A square ABCD is given. A point P is chosen inside the triangle ABC such that ∠CAP=15∘=∠BCP. A point Q is chosen such that APCQ is an isosceles trapezoid: PC∥AQ, and AP=CQ,AP∦. Denote by N the midpoint of PQ. Find the angles of the triangle CAN.
Problem
Source: Israel Autumn 2016 TST2/1
Tags: geometry, trapezoid
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}
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°.