"Things You Wouldn't Know If We Didn't Blog Intermittently."
12 July 2026
Math puzzle
I found this on Facebook without an answer posted there. I've put my answer in the comments but am not sure about it. Would appreciate insight from readers.
I get an area of 4. The square is 2x2. A radius drawn from the center of a circle to the point touching the square forms the hypotenuse of a pair of 3x4x5 right triangles whose legs are formed by extending the horizontal and vertical edges of the square to meet the horizontal and vertical lines passing through the center of the circle. In other words, if the length of the square is X, the length of the legs of the triangles described previously are 5-x and 5-x/2. Solving for X where the hypotenuse/radius is 5 yields a value of X=2, giving the square an area of 4.
I don't think the radius passes through that point. My (inelegant) solution notes that if a side of the square is 2x units, then the corner where it meets the circle is at offset (5-x, -(5-2x)) from the circle's center, which means (5-x)^2 + (5-2x)^2 = 5^2. The only nontrivial solution is x=1, meaning the square has area 4.
Catriona Agg (née Shearer) used to post beautiful geometry puzzles of this sort to pre-acquisition Twitter and I guess now Instagram and Bluesky. The 2018 puzzles were collected in a book (https://www.amazon.com/dp/1095672460); I see that Andrew Stacey has collected all the puzzles, along with solutions: https://notes.mathforge.org/notes/published/HomePage
I'm also not sure, but I'm thinking 6.25 (2.5 squared). My theory: If the square intersects the circles at halfway through the arc between the line and where the circles meet, then the height of the square is half the radius.
Anon and hovav are each correct, so I've erased my initial incorrect calculation. There's also a video here that explains it in much more complicated ways -https://www.youtube.com/watch?v=qOOnBTaHG_Q
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ReplyDeleteI get an area of 4. The square is 2x2. A radius drawn from the center of a circle to the point touching the square forms the hypotenuse of a pair of 3x4x5 right triangles whose legs are formed by extending the horizontal and vertical edges of the square to meet the horizontal and vertical lines passing through the center of the circle. In other words, if the length of the square is X, the length of the legs of the triangles described previously are 5-x and 5-x/2. Solving for X where the hypotenuse/radius is 5 yields a value of X=2, giving the square an area of 4.
ReplyDeleteI don't think the radius passes through that point. My (inelegant) solution notes that if a side of the square is 2x units, then the corner where it meets the circle is at offset (5-x, -(5-2x)) from the circle's center, which means (5-x)^2 + (5-2x)^2 = 5^2. The only nontrivial solution is x=1, meaning the square has area 4.
ReplyDeleteCatriona Agg (née Shearer) used to post beautiful geometry puzzles of this sort to pre-acquisition Twitter and I guess now Instagram and Bluesky. The 2018 puzzles were collected in a book (https://www.amazon.com/dp/1095672460); I see that Andrew Stacey has collected all the puzzles, along with solutions: https://notes.mathforge.org/notes/published/HomePage
I'm also not sure, but I'm thinking 6.25 (2.5 squared). My theory: If the square intersects the circles at halfway through the arc between the line and where the circles meet, then the height of the square is half the radius.
ReplyDeleteAnon and hovav are each correct, so I've erased my initial incorrect calculation. There's also a video here that explains it in much more complicated ways -https://www.youtube.com/watch?v=qOOnBTaHG_Q
ReplyDelete