The triangle $ABC$ has $CA = CB$. $P$ is a point on the circumcircle between $A$ and $B$ (and on the opposite side of the line $AB$ to $C$). $D$ is the foot of the perpendicular from $C$ to $PB$. Show that $PA + PB = 2 \cdot PD$.
Problem
Source: JBMO 2002, Problem 1
Tags: geometry, circumcircle, trigonometry, trig identities, Law of Cosines
socrates
30.10.2005 16:08
Extend $PB$ to $A'$ such that $BA'=PA$ Then compare triangles $CAP$ and $CBA'$ In the contest I solved it by aid of trig!
Andreas
30.10.2005 18:01
$CA = CB = x$
Since $APBC$ is cyclic we have $AB \cdot CP = AP \cdot CB + PB \cdot CA$ $\Rightarrow$ $AB \cdot CP = (PA + PB)x$.
$PA + PB = \frac{2AH \cdot CP}{x}$ where $H$ is the midpoint of $AB$.
Also $\angle BPC = \angle BAC$ and for costruction $\angle AHC = \angle PDC$.
Hence $\Delta AHC \sim \Delta PDC$.
So $\frac{AH}{x} = \frac{PD}{CP}$ $\Rightarrow$ $\frac{2AH \cdot CP}{x} = 2PD = PA +PB$.
Virgil Nicula
30.10.2005 23:26
I note the middlepoint $M$ of the side $AB$. $\widehat {CAB}\equiv \widehat {CPB}\equiv \widehat {CPD}\Longrightarrow \triangle CAM\sim\triangle CPD\Longrightarrow$ $\frac{CA}{AM}=\frac{CP}{PD}\Longrightarrow PC=\frac{2a}{c}\cdot PD$. From the Ptolemeu's theorem results the relation $PA\cdot BC+PB\cdot AC=PC\cdot AB\Longrightarrow a\cdot (PA+PB)=c\cdot PC\Longrightarrow$ $PA+PB=\frac ca\cdot PC=\frac ca\cdot \frac{2a}{c}\cdot PD\Longrightarrow PA+PB=2PD$.
aodeath
27.02.2007 20:54
Well, first note that we can reformulate the problem using an arithmetic trick, writing, then: $PA+PB = PA+PD+DB=2PD \Leftrightarrow PA+DB = PD$ This is nice, because it leads us to another interpretation (which I think is geometric meaningfull). Geomtrically, use the compass on the point P and make a circle with radius $\overline{PA}$. Ok, name the point of intersectiojn between the circle and the segment $\overline{PB}$, M. Draw the line $PC$. Observe that, because of the property of the triangle and other properties on angles and circles, we garantee that $\angle{APC}=\angle{CPB}$ (actually, because they "look" to the same arc). It's wasy to see that our nice $PC$ is notheing less then the mediatrice of the segment $AM$. Now, draw $CM$ and observe the following equality: $CA=CM=CB$. Bingo! $CD$ is mediatrice of the segment $MB$ which completes the proof =]]]
dgreenb801
11.04.2009 05:14
Nice! Let X be the foot of the altitude from C to AP. Then $ \triangle CAX$ and $ \triangle CBD$ are congruent since $ \angle CAX=\angle CBD$ and $ CA=CB$, and $ \triangle CPX$ is congruent to $ \triangle CPD$ since $ CX=CD$. So $ PX=PD$, $ AX=BD$, and it is obvious since $ PA+PB=PA-AX+PB+BD=2PD$
AwesomeToad
21.10.2011 19:28
Since $APBC$ is cyclic, Ptolemy gives us $CA\cdot PB+CB\cdot PA=PC\cdot AB$ and since $CA=CB$, $ CA(PA+PB)=PC\cdot AB$. Thus $PA+PB=\frac{PC\cdot AB}{CA}$.
Since $\angle{CAB}$ and $\angle{BPC}$ are inscribed in the same arc, $\angle{CAB}=\angle{BPC}$. Then Law of Cosines gives us $CA^2+AB^2-2\cdot CA\cdot AB=CB^2=CA^2$, so $AB^2=2\cdot CA \cdot AB$ and $\cos{BPC}=\frac{AB}{2\cdot AC}$.
However, we have $PD=PC\cdot\cos{BPC}=PC\cdot\frac{AB}{2\cdot AC}=\frac{PC\cdot AB}{2\cdot AC}=2(PA+PB)$, so we are done.
Virgil Nicula
22.10.2011 09:37
An equivalent enunciation. Let $ABC$ be a triangle with the circumcircle $w$ . Denote the midpoint $M$ of the side $[BC]$ and the diameter $[NS]$ of $w$ so that line $BC$ separates $A$ and $S$ . Denote $X\in AB$ so that $SX\perp AB$ . Prove that $MX\perp AS$ and $AX=\frac {b+c}{2}$ . Proof 1. Apply Ptolemy's theorem to $ABSC\ :\ (b+c)\cdot SC=a\cdot SA$ . Prove easily $\triangle AXS\sim\triangle CMS$ , i.e. $\frac {AX}{CM}=\frac {AS}{SC}$ . Thus, $\frac {b+c}{a}=\frac {SA}{SC}=\frac 2a\cdot AX$ $\implies$ $AX=\frac {b+c}{2}$ . Since $MXBS$ is cyclically, obtain that $\widehat{MXS}\equiv$ $\widehat{MBS}\equiv$ $\widehat{SAX}$ , i.e.in the $X$-right triangle $AXS$ we have $MX\perp AS$ . Proof 2. Denote $D\in BC\cap BC$ . Well-known relations $\left\{\begin{array}{c} AD\cdot AS=bc\\\\ AD=\frac {2bc\cos\frac A2}{b+c}\end{array}\right|\implies$ $AX=AS\cdot \cos\frac A2=$ $\frac {bc\cdot\cos\frac A2}{AD}=$ $\frac {b+c}{2}$ .
iarnab_kundu
11.06.2014 13:17
Let $\begin{cases}\angle ACP=\theta\\ \angle BCP=\phi\end{cases}$. Then $\begin{cases}PA = \sin\theta \dfrac{PC}{\sin(A+\phi)} \\ PB=\sin\phi \dfrac{PC}{\sin(A+\theta)}\\ PD=2PC\cos A\end{cases}$ To prove $\dfrac{\sin\theta}{\sin(A+\phi)}+\dfrac{\sin\phi}{\sin(A+\theta)}=2\cos A$, which is easy to prove given that $\theta+\phi = \pi-2A$.
jayme
11.06.2014 15:35
Dear Mathlinkers, this problem is immediat if we think to the Archimede broken chord theorem in the opposite version. Sincerely Jean-Louis
Batman007
13.07.2015 15:27
On $PB$ take $K$ such that $PK=PA$ now $ \angle APC= \angle KPC$, $PA=PK$ and & $PC=PC$ so triangles $APC$ and $KPC$ are congruent. So $AC=DC=BC$ or $KD=DB$ so $PD=PK+KD=PA+BD $ adding $PD$ on both sides we get the answer.
sturdyoak2012
28.11.2015 14:41
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Since $ABC$ is isosceles, $\angle CAB = \angle ABC.$ $\angle CPB = \angle CAB$ because they inscribe the same arc. Similarly, $\angle ABC = \angle APC.$ Therefore, $\angle APC = \angle BPC.$
Applying the law of cosines to triangles $APC$ and $BPC,$ we get the equations
\begin{align*}
AC^2 - PC^2 &= AP^2 - 2\cdot AP\cdot PC \cdot \cos \angle P\\
BC^2 - PC^2 &= BP^2 - 2\cdot BP\cdot PC \cdot \cos \angle P.
\end{align*}However, since $AC = BC,$ the right hand sides are equivalent. Therefore,
\[AP^2 - 2\cdot AP\cdot PC \cdot \cos \angle P = BP^2 - 2\cdot BP\cdot PC \cdot \cos \angle P.\]Rearranging and using difference of squares, this becomes
\[ (BP - AP)(BP + AP) = 2\cdot PC \cdot \cos \angle P (BP - AP).\]Thus,
\[BP + AP = 2\cdot PC \cdot \cos \angle P.\]However, $PD = PC\cdot \cos \angle P,$ so $PA+PB=2\cdot PD.\quad \Box$
jayme
23.11.2018 08:50
Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Regard%202.pdf p. 21-22. Sincerely Jean-Louis
Learner13
23.11.2018 11:32
Extend $PB$ to $Q$ such that $BQ=PA$. Then $PQ=PB+BQ=PB+PA$, so we just have to prove that $D$ is the midpoint of $PQ$, or since $CD \perp PQ$ (and using Pythagoras’ Theorem), that $CP=CQ$.
ProofConsider $\triangle CAP, \triangle CBQ$. We will show that they are congruent.
$$CA=CB,AP=BQ$$by definition and construction respectively.
Also $\angle CAP=\angle CBQ$ because $CAPB$ is cyclic ($BC$ separates $C,P$ avoids configuration issues here).
Thus the triangles are congruent by SAS and their corresponding sides $CP,CQ$ are equal.
$\blacksquare$
Let us reformulate the problem to start with $P$ and construct $C$. It becomes:
In a $\triangle PAB$, let the midpoint of arc $AB$ of $\odot PAB$ not containing $P$ be $C$ and let $CD \perp PB,D \in PB$. Prove that $$PD=\frac{1}{2} \cdot (PA+PB)$$ProofAssume $PA,PB$ are unequal. Drop $CE \perp PA,E \in PA$.
Since arcs $CA,CB$ are equal, $\angle APC=\angle CPB$. Thus $C$ is on the internal angle bisector of $\angle APB$ —(1),
so $CD=CE$—(2).
We will show that $PD=PE$ and $BD=AE$. Then since $PA$ is not equal to $PB$, this implies $\{PA,PB\}=\{PD+BD,PD-BD\}$, so that $PD=\frac{1}{2} \cdot (PA+PB)$ as required.
Since $$\angle CDP=\angle CEP= \frac{\pi}{2}$$$$CP=CP$$$$\angle CPD=\angle CPE (1)$$$\triangle CPD,\triangle CPE$ are congruent, so $PD=PE$.
Since $$\angle CBD=\angle CAE$$(because $CAPB$ is cyclic, $AE=AP,BD=BP$ and both angles are almost $\frac{\pi}{2}$)$$\angle CDB=CEA=\frac{\pi}{2}$$$$CD=CE(2)$$$\triangle CBD,\triangle CAE$ are congruent, so $BD=AE$. Thus we are done.
If PAB is isosceles:Since arcs $CA=CB$ and $PA=PB$ of $\odot PAB$, so $C,P$ are diametrically opposite on the circle, so $D=B$. Thus $PD=PB=PA=\frac{1}{2} \cdot (PA+PB)$.
$\blacksquare$
NoteThis very construction actually features in a “spot-the-mistake” proof that every triangle is equilateral by glossing over the possibility that one of $PA,PB$ is the sum of $PD,DB$ and the other their difference.
Aimingformygoal
21.04.2021 18:29
Might be similar to others, but posting for storage
JBMO 2002 wrote:
$ABC$ is an isosceles triangle $(\overline{CA} = \overline{CB})$. Let $P$ be a point on the arc $\overarc{AB}$ on $(ABC)$ that doesn’t contain $C$. Let $D$ be the foot of the perpendicular from $C$ to $PB$. Show that $\overline{PA} + \overline{PB} = 2 \cdot \overline{PD}$.
Apply Ptolemy Theorem to cyclic quadrilateral $ACBP$, we get :
$AB.PC=AC.BP+AP.BC$
$AB.PC=AC(PA+PB)$
$PA+PB=\frac{AB.PC}{AC}$
So, it suffices to show that :.
$AB.PC=2PD.AC$
Claim : $\cos\angle BAC=\frac{AB}{2AC}$
Drop a perpendicular $CD$ onto $AB$ hence :
$\cos\angle BAC=\frac{AD}{AC}\\ \implies \cos\angle BAC=\frac{AB}{2AC} \qquad \square$
Now, since $\angle CPD = \angle BAC$
Hence, $\cos\angle CPD=\frac{AB}{2AC}$.
However, $\cos\angle CPD = \frac{PD}{PC}$.
Thus, $\frac{AB}{2AC}= \frac{PD}{PC}\implies AB.PC=2PD.AC \qquad \blacksquare$
parmenides51
14.08.2024 10:13
Not Different than above
Ptolemy theorem on ACBP :
AP*BC+PB*AC= AB*PC (1)
and
cos<PCD = cos <ACD :
PD/PC= AB/2AC (2)
Combiing (1), (2) gives what is wanted
LeYohan
28.12.2024 00:15
Synthetic-only solution : Apply Ptolemy's theorem to $ACBP$: $PB \cdot CA + CB \cdot PA = PC \cdot AB \implies CA (PB + PA) = PC \cdot AB \implies PB + PA = \frac{PC \cdot AB}{CA}$. It now suffices to show that $\frac{PC \cdot AB}{CA} = 2 \cdot PD$. Let $M$ be the midpoint of $AB$, note that $\triangle ACM \sim \triangle PCD \implies$ $\frac{AM}{PD} = \frac{CA}{PC} \implies PC \cdot AM = CA \cdot PD$. Multiply by $2 \implies$ $PC \cdot AB = CA \cdot 2PD \implies 2PD = \frac{PC \cdot AB}{CA}$, as desired. $\square$