Our next Catriona Shearer puzzle.

**12.
Spike in the Hive**

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

Spoiler!

Continue reading “Spike in the Hive”

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# Tag: geometry

## Spike in the Hive

## All in the square

## Power chords

## A couple more

## Semicircle Turducken

## The Garden of Clocks

## Easy foorp

Our next Catriona Shearer puzzle.

**12.
Spike in the Hive**

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

Spoiler!

Continue reading “Spike in the Hive”

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Another Catriona Shearer puzzle.

**11.
All in the Square**

Spoiler!

Continue reading “All in the square”

Here’s our next Catriona Shearer puzzle.

**8.
Power Chords**

**What’s the area of the circle?**

Spoiler!

Continue reading “Power chords”

Another two of those Catriona Shearer puzzles.

**5.
Shear Beauty**

**The area of the bottom left square is 5. What’s the area of the blue triangle?**

**6.
All Men are Created Equilateral**

Spoilers!

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

**7.
Semicircle Turducken**

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

Spoiler!

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!

**1.
The Garden of Clocks**

Spoilers!

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 , , and . Draw perpendiculars to the hypotenuse at each end and a parallel line through the third vertex to make a rectangle .

Angles and are congruent (alternate interior angles). So are and since both are complementary with . So triangles and are similar; the hypotenuses are in the ratio , so the vertical leg and the horizontal leg . Likewise we find (as it should be) and . But then implies or, multiplying though by , .

(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 through .)

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