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 120o, so XOY is 60o.
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 120o, 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