Tag Archives: plane geometry

Alberti’s Perspective Construction

I was reading yet another book on the Scientific Revolution when I came across a discussion of the mathematical significance of the invention of perspective for painting in the 15th century Italian Renaissance. The main player in the saga was Leon Battista Alberti (1404 – 1472) and his tome De Pictura (On Painting) (1435-6), which contained the first mathematical presentation of perspective. Even though mathematics was advertised, it was not at the level of trigonometry I used in my post “The Perspective Map”, but rather entailed simple Euclidean plane geometry. So the discussion was largely historical rather than mathematical. Nevertheless, I became curious to learn how much Alberti was able to discover about perspective without a lot of math. This essay is the result.

See Alberti’s Perspective Construction

(Update 7/29/2019)  I got a response! Continue reading

Star Polygon Problem

This is problem #25 from the UKMT 2014 Senior Challenge.

“Figure 1 shows a tile in the form of a trapezium [trapezoid], where a = 83⅓°. Several copies of the tile placed together form a symmetrical pattern, part of which is shown in Figure 2. The outer border of the complete pattern is a regular ‘star polygon’. Figure 3 shows an example of a regular ‘star polygon’.
How many tiles are there in the complete pattern?

See the Star Polygon Problem.

Rising Sun

Here is a problem from the UKMT Senior (17-18 year-old) Mathematics Challenge for 2012:

“A semicircle of radius r is drawn with centre V and diameter UW. The line UW is then extended to the point X, such that UW and WX are of equal length. An arc of the circle with centre X and radius 4r is then drawn so that the line XY is tangent to the semicircle at Z, as shown. What, in terms of r, is the area of triangle YVW?”

See the Rising Sun

Six Squares Problem

This is a problem from the UKMT Senior Challenge for 2001. (It has been slightly edited to reflect the colors I added to the diagram.)

“The [arbitrary] blue triangle is drawn, and a square is drawn on each of its edges. The three green triangles are then formed by drawing their lines which join vertices of the squares and a square is now drawn on each of these three lines. The total area of the original three squares is A1, and the total area of the three new squares is A2. Given that A2 = k A1, then

_____A_ k = 1_____B_ k = 3/2_____C_ k = 2_____D_ k = 3_____E_ more information is needed.”

I solved this problem using a Polya principle to simplify the situation, but UKMT’s solution was direct (and more complicated). See the Six Squares Problem.

Parallelogram Cosine Problem

Another challenging problem from Presh Talwalkar. I certainly could not have solved it on a timed test at the age of 16.

One Of The Hardest GCSE Test Questions – How To Solve The Cosine Problem

Construct a hexagon from two congruent parallelograms as shown. Given BP = BQ = 10, solve for the cosine of PBQ in terms of x.

This comes from the 2017 GCSE exam, and it confused many people. I received many requests to solve this problem, and I thank Tom, Ben, and James for suggesting it to me.”

See the Parallelogram Cosine Problem

Kissing Angles

I really was trying to stop including Catriona Shearer’s problems, since they are probably all well-known and popular by now. But this is another virtually one-step-solution problem that again seems impossible at first. Many of her problems entail more steps, but I am especially intrigued by the one-step problems.

“What’s the sum of the two marked angles?”

See Kissing Angles.

Two Block Incline Puzzle

Since everyone by now who has any interest has gone directly to Catriona Shearer’s Twitter account for geometric puzzles, I was not going to include any more. But this one with its one-step solution is too fine to ignore and belongs with the “5 Problem” as one of the most elegant.

“Two squares sit on the hypotenuse of a right-angled triangle. What’s the angle?”

See the Two Block Incline Puzzle

(Update 4/26/2019) Continue reading

Magic Parallelogram

I came across this problem in Alfred Posamentier’s book, but I remember I had seen it a couple of places before and had never thought to solve it. At first, it seems like magic.

In any convex quadrilateral (line between any two points in the quadrilateral lies entirely inside the quadrilateral) inscribe a second convex quadrilateral with its vertices on the midpoints of the sides of the first quadrilateral. Show that the inscribed quadrilateral must be a parallelogram.

See the Magic Parallelogram.

(Update 5/15/2020) Continue reading