Tag Archives: analytic geometry

Geometric Puzzle Munificence

Having fallen under the spell of Catriona Shearer’s geometric puzzles again, I thought I would present the latest group assembled by Ben Orlin, which he dubs “Felt Tip Geometry”, along with a bonus of two more recent ones that caught my fancy as being fine examples of Shearer’s laconic style. Orlin added his own names to the four he assembled and I added names to my two, again ordered from easier to harder.

See Geometric Puzzle Munificence.

A Tidy Theorem

This is another fairly simple puzzle from Futility Closet.

“If an equilateral triangle is inscribed in a circle, then the distance from any point on the circle to the triangle’s farthest vertex is equal to the sum of its distances to the two nearer vertices (q = p + r).

(A corollary of Ptolemy’s theorem.)”

See A Tidy Theorem

Amazing Triangle Problem

Here is another simply amazing problem from Five Hundred Mathematical Challenges:

Problem 154. Show that three solutions, (x1,.y1), (x2,.y2), (x3, y3), of the four solutions of the simultaneous equations
____________(x – h)² + (y – k)² = 4(h² + k²)
______________________xy = hk
are vertices of an equilateral triangle. Give a geometrical interpretation.”

Again, I don’t see how anyone could have discovered this property involving a circle, a hyperbola, and an equilateral triangle. It seems plausible when h.=.k, but it is not at all obvious for h..k. For some reason, I had difficulty getting a start on a solution, until the obvious approach dawned on me. I don’t know why it took me so long.

See the Amazing Triangle Problem.

Magic Hexagons

This is truly an amazing result from Five Hundred Mathematical Challenges.

Problem 119. Two unequal regular hexagons ABCDEF and CGHJKL touch each other at C and are so situated that F, C, and J are collinear.

Show that

(i) the circumcircle of BCG bisects FJ (at O say);
(ii) ΔBOG is equilateral.”

I wonder how anyone ever discovered this.

See the Magic Hexagons

Flipping Parabolas

This is a stimulating problem from the UKMT Senior Math Challenge for 2017. The additional problem “for investigation” is particularly challenging. (I have edited the problem slightly for clarity.)

“The parabola with equation y = x² is reflected about the line with equation y = x + 2. Which of the following is the equation of the reflected parabola?

A_x = y² + 4y + 2_____B_x = y² + 4y – 2_____C_x = y² – 4y + 2
D_x = y² – 4y – 2_____E_x = y² + 2

For investigation: Find the coordinates of the point that is obtained when the point with coordinates (x, y) is reflected about the line with equation y = mx + b.”

See Flipping Parabolas.

Hyperboloid as Ruled Surface

When our daughter-in-law made wheat shocks as center-pieces for hers and our son’s fall-themed wedding reception, I naturally could not help pointing out the age-old observation that they represented a hyperboloid of one sheet. This was naturally greeted with the usual groans, but the thought stayed with me as I realized I had never proved this mathematically to myself. And so I did. See the Hyperboloid as Ruled Surface.

Perspective Map

A number of recent puzzles have involved perspective views of objects. I had never really explored the idea of a perspective map in detail. So some of the properties associated with it always seemed a bit vague to me. I decided I would derive the mathematical equations for the perspective or projective map and see how its properties fell out from the equations. With this information in hand I then addressed some questions I had about the article “Dürer: Disguise, Distance, Disagreements, and Diagonals!” by Annalisa Crannell, Marc Frantz, and Fumiko Futamura concerning a controversy over Albrecht Dürer’s woodcut St. Jerome in His Study (1514). And finally, I read somewhere that a parabola under a perspective map becomes an ellipse, so I was able to show that as well. See the Perspective Map.

(Update 7/1/2019) Continue reading