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.
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
Here is another logic problem from Ian Stewart.
- No cat that wears a heron suit is unsociable.
- No cat without a tail will play with a gorilla.
- Cats with whiskers always wear heron suits.
- No sociable cat has blunt claws.
- No cats have tails unless they have whiskers.
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
This is a most interesting problem proposed by Mirangu and retweeted by Catriona Agg:
“Two equilateral triangles share a vertex. What is the proportion red : green?”
See the Two Equilateral Triangles
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
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
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
Here is another sum problem, this time from the 2021 Math Calendar.
As before, recall that all the answers are integer days of the month. And the solution employs a technique familiar to these pages.
See the Winter Sum
This is another doable puzzle from Sam Loyd.
“BACK OF THE OLDTIME song of “Grandfather’s clock was too tall for the shelf, so it stood for ninety years on the floor,” there was a legend of a pestiferous grand-father and a cantankerous old clock which, from the fitful time when “it was bought on the morn, when the old man was born,” it had made his whole life miserable, owing to an incurable habit which the clock had acquired of getting the hands tangled up whenever they attempted to pass.
These semi-occasional stoppages became of more frequent occurrence as advancing age made the old gentleman more irritable and his feeble hands more incapable of correcting the cranky antics of the balky old timepiece.
Once when the hands came together again and stopped the clock the old man flew into such an ungovernable passion that he fell down in a fit, stone dead, and it was then that
“The clock stopped short,
Never to go again,
When the old man died.”
A photograph of the clock was presented to me, showing the classical figure of a female representing time, and it struck me as remarkable that with the knowledge of the hour and minute hands being together that it should be possible to figure out the exact time at which “the old man died,” from the position of the second hand as shown, without having to see the face of the clock. The idea of being able to figure out the exact time of day from seeing the second hand alone is very odd, although not so difficult a puzzle as one would imagine.”
See the Grandfather Clock Puzzle
This is another nice puzzle from the Scottish Mathematical Council (SMC) Senior Mathematical Challenge of 2008.
“The triangle ABC is inscribed in a circle of radius 1. Show that the length of the side AB is given by 2 sin c°, where c° is the size of the interior angle of the triangle at C.”
The diagram shows the case where C is on the same side of the chord AB as the center of the circle. There is a second case to consider where C is on the other side of the chord from the center.
See the Circle Chord Problem