I solved this question few days ago and found interesting. In a convex quadrilateral \(PQRS\), \(PQ=RS,(\sqrt {3}+1)QR=SP\) and \(\angle RSP- \angle SPQ=30^{\circ}.\)Find \(\angle PQR-\angle QRS.\)This is also a quite easy problem, but let's see who give the shortest solution.Then I will give my solution.

I am observing since few weeks that the geometry problems in the combinatorics and geometry section are very low.This time only 2.They are also quite easy,.What do u think.?

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TopNewestlet PQ=a;QR=b;SR=c and PS=d then a=c and \[d=(\sqrt{3}+1)b\] we have \[PR^{2}=c^{2}+d^{2}-2cdcos\angle S=a^{2}+b^{2}-2abcos\angle Q\] FROM THIS \[b^{2}-d^{2}=2abcos\angle Q-2cdcos\angle S\] similarly \[b^{2}-d^{2}=2bccos\angle R-2adcos\angle P\] comparing we get \[2abcos\angle Q-2cdcos\angle S=2bccos\angle R-2adcos\angle P\] or simflifying it we get [putting the values a=c and \[d=(\sqrt{3}+1)b\]]=== \[cosQ-(\sqrt{3}+1)cosS=cosR-(\sqrt{3}+1)cosP\] OR \[cosQ-cosR=(\sqrt{3}+1)(cosS-cosP)\] OR \[sin\frac{Q+R}{2}sin\frac{Q-R}{2}=(\sqrt{3}+1)sin\frac{S+P}{2}sin\frac{S-P}{2}\] we see \[sin\frac{Q+R}{2}=sin\frac{S+P}{2}\] putting this value in the above equation we get \[sin\frac{Q-R}{2}=(\sqrt{3}+1)sin\frac{30}{2}\][as S-P=30]...... putting the value of sin15 we get \[sin\frac{Q-R}{2}=\frac{1}{\sqrt{2}}\] so \[\frac{Q-R}{2}=45\] or Q-R=90. so we have done now. now you give me your solutin;kishan. i know this is not the shortest solution;nor the best!!!

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My solution is slightly shorter than yours.I would like you to do it.I would just give you a path to my solution.Try to get the area of the quadrilateral in 2 different ways.\([PQS]+[QRS]=[PQR]+[PSR]\) and try to use the sine rule area formula.After few simplifications,you will get your answer.

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thanks!! i will be waiting for your next beautiful problems;kishan

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