perplexus.info :: Geometry : Square in a Sector

Square in a Sector (Posted on 2007-06-06)

Let P and Q be points on a unit circle with center O such that angle POQ is x degrees ( x ≤ 270 ).
Let the bisector of angle POQ intersect the circle at point A.
Let KLMN be a square with vertices K and L on line segments OP and OQ and vertices M and N on the arc PAQ.

Give the area of the square in terms of x.

If x = 45, write the area of the square as (a+b√c)/d where a, b, c, and d are integers.

Submitted by Bractals

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Solution: (Hide)

Let the line OA intersect sides KL and MN in points B and C respectively.
tan(x/2) = |LB|/|OB| = |CM|/|OB| or |OB| = |CM|/tan(x/2) |OA|² + |CM|² = |OM|² (|OB| + |BC|)² + |CM|² = |OM|² (|OB| + 2|CM|)² + |CM|² = 1 |OB|² + 4|OB||CM| + 4|CM|² + |CM|² = 1 (|CM|/tan(x/2))² + 4(|CM|/tan(x/2))|CM| + 5|CM|² = 1 Area of square = 4|CM|² = 4tan²(x/2)/(5tan²(x/2) + 4 tan(x/2) + 1) (1) or = 4/(5 + 4cot(x/2) + cot²(x/2)) (2) If x = 180, then use (2); otherwise, use (1). If x = 45, then tan(x/2) = (√2 - 1) and Area of square = (2 - √2)/3

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Answer	K Sengupta	2009-01-09 15:39:32
Non conforming but interesting	brianjn	2007-06-08 09:58:21
Solution	Jer	2007-06-07 12:06:23
re: No formula but:	Charlie	2007-06-07 10:15:38
No formula but:	Charlie	2007-06-06 16:55:57

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