## GEOMETRY MARATHON: SEASON 2

For discussing Olympiad level Geometry Problems
nahin munkar
Posts: 81
Joined: Mon Aug 17, 2015 6:51 pm
Location: banasree,dhaka

### Re: GEOMETRY MARATHON: SEASON 2

*Mahi* wrote:New Problem!
Let $BE$ and $CF$ be the altitudes in an acute triangle $ABC$. Two circles passing through the points $A$ and $F$ are tangent to the line $BC$ at the points $P$ and $Q$ so that $B$ lies between $C$ and $Q$. Prove that the lines $PE$ and $QF$ intersect on the circumcircle of $\triangle AEF$.
(Intended to those who haven't solved it yet.)
We let $AD$ is another altitude & $H$ be the orthocentre. Suppose, $PE \cap QF = K$ .

$\clubsuit$ Claim 1 : $\angle KPQ = \angle BHP$

proof:

AS, $B$ lies on the radical axis $AF$ of two circles,
$BP^2$
$= BQ^2$
$= BF \cdot BA$
$= BH \cdot BE$
$\Longrightarrow \odot PHE$ is tangent to $BC$ at $P$.

Here, $\triangle BPE$ ~$\triangle PHE$

$.^..$ $\angle KPQ$
$= \angle EPB$
$= \angle BHP$ $\clubsuit$

$\spadesuit$ Claim 2 : $\angle KQP =\angle PHC$

proof :

$CP \cdot CQ$
$= (CB+BP)(CB- BP)$
 $= CB^2 - BP^2$
$= CB^2 - BF \cdot BA$
$= CB^2 - BD \cdot CB$
$= CB(CB-BD)$
$= CB \cdot CD$
$= CH \cdot CF$

So, $FHPQ$ is cyclic .

$\Longrightarrow \angle KQP$
$= \angle FQP$
$= \angle PHC$ $\spadesuit$

Now,
$\angle PKQ = 180^0 - (\angle KPQ + \angle KQP)$
$= 180^0 -(\angle BHP + \angle PHC)$
$= 180^0 - \angle BHC$
$= 180^0 - \angle FHE$
$= \angle FAE$
$\Longrightarrow$ $K$ lies on $\odot AFE$ . $\blacksquare$
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss

nahin munkar
Posts: 81
Joined: Mon Aug 17, 2015 6:51 pm
Location: banasree,dhaka

### Re: GEOMETRY MARATHON

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# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss