Secondary Special Camp 2011: Geometry P 4

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Moon
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Secondary Special Camp 2011: Geometry P 4

Unread post by Moon » Fri Apr 22, 2011 10:34 am

Problem 4: Let $ABC$ be an acute triangle and $D, E, F$ the feet of its altitudes from $A, B, C$, respectively. The line through $D$ parallel to $EF$ meets line $AC$ and line $AB$ at $Q$ and $R$, respectively. Let $P$ be the intersection of line $BC$ and line $EF$.
Prove that the circumcircle of $PQR$ passes through the midpoint of $BC$.
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Zzzz
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Re: Secondary Special Camp 2011: Geometry P 4

Unread post by Zzzz » Wed May 11, 2011 8:21 pm

$M$ is the midpoint of side $BC$.

$\angle AQD=\angle AEF=\angle B$ and $\angle ARD=\angle AFE=\angle C$
\[\angle EPB=180\circ - \angle PEB-\angle EBP=180\circ-(\angle AEB-\angle AEF)-(\angle EBA+\angle ABP)\]
\[=180\circ-(90\circ-\angle B)-\{(90\circ-\angle A)+(180\circ-\angle B)\}=2\angle B+\angle A-180\circ=\angle B-\angle C\]

From $\Delta EBP$ \[\frac{PB}{sin\angle PEB}=\frac{BE}{sin\angle EPB}\Rightarrow \frac{PB}{sin(90\circ - \angle B)}=\frac{c\ sin \angle A}{sin (\angle B-\angle C)}\Rightarrow PB=\frac{c\ sin\angle A cos\angle B}{sin(\angle B-\angle C)}\]
$BD=c\ cos\angle B$
$\therefore PD=PB+BD=c\ cos\angle B\{\frac{sin(\angle B-\angle C)+sin\angle A}{sin(\angle B-\angle C)}\}$

From $\Delta ARD$ \[ \frac{RD}{sin\angle RAD}=\frac{AD}{sin\angle ARD}\Rightarrow \frac{RD}{sin(90\circ-\angle B)}=\frac{c\ sinB}{sin\angle C}\Rightarrow RD=\frac{c\ sin\angle B\ cos\angle B}{sin\angle C}\]
\[\therefore \frac{PD}{RD}=\frac{\{sin(\angle B-\angle C)+sin\angle A\}sin\angle C}{sin(\angle B-\angle C)sin \angle B}\]

From $\Delta ADQ$\[\ \frac{QD}{sin\angle DAQ}=\frac{AD}{sin\angle AQD}\Rightarrow \frac{QD}{sin(90\circ-\angle C)}=\frac{b\ sin\angle C}{sin\angle B}\]\[\Rightarrow QD=\frac{b\ cos\angle C sin\angle C}{sin\angle B}=\frac{a\ cos\angle C sin\angle C}{sin\angle A}\]
\[DM=BM-BD=\frac{a}{2}-c\ cos\angle B=\frac{a}{2}-\frac{a\ sin\angle C cos\angle B}{sin\angle A}=a\{\frac{sin\angle A-2sin\angle C cos\angle B}{2sin\angle A}\}\]
\[\therefore \frac{QD}{DM}=\frac{2sin\angle C cos\angle C}{sin\angle A-2sin\angle C cos\angle B}\]

Now,
\[\frac{PD}{RD}=\frac{QD}{DM}\]
\[\Leftrightarrow \frac{\{sin(\angle B-\angle C)+sin\angle A\}sin\angle C}{sin(\angle B-\angle C)sin \angle B}=\frac{2sin\angle C cos\angle C}{sin\angle A-2sin\angle C cos\angle B}\]
\[\Leftrightarrow \frac{sin(\angle B-\angle C)+sin\angle A}{sin(\angle B-\angle C)sin \angle B}=\frac{2cos\angle C}{sin\angle A-2sin\angle C cos\angle B}\]
\[\Leftrightarrow \frac{sin(\angle B-\angle C)+sin\angle A}{sin(\angle B-\angle C)sin \angle B}=\frac{2cos\angle C}{sin(\angle B+\angle C) -2sin\angle C cos\angle B}\]
\[\Leftrightarrow \frac{sin(\angle B-\angle C)+sin\angle A}{sin(\angle B-\angle C)sin \angle B}=\frac{2cos\angle C}{sin(\angle B-\angle C)}[\text {as} EF \text{intersects} BC, \angle B \ne \angle C\]
\[\Leftrightarrow \frac{sin(\angle B-\angle C)+sin\angle A}{sin \angle B}=2cos\angle C\]
\[\Leftrightarrow sin(\angle B-\angle C)+sin\angle (\angle B+\angle C)=2cos\angle C sin \angle B\]

Which is true.
$\therefore \frac{PD}{RD}=\frac{QD}{DM}\Rightarrow PD\cdot DM=QD\cdot RD$

So, $P,R,M,Q$ cyclic :)
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Tahmid Hasan
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Re: Secondary Special Camp 2011: Geometry P 4

Unread post by Tahmid Hasan » Thu Jun 23, 2011 7:47 pm

for $P,Q,M,R$ to be cyclic we need to prove that $RD.DQ=PD.DM$.it can be easily deduced that $FD=RD=bcosB$
and $DE=DQ=ccosC$.we can also get $DM=\frac{1}{2}a-BD=bcosC-ccosB$.
so it suffices to prove that $PD=\frac{2bccosBcosC}{bcosC-ccosB}$.
applying sine law on $\triangle PDF$ yields us the desired figure.
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tanmoy
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Re: Secondary Special Camp 2011: Geometry P 4

Unread post by tanmoy » Sat Oct 19, 2013 11:54 am

why $\angle AEF=\angle B$ -tanmoy

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Fm Jakaria
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Re: Secondary Special Camp 2011: Geometry P 4

Unread post by Fm Jakaria » Sun Nov 24, 2013 7:43 pm

tanmoy wrote:why $\angle AEF=\angle B$ -tanmoy
Note that the quadrilateral EFBC is cyclic, because <BEC = 90 = <BFC. Hence the equality.
I do mathematics, but still I have no evidence that absence of evidence is not evidence of absence.

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