Presh Talwalkar had another interesting problem.
“A triangle is drawn inside a square with sides 4, 3, and 5, as shown. What is the length of the square’s side?”
The problem looks simple at first, but it takes some care to avoid some hideous quartic equations.
See Tipsy 3-4-5 Triangle
This problem from the 1987 Discover magazine’s Brain Bogglers by Michael Stueben apparently traces back to 1770, though the exact reference is not given.
“Here’s an arithmetic problem taken from a textbook published in Germany in 1770. Three people are gambling. In the first game, Player A loses to each of the others as much money as each of them had when the game started. In the next game, B loses to each of the others as much money as each had when that game began. In the third game, A and B each win from C as much money as each had at the start of that game. The players now find that each has the same sum, 24 guineas. How much money did each have when play began?”
See the 1770 Card Game Problem
I found this problem from the Math Challenges section of the 2002 Pi in the Sky Canadian math magazine for high school students to be truly astonishing.
“Problem 4. Inside of the square ABCD, take any point P. Prove that the perpendiculars from A on BP, from B on CP, from C on DP, and from D on AP are concurrent (i.e. they meet at one point).”
How could such a complicated arrangement produce such an amazing result? I didn’t know where to begin to try to prove it. My wandering path to discovery produced one of my most satisfying “aha!” moments.
See the Mysterious Doppelgänger Problem
Update (12/27/2019) I goofed. I had plotted the original figure incorrectly. (No figure was given in the Pi in the Sky statement of the problem.) Fortunately, the original solution idea still worked.
Here is yet another surprising result from Colin Hughes at Maths Challenge.
It can be shown that a unique circle passes through three given points. In triangle ABC three points A’, B’, and C’ lie on the edges opposite A, B, and C respectively. Given that the circle AB’C’ intersects circle BA’C’ inside the triangle at point P, prove that circle CA’B’ will be concurrent with P.”
I have to admit it took me a while to arrive at the final version of my proof. My original approach had some complicated expressions using various angles, and then I realized I had not used one of my assumptions. Once I did, all the complications faded away and the result became clear.
See Circular Rendezvous Mystery.
This is another train puzzle by H. E. Dudeney. This one has some hairy arithmetic.
“Two trains, A and B, leave Pickleminster for Quickville at the same time as two trains, C and D, leave Quickville for Pickleminster. A passes C 120 miles from Pickleminster and D 140 miles from Pickleminster. B passes C 126 miles from Quickville and D half way between Pickleminster and Quickville. Now, what is the distance from Pickleminster to Quickville? Every train runs uniformly at an ordinary rate.”
See Trains – Pickleminster to Quickville
It is always fascinating to look at problems from the past. This one, given by Thomas Whiting himself, is over 200 years old from Whiting’s 1798 Mathematical, Geometrical, and Philosophical Delights:
“Question 2, by T. W. from Davison’s Repository.
There are two houses, one at the top of a lofty mountain, and the other at the bottom; they are both in the latitude of 45°, and the inhabitants of the summit of the mountain, are carried by the earth’s diurnal rotation, one mile an hour more than those at the foot.
Required the height of the mountain, supposing the earth a sphere, whose radius is 3982 miles.”
See the Mountain Houses Problem
This problem from Colin Hughes at Maths Challenge is a most surprising result that takes a bit of tinkering to solve.
We can see that 3 x 4 x 5 x 6 = 360 = 19² – 1. Prove that the product of four consecutive integers is always one less than a perfect square.”
The result is so mysterious at first that you begin to understand why the ancient Pythagoreans had a mystical relationship with mathematics.
See the Consecutive Product Square.
(Update 11/12/2020) Generalization and Visual Proof
This is another intriguing problem from Presh Talwalkar.
“A car travels 75 miles per hour (mph) downhill, 60 mph on flat roads, and 50 mph uphill. It takes 3 hours to go from town A to B, and it takes 3 hours and 30 minutes for return journey by the same route. What is the distance in miles between towns A and B?”
See the Impossible Car Riddle
This interesting problem comes from Colin Hughes at the Maths Challenge website.
Prove that for any number that is not a multiple of seven, then its cube will be one more or one less than a multiple of 7.”
See Lucky 7 Problem
A glutton for punishment I considered another Sam Loyd puzzle:
“Three men had a tandem and wished to go just forty miles. It could complete the journey with two passengers in one hour, but could not carry the three persons at one time. Well, one who was a good pedestrian, could walk at the rate of a mile in ten minutes; another could walk in fifteen minutes, and the other in twenty. What would be the best possible time in which all three could get to the end of their journey?”
See the Tandem Bicycle Puzzle.