Harder Temple Geometry Problems
We can, however, make an easy problem more difficult. The two integers 45 and 56 have no common divisor except 1, and so they are called “relatively prime.” In general, dividing a number a by a number b gives a quotient and a remainder, in other words, a = q × b + r. In our case 56 = 1 × 45 + 11,
where q = 1 and r = 11. Similarly, 45 = 4 × 11 + 1
and 11 = 11 × 1 + 0.
This sequence of successive divisions is an example of Euclid’s algorithm.3 One can easily show that the last nonzero remainder is always the greatest common divisor of a and b, in our case 1. We can also reverse the procedure and solve for the remainders: 1 = 45 − 4 × 11 = 45 − 4(56 − 45) = 5 × 45 − 4 × 56, which is the same as 15 × 45 − 12 × 56 = 3. Notice that 15 and 12 provide the answer to the problem. The moral of this shaggy- dog story is that, any time the greatest common divisor of a and b is 1, one can write a × c − b × d = 1, where c and d are the given coefficients to a and b. The Japanese learned of this algorithm through the Chinese, independently of the West.
Problem 2 The area of the original trapezoid is S = [(a + b)/2]h, and so the area of the smallest trapezoid is S/n = [(a + k)/2]h1, where h1 is its height. Dropping the dashed perpendicular as shown in figure 5.24, we then have by similar triangles ⎛k −a⎞ h1 = ⎜ ⎟h. ⎝b −a⎠
Solving these equations for n in terms of h, a and b gives the result ⎛ a + b ⎞ ⎛ h ⎞ b2 − a2 n=⎜ . = ⎟ ⎝ a + k ⎠ ⎜⎝ h1 ⎟⎠ k 2 − a 2
3 For more on Euclid’s algorithm, see, for example, Oystein Ore, Number Theory and Its History (Dover, New York, 1988).
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