It used to be a standard and well-known geometry fact that given a point and a circle, all lines through the point intersect the circle at two distances that have the same geometric mean = the length of a tangent. This is easy to prove using similar triangles, given the fact that all arcs of circles subtend the same angle as seen from any point on the circle outside the arc. Apply this to the bottom of the wall. Bill On Mar 11, 2010, at 5:45 PM, Allan Wechsler wrote:
Oh, mercy. Dan, did you know that this remains true when the wall is not vertical? It can lean toward or away from you at any angle, unless I've made a big error. (A and B are still measured along the wall, not by dropping perpendiculars to the ground.)
On Thu, Mar 11, 2010 at 5:36 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Yes. Now write equations and solve for x in terms of A and B. Can you guess the answer beforehand?
On Thu, Mar 11, 2010 at 5:29 PM, Andy Latto <andy.latto@pobox.com> wrote:
Just to check:
I think the geometric answer is "draw a circle through the top and bottom of the window, tangent to the ground; the answer is the point of tangency". But I'm guessing from what both you and Dan have said that there's an even simpler characterization of the point. Is that right?
Andy
On Thu, Mar 11, 2010 at 5:02 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I did this with a geometric construction, but even though the algebra was easy, I wouldn't have been able to predict the answer in the form that Dan obviously intends. And I agree with him; it sort of calls out for something that qualifies as an explanation, rather than a mere proof.
On Thu, Mar 11, 2010 at 4:05 PM, Dan Asimov <dasimov@earthlink.net> wrote:
This is the window puzzle:
Given a window -- on the front of a building -- whose lowest point is height A and whose highest point is height B, how far from the building should a ground-level observer be so that the angle subtended by the window is maximum?
(Let's assume the window is the interval from (0,A) to (0,B) on the y-axis, and the observer is at (x,0) for x > 0.)
This is an easy enough calculus problem, and with a bit of thought can also be solved rigorously without calculus.
But the answer is a very simple function of A and B, and ideally there would be a solution that sheds light on why this should be.
So that's the real puzzle: Find an elegant solution that illuminates why the answer is in the simple form it is.
--Dan
"Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx
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