A square $ABCD$ is given. A point $P$ is chosen inside the triangle $ABC$ such that $\angle CAP = 15^\circ = \angle BCP$. A point $Q$ is chosen such that $APCQ$ is an isosceles trapezoid: $PC \parallel AQ$, and $AP=CQ, AP\nparallel CQ$. Denote by $N$ the midpoint of $PQ$. Find the angles of the triangle $CAN$.
Problem
Source: Israel Autumn 2016 TST2/1
Tags: geometry, trapezoid
ThE-dArK-lOrD
29.09.2016 19:04
Note that $\triangle{PAD}$ is equilateral, assume that coordinate of $A,B,C,D$ are $A(0,0),B(0,1),C(1,1),D(1,0)$ We get $P(\frac{1}{2},\frac{\sqrt{3}}{2})$ and $Q$ lie on the line $y=\frac{1-\frac{\sqrt{3}}{2}}{1-\frac{1}{2}} x$ and circle center at $C(1,1)$ with radius $PA=AD=1$ So we get $Q(\frac{2+\sqrt{3}}{2},\frac{1}{2})\rightarrow N(\frac{3+\sqrt{3}}{4},\frac{\sqrt{3}+1}{4})$, so $tan(\angle{NAD})=\frac{\frac{\sqrt{3}+1}{4}}{\frac{3+\sqrt{3}}{4}} =\frac{1}{\sqrt{3}}$ We get $\angle{NAD}=30^{\circ}\rightarrow \angle{CAD}=15^{\circ}$ and we easily prove that $\angle{CNA}=90^{\circ}$ Finally we get $\angle{CAD}=15^{\circ},\angle{CNA}=90^{\circ},\angle{NCA}=180-15-90=75^{\circ}$
Tumon2001
12.03.2017 10:50
Solution: Let $BD\cap AC = M $. Join $MN $ and $DQ $. Let $a $ be the length of the side of the square. Simple length and angle chasing proves that $D $ is the circumcenter of cyclic $APCQ $. This gives $AB = BC = CD = DA = DQ = CQ = AP = a $. Since, $AP = AB $, so, $\angle APB$ = $ 75$°. Also, $\angle APQ$ = $105$°. Thus, $B $, $P$ and $Q $ are collinear. So, $\angle QBC$ = $15$°. Now, $MN || AQ $. So, $\angle MNB $ = $30$° = $\angle MBN $. This gives $MN = MB = MC = MD = MA $. Thus, $ADNCB $ is cyclic. Trivial angle chasing gives $\angle ANC $ = $90$°, $\angle NAC $ = $15$° and $\angle NCA $ = $75$°.