[math-fun] Window puzzle, sort of
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
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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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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-- Andy.Latto@pobox.com
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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-- Andy.Latto@pobox.com
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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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-- Andy.Latto@pobox.com
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On 3/11/10, Allan Wechsler <acwacw@gmail.com> 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.)
I'm sure Buster Keaton would have found that good to know ... WFL
No-one seems to have mentioned the Euclid theorem T^2=AB explicitly. Isn't that a complete solution? R. On Thu, 11 Mar 2010, 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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-- Andy.Latto@pobox.com
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On 3/11/10, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
No-one seems to have mentioned the Euclid theorem T^2=AB explicitly. Isn't that a complete solution? R.
Not entirely --- you also need Andy's observation that the tangent circle minimises the radius, so maximising the angle. WFL
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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-- Andy.Latto@pobox.com
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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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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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On 3/12/10, Allan Wechsler <acwacw@gmail.com> wrote:
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.
It generalises to a pair of (oriented) circles (or spheres in n-space); depending on how it is normalised, the same bilinear form gives the tangential distance, the incidence angle, or the perpendicular distance from point to line. Followed to its logical conclusion, this leads to (in the plane) tetracyclic coordinates for conformal / inversive / Moebius group geometry, and pentacyclic coordinates for equilong / Laguerre group and contact / Lie-sphere group geometry. Not many people know that ... WFL
Here's another argument that's less geometric and more calculusy. I'll also use A and B to denote the bottom and top of the window and O and X to denote the bottom of the wall and the location of the person. You choose X so that when you walk right at unit speed, say, angle AXB has an inflection point, which means OXB is decreasing at the same rate as OXA. Now having X move right at unit speed has the same effect on OXA as moving A down at speed a/x. So moving A _up_ at speed a/x will decrease the complementary angle OAX at the same rate that moving X right will decrease OXB. Thus triangles OAX and OXB are similar, giving x^2 = ab. David P. Moulton
This problem is discussed and elementary solutions are given on p46-48 of the book "Trigonometric Delights" by Eli Maor, where it is referred to as Regiomontanus's maximum problem. It was proposed by Johann Müller, alias Regiomontanus, in 1471 in a letter to Christian Roder. The elementary solutions given in Eli Maor's book seem similar to some given on this list by others. --Jim On Thu, Mar 11, 2010 at 3:05 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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participants (9)
-
Allan Wechsler -
Andy Latto -
Dan Asimov -
David P. Moulton -
Fred lunnon -
James Buddenhagen -
Joshua Zucker -
Richard Guy -
William Thurston