angle chasing inside a square (Kharkiv City X 2014 - Ukr)

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angle chasing inside a square (Kharkiv City X 2014 - Ukr)

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#1 h Mar 14, 2020, 6:53 AM • 2 Y

Y by Mango247, Rounak_iitr

Let $ABCD$ be a square. The points $N$ and $P$ are chosen on the sides $AB$ and $AD$ respectively, such that $NP=NC$ . The point $Q$ on the segment $AN$ is such that that $\angle QPN=\angle NCB$ . Prove that $\angle BCQ=\frac{1}{2}\angle AQP$ .

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#3 h Mar 14, 2020, 1:45 PM

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Let $\angle NCB=\alpha\ ; \angle CNP=\beta$
$\angle BCQ=\frac{1}{2}\angle AQP \leftrightarrow 180^0-\angle BQP=2(90^o-\angle BQC) \leftrightarrow \angle NQP=2\angle NQC$
To $\triangle NQP$ and point $C$ we use Ceva's: $\frac{sin \angle CQP}{sin \angle CQN} \cdot \frac{cos \alpha}{sin \beta} \cdot \frac{cos \frac{\beta}{2}}{cos(\alpha -\frac{\beta}{2})}=1$
and $\frac{cos \alpha}{sin \beta} \cdot \frac{cos \frac{\beta}{2}}{cos\alpha -\frac{\beta}{2}}=\frac{cos\alpha}{2sin\frac{\beta}{2}cos(\alpha -\frac{\beta}{2})}=\frac{cos\alpha}{sin(\beta -\alpha)+sin\alpha}=\frac{1}{tan\alpha+\frac{sin(\beta -\alpha)}{cos\alpha}}$
Let point $H$ we choose satisfy $HN \perp AB ; HP \perp AD$ and let the length of $AB$ =1
So $tan\alpha+\frac{sin(\beta -\alpha)}{cos\alpha}=NB+\frac{\frac{PH}{NC}}{\frac{1}{NC}}=1 \rightarrow \frac{sin \angle CQP}{sin \angle CQN}=1 \rightarrow \angle CQP=\angle NQP \rightarrow \angle NQP=2\angle NQC$
$Q.E.D$

This post has been edited 2 times. Last edited by ironball, Mar 14, 2020, 1:51 PM

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#4 h Mar 14, 2020, 1:48 PM • 1 Y

Y by Mango247

I haven't found a graceful proof

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#5 h Mar 16, 2020, 6:00 PM

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Just notice $PQ$ is tangent to the circle $(C,CB)$ .

Best regards,
sunken rock

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jayme

#6 h Apr 13, 2020, 1:57 PM

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Dear Mathlinkers,

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=584122

http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20II.pdf p. 18-20.

Sincerely
Jean-Louis

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