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; xa = Pi/3; yb = 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 20100626 08:18:26 