This is a problem from the UKMT Senior Challenge for 2001. (It has been slightly edited to reflect the colors I added to the diagram.)
“The [arbitrary] blue triangle is drawn, and a square is drawn on each of its edges. The three green triangles are then formed by drawing their lines which join vertices of the squares and a square is now drawn on each of these three lines. The total area of the original three squares is A1, and the total area of the three new squares is A2. Given that A2 = k A1, then
_____A_ k = 1_____B_ k = 3/2_____C_ k = 2_____D_ k = 3_____E_ more information is needed.”
I solved this problem using a Polya principle to simplify the situation, but UKMT’s solution was direct (and more complicated). See the Six Squares Problem.
Yet another train problem from H. E. Dudeney.
“We were going by train from Anglechester to Clinkerton, and an hour after starting an accident happened to the engine. We had to continue the journey at three-fifths of the former speed. It made us two hours late at Clinkerton, and the driver said that if only the accident had happened fifty miles farther on the train would have arrived forty minutes sooner. Can you tell from that statement just how far it is from Anglechester to Clinkerton?”
See the Damaged Engine.
Catriona Shearer retweeted the following problem from Antonio Rinaldi @rinaldi6109
“My little contribution to @Cshearer41 October 7, 2018
A point D is randomly chosen inside the equilateral triangle ABC. Determine the probability that the triangle ABD is acute-angled.”
See Triangle Acute-Angle Problem
Another challenging problem from Presh Talwalkar. I certainly could not have solved it on a timed test at the age of 16.
“One Of The Hardest GCSE Test Questions – How To Solve The Cosine Problem
Construct a hexagon from two congruent parallelograms as shown. Given BP = BQ = 10, solve for the cosine of PBQ in terms of x.
This comes from the 2017 GCSE exam, and it confused many people. I received many requests to solve this problem, and I thank Tom, Ben, and James for suggesting it to me.”
See the Parallelogram Cosine Problem
Here is a problem from the UKMT Senior (17-18 year-old) Mathematics Challenge for 2012:
“Tom and Geri have a competition. Initially, each player has one attempt at hitting a target. If one player hits the target and the other does not then the successful player wins. If both players hit the target, or if both players miss the target, then each has another attempt, with the same rules applying. If the probability of Tom hitting the target is always 4/5 and the probability of Geri hitting the target is always 2/3, what is the probability that Tom wins the competition?
______A 4/15______B 8/15______C 2/3______D 4/5______E 13/15”
See Hitting the Target
I really was trying to stop including Catriona Shearer’s problems, since they are probably all well-known and popular by now. But this is another virtually one-step-solution problem that again seems impossible at first. Many of her problems entail more steps, but I am especially intrigued by the one-step problems.
“What’s the sum of the two marked angles?”
See Kissing Angles.
An amazing publication was conceived primarily for women at the beginning of the 18th century in 1704 and was called The Ladies’ Diary or Woman’s Almanack. What made it even more remarkable was that each issue contained mathematical problems whose solutions from the readers were provided in the next issue. One particularly sharp woman was Mary Wright (Mrs. Mary Nelson). This is one of her problems:
“VIII. Question 72 by Mrs. Mary Nelson
(proposed in 1719, answered in 1720)
A prize was divided by a captain among his crew in the following manner: the first took 1 pound and one hundredth part of the remainder; the second 2 pounds and one hundredth part of the remainder; the third 3 pounds and one hundredth part of the remainder; and they proceeded in this manner to the last, who took all that was left, and it was then found that the prize had by this means been equally divided amongst the crew. Now if the number of men of which the crew consisted be added to the number of pounds in each share, the square of that sum will be four times the number of pounds in the chest: How many men did the crew consist of, and what was each share?”
What makes this problem nice is that it does have a clean answer, contrary to most of the problems in The Ladies’ Diary. See the Ladies’ Diary Problem.
(Update 5/6/2019) Continue reading
This is another train puzzle from H. E. Dudeney, which is fairly straight-forward.
“I put this little question to a stationmaster, and his correct answer was so prompt that I am convinced there is no necessity to seek talented railway officials in America or elsewhere. Two trains start at the same time, one from London to Liverpool, the other from Liverpool to London. If they arrive at their destinations one hour and four hours respectively after passing one another, how much faster is one train running than the other?”
See Two Trains – London to Liverpool
The Columbus story shows the intervention of chance in history at its most capricious. The following tale has its own logic, but the confluence of serendipitous events makes it marvelous and uplifting, especially in our current dark times. It was first brought to my attention by my father back in the early 1960s at the height of America’s role as wheat breadbasket of the world. America, and especially Kansas, was supplying essential wheat to the recently independent country of India and to the Soviet Union, whose long struggle with collective farming (and other factors), especially in the Ukraine, had led to its dependency on imports.
I will not try to narrate the story O’Henry-like with a surprise ending, but announce the amazing coincidence from the start—America was supplying the USSR its own wheat! The Kansas wheat was derived from a special hardy winter variety called Turkey Red that had originated in the Ukraine and was brought to America by Mennonites. So the story is how this all came about. See Turkey Red.
Here is a problem from the UKMT Senior (17-18 year-old) Mathematics Challenge for 2009:
“Four positive integers a, b, c, and d are such that
_________abcd + abc + bcd + cda + dab + ab + bc + cd + da + ac + bd + a + b + c + d = 2009.
What is the value of a + b + c + d?
_________A 73_________B 75_________C 77_________D 79_________E 81”
See the Challenging Sum
(Update 4/17/2019) Continue reading