Solution to January’s Puzzle: Counting Triangles
Can you figure out how many triangles Figure A contains?
Can you think of a methodical approach or formula to calculate this
number?
To follow the explanation of this puzzle’s solution,
examine Figure B, in which the points in the diagram
are marked with letters. To find a methodical
approach for solving the puzzle, you must identify
a repeating pattern in the diagram. Note that the
diagram contains a repeating pattern of floors or
levels. Each floor except the top one consists of
two lines crossing each other, as well as a
ceiling.
To create a formula for counting the triangles,
you must determine the effect of
adding each floor. You can start by drawing
only the outermost triangle (ABC). So far,
your count is 1. Add the ceiling of the first
floor (DE), and the triangle ADE adds 1
to the count. Next, add the lines crossing
each other within the first floor (DC, BE).
The new triangles formed as a result of adding
these two lines include 4 one-celled triangles
(DBF, FBC, EFC, DFE), 4 two-celled triangles
(DBC, BCE, CED, EDB) and 2 three-celled
triangles (ABE, ADC), thus adding 10 new
triangles to the count. So adding the first
floor (including the ceiling and the two
lines crossing each other) adds 11 to the
original count of 1.
Add another floor by marking the
lines GH, DH, and GE. This adds 11
new triangles (as the first floor also
added), plus 2 new four-celled triangles
(GBE, HDC). In other words,
the first floor adds 11 to the count,
and every additional floor beyond the
first adds 13 to the count.
Although the top floor doesn’t have a ceiling
(no line exists between points J and K), you can imagine the floor as if
there were a ceiling (namely, add 13 to the count), then subtract the triangles
that are eliminated by removing the floor. Four triangles are eliminated
(AJK, JLK, JGK, JHK). So the total number of triangles you get is 1 + 11 +
13 + 13 - 4 = 34.
The general formula for n floors when the top floor has no ceiling is:
1 + 11 + [(n - 1) × 13] - 4. If you simplify the formula by expanding the
parentheses (i.e., 1 + 11 + (n × 13) - 13 - 4), you get (n × 13) - 5. So
for 3 floors you get (3 × 13) - 5 = 34. Now you can easily calculate the
number of triangles for any given number of floors.
February’s Puzzle: Counterfeit Coins
This puzzle is from Clifford Jensen. Suppose you have 10 stacks of coins,
with 10 coins in each stack. One stack consists of 10 counterfeit coins
and the other 9 stacks each consist of 10 legitimate coins. Each legitimate
coin weighs exactly 1 gram. Each counterfeit coin weighs exactly
0.9 grams. You have a digital scale that’s graduated in tenths of grams.
Using the scale to take only one reading, determine which stack has the
10 counterfeit coins. You can weigh any number of coins from any number
of stacks, but must you weigh them all together (i.e., you can take
only one reading from the scale).