## Wrong Triangle

1. In the amazing book 1000 PlayThinks by Ivan Moscovich (I highly recommend it; order it from Amazon.com), we see the diagram on the left, with this explanation:

Hidden Triangle
Trisect the angles of a triangle, as shown. Note that three points within the triangle form an equilateral triangle. Does such an equilateral triangle appear in every trisected triangle?

Answer: The answer, of course, is yes. But the diagram is wrong. The small triangle in the diagram is not an equilateral triangle. The diagram on the right shows the real equilateral triangle. This was discovered by Frank Morley in 1899. See Morley's Theorem.

2. On this puzzle, the same book misleads slightly:

How many diagonals are needed to divide a heptagon (7-gon), a nonagon (9-gon), and an undecagon (11-gon) into triangles?

In other words, how few lines do you need to draw, in order to divide each polygon into triangles. The book shows three seemingly regular polygons (although the text did not say "regular"). The answer that the book gives is, "In general, a convex polygon of n sides requires n-3 diagonals to triangulate it." That sounds true, in general. But many special n-gons require fewer lines. A regular octagon requires only four lines. Above left is an irregular nonagon which requires only five lines.

3. The same book shows these squares: 5x5, 4x4, and 3x3. It asks:

Can you dissect the 5x5 square into the fewest number of pieces needed to make both a 4x4 square and a 3x3 square?

Answer: The solution given in the book is five pieces. On the right is a solution with four pieces.

4. Another puzzle in the same book is this question: "What three numbers have a sum equal to their product?" The book's answer is 1+2+3=1x2x3. But there are "trivial" solutions like 0, 0, 0 or -17, 0, 17, or more interesting solutions like -1, 1/2, 1/3. That may seem picky of me to mention that, but this is the kind of criticism (constructive, of course) that I get in my email.

5. Yet another puzzle in the same book is this question: "Can you find a square and a rectangle that have perimeters equal to their areas?" You might want to work on that.

The square is 4x4 and one of the infinitely many rectangles is 3x6. The book says that those are the only solutions. Well, those are the only solutions in integers. But 2.5x10 works, as does 3.5x14/3, and infinitely many other triangles.

6. Another puzzle in the same book may give a mistaken impression. The author shows four isosceles right triangles much like those on the left, and asks, "Can you tell in which examples the shapes [rectangles] cover the most area?"

Answer: The book says that the first two have the greatest area. But all have very nearly the same area. The way I have drawn the third one makes it the same area as the first two; the book has the small square slightly smaller which makes the total area slightly smaller. And the fourth one is just the first one (same green area, different blue lines), slightly distorted to make the area slightly smaller.

The mistaken impression that I mentioned is this: One might think from the book that it matters greatly how the rectangles are oriented within the triangle. But this is not true. As long as the sides of the rectangles fit against sides of the triangle or against other rectangles, almost any reasonable orientation can produce a maximal area, as long as the sizes of the rectangles are adjusted to just the right size. Assuming that the leg of the triangle is 1, then (to maximize the areas) the vertices of the rectangles are all 1/4, 1/2, and 3/4 to the right of the left leg of the triangle, and 1/4, 1/2, and 3/4 from above the bottom leg of the triangle. For most of these figures, some elementary calculus can be used to verify that these are the largest possible areas, for those orientations. In the first three triangles that I have drawn above, you can either total up the areas of the rectangles and be surprised that they all total up to 3/8 (the area of the triangle is 1/2), or you can notice that there are many identical yellow triangles (some very small, and others twice as big), and these obviously add up to the same area in each of these three large triangles.

7. Yet another puzzle from the same book. Here we want the longest line segment that goes through A, and has endpoints on the two intersecting circles. The author finds that segment BC is longest when BD and CD are diameters of the two circles, which seems to be right for many intersecting circles. But look at the diagram on the right. AE seems to be the longest possible line segment that meets these conditions, and it is not the one described by the author (BC is). Maybe we have to specify that B and A cannot be the same point, and we are searching for a longest segment BC; then segment BC still seems to work. The problem becomes slightly messy with the extra conditions.

8. Yet another puzzle from the same book is this: "Can you find a way to express the number 100 using six 9's?" The book gives one correct solution: 99+99/99 = 100. Here are my solutions (there are others):

• (999-99)/9
• 9(9)+9+9+9/9
• 99+9/9+9-9
• 9(99+9/9)/9
• (9+9/9)(9+9/9)
• 9/.9^(9/9+9/9)
• 999/9.99
• 99.9999...
• (9/.9)(9/.9)+9-9
• 9(9/.9)(9/.9)/9
• 9(99/.99)/.9