A circle has a radius of 1 unit with its center located at O and P is an arbitrary point inside the circle. Three straight lines are drawn from P to meet the circle’s circumference respectively at the points A, B and C.

It is known that:

Angle APB = Angle BPC = Angle APC = 120 degrees, and:

Area of sector APB = pi/3 square units, and:

Area of sector BPC = pi/4 square units.

Determine the length of the straight line OP.

In fact, P is not an 'arbitary' point; its location is fixed by the size of the sectors. Hence OP can be  computed, as follows:

 

1.       Extend AP to E on the circumference, and do the same with CP,to D, creating 2 chords, AE, CD.

2.        Draw the perpendicular bisector of AE, crossing AE at X, and do the same with CD, crossing at Y.

3.       The angle APC is given as 120^o, so XOY is 60^o.

4.       Although (for example) PE and PD are different lengths, we know that AP*PE = CP*PD, and also that PB bisects DPE.

5.       Call the area BPE, a, and the area BPD, b. Chord area AE=x; Chord area CD=y; x-a = Pi/3; y-b = pi/4; b is to a as x is to y.

6.       From these equalities, calculate the areas x and y, and thereby determine the height of the apothegms OX, OY.

7.       Construct the line XY; compute its length using the law of cosines.

8.       XPY is known to be 120^o, and once XY is known, the values of XP and YP can be computed.

9.       OXP and OYP are right angles. Hence OX^2+XP^2=OY^2+YP^2=OP^2.

 

Edited on June 26, 2010, 8:19 am

Edited on June 26, 2010, 8:23 am

Edited on June 27, 2010, 9:39 am

Posted by broll on 2010-06-26 08:18:26