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Hexagon Hinder 2 (Posted on 2016-05-06) Difficulty: 3 of 5
The hexagon PQRSTU is such that:

(1) All its vertices lie on a circle, and:
(2) PQ = RS = TU, and:
(3) the diagonals PS, QT, RU meet at a point.

N is the intersection of PS and RT


No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution Solution Comment 2 of 2 |

All angles subtended at the circumference by the three equal arcs
will be equal:
            Let angles PTQ = PRQ = RTS = RPS = TPU = TRU = a
Similarly for angles subtended by arc QR:             QPR = QTR = b
and       for angles subtended by arc ST:               SPT = SRT = c
and       for angles subtended by arc UP:               URP = UTP = d
and       for angles subtended by arc PR:         PSR = PTR = a + b
and       for angles subtended by arc PT:         PST = PRT = a + d

With these angles marked, and the intersection Of PS, QT and RT
denoted by O, it is easy to see that triangles ORS, SOT and RPT
are similar since they each have the angles {a+b, a+c, a+d}.

    Thus                         OR/OS = OS/ST = PR/RT                        (1)

Also triangles RNO and TNS are similar since they each have an
angle a, and the angles RNO and TNS are equal (vert. opp.).

    Thus             RN/NT   = OR/ST

                                    = OR/OS x OS/ST

                                    = (PR/RT)2         using (1)

Therefore          (RN/NT)/(PR/RT)2 = 1

  Posted by Harry on 2016-05-08 15:31:37
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