Rightarrow \angle XPY = 180^{\circ} - \angle XQY \Rightarrow \angle XPY + \angle XQY = 180^{\circ}$ As desired.

Now, note that $E$ is the center of perspective of $\triangle APB$ and $\triangle DQC$ .
By the special case of Desargues Theorem which is when $AB \parallel CD$ , we have $XY \parallel AB,CD$ .

Since $Q,X,P,Y$ are concyclic and $XY \parallel CD$ , we have
$\angle CDQ = \angle CDY = \angle CYX = \angle QYX = \angle QPX = 180^{\circ} - \angle QPB$ $\angle CDQ = 180^{\circ} - \angle QPB \Rightarrow \angle QPB = 180^{\circ} - \angle CDQ$ Since $AB\parallel CD$ , we have $\angle CQD = \angle ABC = \angle ECD \Rightarrow \angle ECD = \angle CQD$ . And then we have,
$\angle QCB = 180^{\circ} - \angle ECD - \angle DCQ = 180^{\circ} - \angle CQD - \angle DCQ = \angle CDQ$ $\Rightarrow \angle QCB = \angle CDQ \Rightarrow \angle QCB + \angle QPB = \angle CDQ + 180^{\circ} - \angle CDQ= 180^{\circ}$ $\Rightarrow \angle QCB + \angle QPB = 180^{\circ}$ Note that $Q$ and $P$ are on the same side wrt $BC$ , thus $P,Q,C,B$ are concyclic. As desired.

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TestX01

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TestX01

#71 h May 9, 2024, 10:02 PM

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Nice geo

It is easily seen by Alternate Segment Theorem that $(APB)$ and $(CQD)$ are tangent to $BC$ . The ratio condition gives $KL$ , $BC$ , $AD$ , concurrent at a point $E$ (This is well known). Let $KL$ intersect $(APB)$ at $X$ and $(CQD)$ at $Y$ . By Power of a Point, $EP\times EX=EB^2$ , and $EQ\times EY=EC^2$ . Multiplying these together, $EP\times EQ\times EX\times EY=EB^2\times EC^2$ .

We claim $EX\times EY=EB\times EC$ . But this is true by Power of a Point, as a homothety taking $(APB)$ to $(CQD)$ (Possible by Parallel lines and tangency), gives $CQ \parallel BX$ , hence
$\angle BXE=\angle CQE=\angle ECY$ by Alternate Segment Theorem, so $BCYX$ is cyclic.

This gives $EP\times EQ=EB\times EC$ so we conclude by converse of Power of a Point.

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dudade

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dudade

#72 h Jul 28, 2024, 8:41 PM

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Let $M = AD \cap BC$ . Then, suppose $R = MK \cap (CQD)$ . Notice that $\angle CQD = \angle MCD$ , so $(CQD)$ and $(APB)$ are tangent to $MB$ at $C$ and $B$ , respectively. Thus, $\triangle MCR \sim \triangle MBP$ . Now, note that $MC \cdot MB = MC^2 \cdot \dfrac{MB}{MC} = \left(MR \cdot MQ\right) \cdot \dfrac{MP}{MR} = MQ \cdot MP.$ So, $P$ , $Q$ , $B$ , $C$ are concyclic points, as desired.

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Likeminded2017

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Likeminded2017

#73 h Jul 31, 2024, 2:08 AM

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Let F be the intersection of AP and DQ and E be the intersection of PB and QC. Claim: $FE$ is parallel to $AB,DC.$
Proof:
Suppose not, let $E$ be higher than $F.$ Draw the points $F',E'$ such that $E'F'$ is parallel to $AB$ and goes through $G.$ Observe as $F$ lies on $PA$ and $E$ lies on $PB$ we have:
$\dfrac{FG}{EG}=\dfrac{[FPG]}{[EPG]}>\dfrac{[F'PG]}{[E'PG]}=\dfrac{F'G}{E'G}=\dfrac{AK}{KB}.$ Analogously, $\dfrac{FG}{EG}<\dfrac{DL}{LC}=\dfrac{AK}{KB}$ forming a contradiction so $EF$ is parallel to $AB.$

Let $\angle DCB= \angle 1,$ $\angle DQC=\angle 2.$ Then observe $PEFQ$ is cyclic as $P,Q$ sum to $180^\circ.$ Then we have
$\angle QCB=\angle 1 - \angle DCQ = \angle 1 - \angle FEQ = \angle 1 - \angle FPQ = \angle KPB = \angle QPB$ so $QPCB$ is cyclic.

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eg4334

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eg4334

#74 h Aug 11, 2024, 4:21 PM

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Notice that the length condition translates to $AD$ , $KL$ , and $BC$ being concurrent as some point $E$ . Additionally, the angle conditions imply that $(DQC)$ and $(ABE)$ are tangent to $EB$ at $C$ and $B$ , respectively. Denote $EP \cap (DQC) = X$ .Therefore, for $PQBC$ to be cyclic, we must prove that $\angle PQC = \angle CBP$ . But $\angle PQC = \angle XDC$ by inscribed angles and similarly $\angle CBP = \angle PAB$ . Therefore, we must prove that $\angle XDC = \angle PAB$ , which is clear from a homothety centered at $E$ which takes $(RDC)$ to $(PAB)$ so we are done.

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happypi31415

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happypi31415

#75 h Aug 22, 2024, 3:50 PM

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Feel like i fakesolved cuz i never realized BC tangent to the circumcircles? :skull:

Construct point $X$ on line $LK$ outside of $ABCD$ such that $\angle DXC \cong \angle APB$ . Note that because of the ratio condition, we have a homothety between $\triangle DXC$ and $\triangle APB$ . Then, we have that $DXCM$ is cyclic, so $\angle DMX \cong \angle DCX \cong \angle PBA$ . However, $\angle DMC \cong \angle CBA$ so $\angle CBP = \angle CBA - \angle PBA \implies \angle CBP \cong \angle CML$ . Therefore, $\angle PBC = 180- \angle PMC$ so we are done.

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dolphinday

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dolphinday

#76 h Sep 1, 2024, 5:33 AM

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Let $BC \cap AD = F$ . Then $F = BC \cap AD$ is the center of the homothety $\Psi$ sending $CD$ to $AB$ . Then let $P' = \Psi^{-1}(P)$ and $\Psi(Q) = Q'$ .
Notice that since $\angle BPA = \angle CP'D = \angle BCD = 180^\circ - \angle CBA = 180^{\circ} - \angle CQD$ which implies that $P' \in (CQD)$ . Similarly, we get that $Q' \in (APB)$ .
Also, since $\angle FCD = \angle CQD$ we have that $FC$ is tangent to $(CQD)$ and similarly $FB$ tangent to $(ABP)$ . Then by PoP we get that $FC^2 = FP' \cdot FQ$ and $FB^2 = FP \cdot FQ' = FB^2$ . Notice that $FP' \cdot FQ' = FP \cdot FQ$ by the homothety so by multiplying our previous two equations together we get that $(FB \cdot FC)^2 = (FP \cdot FQ)^2$ which implies that $FP \cdot FQ = FC \cdot FB$ . So $CPQB$ cyclic by PoP, so we are done.

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L13832

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L13832

#77 h Sep 9, 2024, 3:58 PM

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Similar to everyone else's, also not too hard imo.
$\textbf{Construction:}$ Let $AD \cap BC=E$ .
$\textbf{Angle Chasing I:}$ $(APB)$ and $(CQD)$ are tangent to line $BC$ , this is because $\measuredangle APB=\measuredangle ABE$ and $\measuredangle CQD = \measuredangle ECD.$
$\textbf{Homothety:}$ A homothety with center $E$ sending $K$ to $L$ .
Let $(ABP)\cap KL=R\neq P$ . The same homothetic map $\triangle QDC\mapsto \triangle RAB$ .
$\textbf{Angle Chasing II:}$ $\angle BCQ=\angle CDQ =\angle BAR=\angle BPR$ .
So $\odot(PQBC)$ and we are done!

p.s.

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fearsum_fyz

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fearsum_fyz

#78 h Oct 16, 2024, 9:44 AM • 1 Y

Y by Funcshun840

The first step is to find a more natural way to think about the angle conditions.

Let $AD$ and $BC$ intersect at $E$ . Note that the $\frac{AK}{KB} = \frac{DL}{LC}$ condition is equivalent to $E, L, K$ being collinear by the homothety centered at $E$ taking $DC$ to $AB$ .

Further, a simple angle chase shows that the angle conditions for $P$ and $Q$ are equivalent to $BC$ being tangent to $(APB)$ and $(CQD)$ .

Then, power of a point seems like the most hopeful way to show that $P, Q, B, C$ are concyclic (because we have tangents, proportional lengths, and we would like to use the $E-L-P-Q-K$ collinearity somehow).
Note that $EC^2 = EF \cdot EQ$ and $EB^2 = EP \cdot EG$ .
Hence $EC \cdot EB = \sqrt{EF \cdot EQ \cdot EP \cdot EG} = \sqrt{(\frac{1}{\cancel{k}} \cdot EP) \cdot EQ \cdot EP \cdot (\cancel{k} \cdot EQ)} = EP \cdot EQ$ where $k$ is the scale factor of the homothety.
We are done.

This post has been edited 1 time. Last edited by fearsum_fyz, Oct 16, 2024, 9:52 AM
Reason: readability

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Maximilian113

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Maximilian113

#79 h Jan 12, 2025, 11:41 PM

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By homothety, observe that $AD, KL, BC$ concur at a point $R.$ In addition, draw $(APB)$ and $(DQC).$ Notice that $\angle APB = \angle DCB = 180^\circ - \angle DBP,$ so it follows that $BC$ is tangent to $(APB).$ Also, $\angle DQC = \angle ABR = \angle DCR$ so $BC$ is also tangent to $(DQC).$ Now, let $RK$ intersect $DC$ at $P'.$ By homothety it follows that $PB \parallel P'C,$ so $\angle PBC = \angle P'CR = \angle P'QC = \angle PQC,$ and thus the desired result follows. QED

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Bonime

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Bonime

#80 h Apr 12, 2025, 6:41 PM

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$\textbf{G2 IMOSL 2006}$

Let $AP \cap DC=U$ . Then $\measuredangle UCB = \measuredangle UPB \Rightarrow ABUC \ \text{is cyclic}$ $\Rightarrow \measuredangle CBP= \measuredangle CUP = \measuredangle UPB$ And then, we get that $BC$ is tangent to $(APB)$ . Similarly, define $DQ \cap AB=V$ , hence we have $\measuredangle TQC= \measuredangle TBC \Rightarrow CQTB \ \text{is cyclic}$ $\Rightarrow \measuredangle BCQ= \measuredangle BTQ= \measuredangle QDC$ So $BC$ is tangent to $(DQC)$ too.

Therefore, let $AD \cap BC=X$ , $XK\cap (DQC)=M$ and $XK\cap (APB)= N$ . Then, from $DC//AB$ , we get that there's a homothety $\phi(P)=P'$ which makes $DC \mapsto AB$ . But, from the tangency, $(DQC)'=(APB)$ , so $P=M'$ and $Q=N'$ . Thus, we're done, because by PoP, we get that $\frac{XM}{XP}=\frac{XQ}{XN} \ \text{(By the homothety)}$ $\Rightarrow \boxed{XP\cdot XQ =XM \cdot XN}$ But, by PoP, we know that $XC^2 = XM \cdot XQ \ \text{and} \ XB^2=XP\cdot XN$ $\Rightarrow (XC\cdot XB)^2 = (XP\cdot XQ)(XM\cdot XN)=(XP\cdot XQ)^2$ $\boxed{XC\cdot XB=XP \cdot XQ}$ $\blacksquare$

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