This is another intimidating puzzle from Presh Talwalkar:
“Thanks to Eric from Miami for suggesting this problem and sending a solution!
From a 5th grade Chinese textbook: In the quadrilateral ABCD, angle A = 90°, angle ABD = 40°, angle BDC = 5°, angle C = 45°, and the length of AB is 6. Find the area of the quadrilateral ABCD.”
See the Chinese Quadrilateral Puzzle for solutions.
This is an interesting problem from the Canadian Mathematical Society’s 2001 Olymon.
“Suppose that XTY is a straight line and that TU and TV are two rays emanating from T for which XTU = UTV = VTY = 60º. Suppose that P, Q and R are respective points on the rays TY, TU and TV for which PQ = PR. Prove that QPR = 60º.”
See the Ubiquitous 60 Degree Problem
This is a fairly straight-forward problem from the 1999 AIME problems.
“The two squares shown share the same center O and have sides of length 1. The length of AB is 43/99 and the area of octagon ABCDEFGH is m/n where m and n are relatively prime positive integers. Find m + n.”
See the Another Octagonal Area Problem for solutions.
Since Twitter (now X) is no longer public, I was afraid I would not have access to new Catriona Agg puzzles, but she has put them up on Instagram, which is partially available to the public. I managed to find a half dozen interesting new brain ticklers.
Coming across this classic geometric paradox recently in Futility Closet motivated me to write down its solution in detail.
“Where did the empty square come from?”
In any case, this is the canonical example for why I avoid visual geometric proofs—you can be so easily fooled. Real proofs require plane or analytic geometry arguments.
See the Classic Geometry Paradox
(Update 9/14/2024) Penn & Teller – Fool Us – Magic Trick
Continue reading
James Tanton asked to prove the following surprising property of a right triangle and its circumscribed and inscribed circles.
“Every triangle is circumscribed by some circle of diameter D, say, and circumscribes another circle of smaller diameter d. For a right triangle, d + D equals the sum of two side lengths of the triangle. Why?”
Here is another problem from the “Challenges” section of the Quantum magazine.
“Inside a circle there are two intersecting circles. One of them touches the big circle in point A, the other in point B. Prove that if segment AB meets the smaller circles at one of their common points, then the sum of their radii equals the radius of the big circle. Is the converse true? (A. Vesyolov)”
This is a tantalizing problem from the 1977 Crux Mathematicorum.
“278. Proposed by W.A. McWorter, Jr., The Ohio State University.
If each of the medians of a triangle is extended beyond the sides of the triangle to 4/3 its length, show that the three new points formed and the vertices of the triangle all lie on an ellipse.”
See the Elliptical Medians Problem
This is a nifty problem from Presh Talwalkar.
“This is from a Manga called Q.E.D. I thank Sparky from the Philippines for the suggestion!
A string of beads is formed from 25 circles of the same size. The string passes through the center of each circle. The area enclosed by the string inside each circle is shaded in blue, and the remaining areas of the circles are shaded in orange. What is the value of the orange area minus the blue area? Calculate the area in terms of r, the radius of each circle.”
See the String of Beads Puzzle for solutions.
This is another puzzle from the Maths Masters team, Burkard Polster (aka Mathologer) and Marty Ross as part of their “Summer Quizzes” offerings.
“A mysterious square has materialized in the middle of the MCG, hovering in mid-air. The heights above the ground of three of its corners are 13, 21 and 34 metres. The fourth corner is higher still. How high?”
See the Floating Square Puzzle for solutions.
(Update 8/13/2023) Alternative Solution Continue reading
