# Geometry

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.?

Note by Kishan K
4 years, 10 months ago

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let 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!!!

- 4 years, 6 months ago

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.

- 4 years, 6 months ago

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

- 4 years, 6 months ago