GEOMETRY MARATHON: SEASON 2

For discussing Olympiad level Geometry Problems
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nahin munkar
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Re: GEOMETRY MARATHON: SEASON 2

Unread post by nahin munkar » Thu Jan 05, 2017 7:29 pm

*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

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

Re: GEOMETRY MARATHON

Unread post by nahin munkar » Thu Jan 05, 2017 10:16 pm

Next thread goes to the following page :

viewtopic.php?f=25&t=3802
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss

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