Tag Archives: plane geometry

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 →

Putnam Octagon Problem

Here is a problem from the famous (infamous?) Putnam exam, presented by Presh Talwalkar. Needless to say, I did not solve it in 30 minutes—but at least I solved it (after making a blizzard of arithmetic and trigonometric errors).

“Today’s problem is from the 1978 test, problem B1 (the easiest of the second set of problems). A convex octagon inscribed in a circle has four consecutive sides of length 3 and four consecutive sides of length 2. Find the area of the octagon.”

My solution is horribly pedestrian and fraught with numerous chances for arithmetic mistakes to derail it, which happened in spades. As I suspected, there was an elegant, “easy” solution (as demonstrated by Talwalkar)—once you thought of it! Again, this is like a Coffin Problem.

Answer.

See the Putnam Octagon Problem for solutions.

Conical Bottle Problem

I was astonished that this problem was suitable for 8th graders. First of all the formula for the volume of a cone is one of the least-remembered of formulas, and I certainly never remember it. So my only viable approach was calculus, which is probably not a suitable solution for an 8th grader.

Presh Talwalkar: “This was sent to me as a competition problem for 8th graders, so it would be a challenge problem for students aged 12 to 13. When a conical bottle rests on its flat base, the water in the bottle is 8 cm from its vertex. When the same conical bottle is turned upside down, the water level is 2 cm from its base. What is the height of the bottle? (Note “conical” refers to a right circular cone as is common usage.) I at first thought this problem was impossible. But it actually can be solved.”

Answer.

See the Conical Bottle Problem for solutions.

Regiomontanus 1471 Problem

This is an old problem I had seen before. Here is David Wells’s rendition:

“Johannes Müller, named Regiomontanus after the Latin translation of Körnigsberg, his city of birth, later made famous by Euler, proposed this problem in 1471. … it is usually put in this form …: From what distance will a statue on a plinth appear largest to the eye [of a mouse!]? If we approach too close, the statue appears foreshortened, but from a distance it is simply small.”

I have added height numbers in feet for concreteness (as well as the mouse qualification, since the angles are measured from ground level). So the problem is to find the distance x such that the angle is maximal.

Answer.

See the Regiomontanus 1471 Problem for solutions.

Parallelogram Problem

Catriona Shearer has come up with another challenging but elegant geometric problem. In some ways, it is similar to the famous Russian Coffin Problems that have an obvious solution—once you see it—but initially seem impenetrable. I really marvel at Catriona Shearer’s ability to come up with these problems.

“What’s the area of the parallelogram?”

Answer.

See the Parallelogram Problem for a solution.

Straight and Narrow Problem

The following interesting behavior was found at the Futility Closet website:

“A pleasing fact from David Wells’ Archimedes Mathematics Education Newsletter: Draw two parallel lines. Fix a point A on one line and move a second point B along the other line. If an equilateral triangle is constructed with these two points as two of its vertices, then as the second point moves, the third vertex C of the triangle will trace out a straight line. Thanks to reader Matthew Scroggs for the tip and the GIF.”

This is rather amazing and cries out for a proof. It also raises the question of how anyone noticed this behavior in the first place. I proved the result with calculus, but I wonder if there is a slicker way that makes it more obvious. See the Straight and Narrow Problem.

(Update 3/25/2019) Continue reading →

River Crossing

This is a riff on a classic problem, given in Challenging Problems in Algebra.

“N. Bank and S. Bank are, respectively, the north and south banks of a river with a uniform width of one mile. Town A is 3 miles north of N. Bank, town B is 5 miles south of S. Bank and 15 miles east of A. If crossing at the river banks is only at right angles to the banks, find the length of the shortest path from A to B.

Challenge. If the rate of land travel is uniformly 8 mph, and the rowing rate on the river is 1 2/3 mph (in still water) with a west to east current of 1 1/3 mph, find the shortest time it takes to go from A to B. [The path across the river must still be perpendicular to the banks.]”

Answer.

See the River Crossing for a solution.

Chalkdust Triangle Problem

The issue 7 of the Chalkdust mathematics magazine had an interesting geometric problem presented by Matthew Scroggs.

“In the diagram, ABDC is a square. Angles ACE and BDE are both 75°. Is triangle ABE equilateral? Why/why not?”

I had a solution, but alas, the Scroggs’s solution was far more elegant.

Answer.

See the Chalkdust Triangle Problem for solutions.

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