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A square $P_1P_2P_3P_4$ has points $X$ on side $P_2P_3$ and $Y$ on side $P_3P_4$ chosen such that angle $XP_1Y$ equals forty-five degrees. The lines $P_1X$ and $P_1Y$ intersect the circumcircle of the square at points $R$ and $S$, respectively.
How can one show that the lines $XY$ and $RS$ run parallel? enter image description here

A geometric-algebraic solution using trigonometric addition properties is fairly straightforward and gives a solution. Yet here I´m looking for a more elegant elementary geometric solution. The intersections with the circumcircle, points $R$ and $S$ can be construed as corners of another square of equal size as the initial one using the Pythagorean theorem.
Here´s where I think a line or two or a smart angle hunt could solve it but got stuck. Any suggestion?

user376343
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1 Answers1

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Consider the quadrilateral RSYX and show that it is an isosceles trapezoid. You can show it by proving that SY and RX have the same length and SYX and YXR form the same angle.

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JoseSquare
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  • Could you please provide some intermediate steps? Only thing I see is that angles TSO = URO = 45° –  Feb 09 '19 at 16:11
  • Since P1XY is a isosceles triangle then the angles P1YX and YXP1 are the same. This implies that SYX and YXR are the same because SYX = 180º - P1YX and YXR = 180º - YXP1. Now to show that SY and RX lengths are equal just use that SP1P3 = RP1P3 by simmetry. – JoseSquare Feb 09 '19 at 17:30
  • @JoseManuelSanchezCuadrado: The question has been locked for being part of an on-going contest. Perhaps you should delete your question, and "undelete" it after 4 March, 2019. – Blue Feb 10 '19 at 04:52