Five Year Anniversary

So I managed to make it five years.  Again, I thought I would present the statistical pattern of interaction with the website in the absence of any explicit feedback.

But as the summary shows, the fall-off of visitors that began in the middle of last year has persisted throughout 2023.  I have also run out of much new material, so I am basically going to wrap it up.  I have a few things in the hopper, but they are mostly similar to puzzles already presented.  I have one or two essay ideas left, but again I have mostly said what I have to say, and the world of math has moved on.

Anyway, here is the summary for what it’s worth.

See Five Year Anniversary

Distance to Flag Problem

The following puzzle is from the Irishman Owen O’Shea.

“The figure shows the location of three flags [at A, B, and C] in one of the fields on a neighbor’s farm.  The angle ABC is a right angle.  Flag A is 40 yards from Flag B.  Flag B is 120 yards from flag C.  Thus, if one was to walk from A to B and then on to C, one would walk a total of 160 yards.

Now there is a point, marked by flag D, [directly] to the left of flag A.  Curiously, if one were to walk from flag A to flag D and then diagonally across to flag C, one would walk a total distance of 160 yards.

The question for our puzzlers is this: how far is it from flag D to flag A?”

This problem has a simple solution.  But it also suggests a more advanced alternative approach.

Answer.

See the Distance to Flag Problem for a solution.

More Right Triangle Magic

James Tanton asked to prove the following surprising property of a right triangle and its circumscribed and inscribed circles.

“Every triangle is circumscribed by some circle of diameter D, say, and circumscribes another circle of smaller diameter d. For a right triangle, d + D equals the sum of two side lengths of the triangle. Why?”

See More Right Triangle Magic

Two Containers Mixing Puzzle

This is a slightly different type of a mixture problem from Dan Griller.

“Two containers A and B sit on a table, partially filled with water.  First, 40% of the water in A is poured into B, which completely fills it.  Then 75% of the water in B is poured into A, which completely fills it.  80% of the water in A is poured into B, which completely fills it.  Calculate the ratio of the capacity of container A to the capacity of container B, and the fraction of container A that was occupied by water at the start.”

Answer.

See the Two Containers Mixing Puzzle for solution.

Yet Another Race

This is another race puzzle from the Maths Masters team, Burkard Polster (aka Mathologer) and Marty Ross as part of their “Summer Quizzes” offerings for 2013.

“In a 100 meter race, Jacob can beat Johann by 5 meters, and Johann can beat Nicolaus by 10 meters. By how much can Jacob beat Nicolaus?”

Answer.

See Yet Another Race for a solution.

Circles in Circles

Here is another problem from the “Challenges” section of the Quantum magazine.

“Inside a circle there are two intersecting circles. One of them touches the big circle in point A, the other in point B. Prove that if segment AB meets the smaller circles at one of their common points, then the sum of their radii equals the radius of the big circle. Is the converse true?  (A. Vesyolov)”

See Circles in Circles

Elliptic Circles

Here is another UKMT Senior Challenge problem for 2017.

“The diagram shows a square PQRS with edges of length 1, and four arcs, each of which is a quarter of a circle. Arc TRU has centre P; arc VPW has centre R; arc UV has centre S; and arc WT has centre Q.

What is the length of the perimeter of the shaded region?

A_6___B_(2√2 – 1)π___C_(√2 – 1/2 ___D_2___E_(3√2 – 2)π”

Answer.

See Elliptic Circles for a solution.

Maximized Box Problem

This problem is from Colin Hughes’s Maths Challenge website (mathschallenge.net).

“Four corners measuring x by x are removed from a sheet of material that measures a by a to make a square based open-top box.  Prove that the volume of the box is maximised iff the area of the base is equal to the area of the four sides.”

See the Maximized Box Problem