mrios@ihlpg.UUCP (Michael Rios) (10/21/86)
(Is the line eater now an imaginary function?) I recently have been thinking about a problem I'd created out of boredom, but I haven't come up with a satisfactory answer. Will someone please help me put this to rest? Given a square of side 10 (or whatever you want...I use 10 for concreteness) find the length of the side of the largest regular pentagon that may be inscribed within it. Prove it is the largest. Thanx... -- Michael Rios ihnp4!ihlpg!mrios "Gun. No license. I checked. Very bad." -Rorschach, _Watchmen_
crocker@ihwpt.UUCP (ron crocker) (10/24/86)
> Given a square of side 10 (or whatever you want...I use > 10 for concreteness) find the length of the side of the > largest regular pentagon that may be inscribed within it. > Prove it is the largest. 10 sin (36 degrees). Proof: (by construction - you may need a pencil to follow this. It is difficult to describe (I think...)) Let a be the length of the side of the square, and L be the length of the sides of the regular pentagon. First, inscribe the largest circle possible within the square. It will touch the square at the 4 midpoints of the sides of the square, and therefore will have radius r = 5 (a/2). Now, draw your regular pentagon. It will have 5 sides, of equal length, with 108 degree angles between the sides. Connect the center of the circle to each of the 5 "points" of the pentagon. You now have 5 isosceles(sp?) triangles with a 72 degree angle between the radii edges. Note that these line segments bisect the angles of the edges of the pentagon, so the other two angles are both 54 degrees (This does add to 180 degrees). Take any of these triangles, and draw a line segment from the center of the circle to the midpoint of the pentagon edge. This creates a right triangle, where the hypotenuse is r (= 5), half the pentagon edge (now length L/2) is opposite of the 72/2 = 36 degree angle. (See figure below) L/2 ----------- \ |_| \ | \ | \ __|_____ 36 degrees hypo- \ | | tenuse \ v | length r \ | \ | \ | \| . -- center of circle Therefore, using the simple trignometric identity, length of opposite side sin t = ----------------------- length of hypotenuse we see in our triangle that L/2 sin 36 = --- 5 Solving this for L, we see that L = 2 * 5 * sin 36 = 10 sin 36 degrees QED.
desj@brahms (David desJardins) (10/25/86)
In article <1179@ihwpt.UUCP> crocker@ihwpt.UUCP (ron crocker) writes: >> [ What is the area of the largest regular pentagon which can be enclosed >> within a square of side 10? ] > >[ Answer derived from the pentagon inscribed within a circle inscribed in > the square. ] To see that this is not the largest possible regular pentagon, choose a rotation of this inscribed pentagon such that none of its vertices lie at any of the four points of intersection of the circle with the square. Then it is clear that the pentagon can be slightly expanded about its center without extending outside of the square. Any maximal pentagon within the square must necessarily have four of its vertices lie on the sides of the square. This is because any pentagon with only three vertices on the square either can be expanded in the direction of the fourth side of the square, or can be rotated to bring only two vertices into contact with the square, and then expanded. To find the correct solution, we consider the area of the pentagon as a function of the angle of rotation of its main axis with respect to the axis of the square. Since the square has 90-degree rotational symmetry, and the pentagon has 72-degree rotational symmetry, the possible orientations of the pentagon are parametrized by an angle ranging from 0 to 18 degrees. Or, to look at it differently, rotating the pentagon by 18 degrees within the square gives the same orientation, with the vertices relabeled. If you need to, you can draw a picture to see this. In fact, there is an additional symmetry given by reflection, and we can thus reduce the range of angles in question to the range from 0 to 9 degrees. Another way to see this is to see that every possible orientation of the pentagon within the square will have some side of the pentagon intersect some side of the square at an angle of 9 degrees or less. Again, draw a picture if necessary to see this. We can now proceed to maximize the area using the calculus. Let x be the smallest angle that a side of the pentagon makes with respect to an edge of the square; as we have seen, this will be in the interval from 0 to 9 degrees. WLOG assume that the edge in question is on the bottom of the square, and that the side in question actually intersects it (simple translation and/or reflec- tion will accomplish this). Let S be the length of a side of the pentagon. Then simple trigonometry tells us that the distance from the center of the pentagon to a vertex is S/(2 sin36), the distance from the center of the pentagon to a side is S/(2 tan36), and the distance between two nonadjacent vertices is 2Scos36. The height of the rotated pentagon (from the bottom vertex to the top vertex) is (S/(2 sin36) + S/(2 tan36)) (cos x) + (S/2) (sin x) and the width (from the leftmost to the rightmost vertex) is (2 S cos36) (cos x). For the pentagon to be maximal one of these values must be equal to 10. To maximize S, we will minimize the larger of the factors by which it is multiplied in the above expressions. It turns out that this minimum comes at the value x = 9 degrees, at which the two expressions become equal (for all smaller value of x the second expression is the larger). So, the answer is that the maximum side length is S = 10 / (2 cos36 cos9) = 6.2574 and the corresponding area is A = 5/(4 tan36) S^2 = 67.3649. If you want to draw a picture of this orientation, one of the sides of the pentagon intersects two of the edges of the square at 45-degree angles. This actually wasn't all that hard -- I was pretty sure that the above was the correct answer from the beginning -- but I wanted to go through it in enough detail that everyone who is interested can understand it. I hope I succeeded. -- David desJardins
avg@navajo.UUCP (10/25/86)
Put the pentagon like a house, flat side down. Now put the square around it like a diamond. The pentagon touches all 4 sides of the square and is held rigidly, thus is locally maximum. I.e., any slightly different orientation of the same sized square would force the pentagon to be smaller. Now convince yourself that there is no other local maximum, i.e., rigid orientation, and move on to something else.
pash@sjuvax.UUCP (10/31/86)
In article <2595@ihlpg.UUCP> mrios@ihlpg.UUCP (Michael Rios) writes: >(Is the line eater now an imaginary function?) > > I recently have been thinking about a problem I'd created >out of boredom, but I haven't come up with a satisfactory answer. >Will someone please help me put this to rest? > > Given a square of side 10 (or whatever you want...I use > 10 for concreteness) find the length of the side of the > largest regular pentagon that may be inscribed within it. > Prove it is the largest. You can't inscribe a regular pentagon in a square. For, suppose the five vertices of a regular pentagon were each on the square. Then two would have to be on one side of the square, say on AB, where the square's vertices are ABCD. Call the pentagon EFGHI, where EF lies on AB. Then it's clear that |IG| = |AB| = |BC| = |HJ|, where J is on EF and HJ is perpendicular to EF. (i.e., HJ is an altitude of the pentagon.) But since the length of all diagonals of a regular pentagon are equal, we also have |HF| = |IG|, so that |HF| = |HJ|. But this is impossible, because HF is the hypotenuse of right triangle HJF and HJ is a leg. If you remove the restriction "regular", the problem still has no answer as you can inscribe a pentagon of area 100 - e, where e is any positive integer, by lopping off a triangular corner of the square of area e, so there is no maximal pentagon. --Peter Ash St. Joseph's University