A16. Treasure search

A treasure hunter is on an island which is in the shape of the unit disc, and they are located on the boundary at (1,0). The treasure hunter has a treasure-detector, which will light up if the detector is within 1/2 unit of any buried treasure. The treasure hunter can move along any path (say, continuous and piecewise infinitely differentiable) that starts at (1,0) and stays within the closed unit disk. The treasure hunter claims that they can discover whether or not there is treasure buried on the island (or in other words, that the union of all disks of radius 1/2 centered on points along the path covers the unit disk) by traveling along a path of length π. Prove or disprove the claim.

This is my revision of a problem sent to me by Apratim Roy.

E35 - Three Triangles in a Regular Hexagon

Here is another problem from the Catriona Shearer collection. ABCDEF is a regular hexagon. Lines have been drawn to create three triangles, whose interiors have been shaded. What fraction of the hexagon is shaded?

The points labeled M_i are the midpoints of  their sides, and the unlabeled point marks the center of the hexagon.

E34. Tangent circles in a rectangle

This problem was called to my attention to Apratim Roy. It comes from a collection, on Twitter, of many dozens of interesting geometry problems, at the level of high school geometry, by Catriona Shearer, a math teacher in the UK, which she has posted on Twitter, @Cshearer41.

In the diagram below. you are given circles O and P, externally tangent at B. Circle O is tangent to two sides of rectangle DEFG, and circle P is tangent to three sides of the rectangle, as shown. Find the measure of angle ABC.

It is quite surprising that you don’t need to be given any measurements to answer the problem. (But no fair using that fact in your proof; that is, you can’t just prove a special case such as two circles of equal size.) I also like this problem because it is hard, but not too hard, and because there are several different nice ways to solve it.

E33. Tiling a checkerboard

This problem was shown to me by my student, Apratim Roy. Though it involves only elementary concepts, I found it rather difficult, and thought the solution was very surprising and elegant.

You are given a standard 8 x 8 checkerboard, with one square removed, and 21 3 x 1 tiles. In other words, there are exactly enough tiles to cover the modified board. Your task is to find a way to do this, without cutting any tile.

(a) Find out what square must be removed for the task to be possible. (4 possible answers).
(b) Describe the tiling. (Many possible answers).

You might guess that the removed square needs to be one of the four corners of the checkerboard, but you would be wrong.

A13. Points of tangency of an ellipse and a circle

Let E be the ellipse with equation x²/4 + y² = 1 and C(r) be the circle with center (1,0) and radius r. For which values of r do the curves E and C(r) have point(s) of tangency?

This is fairly routine, but still a bit challenging to find all solutions.

Hippasus & The Discovery of Irrational Numbers

In November I gave a speculative talk to the New England Section of the Mathematical Association of America on the discovery of irrational numbers by the Pythagorean Hippasus through an examination of the mystic pentagram, the sacred symbol of the Pythagoreans. I have expanded it a bit, adding a method of recursively computing the golden ratio, phi. If you want to see how these topics are related, please check my paper on Scribd: https://www.scribd.com/doc/294006886/Irrational-Numbers-the-Mystic-Pentagram-and-Eigenvectors.

A11. Splitting a triangle and a tetrahedron

János Kurdics published this interesting problem in The Math Connection Linkedin group:
Given a triangle, find the shortest line segment that divides the triangle into two regions of equal area. A solution that does not involve calculus is preferred.
The solution that I know is very nice, and gives quite a workout in elementary trigonometry.

János says that he is working on three-dimensional analog (plane that divides a tetrahedron into two regions of equal volume and gives smallest possible cross-sectional area with the tetrahedron) but has only solved it for a regular tetrahedron.

A10. Triangle with sides in arithmetic sequence.

Find a number n such that there is a triangle with sides n, n + 1, and n + 2 in which the largest angle is twice the smallest angle. How many such numbers n are there? Note that n does not have to be an integer.

This problem came up in a high school textbook during a tutoring session, except there were several hints given that made the problem much easier. See if you can do it without hints.

E28. A balancing cube

A sculpture is in the form of a cube, one meter on a side, balancing unstably on one vertex. Let A be the bottom vertex (on the ground), B one of the three adjacent vertices, and C the vertex at the other end of the space diagonal of the cube from A. Assume the ground is a horizontal plane, and AC is perpendicular to to the ground. Find the distance from B to the ground, with answer in exact form.

A9. Squares erected on a triangle.

Here is a nice problem from Coxeter and Greitzer’s classic book, Geometry Revisited, where an elegant solution is given.

Let ABC be a triangle, and construct squares externally on the sides. Let O₁ be the center of the square on AB, O₂ the center of the square on BC, and O₃ the center of the square on AC, as in the following diagram:

Prove that the segments O₁O₃ and O₂A have the same length and are perpendicular to each other.

A8. Acute triangles with given side lengths

This is a neat problem from last year's Putnam Examination. It was published (with answer given) in the most recent MAA Monthly.

Given 12 real numbers d₁, ... , d₁₂ on the interval (1, 12), show that there exist distinct indices i, j, k such that there is an acute triangle with side lengths d_i, d_j, d_k.

I will post a hint as a comment in a few days.

E 24. Egyptian Area Formula

Let ABCD be a quadrilateral with area K and let w = AB, x = BC, y = CD, z = DA. The ancient Egyptians used the expression K' = [(w + y)/2][(x + z)/2] for the area. Since they used this formula to compute the area of fields for taxing purposes, the government probably didn't mind that this formula overestimates the area for all quadrilaterals except rectangles. Prove this fact, that is K <= K', with K = K' iff ABCD is a rectangle. The proof is surprisingly easy, and only requires high school mathematics.

A7. Comparing areas

In the diagram below, ABCD is a square, DCE is an equilateral triangle, F is the intersection of AE with CD, G is the intersection of BE with CD, IFJ and HGK are perpendicular to CD, and L is on CB so CF = CL. Prove that the area of triangle AFL is equal to the area of rectangle FGKJ.

Sketchpad - Jacobs problems now available

I have finished writing 23 pages of problems for Jacobs' high school geometry book that use GSP. It is posted on Scribd as a pdf.

http://www.scribd.com/doc/146440260/Extra-Problems-for-Jacobs-Using-GSP

My Jacobs - GSP Project

I am using the textbook Geometry: Seeing, Doing, Understanding (Third Edition) by Harold R. Jacobs in my course for prospective middle school math teachers. I love many things about this book, but I also am a strong believer in having my students use Geometer's Sketchpad (GSP) software. Jacobs has lots of hands-on work, but he made a decision not to use geometry software, so I have been supplementing his book with Sketchpad work. Mostly, I have taken a number of exercises from Jacobs and made very similar Sketchpad exercises out of them.

So far, I have written out seven pages of exercises, and I imagine I will end up with 20 - 30 pages. I key each of my exercise sets to section of Jacobs. I intend to class-test my exercises, and once they are complete to post them on Scribd. In the meantime, I am happy to send my work-in-progress to anyone who wants to send me an email for it. My address is peter.ash@MathForTheRestOfUs.com, and I appreciate feedback.

Sangaku, Harold Jacobs, and Geometer's Sketchpad

As I mentioned at the time, I delivered a talk at the New England Section of the Mathematical Association of America meeting in Bridgewater, Massachusetts, in November. I decided that it would be good to make the talk available online. The talk was about my adventures in trying to prove a difficult theorem mentioned in Harold Jacobs' Geometry. After finding a proof with the aid of Geometer's Sketchpad I happened to discover through Wikipedia that the theorem has a name: The Japanese Theorem For Quadrilaterals. Then Peter Renz, one of Jacobs' editors, suggested I look at the book Sacred Geometry: Japanese Temple Geometry by Fukagawa Hidetoshi and Tony Rothman, which allowed me to place the theorem in a rich cultural and mathematical context.

The talk is at http://www.scribd.com/doc/129968316/NES-MAA-Presentation. This consists of the slides that I used, put in portrait page orientation, but otherwise unchanged. It is a bit terse, but I hope some find it interesting.

Constructible Angles

A student asked what angles are constructible, considering only angles that measure a whole number of degrees (which I'll call integral angles). The answer is very simple: The only constructible integral angles are those which measure 3n degrees, where n is any natural number.

To see this, first note that angles of measure 60 and 72 (vertex angles in a regular pentagon) are constructible. Since angles may be bisected repeatedly by construction, this means that angles of measure 15 = 60/4 and 9 = 72/8 degrees may be constructed. Since 3 = 9*2 – 15, an angle of 3 degrees may be constructed by constructing two adjacent 9 degree angles, and then a 15 degree angle inside the resulting 18 degree angle. By adding n 3-degree angles together, a 3n degree angle can be constructed.

The proof that only these integral angles can be constructed seems to require the more advanced result that there is at least one integral angle that is not constructible. For example, an angle of 20 degrees is non-constructible. (This is the standard proof, using Galois theory, that demonstrates the impossibility of angle trisection.) If there were an integral angle of measure n that is not a multiple of 3 that is constructible, then since (n,3) = 1, there is a linear combination with integral coefficients of n and 3 that gives 1, and so an angle of 1 degree would be constructible, and hence an angle of 20 degrees would be constructible, which it is not..

A Sangaku Problem

I will be giving a talk Saturday at the NES-MAA meeting in Bridgewater MA in which I will talk about sangaku, or Japanese temple geometry problems. These problems, created by people from a wide walk of life, were beautifully drawn on wooden tablets which were then placed in Buddhist temples or Shinto shrines. Hundreds have been discovered, and probably thousands existed at one time. I became involved in this when I was challenged to solve one of the difficult sangaku. Here I present one of the easier ones, from the delightful book Sacred Geometry: Japanese Temple Geometry by Fukagawa Hidetoshi and Tony Rothman.

I chose this problem because the diagram is so beautiful, the solution is fairly simple, yet satisfying, and it is one of the few sangaku created by a woman (Okuda Tsume).

In a circle of diameter AB = 2R, draw two arcs of radius R with centers A and B respectively, and 10 inscribed circles, two green circles of diameter R, four red circles of radius t, and four blue circles of radius t'. Show that t = t' = R/6.

In the diagram below, we follow the convention of labeling the center of a circle with the radius of that circle.

A6. Counting Triangulations

Here is a counting problem that was solved a long time ago. Feel free to try your hand at it.

Given P, a convex n-gon, a triangulation of P is a subdivision of P into n - 2 non-overlapping triangles. A triangulation is obtained by drawing n - 3 non-intersecting diagonals. Let f(n) be the number of different triangulations. Clearly, f(3) = 1, f(4) = 2, and f(5) = 5. Careful counting shows f(6) = 14. Find an expression for f(n).

E21. Equiangular and Equilateral Polygons

A polygon is equiangular if all of its angles are equal. In particular, if the polygon has n sides, each angle measures (n - 2) * 180 / n degrees. A polygon is equilateral if all of its sides have the same length. It can be shown very easily that every equiangular triangle is equilateral. Of course, it is not true that every equiangular quadrilateral is equilateral. Any rectangle that is not a square provides a counterexample. Show that for every n > 3 there exists an equiangular n-gon that is not equilateral.