Tag Archives: Presh Talwalkar

Sum Of Squares Puzzle

Yet another interesting problem from Presh Talwalkar.

“Two side-by-side squares are inscribed in a semicircle.  If the semicircle has a radius of 10, can you solve for the total area of the two squares? If no, demonstrate why not. If yes, calculate the answer.”

This puzzle shares the characteristics of all good problems where the information provided seems insufficient.

See the Sum of Squares Puzzle.

Hard Geometric Problem

This is another problem from the indefatigable Presh Talwalkar.

_    _____Hard Geometry Problem
“In triangle ABC above, angle A is bisected into two 60° angles. If AD = 100, and AB = 2(AC), what is the length of BC?”

See Hard Geometric Problem

(Update 7/18/2020, 7/20/2020)  Alternative Solution Continue reading

Impossible Car Riddle

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

Parallelogram Cosine 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

Putnam Octagon Problem

Here is a problem from the famous (infamous?) Putnam exam, presented by Presh Talwalkar. Needless to say, I did not solve it in 30 minutes—but at least I solved it (after making a blizzard of arithmetic and trigonometric errors).

“Today’s problem is from the 1978 test, problem B1 (the easiest of the second set of problems). A convex octagon inscribed in a circle has four consecutive sides of length 3 and four consecutive sides of length 2. Find the area of the octagon.”

My solution is horribly pedestrian and fraught with numerous chances for arithmetic mistakes to derail it, which happened in spades. As I suspected, there was an elegant, “easy” solution (as demonstrated by Talwalkar)—once you thought of it! Again, this is like a Coffin Problem. See the Putnam Octagon Problem.

Conical Bottle Problem

I was astonished that this problem was suitable for 8th graders. First of all the formula for the volume of a cone is one of the least-remembered of formulas, and I certainly never remember it. So my only viable approach was calculus, which is probably not a suitable solution for an 8th grader.

Presh Talwalkar: “This was sent to me as a competition problem for 8th graders, so it would be a challenge problem for students aged 12 to 13. When a conical bottle rests on its flat base, the water in the bottle is 8 cm from its vertex. When the same conical bottle is turned upside down, the water level is 2 cm from its base. What is the height of the bottle? (Note “conical” refers to a right circular cone as is common usage.) I at first thought this problem was impossible. But it actually can be solved. Give it a try and then watch the video for a solution.”

See the Conical Bottle Problem.