Spike in the Hive

Our next Catriona Shearer puzzle.
Spike in the Hive

Two of the regular hexagons are identical; the third has area 10. What’s the area of the red triangle?

Continue reading “Spike in the Hive”


Semicircle Turducken

Here’s another of those Catriona Shearer puzzles that had me stumped for a while.

Semicircle Turducken

This looks ridiculous. There are no lengths, areas, or angles given, but you’re supposed to figure out an angle?


Continue reading “Semicircle Turducken”

The Garden of Clocks

Last week Math With Bad Drawings published a bunch of geometry puzzles by Catriona Shearer. I’m about ready to talk about the first one!

The Garden of Clocks

What fraction of each circle is shaded? (The 12 dots are equally spaced; the only point used inside the circle is the centre.)


Continue reading “The Garden of Clocks”

Easy foorp

That proof can also be done sort of backwards and is arguably easier that way. Instead of using three triangles of known dimensions and proving they can be assembled into a rectangle, start with the rectangle and find the dimensions of the triangles.

Take a right triangle with sides ab, and c. Draw perpendiculars to the hypotenuse at each end and a parallel line through the third vertex to make a rectangle BCDE.

Angles CDA and EAD are congruent (alternate interior angles). So are ADE and ACD since both are complementary with ADC. So triangles ADE and ACD are similar; the hypotenuses are in the ratio b/c, so the vertical leg DE = a(b/c) and the horizontal leg AE = b(b/c). Likewise we find BC = b(a/c) (as it should be) and AB = a(a/c). But then BE = CD implies a(a/c)+b(b/c) = c or, multiplying though by ca^2+b^2=c^c.

(Right away you know this proof fails in non Euclidean geometry, because in the first steps it relies on there being one and only one line parallel to CD through A.)