Denote $Z \in BJ \cap AC$. We have that $\frac{BJ}{JZ}=\frac{a+c}{CD}$, becuase $BJ$ is the bisector of $\angle{CBD}$. Thus $ZJ=\frac{BJ \cdot CD}{a+c}$, and $BJ=\frac{BJ \cdot (CD+a+c)}{a+c}$. Now the main idea is to use the transversal theorem in $\triangle{ABZ}$, for $KL$ and the cevian $AJ$. So let's denote $T \in AJ \cap KL$. Therefore $TA=TJ \Longleftrightarrow \frac{BK}{KA}\cdot ZJ+\frac{ZL}{LA}\cdot BJ = BZ$ $\iff \frac{a+c+CD}{a+c}-\frac{p-b}{p-a}\cdot \frac{CD}{a+c}= \frac{ZL}{LA}$. ($\star$) Now from the bisector theorem $\frac{ZD}{ZC}=\frac{c}{a}\iff ZD=\frac{c \cdot CD}{a+c}$. And from $AD=b-CD$, we have that $AZ = b-CD+\frac{c \cdot CD}{a+c}$, so: $ZL = \frac{(a+b-p-CD)(a+c)+c \cdot CD}{a+c}$, and $AL=p-a$. Thus $(\star)$ is equivalent with: $\frac{a+c+CD}{a+c}-\frac{p-b}{p-a}\cdot \frac{CD}{a+c}= \frac{(a+b-p-CD)(a+c)+c \cdot CD}{(a+c)(p-a)}$, which after clearing the denominator is true. [EDIT:typing errors]

PP13. Let $\triangle ABC$ such that $C<A<\frac{\pi}{2}$. Let $D\in [AC]$ such that $BD=BA$. The incircle $w=C(I,r)$ of $\triangle ABC$ touches its sides in $K\in [AB]$ and $L\in [AC]$. Let $w_{1}=C(J,r_{1})$ be the incircle of $\triangle BCD$. Denote $T\in KL\cap AJ$. Prove that $TA=TJ$. Proof.. The incircle $w_{1}$ touches $\triangle BCD$ in $U\in BD$, $V\in BC$ and $W\in BC$. Observe that $DC=AC-AD=b-2c\cdot\cos A=$ $\frac{b^{2}-(b^{2}+c^{2}-a^{2})}{b}$ $\implies$ $DC=\frac{a^{2}-c^{2}}{b}$. Thus, $VC=\frac{1}{2}\cdot (BC+CD-BD)=\frac{1}{2}\cdot(a+\frac{a^{2}-c^{2}}{b}-c)\implies VC=\frac{p(a-c)}{b}$ . Since $VC=JC\cdot\cos \frac{C}{2}$ obtain that $\boxed{JC=\frac{p(a-c)}{b\cdot\cos\frac{C}2}}$. Denote $P\in KL\cap CI$. The property $PC\perp PB$ is well-known (see the lower remark). Thus, $APJW$ is cyclically. Analogously $P\in UV\cap CJ$, i.e. $\boxed{\ P\in KL\cap UV\ }$. Observe that $\boxed{PC=a\cdot \cos \frac{C}{2}}\implies$ $\frac{PC}{JC}=\frac{ab\cdot\cos^{2}\frac{C}{2}}{p(a-c)}=\frac{p-c}{a-c}$ $\implies$ $\frac{PC}{p-c}=\frac{JC}{a-c}=\frac{PJ}{p-a}$ $\implies$ $\boxed{\ \frac{PJ}{PC}=\frac{p-a}{p-c}\ }$. Apply the Menelaus' theorem to transversal $\overline{PTL}$ in $\triangle AJC$ : $\frac{PJ}{PC}\cdot\frac{LC}{LA}\cdot\frac{TA}{TJ}=1$ $\implies$ $\frac{p-a}{p-c}\cdot\frac{p-c}{p-a}\cdot\frac{TA}{TJ}=1$ $\implies TA=TJ$. So $PC\perp PB$ $\Longleftrightarrow$ the quad. $PKIB$ is cyclically $\Longleftrightarrow$ $m(\widehat{PKB})=m(\widehat{PIB})=\frac{B+C}{2}$ or $m(\widehat{PKA})=m(\widehat{PIB})=\frac{B+C}{2}$

Invalid URL I have a synthetic solution: Let S be the foot of the perpendicular from B, M the point where the incircle of BCD touches AC, F the midpoint of SL. Let Q be the point where the perpendicular bisector of SL meets KL. Now, the triangle SQL is isosceles, and by angle chasing we get $\angle QSL = \angle SLQ = 90-\frac{\alpha}2$. With angle chasing you can also find that $\angle JDM = 90-\frac{\alpha}2$, therefore $SQ \parallel JD$. Now we calculate $2LM = 2CL-2CM = (AC+BC-AB)-(DC+BC-BD) = AC-DC = AD$. Therefore $AS = SD = LM$, $SL = DM$, $2SF = DM$. Consider the homotety with centre A and factor 2: - it sends S to D - therefore it sends the line QS to the line JD - it sends F to M - therefore it sends the line QF to the line JM - therefore it sends Q to J Finally, Q is on the lines AJ and KL, then Q is P! But the homotety had a factor 2; we conclude $JP = PA$.

This problema was in the first Brazilian TST!

That's not so suprising, this is a Shortlist problem.

Arne wrote: That's not so suprising, this is a Shortlist problem. by the way I would like to know if you have the last IMO short list.

I do...

Let $P$ be the point of intersection of $JD$ and $KL$, $I$ be the incentre of $\triangle ABC$, $Q$ be the excentre of $\triangle BAD$ opposite to $B$ and $M$ be the midpoint of $AD$. A little angle-chasing shows that $\angle PDL=\angle DAQ =\angle ALK=\angle AKL = 90^\circ-\alpha/2$, $\angle AQM=\alpha/2$ and $\angle JBQ=\beta/2$. From $\angle ALK=\angle QAD$ we conclude that $KL\parallel AQ$, so it is enough to prove that $QJ=2\cdot QP$. It follows from $\angle PDQ=\angle PDL=\angle AKL=\angle AKP$ that $AKDP$ is cyclic, hence $\angle QPA=\angle DKA$, so the triangles $QPA$ and $AKD$ are similar. Hence \[\frac{QP}{AK}=\frac{QA}{AD}= \frac{1}{2}\cdot\frac{QA}{AM}=\frac{1}{2\sin\frac{\alpha}{2}},\ \ \ \ \ \ \ \ (1) \] the latter following from $\QMA=90^\circ$. Also, since $\angle JBQ=\angle IBA=\beta/2$ and $\angle JQB=\alpha/2=\angle IAB$, the triangles $QJB$ and $AIB$ are similar, so \[\frac{QJ}{AI}=\frac{BQ}{BA}=\frac{\sin(90^\circ-\frac{\alpha}{2})}{\sin\frac{\alpha}{2}}= \frac{\cos\frac{\alpha}{2}}{\sin\frac{\alpha}{2}}.\ \ \ \ \ \ \ \ (2) \] From (1) and (2), we conclude that \[\frac{QJ}{QP}=2\cdot\sin\frac{\alpha}{2}\cdot\frac{\cos\frac{\alpha}{2}}{\sin\frac{\alpha}{2}}\cdot \frac{AI}{AK}= 2\cdot\cos\frac{\alpha}{2}\cdot\frac{AI}{AK}.\ \ \ \ \ \ \ \ (3) \] However, since $AKI=90^\circ$, we have \[\frac{AI}{AK}=\frac{1}{\cos\frac{\alpha}{2}}\ \ \ \ \ \ \ \ (4) \] so combining (3) and (4), we obtain \[\frac{QJ}{QP}=2, \] as desired. $\Box$

http://www.mathlinks.ro/Forum/viewtopic.php?t=106306 How did this succeed to get disclosed in August 2006? Darij

Let $ S$ be the midpoint of $ AJ$ and let $ T$ be the intersection of lines $ JD$ and $ KL$. Moreover, let $ M$ be the foot of the perpendicular of $ J$ on $ CD$. It's our aim to show that $ K$, $ S$ and $ T$ are collinear. First note that triangles $ \triangle TLD$ and $ \triangle AKL$ are similar. This follows from a short angle-chase. Hence $ DT/DL = LA/LK$. Moreover, $ LA/LK = 1/(2\sin\angle CAB/2)$ $ (*)$. This follows from the sine law and the addition theorem. Another consequence of the similarity is that $ AKDT$ is cyclic. Additionally, $ AL - LD = 2MD$ $ (**)$. This is easily seen, as $ AL - LD = 2AL - AD = AB + AC - BC - AD = AB + CD - BC$ and $ 2MD = BD + CD - BC$, hence the conclusion. We have $ MD/JD = \sin \angle CAB/2$ $ (***)$. Now, applying Menelaos' theorem in triangle $ \triangle AJD$ yields: \[ \frac {AS}{SJ} \cdot \frac {JT}{DT} \cdot \frac {DL}{LA} = 1 \\ \Leftrightarrow JT = LA \cdot \frac { DT}{DL} = \frac{LA^2}{LK}.\] This is equivalent to $ JT \cdot LK = (JD + DT)LK = LA^2 \Leftrightarrow JD \cdot LK = LA^2 - DT \cdot LK = LA(LA - DL)$, where $ DT \cdot LK = LA \cdot DL$ follows from the power of a point wrt quadrilateral $ AKDT$. The conclusion follows now from combining $ (*)$, $ (**)$ and $ (***)$.

Dear Mathlinkers, can we imagine a proof with trigonometry, ratio, angle, homothetie, Menelaüs and case of egality of triangles? I think so A new challenge... Sincerely Jean-Louis

I have another idea. let $m(JAD)=b,m(JAI)=a$ $\implies$ $m(IAK)=a+b$ $m(DCJ)=m(JCB)=c$ $\implies$ $m(JBC)=m(JBD)=a+b-c,$ $m(DBI)=90-3(a+b)-c,m(IBK)=90-a-b-c$ $AJ\cap KL=\{P\}$ $AP=BJ{\frac{cos(a+b+c).cos(a+b)}{cosa}}$ $AJ=BJ{\frac{cos(2a+2b)sin(a+b+c)}{cos(a+b)sina}}$, in $\triangle ABJ$ $sin(b+c)cos(2a+2b)=cos(a+b+c)sina\cdots (1)$ $2cos(a+b+c)cos(a+b)cos(a+b)sina=cosacos(2a+2b)sin(a+b+c)$ $\iff$ $2cos^{2}(a+b)sin(b+c)=cosasin(a+b+c)$ $\iff$ $(cos(2a+2b)+1)sin(b+c)=cosasin(a+b+c)$ $\iff$ $(cos(2a+2b)+1)sin(b+c)=sin(2a+b+c)-sin(b+c)cos(2a+2b)$ $\iff$ $sin(b+c)(2cos(2a+2b)+1)=sin(2a+b+c)$ $\iff$ $2cos(2a+2b)sin(b+c)=sin(2a+b+c)-sin(b+c)=2sinasin(a+b+c)$ Which is obviously true Q.E.D

Yimin Ge wrote: http://www.artofproblemsolving.com/Forum/download/file.php?id=9462&mode=view How can you draw geometries shapes?

Dear Mathlinkers, I come back with the same idea where I made a typo: can we imagine a proof without trigonometry, ratio, angle, homothetie, Menelaüs and case of egality of triangles? I think so. Sincerely Jean-Louis

We can also imagine no solutions, it's equivalent...

Dear Joao and Mathlinkers, yes, for this nice remark but I have a proof which you will see next on my site. Sincerely Jean-Louis

Dear Mathlinkers, an article concerning this problem with an original proofs has been put on my website http://perso.orange.fr/jl.ayme vol. 5 (last article) Use Google translator Merry Christmas Sincerely Jean-Louis

Dear Mathlinkers,

Dear Armpist (M) and Mathlinkers, your idea is very brilliant because you made a link with a San Gaku and have a nice chain history connected the figures... I have connected your point of view with the Thébault's result also... How do you prove the San Galu in question? Sincerely Jean-Louis

Why did I try to length chase/PoP the first claim for so long... Let $K'$ and $L'$ be the reflections of $A$ in $K$ and $L,$ respectively. Claim: $BIJK'$ is cyclic. Proof. \begin{align*}\angle IBJ&=180-\angle BJC=180-(90+\tfrac{1}{2}\angle BDC)\\&=90-\tfrac{1}{2}(180-\angle ADB)=\tfrac{1}{2}\angle A=\angle K'AI=\angle IK'B\end{align*}since $AI=IK'.$ $\blacksquare$ Then, $$\angle JK'A=180-\angle BIC=180-(90+\tfrac{1}{2}\angle A)=90-\tfrac{1}{2}\angle A=\angle LKA=\angle L'K'A$$so $K',J,$ and $L'$ are collinear. By homothety at $A$ with scale factor $\tfrac{1}{2},$ $\overline{AJ}\cap\overline{KL}$ is the midpoint of $\overline{AJ}.$ $\square$

Let $KL\cap JD=P$ and $AJ\cap KL=E$ and $\angle BAC=a$. Note that $\angle LDP=90^{\circ}-\frac{a}{2}=\angle DLP$. By Menelaus's theorem on $\triangle AJD$ with collinear points $E,L,P$, we have that $$\frac{AE}{EJ}\cdot \frac{JP}{DP}\cdot\frac{DL}{LA}=1 \implies \frac{AE}{EJ}=\frac{DP}{LD}\cdot\frac{AL}{JP}$$Note that by Law of sines in $\triangle LDP$, we have that $\frac{DP}{LD}=\frac{\sin{(90^{\circ}-a/2)}}{\sin{a}}=\frac{1}{2\sin(a/2)}$. Now $$JP=DP+JD=\frac{DL}{2\sin{(a/2)}}+\frac{BD+DC-BC}{2\sin{(a/2)}}=\frac{DL+BA+DC-BC}{2\sin{(a/2)}}=\frac{BA+AC-BC-AL}{2\sin{(a/2)}}=\frac{2AL-AL}{2\sin{(a/2)}}=\frac{AL}{2\sin{(a/2)}}\implies \frac{AL}{JP}=2\sin{(a/2)}$$Hence $\frac{AE}{EJ}=\frac{1}{2\sin{(a/2)}}\cdot 2\sin{(a/2)}=1$ as desired.

Let $X$ be the projection of $B$ onto $AC$, the incenter of $ABC$ be $I$, $T = BX \cap KL$, $M = AI \cap KL$, the incircle of $BCD$ touch $\overline{ADC}$ at $P$, and $S_{\delta}$ denote the semiperimeter of triangle $\delta$. It's easy to see $XA = XD$ and $AI \perp KL$. Clearly, we have $$S_{ABC} - S_{BCD} = \frac{CA - CD}{2} = \frac{AD}{2} = XA$$so $$LP = CL - CP = (S_{ABC} - BA) - (S_{BCD} - BD) = XA.$$Thus, basic length chasing yields $AL = XP$ and $XL = DP$. Now, because $$\angle TLX = 90^{\circ} - \angle MAL = 90^{\circ} - \frac{\angle A}{2} = 90^{\circ} - \frac{\angle BDA}{2} = \angle JDP$$follows from properties of external bisectors, we know $TLX \cong JDP$. This implies that $JPXT$ is a rectangle, so $$\overrightarrow{TJ} = \overrightarrow{XP} = \overrightarrow{AL}$$which means $ALJT$ is a parallelogram. Hence, we know $\overline{KTL}$ bisects $AJ$, as required. $\blacksquare$ Remark: I almost missed the parallelogram argument completely. In fact, I was prepared to perform a full out length bash.

Let $M$ be the midpoint of $AD$, $N$ be the perpendicular from $J$ to $CD$, and $P$ be the point such that $L$ is the midpoint of $AP$. Note that $2AL-2ND=(AB+AC-BC)-(BD+CD-BC)=AD$ so we have \[ND=2AL-ND-AD=2AL-AN=AL-(AN-AL)=PL-NL=PN\]Since $JN\perp PD$, \[\angle JPA=\angle JDC=\frac12 (180^\circ-\angle BDA)=90^\circ-\frac12\angle BAC=\angle KLA\]so $JP\parallel KL$. Since $KL$ bisects $AP$, it must also bisect $AJ$.

This was also Serbian MO 2007 P1.

Best solution ! https://www.google.com/search?ie=UTF-8&client=tablet-android-samsung-rev2&source=android-browser&q=shortlist+imo+2007

nice, another one from... 18 years ago

Let $CJ$ meet $KL$ at $X$ and the external $A$-bisector at $Y$. Since $KL\parallel AY$, we have the condition is equivalent to $X$ being the midpoint of $JY$. Now $X$ is the iran point, so $BX\perp CX$ and we now need to show that $\angle JBX=\angle XBY$. But since $Y$ is the $C$-excentre it is not hard to chase that both angles equal $90^\circ-\frac{A}{2}$. So we are done.

easy solve without diagram We use barycentric coordinates with $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$. Let $D=:(d,0,1-d)$ so \[ db=AD=2c\cos A=\frac{b^2+c^2-a^2}{b}. \]Also $K=(s-b:s-a:0)$ and $L=(s-c:0:s-a)$. The angle bisector of $\angle BCD$ is given by $(a:b:t)$ where $t$ varies. Since $BD=c$, by the Angle Bisector Theorem we have $CD=\frac{a}{a+c}(b-db)=\frac{ab(1-d)}{a+c}$. Thus the angle bisector of $\angle CBD$ is given by $(a(1-d):t:c+ad)$ where $t$ varies. It follows that \[ J=(a(1-d):b(1-d):c+ad)=\left(\frac{a(1-d)}{a+b+c-db},\frac{b(1-d)}{a+b+c-db},\frac{c+ad}{a+b+c-db}\right). \]Line $KL$ is given by \[ \begin{vmatrix} x & y & x\\ s-b & s-a & 0\\ s-c & 0 & s-a \end{vmatrix} =0 \]which simplifies to $(s-a)x-(s-b)y-(s-c)z=0$. The midpoint of $\overline{AJ}$ is \[ \frac{1}{2}\left(1+\frac{a(1-d)}{a+b+c-db},\frac{b(1-d)}{a+b+c-db},\frac{c+ad}{a+b+c-db}\right). \]Now we just need to check that it lies on line $KL$. This is easy: \begin{align*} &(s-a)(2a+b+c-ad-db)-(s-b)b(1-d)-(s-c)(c+ad)\\ =\ &2as+sb+sc-sad-sdb-2a^2-ab-ac+a^2d+adb-sb+sdb+b^2-b^2d-sc-sad+c^2+cad\\ =\ &a(2s-b-c-2a)-ad(2s-a-b-c)+b^2+c^2-b^2d\\ =\ &-a^2+b^2+c^2-(b^2+c^2-a^2)\\ =\ &0. \end{align*}We are done. $\square$

Construct line through $J$ parallel to $KL$. Let it meet $AC,AB$ at $X,Y$. From $$\measuredangle CJX = \measuredangle JXA - \measuredangle JCX = 90 - \frac{\measuredangle A}{2} - \frac{\measuredangle C}{2} = \frac{\measuredangle B}{2} = \measuredangle IBA$$ we have $I,J,B,Y$ cyclic. Now $\measuredangle JBI = \frac{\measuredangle DBA}{2} = \measuredangle JYI$ which give us $\measuredangle IYA = \measuredangle YAI$ . As $I$ is center, we have $K$ is midpoint of $AY$

Wizard0001 wrote: Let $KL\cap JD=P$ and $AJ\cap KL=E$ and $\angle BAC=a$. Note that $\angle LDP=90^{\circ}-\frac{a}{2}=\angle DLP$. By Menelaus's theorem on $\triangle AJD$ with collinear points $E,L,P$, we have that $$\frac{AE}{EJ}\cdot \frac{JP}{DP}\cdot\frac{DL}{LA}=1 \implies \frac{AE}{EJ}=\frac{DP}{LD}\cdot\frac{AL}{JP}$$Note that by Law of sines in $\triangle LDP$, we have that $\frac{DP}{LD}=\frac{\sin{(90^{\circ}-a/2)}}{\sin{a}}=\frac{1}{2\sin(a/2)}$. Now $$JP=DP+JD=\frac{DL}{2\sin{(a/2)}}+\frac{BD+DC-BC}{2\sin{(a/2)}}=\frac{DL+BA+DC-BC}{2\sin{(a/2)}}=\frac{BA+AC-BC-AL}{2\sin{(a/2)}}=\frac{2AL-AL}{2\sin{(a/2)}}=\frac{AL}{2\sin{(a/2)}}\implies \frac{AL}{JP}=2\sin{(a/2)}$$Hence $\frac{AE}{EJ}=\frac{1}{2\sin{(a/2)}}\cdot 2\sin{(a/2)}=1$ as desired. Had the same solution as this one, just my trig bash was a little different.

I did this problem while working on Lemmas in OG on the chapter on Menelaus so adding the intersection of $KL$ and $JD$ was motivated. Else how can you motivate it?

Let $ AJ $ and $ KL $ intersected at point $ M $ . Let $ I $ is the incenter of triangle $ ABC $. We know $ C, J, I $ collinear. Let $ CJ $ and $ KL $ intersect at point $ T $ . By lemma 255 ( or by Iran Lemma ), We know $ \measuredangle CTB=90 $ . By angle chasing, we can find easily $ CT/TJ=Cot( \measuredangle C/2 )/tan( \measuredangle A/2 )=CL/AL $ . So By Menelaus theorem for triangle $ CTL $ and points $ A, M, J $ we can find $ MA=MJ $ . Proved

Let $I_C$ be the $C$-excenter of $\bigtriangleup ABC, X = JA \cap KL, Y = KL \cap CI_C$ Firstly, notice $LK$ and $AI_C$ is perpendicular to $AJ$, thus $KL \parallel AI_C$ and our new RTP is to prove $Y$ is the midpoint of $JI_C$. Next, $\angle BJI_C = \frac{\angle BCA}{2} + \frac{\angle BAC - \angle ACB}{2} = \frac{\angle CAB}{2} = \angle BI_CC$, thus $\bigtriangleup BI_CJ$ is isosceles and $BJ = BI_C$. Now also notice that $\angle IYK = \angle CI_CA = \angle IBA = \angle IBK$ thus $IBYK$ is cyclic and $\angle IYB = \angle IKB = 90$ Since we have $BY \perp JI_C$ and $BJ = BI_C$, we have $JY = YI_C$ thus the question is done.