Problem: In the unit circle shown in the figure to the right, chords PQ and MN are parallel to the unit radius OR of the circle with center at 0. Chords MP,PQ and NR are each s units long and chord MN is d units long. Of the three equations
I. d−s=1,
II. ds =1,
III. d2−s2=5​
those which are necessarily true are
Answer Choices:
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
Solution:
Chord MN has length (See figure) and each chord of length a subtends an arc of 5180∘​=36∘. Draw a radius OP meeting MN at T. Also draw perpendicular OV and QU to MN from O and Q reapectively. Triangles OVN and QUN are similar so that
QNUN​=1VN​ or 2d​−2s​=s2d​, Therefore d−s=ds
To see thet d−s=1, note that MT=s because ∠MPT=∠MTP=72∘. Also ∠OTN=∠ORN so that ORNT is a parallelogram with opposite sides TN=OR=1. MN−MT=d−s=1. Now (d−s)2+4ds=1+4=5 or d2+2ds+s2=5 and d+s=5​ .∴d2−s2=(d+s)(d−s)=5​×1. All three of the equations I, II, and III are necessarily true as stated in choice (E).
or
Alternately, d−s=1 follows from MT=s and MN=d because OTNR is a parallelogram with side TN=MN−MT=d−s=1. Now let P′ (not shown) danote the other and of the diameter through P. Then chords PP′ and MN intersect at T and the products of their segments are equal.
(PT)⋅(TP′)=(MT)⋅(TN) or (1−s)(1+s)=s⋅1
This quadratic equation iess the pooitive root s=21​(5​−1) from which relations II and III can be verified. Therefore all three equations I , II and III are pecessarily true.
