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BdMO Online Forum • View topic - USA TST 2011/1

USA TST 2011/1

For discussing Olympiad level Geometry Problems
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USA TST 2011/1

Post Number:#1  Unread postby dshasan » Thu Apr 20, 2017 7:50 pm

In acute scalene triangle $ABC$, $D,E,F$ be the feet of the perpendicular from $A,B,C$ to $BC,AC$ and $AB$ respectively.Let $H$ be the orthocenter. Points $P$ a.d $Q$ lie on segment $EF$ such that $AP$ and $HQ$ are perpendicular on $EF$. Lines $DP$ and $QH$ intersects at point $R$. Compute $\dfrac{HQ}{HR}$.
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Re: USA TST 2011/1

Post Number:#2  Unread postby rubab » Thu Apr 20, 2017 10:17 pm

Well known- $A$ and $H$ are the excenter and incenter of
$\bigtriangleup DEF$ respectively. So, $Q$ and $P$ are respectively incircle and excircle touchpoint on EF of $\bigtriangleup DEF$. By a well known lemma $R$ lies on the incircle of $\bigtriangleup DEF$. Since H is the incenter of $\bigtriangleup DEF$ $\frac{HQ}{HR}=1$
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Re: USA TST 2011/1

Post Number:#3  Unread postby Atonu Roy Chowdhury » Thu Apr 20, 2017 10:34 pm

Similarity on $\triangle HQA$ and $\triangle HEA$ implies $\frac {HQ}{HF} = \frac {HE}{HA}$
Similarity on $\triangle HDR$ and $\triangle ADP$, $\triangle ABD$ and $\triangle AEP$, $\triangle BEA$ and $\triangle HEC$ implies $\frac {HD}{HR} = \frac {AD}{AP} = \frac {AB}{AE} = \frac {HC}{HE}$ . These two equations along with
$ \frac {HD}{HF} = \frac {HC}{HA}$ implies $ \frac {HQ}{HR}=1$
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Re: USA TST 2011/1

Post Number:#4  Unread postby Thamim Zahin » Fri Apr 21, 2017 1:37 am

Let the lines $\overleftrightarrow{FE}$ and $\overleftrightarrow{BC}$ intersect at $M$. and let the segment $\overline{FE}$ and $\overline{AD}$ intersect at $N$. And let take $\overleftrightarrow{AP}$ and $\overleftrightarrow{QR}$ meet at $P_{\infty}$.

Now we know that $(M,D;B,C)$ is harmonic. Take the pencil at $E$. We get $(N,D;H,A)$ is harmonic. Now take the pencil at $P$. We get $(Q,R;H,P_{\infty})$ is harmonic.

$Q.E.D$ $\blacksquare$
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