I have never heard the expression "power of a point" before, so I had to look it up. That's a lovely invariant, that is. On Thu, Mar 11, 2010 at 6:23 PM, Joshua Zucker <joshua.zucker@gmail.com>wrote:
The geometric construction I used gives what I feel like is a pretty good explanation.
Spoilers follow!
Let C be the point on the ground. Then the radius of the circumcircle of ABC is minimized exactly when this angle is maximized. For that radius to be minimized, you must have the circle tangent to the ground at that point.
The center of the circle is on the perpendicular bisector of AB, which is always (A+B)/2 above the ground, so the radius of the circle is (A+B)/2. (Sorry, I'm using A and B interchangeably as the measure of the distance and as the name of the points.)
Then to find the horizontal distance from the wall, its square is [(A+B)/2]^2 - [(A-B)/2]^2 = AB, so it's the geometric mean of the two given lengths.
The geometric mean makes me look for an even better construction, and indeed a smarter calculation from the same figure gives it: you have the power of a point at the base of the wall telling you that x^2 = AB. Reminder to self: don't use the Pythagorean theorem when there's power of a point to be had! Similar triangles are always at least as nice. At worst you end up proving the Pythagorean theorem again.
--Joshua Zucker
On Thu, Mar 11, 2010 at 2:02 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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