Prove that BC^2 = 4 PF . AK (CGMO 2010-P6)

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Prove that BC^2 = 4 PF . AK (CGMO 2010-P6)

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#1 h Aug 17, 2010, 4:44 PM • 4 Y

Y by son7, Adventure10, Mango247, and 1 other user

In acute triangle $ABC$ , $AB > AC$ . Let $M$ be the midpoint of side $BC$ . The exterior angle bisector of $\widehat{BAC}$ meet ray $BC$ at $P$ . Point $K$ and $F$ lie on line $PA$ such that $MF \perp BC$ and $MK \perp PA$ . Prove that $BC^2 = 4 PF \cdot AK$ .

$[asy] defaultpen(fontsize(10)); size(7cm); pair A = (4.6,4), B = (0,0), C = (5,0), M = midpoint(B--C), I = incenter(A,B,C), P = extension(A, A+dir(I--A)*dir(-90), B,C), K = foot(M,A,P), F = extension(M, (M.x, M.x+1), A,P); draw(K--M--F--P--B--A--C); pair point = I; pair[] p={A,B,C,M,P,F,K}; string s = "A,B,C,M,P,F,K"; int size = p.length; real[] d; real[] mult; for(int i = 0; i<size; ++i) { d[i] = 0; mult[i] = 1;} string[] k= split(s,","); for(int i = 0;i<p.length;++i) { label("$"+k[i]+"$",p[i],mult[i]*dir(point--p[i])*dir(d[i])); }[/asy]$

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#2 h Aug 17, 2010, 5:41 PM • 2 Y

Y by Adventure10, Mango247

Let $(O)$ be the circumcircle of $\triangle ABC.$ $F$ is obviously the midpoint of arc $BAC,$ thus line $FM$ cuts $(O)$ again at the midpoint $D$ of arc $BC,$ i.e. $AD$ bisects $\angle BAC$ $\Longrightarrow$ $AD \parallel MK.$ Therefore $\frac{AK}{FA}=\frac{DM}{FD},$ but power of $M$ WRT $(O)$ is given by $\frac{BC^2}{4}=DM \cdot FM$ $\Longrightarrow$ $\frac{AK}{FA}=\frac{BC^2}{4FM \cdot FD}.$ Since $FA \cdot FP=FM \cdot FD$ $\Longrightarrow$ $BC^2=4PF \cdot AK.$

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#3 h Aug 17, 2010, 11:31 PM • 2 Y

Y by Adventure10, Mango247

hello

to luis ; can you explain why u say in your first sentense it is obvious that f is the midpoint of arc BAC explain me please

enigma

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#4 h Aug 18, 2010, 4:01 AM • 2 Y

Y by Adventure10, Mango247

Quote:

In acute triangle $ABC$ , $c > b$ . Let $M$ be the midpoint of side $[BC]$ . The exterior angle bisector of $\widehat{BAC}$ meet ray $(BC$
at $P$ . The points $K$ and $F$ lie on $PA$ such that $MF \perp BC$ and $MK \perp PA$ . Prove that $4\cdot PF \cdot AK=BC^2$ .

Proof 1 (similar with luisgeometria's). Denote the circumcircle $w$ of $\triangle ABC$ . Observe that $F\in w$ . Denote the diameter $[FD]$ of $w$ . Observe that $MK\parallel DA$ and $\widehat{MPA}\equiv\widehat{FMK}\equiv\widehat{MDA}$ $\implies$ $\widehat{MPA}\equiv\widehat{MDA}$ , i.e. $AMDP$ is cyclically. Thus,

$\left\|\begin{array}{ccc} MK\parallel DA & \iff & \frac {FA}{KA}=\frac {FD}{MD}\\\\ AMDP\ -\ \mathrm{cyclic} & \iff & FM\cdot FD=FA\cdot FP\\\\ FBDC\ -\ \mathrm{cyclic} & \iff & BC^2=4\cdot MD\cdot FM\end{array}\right\|\ \bigodot$ $\implies\ 4\cdot PF \cdot AK=BC^2$ .

Proof 2 I"ll use same notations from upper proof. Denote $L\in AD$ for which $ML\perp AD$ . Observe that $AK=ML$ and
$\triangle LMD\sim\triangle MFP$ . Therefore, $\frac {LM}{MF}=\frac {MD}{FP}$ $\iff$ $AK\cdot PF=MD\cdot MF=MB^2$ $\iff$ $4\cdot PF \cdot AK=BC^2$ .

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#5 h Aug 18, 2010, 5:03 AM • 2 Y

Y by Adventure10, Mango247

Lemma: Suppose A,C are harmonic conjugates w.r.t B,D. Let M be the midpoint of AC. Then $AM^2=BM*DM$

Proof:
Since the guys are harmonic, there exists a point P s.t. $APC=90$ , and $PC$ bisects $BPD$ . Let Q be on PM s.t. BQ||CP.

Since $\Delta PMC$ is iscoselse, quadrilateral $PCBQ$ is also iscoselese. Thus, $DPC=BPC=ACP$ , so DP||QC.

Now by similar triangles, $MP/MQ=MC/MB=MD/MC$ , as desired. []

Let the A internal bisector meet BC at D, then $AD||KM$ and BDCP is harmonic, so by lemma 1,
$PM*DM=MC^2$
$4PM*DM=BC^2$ (1)
Additionally, by similar triangles,
$MD/KA=PM/PK=PF/PM$ (2)
Combining (1) and (2) we get
$4KA*PF=4PM*MD=BC^2$

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#6 h Aug 18, 2010, 10:53 PM • 1 Y

Y by Adventure10

Here is my solution:
Let $\theta=\angle APB=\frac{1}{2}(C-B)$
It is well-known that $\sin\theta=\frac{c-b}{a}\cos{\frac{A}{2}}$ and $\cos\theta=\frac{c+b}{a}\sin{\frac{A}{2}}$
Sine Rule on $\triangle ACP$ gives: $|CP|=\frac{ab}{c-b}$
Sine Rule on $\triangle ABP$ gives: $|BP|=\frac{ca}{c-b}$
So $|MP|=\frac{|CP|+|BP|}{2}=\frac{a(b+c)}{2(c-b)}$
$|PF|=\frac{|MP|}{\cos\theta}=\frac{a^2}{2(c-b)\sin\frac{A}{2}}$
$|PK|=|MP|\cos\theta=\frac{(b+c)^2\sin\frac{A}{2}}{2(c-b)}$
Sine Rule on $\triangle ACP$ gives: $|PA|=\frac{|CP|\sinh}{\cos\frac{A}{2}}=\frac{4bc\sin\frac{A}{2}}{2(c-b)}$
So $|AK|=|PK|-|PA|=\frac{(c-b)\sin\frac{A}{2}}{2}$
Hence $4\cdot|PF|\cdot|AK|=a^2=|BC|^2$ as required.

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#7 h Aug 19, 2010, 12:11 AM • 2 Y

Y by Adventure10, Mango247

enigmation wrote:

to luis ; can you explain why u say in your first sentense it is obvious that f is the midpoint of arc BAC explain me please

There may be an obvious way that I'm missing. But if you let $O$ be the circumcenter of $\triangle{ABC}$ (which is on line $FM$ ), then it's easy to get that $\angle{AOM}=2\angle{FAO}$ , whence
$\angle{AFO}=\angle{AFM}-\angle{FAO}=\angle{FAO}.$

I have a slightly different solution than the ones above that uses that $ACBF$ is cyclic, and that $PM$ is tangent to the circumcircle of $\triangle{KFM}$ . Note that what we need to show is equivalent to
$\begin{align*} BC^2=4PF\cdot AK=4PF(PK-PA)&=4PF\cdot PK-4PF\cdot PA\\ &=4PM^2-4PF\cdot PA=(PB+PC)^2-4PF\cdot PA, \end{align*}$ or
$4PF\cdot PA=(PB+PC-BC)(PB+PC+BC)=(2PC)(2PB)=4PC\cdot PB,$ which follows from the power of point $P$ with respect to the circle through $ACBF$ .

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#8 h Aug 19, 2010, 9:27 AM • 3 Y

Y by rashah76, Adventure10, Mango247

How about this? $PA\cdot PF=PC\cdot PB=PM^2-MC^2$ and $PK\cdot PF=PM^2$ . Subtract and get the desired result.

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#9 h May 9, 2013, 3:34 AM • 1 Y

Y by Adventure10

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#10 h Apr 15, 2015, 8:53 AM • 2 Y

Y by Adventure10, Mango247

My Solution
Draw the interior angle bisector $AM'$ so we will get that $AM' \parallel KM$
This implies $\frac {PK}{AK}=\frac{PM}{MM'}$ so $\frac {MM'}{AK}=\frac{PM}{PK}=\frac{PF}{PM}$
This implies $PM.MM'=PF.AK$
$PM.MM'= (\frac{a}{2}+\frac{ab}{c-b})(\frac{a}{2}-\frac{ab}{b+c}) =a^2/4$
hence $BC^2=4PF.AK$

This post has been edited 1 time. Last edited by KudouShinichi, Apr 21, 2015, 6:43 PM

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Mahdi_Mashayekhi

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#11 h Jan 27, 2022, 4:48 PM

Y by

Note that BC^2 = (2MC)^2 so we can instead prove MC^2 = PF.AK . we have PM^2 = PA.PF so we can show PM^2 - MC^2 = PA.PF or PC.PB = PA.PF or FACB is cyclic. Let S be reflection of C across FA clearly S lies on DA. we have ∠ACF = ∠ASF and CF = SF = BF so ∠ACF = ∠ABF so FACB is cyclic.
we're Done.

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