Let ABCD be a convex quadrilateral. Also let P, Q, R, S be the respective midpoints of sides AB, BC, CD, DA such that triangles AQR and CSP are both equilateral.
Find the largest angle of quadrilateral ABCD.
From geometry this is a rhombus.
If a/2 is the required angle and b is a/2-30 in degrees. L is a side of the rhombus, l is a side of the triangle equilateral, H and h are the heights in corrispondence with L and l
tan a/2 = H/(2/3h) = L sin (30+b) / 2l cos(30)
in addition L^2=1/4 L^2 + l^2
Then
tan (a/2) = 2 sin (a/2)
a/2=60° and a=120°
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Posted by armando
on 2019-03-08 11:07:58 |
Solution?