All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Independent projections (Posted on 2021-10-11)
Let AB and CD be two diameters of the circle. For an arbitrary point P on the circle, let R and S be the feet of the perpendiculars from P to AB and CD, respectively. Show that the length of RS is independent of the choice of P.

 No Solution Yet Submitted by Danish Ahmed Khan Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution + formula | Comment 1 of 2
First, a little angle chasing shows the angle between the diameters is equal to angle RPS, so this is a fixed angle.  Call this angle O.

Next extend PR to intersect the circle at a second point and form chord PX with midpoint R.
Ans extend PS to intersect the circle at a second point and form chord PY with midpoint S.

The length of chord XY depends only on angle O.

By the midpoint connector theorem, RS = XY/2.

Thus the length of RS depends only on angle O and is independent of P.

Note:  Now that we've shown P doesn't matter, we can move P to A and show length RS = AO sin(O)

 Posted by Jer on 2021-10-11 11:06:30

 Search: Search body:
Forums (0)