Category Archives: Puzzles and Problems

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.

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

Two More Jugs

Here is another classic example of the three jug problem posed in the Mathigon Puzzle Calendars for 2017.

“How can I measure exactly 8 liters of water, using just one 11 liter and one 6 liter bucket?”

It is assumed you have unlimited access to water (the “third jug” of at least 17 liters).  You can only fill or empty the jugs, unless in poring from one jug to another you fill the receiving jug before emptying the poring jug.  (Hint: see the Three Jugs Problem.)

See Two More Jugs.

Elliptical Medians Problem

This is a tantalizing problem from the 1977 Crux Mathematicorum.

“278. Proposed by W.A. McWorter, Jr., The Ohio State University.

If each of the medians of a triangle is extended beyond the sides of the triangle to 4/3 its length, show that the three new points formed and the vertices of the triangle all lie on an ellipse.”

See the Elliptical Medians Problem

Storm Chaser Problem

This is a somewhat challenging problem from the 1997 American Invitational Mathematics Exam (AIME).

“A car travels due east at 2/3 miles per minute on a long, straight road. At the same time, a circular storm, whose radius is 51 miles, moves southeast at √2/2 miles per minute. At time t = 0, the center of the storm is 110 miles due north of the car. At time t = t1 minutes, the car enters the storm circle, and at time t = t2 minutes, the car leaves the storm circle. Find (t1 + t2)/2.”

Answer.

See the Storm Chaser Problem for solutions.

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