Concurrency in a quadrilateral implies equal length

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Source: 2022 China TST, Test 4 P2

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JustPostChinaTST

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JustPostChinaTST

#1 h Apr 30, 2022, 3:48 PM • 3 Y

Y by ImSh95, Rounak_iitr, ys-lg

Let $ABCD$ be a convex quadrilateral, the incenters of $\triangle ABC$ and $\triangle ADC$ are $I,J$ , respectively. It is known that $AC,BD,IJ$ concurrent at a point $P$ . The line perpendicular to $BD$ through $P$ intersects with the outer angle bisector of $\angle BAD$ and the outer angle bisector $\angle BCD$ at $E,F$ , respectively. Show that $PE=PF$ .

This post has been edited 3 times. Last edited by JustPostChinaTST, May 2, 2022, 4:11 PM
Reason: typos

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#2 h May 2, 2022, 12:21 AM • 4 Y

Y by ImSh95, I-love-K2I, aidenkim119, Kingsbane2139

Is the problem right?
Could you please check the original problem?
I think there are some typos somewhere.

Here is a proof that states : $AC,BD,IJ$ can never concurrent.

Let $AC,BD$ intersect at $K$
Let incenter of triangle $ABK, CDK$ be $S,T$ .
Then it is trivial that $S$ lies on segment $BI$ , $T$ lies on segment $CJ$ .
But $ST$ passes $K$ .
So $IJ$ never passes $K$ .

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JG666

#3 h May 2, 2022, 4:26 AM • 1 Y

Y by ImSh95

teddy8732 wrote:

Here is a proof that states : $AC,BD,IJ$ can never concurrent.

Wrong.

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#4 h May 2, 2022, 11:41 AM • 4 Y

Y by ImSh95, I-love-K2I, aidenkim119, Kingsbane2139

JG666 wrote:

teddy8732 wrote:

Here is a proof that states : $AC,BD,IJ$ can never concurrent.

Wrong.

Why is it wrong? Did I understood problem something wrong? ...

JustPostChinaTST wrote:

Let $ABCD$ be a convex quadrilateral, the incenters of $\triangle ABD$ and $\triangle ADC$ are $I,J$ , respectively. It is known that $AC,BD,IJ$ concurrent at a point $P$ . The line perpendicular to $BD$ through $P$ intersects with the inner angle bisector of $\angle ABD$ and the outer angle bisector of $\angle BCD$ at $E,F$ , respectively. Show that $PE=PF$ .

Is $I,J$ are incenters of $\triangle ABD$ and $\triangle ADC$ right?
No typos??

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#5 h May 2, 2022, 1:05 PM • 3 Y

Y by ImSh95, JG666, Mango247

JustPostChinaTST wrote:

the incenters of $\triangle ABD$ and $\triangle ADC$

Well, that should be triangle ABC instead of triangle ABD.

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#6 h May 5, 2022, 8:44 AM • 6 Y

Y by ImSh95, JG666, I-love-K2I, aidenkim119, Mango247, Kingsbane2139

This fact kills the problem.
Hidden

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#7 h May 5, 2022, 12:45 PM • 1 Y

Y by ImSh95

Just paying attention that ABCD has a Tangential Circle outside（the construction of IMO2008）then it is easy to prove by using polar line and polar point(please forgive my poor English)

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#8 h May 5, 2022, 12:47 PM • 2 Y

Y by ImSh95, JG666

This is the solution(without explaination）

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#9 h May 11, 2022, 7:35 AM • 2 Y

Y by JG666, Ru83n05

For completeness, here is the full solution:

Let $\omega_1$ and $\omega_2$ be the incircles of $\triangle ABC$ and $\triangle ADC$ .

Claim: $AB+AD=CB+CD$ .
Proof: Observe that $P$ is the center of negative homothety sending $\omega _1$ to $\omega_2$ . Due to symmetry reasons, we may assume that $BC$ cuts $AD$ on the ray $(DA$ beyond $A$ , in the point $M$ . Consider $\omega$ the excircle of $\triangle DCM$ opposite angle $D$ . Then $D$ is the center of positive homothety sending $\omega_2$ to $\omega$ . Monge's theorem implies that the center of negative homothety sending $\omega_1$ to $\omega$ lies on line $BD$ . However, $BC$ is the internal tangent of $\omega$ and $\omega_1$ , thus $B$ is the center we are looking for. This further implies that $AB$ is tangent to $\omega$ , making $\omega$ tangent to rays $(DA$ beyond $A$ , $(DC$ beyond $C$ , $(AB$ beyond $B$ and $(CB$ beyond $B$ . Calling these tangency points $U,V,W,T$ respectively, we would have that $DU=DA+AU=DA+AW=DA+AB+BW$ and similarly $DV=DC+CV=DC+CT=DC+CB+BT.$ Since $DU=DV$ and $BW=BT$ , we get the desired conclusion.

Call $\Gamma$ the ellipse with foci in $B$ and $D$ passing through $A$ and $C$ . It's well known that the external angle bisectors of angles $BAD$ and $BCD$ are the tangents in $A$ and $C$ to $\Gamma$ . Let $d$ be the line through $P$ perpendicular to $BD$ , and set $d\cap \Gamma = \{R,S\}$ . Since $BD$ is an axis of symmetry in $\Gamma$ , $PR=PS$ . But now, applying Desargues Involution Theorem to the degenerate quadrilateral $AACC$ , conic $\Gamma$ and line $d$ , we get that $(P,P)$ , $(R,S)$ and $(E,F)$ are involution pairs on $d$ , the conclusion following.

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hakN

#10 h May 20, 2022, 6:41 PM • 4 Y

Y by GuvercinciHoca, SerdarBozdag, sevket12, sabkx

Solved with BarisKoyuncu, SerdarBozdag and sevket12.

Let $AB \cap CD = X , AD \cap BC = Y , BI \cap DJ = M$ . Applying Desargues on triangles $\triangle AIB$ and $\triangle CJD$ , we obtain that the points $X , M , G = AI \cap CJ$ are collinear. Now note that since $AI$ is the exterior bisector of $\angle XAC$ and $CJ$ is the bisector of $\angle XCA$ , we get that $G$ is the excenter of $\triangle XAC$ . So, since $G,X,M$ are collinear, this means $M$ lies on the bisector of $\angle (AB,CD)$ , so $d(M,AB)=d(M,CD)$ . Since $M$ also lies on $BI$ and $DJ$ , we similarly obtain that $d(M,AB)=d(M,BC)=d(M,CD)=d(M,AD)$ , implying that the quadrilateral $ABCD$ has an excircle with center $M$ .

Now we will show that $\angle ADE = \angle CDF$ . Note that if we show this, then $A,F$ would be isogonal to $\angle EDC$ , and since $E,C$ are obviously isogonal to $\angle EDC$ , by isogonality lemma we would obtain $P , M$ are isogonal wrt $\angle EDC$ , so this would mean $\angle ADP = \angle JDF$ , and so $\angle EDP = \angle FDP$ , and from here we would be done. So it indeed suffices to show $\angle ADE = \angle CDF$ .

Let $H$ and $N$ be the foots from $E$ and $F$ to $DA$ and $DC$ respectively. Since $EHPD$ and $DPFN$ are cyclic, to show $\angle ADE = \angle CDF$ , it suffices to show $H,P,N$ are collinear. By Desargues theorem on $\triangle AEH$ and $\triangle CFN$ , this collinearity is equivalent to the points $D , M , O = EH \cap FN$ being collinear. Let $S$ be the intersection of the perpendicular from $D$ to $DA$ with $MA$ and let $T$ be the intersection of the perpendicular from $D$ to $DC$ with $MC$ . By Thales, it suffices to show $\frac{MS}{ME} = \frac{MT}{MF}$ which is equivalent to $ST \parallel EF$ , or $BD \perp ST$ . With this, we get rid of points $E,F$ .

Let $K,L$ be the foots from $M$ to $CD,AD$ respectively. Let $Z$ be the foot from $M$ to $BD$ . We will show that $\triangle KZL \sim \triangle SDT$ , and this would imply that $\angle DTS = \angle ZLK = \angle ZDK = 90^\circ - \angle ZDT \implies BD \perp ST$ .
Also note that $\angle KZL = \angle ADK = 90^\circ - \angle KDS = \angle SDT$ , so it suffices to show $\frac{DS}{DT} = \frac{ZK}{ZL}$ .

Now let $\theta = \angle XAM = \angle MAD$ and $\gamma = \angle YCM = \angle MCD$ . We clearly have $DS = DA \cdot \tan{\theta}$ and $DT = DC \cdot \tan{\gamma}$ , so $\frac{DS}{DT} = \frac{DA}{DC} \cdot \frac{\tan{\theta}}{\tan{\gamma}}$ .

By LOS and ratio lemma, we get $\frac{ZK}{ZL} = \frac{\sin{\angle PDC}}{\sin{\angle PDA}} = \frac{PC}{PA} \cdot \frac{DA}{DC}$ , so it suffices to show $\frac{PC}{PA} = \frac{\tan{\theta}}{\tan{\gamma}}$ .

Note that $\frac{PC}{PA} = \frac{\text{area}(BAD)}{\text{area}(BCD)} = \frac{DA \cdot AB \cdot \sin{2\theta}}{DC \cdot CB \cdot \sin{2\gamma}} = \frac{\tan{\theta}}{\tan{\gamma}} \iff DA \cdot AB \cdot \cos^2{\theta} = DC \cdot CB \cdot \cos^2{\gamma}$ .

By law of cosines in $\triangle ABD$ and $\triangle CBD$ , we have $DA^2 + AB^2 + 2 \cdot DA \cdot AB \cdot \cos{2\theta} = DC^2 + CB^2 + 2 \cdot DC \cdot CB \cdot \cos{2\gamma}$ .
Now using the fact that $\cos{2\theta} = 2\cos^2{\theta} - 1$ , this gives that $(DA - AB)^2 + 4\cdot DA \cdot AB \cdot \cos^2{\theta} = (DC - CB)^2 + 4\cdot DC \cdot CB \cdot \cos^2{\gamma}$ . Since $ABCD$ has an excircle, $DA - AB = DC - CB$ and so we get $DA \cdot AB \cdot \cos^2{\theta} = DC \cdot CB \cdot \cos^2{\gamma}$ , and we are done. $\square$

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#11 h May 20, 2022, 6:45 PM • 1 Y

Y by Mango247

hakN wrote:

Solved with BarisKoyuncu, SerdarBozdag and sevket12.

Let $AB \cap CD = X , AD \cap BC = Y , BI \cap DJ = M$ . Applying Desargues on triangles $\triangle AIB$ and $\triangle CJD$ , we obtain that the points $X , M , G = AI \cap CJ$ are collinear. Now note that since $AI$ is the exterior bisector of $\angle XAC$ and $CJ$ is the bisector of $\angle XCA$ , we get that $G$ is the excenter of $\triangle XAC$ . So, since $G,X,M$ are collinear, this means $M$ lies on the bisector of $\angle (AB,CD)$ , so $d(M,AB)=d(M,CD)$ . Since $M$ also lies on $BI$ and $DJ$ , we similarly obtain that $d(M,AB)=d(M,BC)=d(M,CD)=d(M,AD)$ , implying that the quadrilateral $ABCD$ has an excircle with center $M$ .

Now we will show that $\angle ADE = \angle CDF$ . Note that if we show this, then $A,F$ would be isogonal to $\angle EDC$ , and since $E,C$ are obviously isogonal to $\angle EDC$ , by isogonality lemma we would obtain $P , M$ are isogonal wrt $\angle EDC$ , so this would mean $\angle ADP = \angle JDF$ , and so $\angle EDP = \angle FDP$ , and from here we would be done. So it indeed suffices to show $\angle ADE = \angle CDF$ .

Let $H$ and $N$ be the foots from $E$ and $F$ to $DA$ and $DC$ respectively. Since $EHPD$ and $DPFN$ are cyclic, to show $\angle ADE = \angle CDF$ , it suffices to show $H,P,N$ are collinear. By Desargues theorem on $\triangle AEH$ and $\triangle CFN$ , this collinearity is equivalent to the points $D , M , O = EH \cap FN$ being collinear. Let $S$ be the intersection of the perpendicular from $D$ to $DA$ with $MA$ and let $T$ be the intersection of the perpendicular from $D$ to $DC$ with $MC$ . By Thales, it suffices to show $\frac{MS}{ME} = \frac{MT}{MF}$ which is equivalent to $ST \parallel EF$ , or $BD \perp ST$ . With this, we get rid of points $E,F$ .

Let $K,L$ be the foots from $M$ to $CD,AD$ respectively. Let $Z$ be the foot from $M$ to $BD$ . We will show that $\triangle KZL \sim \triangle SDT$ , and this would imply that $\angle DTS = \angle ZLK = \angle ZDK = 90^\circ - \angle ZDT \implies BD \perp ST$ .
Also note that $\angle KZL = \angle ADK = 90^\circ - \angle KDS = \angle SDT$ , so it suffices to show $\frac{DS}{DT} = \frac{ZK}{ZL}$ .

Now let $\theta = \angle XAM = \angle MAD$ and $\gamma = \angle YCM = \angle MCD$ . We clearly have $DS = DA \cdot \tan{\theta}$ and $DT = DC \cdot \tan{\gamma}$ , so $\frac{DS}{DT} = \frac{DA}{DC} \cdot \frac{\tan{\theta}}{\tan{\gamma}}$ .

By LOS and ratio lemma, we get $\frac{ZK}{ZL} = \frac{\sin{\angle PDC}}{\sin{\angle PDA}} = \frac{PC}{PA} \cdot \frac{DA}{DC}$ , so it suffices to show $\frac{PC}{PA} = \frac{\tan{\theta}}{\tan{\gamma}}$ .

Note that $\frac{PC}{PA} = \frac{\text{area}(BAD)}{\text{area}(BCD)} = \frac{DA \cdot AB \cdot \sin{2\theta}}{DC \cdot CB \cdot \sin{2\gamma}} = \frac{\tan{\theta}}{\tan{\gamma}} \iff DA \cdot AB \cdot \cos^2{\theta} = DC \cdot CB \cdot \cos^2{\gamma}$ .

By law of cosines in $\triangle ABD$ and $\triangle CBD$ , we have $DA^2 + AB^2 + 2 \cdot DA \cdot AB \cdot \cos{2\theta} = DC^2 + CB^2 + 2 \cdot DC \cdot CB \cdot \cos{2\gamma}$ .
Now using the fact that $\cos{2\theta} = 2\cos^2{\theta} - 1$ , this gives that $(DA - AB)^2 + 4\cdot DA \cdot AB \cdot \cos^2{\theta} = (DC - CB)^2 + 4\cdot DC \cdot CB \cdot \cos^2{\gamma}$ . Since $ABCD$ has an excircle, $DA - AB = DC - CB$ and so we get $DA \cdot AB \cdot \cos^2{\theta} = DC \cdot CB \cdot \cos^2{\gamma}$ , and we are done. $\square$

$F$

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#12 h Aug 20, 2023, 5:03 PM • 1 Y

Y by GeoKing

Nice problem!

Claim: The two internal bisectors of $\angle ABC$ , $\angle ADC$ and two external of $\angle BAD$ , $\angle BCD$ are concurrent, i.e. there's an excircle.
Proof. Let the first two intersect at $O$ and draw $\omega$ centered at $O$ tangent to $BA$ and $BC$ . By Monge d'Alembert's theorem, the insimilicenter of $\omega$ and the incircle of $\triangle ADC$ lies on $BD$ . But since $D\in JO$ it follows that the insimilicenter is $D$ . So $DA$ and $DC$ are tangent to $\omega$ and the claim follows.

Now, it's well known that $BC+CD=BA+AD$ which means that $A$ and $C$ lie on an ellipse $\mathcal{E}$ with foci $B$ , $D$ . Then $(E,F;P,\infty_{\perp BD})\stackrel{O}{=}(A,C;P,OX\cap AC)$ where $X$ is the projection of $O$ onto $BD$ . It remains to show that $PX$ is the bisector of $\angle AXC$ and we'd be done by the perpendicularity-angle bisector lemma. To this end, note $AO$ and $CO$ are tangents to $\mathcal{E}$ and if $A'$ and $C'$ are reflections of $A$ and $C$ over the major axis, Brokard's theorem on $AA'C'C$ finishes.

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#14 h Aug 23, 2023, 7:23 PM

Y by

Claim: We have $AB+AD=CB+CD$ .
Proof. By Pitot's theorem twice, proving that there is a circle tangent to rays $\overrightarrow{AB}$ , $\overrightarrow{CB}$ , $\overrightarrow{DA}$ , and $\overrightarrow{DC}$ is sufficient. Let $Q = \overline{BC} \cap \overline{AD}$ , and denote $\omega$ by the $D$ -excircle of $\triangle DQM$ and $O$ by the center of $\omega$ . I contend that $\omega$ is the desired circle. Let $\omega_B$ be the incircle of $\triangle ABC$ and $\omega_D$ be the incircle of $\triangle ADC$ . By Monge's theorem, the insimilicenter $\textbf{I}(\omega_B, \omega)$ of $\omega_B$ and $\omega$ lies on $\overline{BD}$ , since $D$ is the insimilicenter of $\omega_D$ and $\omega$ . Since $\textbf{I}(\omega_B, \omega)$ also lies on $\overline{OJ}$ , it must be $D$ . Thus the desired tangencies follow. :yoda:

:yoda:

Note that by the claim, there exists an ellipse $\epsilon$ with foci at $B$ and $D$ passing through $A$ and $C$ . Define $\ell$ as the line through $P$ perpendicular to $\overline{BD}$ , and intersect $\epsilon$ at points $X$ and $Y$ . Clearly we have $PX=PY$ . Applying DIT on $\epsilon$ with respect to $\ell$ , there exists an involution swapping $(P, P)$ , $(E, F)$ , and $(X, Y)$ , but this means that $PE=PF$ , as desired. :starwars:

:starwars:

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#15 h Nov 23, 2023, 10:38 PM

Y by

Claim: $ABCD$ is extangential.
Proof. By Desargues on $AIB$ and $CJD$ it follows that $H = \overline{AI} \cap \overline{CJ}$ , $O = \overline{IB} \cap \overline{JD}$ , and $G = \overline{AB} \cap \overline{CD}$ are collinear.
It follows that $H$ is the $C$ -excenter of $ACG$ , so $HG$ is the external bisector of $AGD$ , so $O$ is the desired excenter. $\blacksquare$
Then, the external angle bisectors of $BAD$ and $BCD$ are $AO$ and $CO$ respectively.
Let $Q$ be the polar of $BD$ wrt to the excircle. Then it remains to show that $(AC;PQ) = -1$ . Polar reciprocating the diagram gives the result with taking $P$ as the intersection of the diagonals.

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#16 h Jan 19, 2024, 4:40 PM • 1 Y

Y by OronSH

Why are so many CTST geos projective spam?
The following is the key claim:
Key Claim: $ABCD$ is extangential quadrilateral.
proof
Now let $\mathcal{E}$ be an ellipse focussed at $B$ and $D$ passing through $A$ and $C$ .Let $\mathcal{l}$ denote the line through $P$ perpendicular to $BD$
Extend line $BD$ to $\mathcal{E}$ at $U,V$ . Apply DIT on $\mathcal{E}$ WRT $\mathcal{l}$ and realise you are done.

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#17 h Sep 11, 2024, 2:42 PM • 2 Y

Y by OronSH, ys-lg

Remark from 数之谜: After $|AB|+|AD|=|CB|+|CD|,$ $A,C$ moves on an ellipse $\Gamma$ with focus $B,D.$ Since $AE$ is the outer angle bisector of $\angle BAD,$ $AE$ is tangent line of $\Gamma,$ similarly $CF$ is also tangent line of $\Gamma.$ Butterfly Theorem gives $|PE|=|PF|.\Box$

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OronSH

#18 h Sep 21, 2024, 3:47 PM • 1 Y

Y by Zhaom

Let $X,Y=AB\cap CD,AD\cap BC$ . Say $X,Y$ lie on same side of $AC$ as $B$ . Construct circle $\Omega$ with center $Z$ tangent to extensions of rays $AB,CB,CX$ . By insimilicenter monge on $\Omega,(I),(J)$ we get $\Omega$ also tangent to $AY$ . Now dual brocard's gives that dual of $BD$ wrt $\Omega$ is $AC\cap XY=Q$ . Thus $-1=(A,C;P,Q)\stackrel Z=(E,F;P,\infty)$ finishes.

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