Problem:
A circle with center O has radius 25. Chord AB of length 30 and chord CD of length 14 intersect at point P. The distance between the midpoints of the two chords is 12. The quantity OP2 can be represented as nm, where m and n are relatively prime positive integers. Find the remainder when m+n is divided by 1000 .
Solution:
Let M and N be the midpoints of chords AB and CD, respectively. By the Pythagorean Theorem, MO2=AO2−AM2 and NO2=CO2−CN2, so MO=20 and NO=24. Applying the Law of Cosines to △OMN results in
cos∠MON=2⋅20⋅24202+242−122=1513,cos∠OMN=2⋅20⋅12202+122−242=−151, and cos∠MNO=2⋅12⋅24122+242−202=95.
The corresponding sines are
sin∠MON=15214,sin∠OMN=15414, and sin∠MNO=9214.
In △MNP,MN=12,∠MNP=∠MNO+90∘, and ∠NMP=∠OMN−90∘. Thus sin∠MNP=cos∠MNO=95, and sin∠NMP=−cos∠OMN=151. The corresponding cosines are cos∠MNP=−9214 and cos∠NMP=15414. Applying the Law of Sines to △MNP yields 5/9MP=sin∠NPM12. Now ∠NPM=180∘−(∠NMP+∠MNP), so sin∠NPM=sin(∠NMP+∠MNP)=151⋅9−214+15414⋅95=15214, and
MP=1521495⋅12=1450
By the Pythagorean Theorem, OP2=MO2+MP2=400+142500=74050. Thus m+n=4057, and the required remainder is 57
OR
Note that ∠OMP and ∠ONP are right angles, and thus △OMP and △ONP are right triangles with the common hypotenuse OP. Therefore the two triangles are inscribed in the same semicircle, and quadrilateral OMNP is cyclic. Hence ∠NMO and ∠NPO are supplementary, and sin∠NMO=sin∠NPO. Applying the Law of Cosines to △MNO yields cos∠NMO=−151,sin∠NMO=sin∠NPO=15414, and the result follows as before.
