In my puzzle blog earlier today I set you the following three questions:
1) In the image of an equilateral triangle above, what fraction of the whole triangle is the red triangle?
You could have shown this in one of at least three ways, illustrated below: by dividing the entire triangle into 8 identical small triangles, and deducing that the red triangle comprises two of them; by constructing an “envelope” shape that must have the same area as the original triangle, and deducing the red triangle is a quarter of this envelope; by dividing the triangle into four identical smaller triangles, and deducing that the red triangle has equal area to one of these smaller triangles, because if two triangles have the same base and height, they have the same area.
2) Four semicircles with radius 2 are constructed in the red square below. What is the area of the square?
Solution: 8(2 + √3)
In order to solve this one you need to be familiar with Pythagoras’s Theorem, which states that for right-angled triangles the square of the hypotenuse is equal to the sum of the squares on the other two sides.
Method A: Look at the triangle in the image below left. It has hypotenuse 4 and one of the other sides is 2. If the third side has length x, Pythagoras’s Theorem tells us that x2 + 22 = 42. So x2 = 16 – 4 = 12, making x = √12 = 2√3. The red square thus has side 2 + 2√3, and area (2 + 2√3)2 = 4 + 8√3 + 12 = 16 + 8√3 = 8(2 +√3)
Method B: Look at the middle image below, with a red square inside the original square. We can rearrange the triangles outside that red square to make a rectangle (below right) which has sides 2√3 and 4. The area of the large square is the area of the smaller square plus the rectangle, which is 42 + 8√3 = 16 + 8√3 = 8(2 +√3)
3) A quarter of a circle with radius 6 is shown below. Inside it are two semicircles and a circle, each of them at a tangent to the others. What are the radii of the black circle and the smaller black semicircle?
Solution: The smallest circle has radius 1, and the smaller semicircle has radius 2.
First, some notation, and basic deduction. Let the smallest circle have radius r, and the smaller semicircle radius R. Since the quarter-circle is of a circle with radius 6, the radius of the larger semi-circle must be 3.
Method 1: Consider the right-angled triangle shown left. The hypotenuse is 3 + R, and the sides have length 3, and r + 3. So, from Pythagoras:
33 + (r + 3)2 = (3 + R)2
We also see that r + R = 3, or r = 3 – R.
33 + (6 – R)2 = (3 + R)2
which works out as 18R – 27 = 9, so R = 2. And then r = 3 – R = 1
Method 2: Consider the right-angled triangle shown left. The hypotenuse is 6 – r, and the sides are length 3 + r and 3. So, from Pythagoras,
(6 – r)2 = (3 + r)2 + 32
or, 27 – 18r = 9, or r = 1.
As above, since r + R = 3, R = 2.
I hope you enjoyed today’s puzzles. I’ll be back in two weeks, on January 1.
I’m the author of several popular maths books, most recently Puzzle Ninja: Pit Your Wits Against the Japanese Puzzle Masters.
I set a puzzle here every two weeks on a Monday. Send me your email if you want me to alert you each time I post a new one. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
Thanks to Ed Southall and Vincent Pantaloni for today’s puzzles. Check out their book Geometry Snacks.