Author Archives: Jim Stevenson

Seven Girls Puzzle

This problem comes from the Scottish Mathematical Council (SMC) Senior Mathematical Challenge of 2007:

“A group of seven girls—Ally, Bev, Chi-chi, Des, Evie, Fi and Grunt—were playing a game in which the counters were beans. Whenever a girl lost a game, from her pile of beans she had to give each of the other girls as many beans as they already had. They had been playing for some time and they all had different numbers of beans. They then had a run of seven games in which each girl lost a game in turn, in the order given above. At the end of this sequence of games, amazingly, they all had the same number of beans—128. How many did each of them have at the start of this sequence of seven games?”

See the Seven Girls Puzzle

Challenging Triangle Problem

This is a challenging problem from the 1986 American Invitational Mathematics Exam (AIME).

“Let triangle ABC be a right triangle in the xy-plane with a right angle at C. Given that the length of the hypotenuse AB is 60, and that the medians through A and B lie along the lines y = x + 3 and y = 2x + 4 respectively, find the area of triangle ABC.”

I have included a sketch to indicate that the sides of the right triangle are not parallel to the Cartesian coordinate axes. 

The AIME (American Invitational Mathematics Examination) is an intermediate examination between the American Mathematics Competitions AMC 10 or AMC 12 and the USAMO (United States of America Mathematical Olympiad). All students who took the AMC 12 (high school 12th grade) and achieved a score of 100 or more out of a possible 150 or were in the top 5% are invited to take the AIME. All students who took the AMC 10 (high school 10th grade and below) and had a score of 120 or more out of a possible 150, or were in the top 2.5% also qualify for the AIME.

See the Challenging Triangle Problem.

Playing with Triangles

Here is another elegant Quantum math magazine Brainteaser from the imaginative V. Proizvolov.

“Two isosceles right triangles are placed one on the other so that the vertices of each of their right angles lie on the hypotenuse of the other triangle (see the figure at left). Their other four vertices form a quadrilateral. Prove that its area is divided in half by the segment joining the right angles. (V. Proizvolov)”

See Playing with Triangles

Heron Suit Problem

Here is another logic problem from Ian Stewart.

  1. No cat that wears a heron suit is unsociable.
  2. No cat without a tail will play with a gorilla.
  3. Cats with whiskers always wear heron suits.
  4. No sociable cat has blunt claws.
  5. No cats have tails unless they have whiskers.

Therefore:

No cat with blunt claws will play with a gorilla.

Is the deduction logically correct?

I confess I don’t know what a heron suit is.  Google showed various garments with herons imprinted on the cloth, so maybe that is what it is.

See the Heron Suit Problem

Ladder Locus Puzzle

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

“A ladder is leaning against a wall. The base of the ladder starts sliding away from the wall, with the top of the ladder sliding down the wall. As the ladder slides, you watch the red point in the middle of the ladder. What figure does the red point trace? What about other points on the ladder?”

See the Ladder Locus Puzzle

Octagonal Area Problem

Here is another problem from the Polish Mathematical Olympiads published in 1960.

“95. In a parallelogram of given area S each vertex has been connected with the mid-points of the opposite two sides.  In this manner the parallelogram has been cut into parts, one of them being an octagon.  Find the area of that octagon.”

See the Octagonal Area Problem

Another Christmas Tree Puzzle

This is a belated Christmas puzzle from December 2019 MathsMonday.

“A Christmas tree is made by stacking successively smaller cones. The largest cone has a base of radius 1 unit and a height of 2 units. Each smaller cone has a radius 3/4 of the previous cone and a height 3/4 of the previous cone. Its base overlaps the previous cone, sitting at a height 3/4 of the way up the previous cone.

What are the dimensions of the smallest cone, by volume, that will contain the whole tree for any number of cones?”

Recall that the volume of a cone is π r2 h/3.

See Another Christmas Tree Puzzle